Download BSE NEET AIPMT 2018 NEET AIPMT 2018 Answers and Solutions Code NN

Download NEET 2018 (National Eligibility cum Entrance Test-Under Graduate) Entrance Test NEET AIPMT 2018 Answers and Solutions Code NN

Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
3. Rough work is to be done on the space provided for this purpose in the Test Booklet only.
4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
5. The CODE for this Booklet is NN.
6. The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on
the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
Regulations of this examination.
11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
FirstRanker.com - FirstRanker's Choice
Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
3. Rough work is to be done on the space provided for this purpose in the Test Booklet only.
4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
5. The CODE for this Booklet is NN.
6. The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on
the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
Regulations of this examination.
11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
FirstRanker.com - FirstRanker's Choice
Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
3. Rough work is to be done on the space provided for this purpose in the Test Booklet only.
4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
5. The CODE for this Booklet is NN.
6. The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on
the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
Regulations of this examination.
11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
FirstRanker.com - FirstRanker's Choice
Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
3. Rough work is to be done on the space provided for this purpose in the Test Booklet only.
4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
5. The CODE for this Booklet is NN.
6. The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on
the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
Regulations of this examination.
11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4
FirstRanker.com - FirstRanker's Choice
Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
3. Rough work is to be done on the space provided for this purpose in the Test Booklet only.
4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
5. The CODE for this Booklet is NN.
6. The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on
the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
Regulations of this examination.
11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
FirstRanker.com - FirstRanker's Choice
Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
3. Rough work is to be done on the space provided for this purpose in the Test Booklet only.
4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
5. The CODE for this Booklet is NN.
6. The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on
the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
Regulations of this examination.
11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
FirstRanker.com - FirstRanker's Choice
Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
3. Rough work is to be done on the space provided for this purpose in the Test Booklet only.
4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
5. The CODE for this Booklet is NN.
6. The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on
the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
Regulations of this examination.
11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
FirstRanker.com - FirstRanker's Choice
Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
3. Rough work is to be done on the space provided for this purpose in the Test Booklet only.
4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
5. The CODE for this Booklet is NN.
6. The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on
the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
Regulations of this examination.
11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
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one mark will be deducted from the total scores. The maximum marks are 720.
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
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4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
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Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
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7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
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12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
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Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
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12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
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12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
14
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Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
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4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
5. The CODE for this Booklet is NN.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
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11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
14
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
15
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Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
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4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
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11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
14
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
15
NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
16
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Regulations of this examination.
11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
14
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
15
NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
16
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
17
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4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
14
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
15
NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
16
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
17
NEET (UG) - 2018 (Code-NN) HLAAC
77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
18
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Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
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4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
5. The CODE for this Booklet is NN.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
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12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
14
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
15
NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
16
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
17
NEET (UG) - 2018 (Code-NN) HLAAC
77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
18
NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
19
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11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
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NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
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NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
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NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
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NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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NEET (UG) - 2018 (Code-NN) HLAAC
77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
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NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
19
NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
20
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Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
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NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
14
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
15
NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
16
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
17
NEET (UG) - 2018 (Code-NN) HLAAC
77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
18
NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
19
NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
20
NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
21
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Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
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7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
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12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
14
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
15
NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
16
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
17
NEET (UG) - 2018 (Code-NN) HLAAC
77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
18
NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
19
NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
20
NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
21
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
22
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Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
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4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
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5. The CODE for this Booklet is NN.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
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9. Use of Electronic/Manual Calculator is prohibited.
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12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
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NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
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NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
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NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
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NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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NEET (UG) - 2018 (Code-NN) HLAAC
77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
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NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
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NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
20
NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
21
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
22
NEET (UG) - 2018 (Code-NN) HLAAC
106. Which one is wrongly matched?
(1) Unicellular organism ? Chlorella
(2) Uniflagellate gametes ? Polysiphonia
(3) Gemma cups ? Marchantia
(4) Biflagellate zoospores ? Brown algae
Answer (2)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (1, 3 & 4) are correctly
matched
107. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Herbarium (i) It is a place having a
collection of preserved
plants and animals
b. Key

(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum

(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue (iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
a b c d
(1) (iii) (iv) (i) (ii)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iv) (iii) (i)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
108. Which one of the following plants shows a
very close relationship with a species of moth,
where none of the two can complete its life
cycle without the other?
(1) Viola (2) Hydrilla
(3) Banana (4) Yucca
Answer (4)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
109. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?160?C
(2) ?120?C
(3) ?196?C
(4) ?80?C
Answer (3)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C
(Cryopreservation)
110. In which of the following forms is iron
absorbed by plants?
(1) Both ferric and ferrous
(2) Ferric
(3) Free element
(4) Ferrous
Answer (2*)
Sol. Iron is absorbed by plants in the form of ferric
ions. (According to NCERT)
*Plants absorb iron in both form i.e. Fe
++
and
Fe
+++
. (Preferably Fe
++
)
111. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Calcium (2) Magnesium
(3) Potassium (4) Sodium
Answer (3)
Sol. Potassium helps in maintaining turgidity of
cells.
112. Double fertilization is
(1) Syngamy and triple fusion
(2) Fusion of two male gametes of a pollen
tube with two different eggs
(3) Fusion of two male gametes with one egg
(4) Fusion of one male gamete with two polar
nuclei
Answer (1)
23
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Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
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4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
5. The CODE for this Booklet is NN.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
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11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
14
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
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NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
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NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
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NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
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NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
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NEET (UG) - 2018 (Code-NN) HLAAC
106. Which one is wrongly matched?
(1) Unicellular organism ? Chlorella
(2) Uniflagellate gametes ? Polysiphonia
(3) Gemma cups ? Marchantia
(4) Biflagellate zoospores ? Brown algae
Answer (2)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (1, 3 & 4) are correctly
matched
107. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Herbarium (i) It is a place having a
collection of preserved
plants and animals
b. Key

(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum

(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue (iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
a b c d
(1) (iii) (iv) (i) (ii)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iv) (iii) (i)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
108. Which one of the following plants shows a
very close relationship with a species of moth,
where none of the two can complete its life
cycle without the other?
(1) Viola (2) Hydrilla
(3) Banana (4) Yucca
Answer (4)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
109. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?160?C
(2) ?120?C
(3) ?196?C
(4) ?80?C
Answer (3)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C
(Cryopreservation)
110. In which of the following forms is iron
absorbed by plants?
(1) Both ferric and ferrous
(2) Ferric
(3) Free element
(4) Ferrous
Answer (2*)
Sol. Iron is absorbed by plants in the form of ferric
ions. (According to NCERT)
*Plants absorb iron in both form i.e. Fe
++
and
Fe
+++
. (Preferably Fe
++
)
111. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Calcium (2) Magnesium
(3) Potassium (4) Sodium
Answer (3)
Sol. Potassium helps in maintaining turgidity of
cells.
112. Double fertilization is
(1) Syngamy and triple fusion
(2) Fusion of two male gametes of a pollen
tube with two different eggs
(3) Fusion of two male gametes with one egg
(4) Fusion of one male gamete with two polar
nuclei
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
113. What is the role of NAD
+
in cellular
respiration?
(1) It is the final electron acceptor for
anaerobic respiration.
(2) It functions as an enzyme.
(3) It is a nucleotide source for ATP synthesis.
(4) It functions as an electron carrier.
Answer (4)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
114. Oxygen is not produced during photosynthesis
by
(1) Chara
(2) Green sulphur bacteria
(3) Cycas
(4) Nostoc
Answer (2)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
115. The Golgi complex participates in
(1) Activation of amino acid
(2) Fatty acid breakdown
(3) Respiration in bacteria
(4) Formation of secretory vesicles
Answer (4)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
116. Stomatal movement is not affected by
(1) CO
2
concentration
(2) Temperature
(3) O
2
concentration
(4) Light
Answer (3)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
117. Stomata in grass leaf are
(1) Barrel shaped
(2) Dumb-bell shaped
(3) Rectangular
(4) Kidney shaped
Answer (2)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
118. Which of the following is true for nucleolus?
(1) It is a site for active ribosomal RNA
synthesis
(2) Larger nucleoli are present in dividing
cells
(3) It takes part in spindle formation
(4) It is a membrane-bound structure
Answer (1)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
119. Which of the following is not a product of light
reaction of photosynthesis?
(1) Oxygen (2) ATP
(3) NADPH (4) NADH
Answer (4)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
120. The stage during which separation of the
paired homologous chromosomes begins is
(1) Zygotene (2) Pachytene
(3) Diakinesis (4) Diplotene
Answer (4)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
121. The two functional groups characteristic of
sugars are
(1) Carbonyl and hydroxyl
(2) Hydroxyl and methyl
(3) Carbonyl and phosphate
(4) Carbonyl and methyl
Answer (1)
Sol. Sugar is a common term used to denote
carbohydrate.
Carbohydrates are polyhydroxy aldehyde,
ketone or their derivatives, which means they
have carbonyl and hydroxyl groups.
122. Which among the following is not a
prokaryote?
(1) Oscillatoria
(2) Saccharomyces
(3) Nostoc
(4) Mycobacterium
Answer (2)
24
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
14
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
15
NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
16
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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NEET (UG) - 2018 (Code-NN) HLAAC
77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
18
NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
19
NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
20
NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
21
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
22
NEET (UG) - 2018 (Code-NN) HLAAC
106. Which one is wrongly matched?
(1) Unicellular organism ? Chlorella
(2) Uniflagellate gametes ? Polysiphonia
(3) Gemma cups ? Marchantia
(4) Biflagellate zoospores ? Brown algae
Answer (2)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (1, 3 & 4) are correctly
matched
107. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Herbarium (i) It is a place having a
collection of preserved
plants and animals
b. Key

(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum

(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue (iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
a b c d
(1) (iii) (iv) (i) (ii)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iv) (iii) (i)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
108. Which one of the following plants shows a
very close relationship with a species of moth,
where none of the two can complete its life
cycle without the other?
(1) Viola (2) Hydrilla
(3) Banana (4) Yucca
Answer (4)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
109. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?160?C
(2) ?120?C
(3) ?196?C
(4) ?80?C
Answer (3)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C
(Cryopreservation)
110. In which of the following forms is iron
absorbed by plants?
(1) Both ferric and ferrous
(2) Ferric
(3) Free element
(4) Ferrous
Answer (2*)
Sol. Iron is absorbed by plants in the form of ferric
ions. (According to NCERT)
*Plants absorb iron in both form i.e. Fe
++
and
Fe
+++
. (Preferably Fe
++
)
111. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Calcium (2) Magnesium
(3) Potassium (4) Sodium
Answer (3)
Sol. Potassium helps in maintaining turgidity of
cells.
112. Double fertilization is
(1) Syngamy and triple fusion
(2) Fusion of two male gametes of a pollen
tube with two different eggs
(3) Fusion of two male gametes with one egg
(4) Fusion of one male gamete with two polar
nuclei
Answer (1)
23
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
113. What is the role of NAD
+
in cellular
respiration?
(1) It is the final electron acceptor for
anaerobic respiration.
(2) It functions as an enzyme.
(3) It is a nucleotide source for ATP synthesis.
(4) It functions as an electron carrier.
Answer (4)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
114. Oxygen is not produced during photosynthesis
by
(1) Chara
(2) Green sulphur bacteria
(3) Cycas
(4) Nostoc
Answer (2)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
115. The Golgi complex participates in
(1) Activation of amino acid
(2) Fatty acid breakdown
(3) Respiration in bacteria
(4) Formation of secretory vesicles
Answer (4)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
116. Stomatal movement is not affected by
(1) CO
2
concentration
(2) Temperature
(3) O
2
concentration
(4) Light
Answer (3)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
117. Stomata in grass leaf are
(1) Barrel shaped
(2) Dumb-bell shaped
(3) Rectangular
(4) Kidney shaped
Answer (2)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
118. Which of the following is true for nucleolus?
(1) It is a site for active ribosomal RNA
synthesis
(2) Larger nucleoli are present in dividing
cells
(3) It takes part in spindle formation
(4) It is a membrane-bound structure
Answer (1)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
119. Which of the following is not a product of light
reaction of photosynthesis?
(1) Oxygen (2) ATP
(3) NADPH (4) NADH
Answer (4)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
120. The stage during which separation of the
paired homologous chromosomes begins is
(1) Zygotene (2) Pachytene
(3) Diakinesis (4) Diplotene
Answer (4)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
121. The two functional groups characteristic of
sugars are
(1) Carbonyl and hydroxyl
(2) Hydroxyl and methyl
(3) Carbonyl and phosphate
(4) Carbonyl and methyl
Answer (1)
Sol. Sugar is a common term used to denote
carbohydrate.
Carbohydrates are polyhydroxy aldehyde,
ketone or their derivatives, which means they
have carbonyl and hydroxyl groups.
122. Which among the following is not a
prokaryote?
(1) Oscillatoria
(2) Saccharomyces
(3) Nostoc
(4) Mycobacterium
Answer (2)
24
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
123. Offsets are produced by
(1) Parthenogenesis (2) Meiotic divisions
(3) Parthenocarpy (4) Mitotic divisions
Answer (4)
Sol. Offset is a vegetative part of a plant, formed
by mitosis.
? Meiotic divisions do not occur in somatic
cells.
? Parthenogenesis is the formation of
embryo from ovum or egg without
fertilisation.
? Parthenocarpy is the fruit formed without
fertilisation, (generally seedless)
124. Select the correct statement
(1) Transduction was discovered by S. Altman
(2) Franklin Stahl coined the term ?linkage?
(3) Spliceosomes take part in translation
(4) Punnett square was developed by a British
scientist
Answer (4)
Sol. Punnett square was developed by a British
geneticist, Reginald C. Punnett.
? Franklin Stahl proved semi-conservative
mode of replication.
? Transduction was discovered by Zinder
and Laderberg.
? Spliceosome formation is part of post-
transcriptional change in Eukaryotes
125. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Sporopollenin (2) Pollenkitt
(3) Oil content (4) Cellulosic intine
Answer (1)
Sol. Sporopollenin cannot be degraded by
enzyme; strong acids and alkali, therefore it is
helpful in preserving pollen as fossil.
Pollenkitt ? Help in insect pollination.
Cellulosic Intine ? Inner sporoderm layer of
pollen grain known as intine made up
cellulose & pectin.
Oil content ? No role is pollen preservation.
126. Which of the following pairs is wrongly
matched?
(1) T.H. Morgan : Linkage
(2) Starch synthesis in pea : Multiple alleles
(3) XO type sex : Grasshopper
determination
(4) ABO blood grouping : Co-dominance
Answer (2)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (1, 3 & 4) are correctly
matched.
127. Select the correct match
(1) Francois Jacob and - Lac operon
Jacques Monod
(2) Alec Jeffreys - Streptococcus
pneumoniae
(3) Matthew Meselson - Pisum sativum
and F. Stahl
(4) Alfred Hershey and - TMV
Martha Chase
Answer (1)
Sol. Francois Jacob and Jacque Monod proposed
model of gene regulation known as operon
model/lac operon.
? Alec Jeffreys ? DNA fingerprinting
technique.
? Matthew Meselson and F. Stahl ? Semi-
conservative DNA replication in E. coli.
? Alfred Hershey and Martha Chase ?
Proved DNA as genetic material not
protein
128. Which of the following flowers only once in its
life-time?
(1) Papaya (2) Bamboo species
(3) Mango (4) Jackfruit
Answer (2)
Sol. Bamboo species are monocarpic i.e., flower
generally only once in its life-time after 50-
100 years.
Jackfruit, papaya and mango are polycarpic
i.e., produce flowers and fruits many times in
their life-time.
129. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Virus (2) Fungus
(3) Plant (4) Bacterium
Answer (4)
25
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Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
3. Rough work is to be done on the space provided for this purpose in the Test Booklet only.
4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
5. The CODE for this Booklet is NN.
6. The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on
the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
Regulations of this examination.
11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
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NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
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NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
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NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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NEET (UG) - 2018 (Code-NN) HLAAC
77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
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NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
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NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
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NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
21
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
22
NEET (UG) - 2018 (Code-NN) HLAAC
106. Which one is wrongly matched?
(1) Unicellular organism ? Chlorella
(2) Uniflagellate gametes ? Polysiphonia
(3) Gemma cups ? Marchantia
(4) Biflagellate zoospores ? Brown algae
Answer (2)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (1, 3 & 4) are correctly
matched
107. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Herbarium (i) It is a place having a
collection of preserved
plants and animals
b. Key

(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum

(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue (iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
a b c d
(1) (iii) (iv) (i) (ii)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iv) (iii) (i)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
108. Which one of the following plants shows a
very close relationship with a species of moth,
where none of the two can complete its life
cycle without the other?
(1) Viola (2) Hydrilla
(3) Banana (4) Yucca
Answer (4)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
109. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?160?C
(2) ?120?C
(3) ?196?C
(4) ?80?C
Answer (3)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C
(Cryopreservation)
110. In which of the following forms is iron
absorbed by plants?
(1) Both ferric and ferrous
(2) Ferric
(3) Free element
(4) Ferrous
Answer (2*)
Sol. Iron is absorbed by plants in the form of ferric
ions. (According to NCERT)
*Plants absorb iron in both form i.e. Fe
++
and
Fe
+++
. (Preferably Fe
++
)
111. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Calcium (2) Magnesium
(3) Potassium (4) Sodium
Answer (3)
Sol. Potassium helps in maintaining turgidity of
cells.
112. Double fertilization is
(1) Syngamy and triple fusion
(2) Fusion of two male gametes of a pollen
tube with two different eggs
(3) Fusion of two male gametes with one egg
(4) Fusion of one male gamete with two polar
nuclei
Answer (1)
23
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
113. What is the role of NAD
+
in cellular
respiration?
(1) It is the final electron acceptor for
anaerobic respiration.
(2) It functions as an enzyme.
(3) It is a nucleotide source for ATP synthesis.
(4) It functions as an electron carrier.
Answer (4)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
114. Oxygen is not produced during photosynthesis
by
(1) Chara
(2) Green sulphur bacteria
(3) Cycas
(4) Nostoc
Answer (2)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
115. The Golgi complex participates in
(1) Activation of amino acid
(2) Fatty acid breakdown
(3) Respiration in bacteria
(4) Formation of secretory vesicles
Answer (4)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
116. Stomatal movement is not affected by
(1) CO
2
concentration
(2) Temperature
(3) O
2
concentration
(4) Light
Answer (3)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
117. Stomata in grass leaf are
(1) Barrel shaped
(2) Dumb-bell shaped
(3) Rectangular
(4) Kidney shaped
Answer (2)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
118. Which of the following is true for nucleolus?
(1) It is a site for active ribosomal RNA
synthesis
(2) Larger nucleoli are present in dividing
cells
(3) It takes part in spindle formation
(4) It is a membrane-bound structure
Answer (1)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
119. Which of the following is not a product of light
reaction of photosynthesis?
(1) Oxygen (2) ATP
(3) NADPH (4) NADH
Answer (4)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
120. The stage during which separation of the
paired homologous chromosomes begins is
(1) Zygotene (2) Pachytene
(3) Diakinesis (4) Diplotene
Answer (4)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
121. The two functional groups characteristic of
sugars are
(1) Carbonyl and hydroxyl
(2) Hydroxyl and methyl
(3) Carbonyl and phosphate
(4) Carbonyl and methyl
Answer (1)
Sol. Sugar is a common term used to denote
carbohydrate.
Carbohydrates are polyhydroxy aldehyde,
ketone or their derivatives, which means they
have carbonyl and hydroxyl groups.
122. Which among the following is not a
prokaryote?
(1) Oscillatoria
(2) Saccharomyces
(3) Nostoc
(4) Mycobacterium
Answer (2)
24
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
123. Offsets are produced by
(1) Parthenogenesis (2) Meiotic divisions
(3) Parthenocarpy (4) Mitotic divisions
Answer (4)
Sol. Offset is a vegetative part of a plant, formed
by mitosis.
? Meiotic divisions do not occur in somatic
cells.
? Parthenogenesis is the formation of
embryo from ovum or egg without
fertilisation.
? Parthenocarpy is the fruit formed without
fertilisation, (generally seedless)
124. Select the correct statement
(1) Transduction was discovered by S. Altman
(2) Franklin Stahl coined the term ?linkage?
(3) Spliceosomes take part in translation
(4) Punnett square was developed by a British
scientist
Answer (4)
Sol. Punnett square was developed by a British
geneticist, Reginald C. Punnett.
? Franklin Stahl proved semi-conservative
mode of replication.
? Transduction was discovered by Zinder
and Laderberg.
? Spliceosome formation is part of post-
transcriptional change in Eukaryotes
125. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Sporopollenin (2) Pollenkitt
(3) Oil content (4) Cellulosic intine
Answer (1)
Sol. Sporopollenin cannot be degraded by
enzyme; strong acids and alkali, therefore it is
helpful in preserving pollen as fossil.
Pollenkitt ? Help in insect pollination.
Cellulosic Intine ? Inner sporoderm layer of
pollen grain known as intine made up
cellulose & pectin.
Oil content ? No role is pollen preservation.
126. Which of the following pairs is wrongly
matched?
(1) T.H. Morgan : Linkage
(2) Starch synthesis in pea : Multiple alleles
(3) XO type sex : Grasshopper
determination
(4) ABO blood grouping : Co-dominance
Answer (2)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (1, 3 & 4) are correctly
matched.
127. Select the correct match
(1) Francois Jacob and - Lac operon
Jacques Monod
(2) Alec Jeffreys - Streptococcus
pneumoniae
(3) Matthew Meselson - Pisum sativum
and F. Stahl
(4) Alfred Hershey and - TMV
Martha Chase
Answer (1)
Sol. Francois Jacob and Jacque Monod proposed
model of gene regulation known as operon
model/lac operon.
? Alec Jeffreys ? DNA fingerprinting
technique.
? Matthew Meselson and F. Stahl ? Semi-
conservative DNA replication in E. coli.
? Alfred Hershey and Martha Chase ?
Proved DNA as genetic material not
protein
128. Which of the following flowers only once in its
life-time?
(1) Papaya (2) Bamboo species
(3) Mango (4) Jackfruit
Answer (2)
Sol. Bamboo species are monocarpic i.e., flower
generally only once in its life-time after 50-
100 years.
Jackfruit, papaya and mango are polycarpic
i.e., produce flowers and fruits many times in
their life-time.
129. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Virus (2) Fungus
(3) Plant (4) Bacterium
Answer (4)
25
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
130. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) pBR 322 (2) Retrovirus
(3) X phage (4) Ti plasmid
Answer (2)
Sol. Retrovirus is commonly used as vector for
introducing a DNA fragment in human
lymphocyte.
Gene therapy : Lymphocyte from blood of
patient are grown in culture outside the body,
a functional gene is introduced by using a
retroviral vector, into these lymphocyte.
131. Select the correct match
(1) G. Mendel - Transformation
(2) Ribozyme - Nucleic acid
(3) T.H. Morgan - Transduction
(4) F
2
? Recessive parent - Dihybrid cross
Answer (2)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
132. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bioexploitation (2) Bio-infringement
(3) Biodegradation (4) Biopiracy
Answer (4)
Sol. Biopiracy is term used for or refer to the use
of bioresources by multinational companies
and other organisation without proper
authorisation from the countries and people
concerned with compensatory payment
(definition of biopiracy given in NCERT).
133. A ?new? variety of rice was patented by a
foreign company, though such varieties have
been present in India for a long time. This is
related to
(1) Basmati (2) Co-667
(3) Lerma Rojo (4) Sharbati Sonora
Answer (1)
Sol. In 1997, an American company got patent
rights on Basmati rice through the US patent
and trademark office that was actually been
derived from Indian farmer?s varieties.
The diversity of rice in India is one of the
richest in the world, 27 documented varieties
of Basmati are grown in India.
Indian basmati was crossed with semi-dwarf
varieties and claimed as an invention or a
novelty.
Sharbati Sonora and Lerma Rojo are varieties
of wheat.
134. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Genetic Engineering Appraisal Committee
(GEAC)
(2) Indian Council of Medical Research (ICMR)
(3) Research Committee on Genetic
Manipulation (RCGM)
(4) Council for Scientific and Industrial
Research (CSIR)
Answer (1)
Sol. Indian Government has setup organisation
such as GEAC (Genetic Engineering Appraisal
Committee) which will make decisions
regarding the validity of GM research and
safety of introducing GM-organism for public
services. (Direct from NCERT).
135. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Denaturation, Annealing, Extension
(2) Extension, Denaturation, Annealing
(3) Denaturation, Extension, Annealing
(4) Annealing, Extension, Denaturation
Answer (1)
Sol. This technique is used for making multiple
copies of gene (or DNA) of interest in vitro.
Each cycle has three steps
(i) Denaturation
(ii) Primer annealing
(iii) Extension of primer
26
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Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
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12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
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NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
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65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
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NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
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NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
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NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
22
NEET (UG) - 2018 (Code-NN) HLAAC
106. Which one is wrongly matched?
(1) Unicellular organism ? Chlorella
(2) Uniflagellate gametes ? Polysiphonia
(3) Gemma cups ? Marchantia
(4) Biflagellate zoospores ? Brown algae
Answer (2)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (1, 3 & 4) are correctly
matched
107. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Herbarium (i) It is a place having a
collection of preserved
plants and animals
b. Key

(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum

(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue (iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
a b c d
(1) (iii) (iv) (i) (ii)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iv) (iii) (i)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
108. Which one of the following plants shows a
very close relationship with a species of moth,
where none of the two can complete its life
cycle without the other?
(1) Viola (2) Hydrilla
(3) Banana (4) Yucca
Answer (4)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
109. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?160?C
(2) ?120?C
(3) ?196?C
(4) ?80?C
Answer (3)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C
(Cryopreservation)
110. In which of the following forms is iron
absorbed by plants?
(1) Both ferric and ferrous
(2) Ferric
(3) Free element
(4) Ferrous
Answer (2*)
Sol. Iron is absorbed by plants in the form of ferric
ions. (According to NCERT)
*Plants absorb iron in both form i.e. Fe
++
and
Fe
+++
. (Preferably Fe
++
)
111. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Calcium (2) Magnesium
(3) Potassium (4) Sodium
Answer (3)
Sol. Potassium helps in maintaining turgidity of
cells.
112. Double fertilization is
(1) Syngamy and triple fusion
(2) Fusion of two male gametes of a pollen
tube with two different eggs
(3) Fusion of two male gametes with one egg
(4) Fusion of one male gamete with two polar
nuclei
Answer (1)
23
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
113. What is the role of NAD
+
in cellular
respiration?
(1) It is the final electron acceptor for
anaerobic respiration.
(2) It functions as an enzyme.
(3) It is a nucleotide source for ATP synthesis.
(4) It functions as an electron carrier.
Answer (4)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
114. Oxygen is not produced during photosynthesis
by
(1) Chara
(2) Green sulphur bacteria
(3) Cycas
(4) Nostoc
Answer (2)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
115. The Golgi complex participates in
(1) Activation of amino acid
(2) Fatty acid breakdown
(3) Respiration in bacteria
(4) Formation of secretory vesicles
Answer (4)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
116. Stomatal movement is not affected by
(1) CO
2
concentration
(2) Temperature
(3) O
2
concentration
(4) Light
Answer (3)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
117. Stomata in grass leaf are
(1) Barrel shaped
(2) Dumb-bell shaped
(3) Rectangular
(4) Kidney shaped
Answer (2)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
118. Which of the following is true for nucleolus?
(1) It is a site for active ribosomal RNA
synthesis
(2) Larger nucleoli are present in dividing
cells
(3) It takes part in spindle formation
(4) It is a membrane-bound structure
Answer (1)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
119. Which of the following is not a product of light
reaction of photosynthesis?
(1) Oxygen (2) ATP
(3) NADPH (4) NADH
Answer (4)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
120. The stage during which separation of the
paired homologous chromosomes begins is
(1) Zygotene (2) Pachytene
(3) Diakinesis (4) Diplotene
Answer (4)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
121. The two functional groups characteristic of
sugars are
(1) Carbonyl and hydroxyl
(2) Hydroxyl and methyl
(3) Carbonyl and phosphate
(4) Carbonyl and methyl
Answer (1)
Sol. Sugar is a common term used to denote
carbohydrate.
Carbohydrates are polyhydroxy aldehyde,
ketone or their derivatives, which means they
have carbonyl and hydroxyl groups.
122. Which among the following is not a
prokaryote?
(1) Oscillatoria
(2) Saccharomyces
(3) Nostoc
(4) Mycobacterium
Answer (2)
24
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
123. Offsets are produced by
(1) Parthenogenesis (2) Meiotic divisions
(3) Parthenocarpy (4) Mitotic divisions
Answer (4)
Sol. Offset is a vegetative part of a plant, formed
by mitosis.
? Meiotic divisions do not occur in somatic
cells.
? Parthenogenesis is the formation of
embryo from ovum or egg without
fertilisation.
? Parthenocarpy is the fruit formed without
fertilisation, (generally seedless)
124. Select the correct statement
(1) Transduction was discovered by S. Altman
(2) Franklin Stahl coined the term ?linkage?
(3) Spliceosomes take part in translation
(4) Punnett square was developed by a British
scientist
Answer (4)
Sol. Punnett square was developed by a British
geneticist, Reginald C. Punnett.
? Franklin Stahl proved semi-conservative
mode of replication.
? Transduction was discovered by Zinder
and Laderberg.
? Spliceosome formation is part of post-
transcriptional change in Eukaryotes
125. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Sporopollenin (2) Pollenkitt
(3) Oil content (4) Cellulosic intine
Answer (1)
Sol. Sporopollenin cannot be degraded by
enzyme; strong acids and alkali, therefore it is
helpful in preserving pollen as fossil.
Pollenkitt ? Help in insect pollination.
Cellulosic Intine ? Inner sporoderm layer of
pollen grain known as intine made up
cellulose & pectin.
Oil content ? No role is pollen preservation.
126. Which of the following pairs is wrongly
matched?
(1) T.H. Morgan : Linkage
(2) Starch synthesis in pea : Multiple alleles
(3) XO type sex : Grasshopper
determination
(4) ABO blood grouping : Co-dominance
Answer (2)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (1, 3 & 4) are correctly
matched.
127. Select the correct match
(1) Francois Jacob and - Lac operon
Jacques Monod
(2) Alec Jeffreys - Streptococcus
pneumoniae
(3) Matthew Meselson - Pisum sativum
and F. Stahl
(4) Alfred Hershey and - TMV
Martha Chase
Answer (1)
Sol. Francois Jacob and Jacque Monod proposed
model of gene regulation known as operon
model/lac operon.
? Alec Jeffreys ? DNA fingerprinting
technique.
? Matthew Meselson and F. Stahl ? Semi-
conservative DNA replication in E. coli.
? Alfred Hershey and Martha Chase ?
Proved DNA as genetic material not
protein
128. Which of the following flowers only once in its
life-time?
(1) Papaya (2) Bamboo species
(3) Mango (4) Jackfruit
Answer (2)
Sol. Bamboo species are monocarpic i.e., flower
generally only once in its life-time after 50-
100 years.
Jackfruit, papaya and mango are polycarpic
i.e., produce flowers and fruits many times in
their life-time.
129. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Virus (2) Fungus
(3) Plant (4) Bacterium
Answer (4)
25
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
130. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) pBR 322 (2) Retrovirus
(3) X phage (4) Ti plasmid
Answer (2)
Sol. Retrovirus is commonly used as vector for
introducing a DNA fragment in human
lymphocyte.
Gene therapy : Lymphocyte from blood of
patient are grown in culture outside the body,
a functional gene is introduced by using a
retroviral vector, into these lymphocyte.
131. Select the correct match
(1) G. Mendel - Transformation
(2) Ribozyme - Nucleic acid
(3) T.H. Morgan - Transduction
(4) F
2
? Recessive parent - Dihybrid cross
Answer (2)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
132. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bioexploitation (2) Bio-infringement
(3) Biodegradation (4) Biopiracy
Answer (4)
Sol. Biopiracy is term used for or refer to the use
of bioresources by multinational companies
and other organisation without proper
authorisation from the countries and people
concerned with compensatory payment
(definition of biopiracy given in NCERT).
133. A ?new? variety of rice was patented by a
foreign company, though such varieties have
been present in India for a long time. This is
related to
(1) Basmati (2) Co-667
(3) Lerma Rojo (4) Sharbati Sonora
Answer (1)
Sol. In 1997, an American company got patent
rights on Basmati rice through the US patent
and trademark office that was actually been
derived from Indian farmer?s varieties.
The diversity of rice in India is one of the
richest in the world, 27 documented varieties
of Basmati are grown in India.
Indian basmati was crossed with semi-dwarf
varieties and claimed as an invention or a
novelty.
Sharbati Sonora and Lerma Rojo are varieties
of wheat.
134. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Genetic Engineering Appraisal Committee
(GEAC)
(2) Indian Council of Medical Research (ICMR)
(3) Research Committee on Genetic
Manipulation (RCGM)
(4) Council for Scientific and Industrial
Research (CSIR)
Answer (1)
Sol. Indian Government has setup organisation
such as GEAC (Genetic Engineering Appraisal
Committee) which will make decisions
regarding the validity of GM research and
safety of introducing GM-organism for public
services. (Direct from NCERT).
135. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Denaturation, Annealing, Extension
(2) Extension, Denaturation, Annealing
(3) Denaturation, Extension, Annealing
(4) Annealing, Extension, Denaturation
Answer (1)
Sol. This technique is used for making multiple
copies of gene (or DNA) of interest in vitro.
Each cycle has three steps
(i) Denaturation
(ii) Primer annealing
(iii) Extension of primer
26
136. Identify the major products P, Q and R in the
following sequence of reactions:
(i) O
2

P Q + R
(ii) H
8
O / A
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

P Q R
137. Which of the following compounds can form a
zwitterion?
(1) Glycine (2) Aniline
(3) Benzoic acid (4) Acetanilide
Answer (1)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
pK
,
= 9.60 pK
,
= 2.34
CH
3

CH ? CH
3

O
2

(P)
CH
3
CH
3
? CH ? CH
3

H /H O
+
2
Hydroperoxide
Rearrangement
OH
O
(R)
(Q)
3
HC?C?O?O?H
+ CH
3
? C ? CH
3

27
NEET (UG) - 2018 (Code-NN) HLAAC
OH
CH(CH
3
)
2
CH
3
? CO ? CH
3
(1)
CH
3
CH
2
? OH
(2)
(3)
(4)
Answer (1)
? ? ? ? ? ? ??
Cl
Sol
Cl Cl
CO
1, 2?H
Shift
6+ 6+ 6?
(Incipient carbocation)
Cl
Ni
CO
6?
OC
AlCl
3

CO
Now,
.
CH
3
CH
2
CH
2
? Cl + Al
CH
3
? CH ? CH
3

CH
2
CH
2
CH
3
CHO
CH
2
CH
2
CH
3
CHO COOH
CH(CH
3
)
2
OH
CH
3
CH
2
CH
2
Cl AlCl
3

, CH
3
CH(OH)CH
3
H
2
N ? CH
2
? COO
?

138. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Tetrahedral geometry and paramagnetic
(2) Square planar geometry and diamagnetic
(3) Square planar geometry and
paramagnetic
(4) Tetrahedral geometry and diamagnetic
Answer (4)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
sp -hybridisation
3
~
H
3
N ? CH
2
? COO
?
139. Iron carbonyl, Fe(CO)
5
is
(1) Dinuclear (2) Tetranuclear
(3) Trinuclear (4) Mononuclear
Answer (4)
Sol. ? Based on the number of metal atoms
present in a complex, they are classified
into mononuclear, dinuclear, trinuclear and
so on.
eg: Fe(CO)
5
: mononuclear
Co
2
(CO)
8
: dinuclear
Fe
3
(CO)
12
: trinuclear
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Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
3. Rough work is to be done on the space provided for this purpose in the Test Booklet only.
4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
5. The CODE for this Booklet is NN.
6. The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on
the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
Regulations of this examination.
11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
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NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
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NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
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NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
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NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
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NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
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NEET (UG) - 2018 (Code-NN) HLAAC
106. Which one is wrongly matched?
(1) Unicellular organism ? Chlorella
(2) Uniflagellate gametes ? Polysiphonia
(3) Gemma cups ? Marchantia
(4) Biflagellate zoospores ? Brown algae
Answer (2)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (1, 3 & 4) are correctly
matched
107. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Herbarium (i) It is a place having a
collection of preserved
plants and animals
b. Key

(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum

(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue (iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
a b c d
(1) (iii) (iv) (i) (ii)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iv) (iii) (i)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
108. Which one of the following plants shows a
very close relationship with a species of moth,
where none of the two can complete its life
cycle without the other?
(1) Viola (2) Hydrilla
(3) Banana (4) Yucca
Answer (4)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
109. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?160?C
(2) ?120?C
(3) ?196?C
(4) ?80?C
Answer (3)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C
(Cryopreservation)
110. In which of the following forms is iron
absorbed by plants?
(1) Both ferric and ferrous
(2) Ferric
(3) Free element
(4) Ferrous
Answer (2*)
Sol. Iron is absorbed by plants in the form of ferric
ions. (According to NCERT)
*Plants absorb iron in both form i.e. Fe
++
and
Fe
+++
. (Preferably Fe
++
)
111. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Calcium (2) Magnesium
(3) Potassium (4) Sodium
Answer (3)
Sol. Potassium helps in maintaining turgidity of
cells.
112. Double fertilization is
(1) Syngamy and triple fusion
(2) Fusion of two male gametes of a pollen
tube with two different eggs
(3) Fusion of two male gametes with one egg
(4) Fusion of one male gamete with two polar
nuclei
Answer (1)
23
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
113. What is the role of NAD
+
in cellular
respiration?
(1) It is the final electron acceptor for
anaerobic respiration.
(2) It functions as an enzyme.
(3) It is a nucleotide source for ATP synthesis.
(4) It functions as an electron carrier.
Answer (4)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
114. Oxygen is not produced during photosynthesis
by
(1) Chara
(2) Green sulphur bacteria
(3) Cycas
(4) Nostoc
Answer (2)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
115. The Golgi complex participates in
(1) Activation of amino acid
(2) Fatty acid breakdown
(3) Respiration in bacteria
(4) Formation of secretory vesicles
Answer (4)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
116. Stomatal movement is not affected by
(1) CO
2
concentration
(2) Temperature
(3) O
2
concentration
(4) Light
Answer (3)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
117. Stomata in grass leaf are
(1) Barrel shaped
(2) Dumb-bell shaped
(3) Rectangular
(4) Kidney shaped
Answer (2)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
118. Which of the following is true for nucleolus?
(1) It is a site for active ribosomal RNA
synthesis
(2) Larger nucleoli are present in dividing
cells
(3) It takes part in spindle formation
(4) It is a membrane-bound structure
Answer (1)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
119. Which of the following is not a product of light
reaction of photosynthesis?
(1) Oxygen (2) ATP
(3) NADPH (4) NADH
Answer (4)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
120. The stage during which separation of the
paired homologous chromosomes begins is
(1) Zygotene (2) Pachytene
(3) Diakinesis (4) Diplotene
Answer (4)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
121. The two functional groups characteristic of
sugars are
(1) Carbonyl and hydroxyl
(2) Hydroxyl and methyl
(3) Carbonyl and phosphate
(4) Carbonyl and methyl
Answer (1)
Sol. Sugar is a common term used to denote
carbohydrate.
Carbohydrates are polyhydroxy aldehyde,
ketone or their derivatives, which means they
have carbonyl and hydroxyl groups.
122. Which among the following is not a
prokaryote?
(1) Oscillatoria
(2) Saccharomyces
(3) Nostoc
(4) Mycobacterium
Answer (2)
24
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
123. Offsets are produced by
(1) Parthenogenesis (2) Meiotic divisions
(3) Parthenocarpy (4) Mitotic divisions
Answer (4)
Sol. Offset is a vegetative part of a plant, formed
by mitosis.
? Meiotic divisions do not occur in somatic
cells.
? Parthenogenesis is the formation of
embryo from ovum or egg without
fertilisation.
? Parthenocarpy is the fruit formed without
fertilisation, (generally seedless)
124. Select the correct statement
(1) Transduction was discovered by S. Altman
(2) Franklin Stahl coined the term ?linkage?
(3) Spliceosomes take part in translation
(4) Punnett square was developed by a British
scientist
Answer (4)
Sol. Punnett square was developed by a British
geneticist, Reginald C. Punnett.
? Franklin Stahl proved semi-conservative
mode of replication.
? Transduction was discovered by Zinder
and Laderberg.
? Spliceosome formation is part of post-
transcriptional change in Eukaryotes
125. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Sporopollenin (2) Pollenkitt
(3) Oil content (4) Cellulosic intine
Answer (1)
Sol. Sporopollenin cannot be degraded by
enzyme; strong acids and alkali, therefore it is
helpful in preserving pollen as fossil.
Pollenkitt ? Help in insect pollination.
Cellulosic Intine ? Inner sporoderm layer of
pollen grain known as intine made up
cellulose & pectin.
Oil content ? No role is pollen preservation.
126. Which of the following pairs is wrongly
matched?
(1) T.H. Morgan : Linkage
(2) Starch synthesis in pea : Multiple alleles
(3) XO type sex : Grasshopper
determination
(4) ABO blood grouping : Co-dominance
Answer (2)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (1, 3 & 4) are correctly
matched.
127. Select the correct match
(1) Francois Jacob and - Lac operon
Jacques Monod
(2) Alec Jeffreys - Streptococcus
pneumoniae
(3) Matthew Meselson - Pisum sativum
and F. Stahl
(4) Alfred Hershey and - TMV
Martha Chase
Answer (1)
Sol. Francois Jacob and Jacque Monod proposed
model of gene regulation known as operon
model/lac operon.
? Alec Jeffreys ? DNA fingerprinting
technique.
? Matthew Meselson and F. Stahl ? Semi-
conservative DNA replication in E. coli.
? Alfred Hershey and Martha Chase ?
Proved DNA as genetic material not
protein
128. Which of the following flowers only once in its
life-time?
(1) Papaya (2) Bamboo species
(3) Mango (4) Jackfruit
Answer (2)
Sol. Bamboo species are monocarpic i.e., flower
generally only once in its life-time after 50-
100 years.
Jackfruit, papaya and mango are polycarpic
i.e., produce flowers and fruits many times in
their life-time.
129. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Virus (2) Fungus
(3) Plant (4) Bacterium
Answer (4)
25
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
130. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) pBR 322 (2) Retrovirus
(3) X phage (4) Ti plasmid
Answer (2)
Sol. Retrovirus is commonly used as vector for
introducing a DNA fragment in human
lymphocyte.
Gene therapy : Lymphocyte from blood of
patient are grown in culture outside the body,
a functional gene is introduced by using a
retroviral vector, into these lymphocyte.
131. Select the correct match
(1) G. Mendel - Transformation
(2) Ribozyme - Nucleic acid
(3) T.H. Morgan - Transduction
(4) F
2
? Recessive parent - Dihybrid cross
Answer (2)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
132. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bioexploitation (2) Bio-infringement
(3) Biodegradation (4) Biopiracy
Answer (4)
Sol. Biopiracy is term used for or refer to the use
of bioresources by multinational companies
and other organisation without proper
authorisation from the countries and people
concerned with compensatory payment
(definition of biopiracy given in NCERT).
133. A ?new? variety of rice was patented by a
foreign company, though such varieties have
been present in India for a long time. This is
related to
(1) Basmati (2) Co-667
(3) Lerma Rojo (4) Sharbati Sonora
Answer (1)
Sol. In 1997, an American company got patent
rights on Basmati rice through the US patent
and trademark office that was actually been
derived from Indian farmer?s varieties.
The diversity of rice in India is one of the
richest in the world, 27 documented varieties
of Basmati are grown in India.
Indian basmati was crossed with semi-dwarf
varieties and claimed as an invention or a
novelty.
Sharbati Sonora and Lerma Rojo are varieties
of wheat.
134. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Genetic Engineering Appraisal Committee
(GEAC)
(2) Indian Council of Medical Research (ICMR)
(3) Research Committee on Genetic
Manipulation (RCGM)
(4) Council for Scientific and Industrial
Research (CSIR)
Answer (1)
Sol. Indian Government has setup organisation
such as GEAC (Genetic Engineering Appraisal
Committee) which will make decisions
regarding the validity of GM research and
safety of introducing GM-organism for public
services. (Direct from NCERT).
135. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Denaturation, Annealing, Extension
(2) Extension, Denaturation, Annealing
(3) Denaturation, Extension, Annealing
(4) Annealing, Extension, Denaturation
Answer (1)
Sol. This technique is used for making multiple
copies of gene (or DNA) of interest in vitro.
Each cycle has three steps
(i) Denaturation
(ii) Primer annealing
(iii) Extension of primer
26
136. Identify the major products P, Q and R in the
following sequence of reactions:
(i) O
2

P Q + R
(ii) H
8
O / A
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

P Q R
137. Which of the following compounds can form a
zwitterion?
(1) Glycine (2) Aniline
(3) Benzoic acid (4) Acetanilide
Answer (1)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
pK
,
= 9.60 pK
,
= 2.34
CH
3

CH ? CH
3

O
2

(P)
CH
3
CH
3
? CH ? CH
3

H /H O
+
2
Hydroperoxide
Rearrangement
OH
O
(R)
(Q)
3
HC?C?O?O?H
+ CH
3
? C ? CH
3

27
NEET (UG) - 2018 (Code-NN) HLAAC
OH
CH(CH
3
)
2
CH
3
? CO ? CH
3
(1)
CH
3
CH
2
? OH
(2)
(3)
(4)
Answer (1)
? ? ? ? ? ? ??
Cl
Sol
Cl Cl
CO
1, 2?H
Shift
6+ 6+ 6?
(Incipient carbocation)
Cl
Ni
CO
6?
OC
AlCl
3

CO
Now,
.
CH
3
CH
2
CH
2
? Cl + Al
CH
3
? CH ? CH
3

CH
2
CH
2
CH
3
CHO
CH
2
CH
2
CH
3
CHO COOH
CH(CH
3
)
2
OH
CH
3
CH
2
CH
2
Cl AlCl
3

, CH
3
CH(OH)CH
3
H
2
N ? CH
2
? COO
?

138. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Tetrahedral geometry and paramagnetic
(2) Square planar geometry and diamagnetic
(3) Square planar geometry and
paramagnetic
(4) Tetrahedral geometry and diamagnetic
Answer (4)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
sp -hybridisation
3
~
H
3
N ? CH
2
? COO
?
139. Iron carbonyl, Fe(CO)
5
is
(1) Dinuclear (2) Tetranuclear
(3) Trinuclear (4) Mononuclear
Answer (4)
Sol. ? Based on the number of metal atoms
present in a complex, they are classified
into mononuclear, dinuclear, trinuclear and
so on.
eg: Fe(CO)
5
: mononuclear
Co
2
(CO)
8
: dinuclear
Fe
3
(CO)
12
: trinuclear
NEET (UG) - 2018 (Code-NN) HLAAC
NO
.
NO
2

H V
~
H
(1)
V
(2)
~
(3)
H
V
~
NO
.
(4)
V H
~
NO
2
140. Match the metal ions given in Column I with
the spin magnetic moments of the ions given
in Column II and assign the correct code :
Column I Column II
a. Co
3+
i.
8
BM
b. Cr
3+
ii.
35
BM
c. Fe
3+
iii.
3
BM
d. Ni
2+
iv.
24
BM
v.
15
BM
c d
i ii
ii i
ii iii
iii iv
Answer (2)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Spin magnetic moment = 4(4 + 2) = 24 BM
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Spin magnetic moment = 3(3 + 2) = 15 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
Spin magnetic moment = 2(2 + 2) = 8 BM
141. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Linkage isomerism
(2) Geometrical isomerism
(3) Ionization isomerism
(4) Coordination isomerism
Answer (2)
Sol. In [CoCl
2
(en)
2
], Coordination number of Co
is 6 and this compound has octahedral
geometry.
? As per given option, type of isomerism is
geometrical isomerism.
142. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) MnO
4
2?
(2) CrO
4
2?

(3) MnO
4
?
(4) Cr
2
O
7
2?
Answer (1)
Sol. CrO
4
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
Cr
2
O
7
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
MnO
4
2?
= Mn
6+
= [Ar] 3d
1

Unpaired electron (n) = 1; Paramagnetic
MnO
4
?
= Mn
7+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
143. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) CH
3
? CH = CH ? CH
3

(2) HC ~ C ? C ~ CH
(3) CH
2
= CH ? CH = CH
2

(4) CH
2
= CH ? C ~ CH
Answer (4)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
144. Which of the following carbocations is
expected to be most stable?
Answer (3)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (3)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
a b
(1) iii v
(2) iv v
(3) iv i
(4) i ii
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
28
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1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
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12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
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NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
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NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
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NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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NEET (UG) - 2018 (Code-NN) HLAAC
77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
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NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
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NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
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NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
21
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
22
NEET (UG) - 2018 (Code-NN) HLAAC
106. Which one is wrongly matched?
(1) Unicellular organism ? Chlorella
(2) Uniflagellate gametes ? Polysiphonia
(3) Gemma cups ? Marchantia
(4) Biflagellate zoospores ? Brown algae
Answer (2)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (1, 3 & 4) are correctly
matched
107. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Herbarium (i) It is a place having a
collection of preserved
plants and animals
b. Key

(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum

(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue (iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
a b c d
(1) (iii) (iv) (i) (ii)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iv) (iii) (i)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
108. Which one of the following plants shows a
very close relationship with a species of moth,
where none of the two can complete its life
cycle without the other?
(1) Viola (2) Hydrilla
(3) Banana (4) Yucca
Answer (4)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
109. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?160?C
(2) ?120?C
(3) ?196?C
(4) ?80?C
Answer (3)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C
(Cryopreservation)
110. In which of the following forms is iron
absorbed by plants?
(1) Both ferric and ferrous
(2) Ferric
(3) Free element
(4) Ferrous
Answer (2*)
Sol. Iron is absorbed by plants in the form of ferric
ions. (According to NCERT)
*Plants absorb iron in both form i.e. Fe
++
and
Fe
+++
. (Preferably Fe
++
)
111. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Calcium (2) Magnesium
(3) Potassium (4) Sodium
Answer (3)
Sol. Potassium helps in maintaining turgidity of
cells.
112. Double fertilization is
(1) Syngamy and triple fusion
(2) Fusion of two male gametes of a pollen
tube with two different eggs
(3) Fusion of two male gametes with one egg
(4) Fusion of one male gamete with two polar
nuclei
Answer (1)
23
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
113. What is the role of NAD
+
in cellular
respiration?
(1) It is the final electron acceptor for
anaerobic respiration.
(2) It functions as an enzyme.
(3) It is a nucleotide source for ATP synthesis.
(4) It functions as an electron carrier.
Answer (4)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
114. Oxygen is not produced during photosynthesis
by
(1) Chara
(2) Green sulphur bacteria
(3) Cycas
(4) Nostoc
Answer (2)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
115. The Golgi complex participates in
(1) Activation of amino acid
(2) Fatty acid breakdown
(3) Respiration in bacteria
(4) Formation of secretory vesicles
Answer (4)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
116. Stomatal movement is not affected by
(1) CO
2
concentration
(2) Temperature
(3) O
2
concentration
(4) Light
Answer (3)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
117. Stomata in grass leaf are
(1) Barrel shaped
(2) Dumb-bell shaped
(3) Rectangular
(4) Kidney shaped
Answer (2)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
118. Which of the following is true for nucleolus?
(1) It is a site for active ribosomal RNA
synthesis
(2) Larger nucleoli are present in dividing
cells
(3) It takes part in spindle formation
(4) It is a membrane-bound structure
Answer (1)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
119. Which of the following is not a product of light
reaction of photosynthesis?
(1) Oxygen (2) ATP
(3) NADPH (4) NADH
Answer (4)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
120. The stage during which separation of the
paired homologous chromosomes begins is
(1) Zygotene (2) Pachytene
(3) Diakinesis (4) Diplotene
Answer (4)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
121. The two functional groups characteristic of
sugars are
(1) Carbonyl and hydroxyl
(2) Hydroxyl and methyl
(3) Carbonyl and phosphate
(4) Carbonyl and methyl
Answer (1)
Sol. Sugar is a common term used to denote
carbohydrate.
Carbohydrates are polyhydroxy aldehyde,
ketone or their derivatives, which means they
have carbonyl and hydroxyl groups.
122. Which among the following is not a
prokaryote?
(1) Oscillatoria
(2) Saccharomyces
(3) Nostoc
(4) Mycobacterium
Answer (2)
24
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
123. Offsets are produced by
(1) Parthenogenesis (2) Meiotic divisions
(3) Parthenocarpy (4) Mitotic divisions
Answer (4)
Sol. Offset is a vegetative part of a plant, formed
by mitosis.
? Meiotic divisions do not occur in somatic
cells.
? Parthenogenesis is the formation of
embryo from ovum or egg without
fertilisation.
? Parthenocarpy is the fruit formed without
fertilisation, (generally seedless)
124. Select the correct statement
(1) Transduction was discovered by S. Altman
(2) Franklin Stahl coined the term ?linkage?
(3) Spliceosomes take part in translation
(4) Punnett square was developed by a British
scientist
Answer (4)
Sol. Punnett square was developed by a British
geneticist, Reginald C. Punnett.
? Franklin Stahl proved semi-conservative
mode of replication.
? Transduction was discovered by Zinder
and Laderberg.
? Spliceosome formation is part of post-
transcriptional change in Eukaryotes
125. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Sporopollenin (2) Pollenkitt
(3) Oil content (4) Cellulosic intine
Answer (1)
Sol. Sporopollenin cannot be degraded by
enzyme; strong acids and alkali, therefore it is
helpful in preserving pollen as fossil.
Pollenkitt ? Help in insect pollination.
Cellulosic Intine ? Inner sporoderm layer of
pollen grain known as intine made up
cellulose & pectin.
Oil content ? No role is pollen preservation.
126. Which of the following pairs is wrongly
matched?
(1) T.H. Morgan : Linkage
(2) Starch synthesis in pea : Multiple alleles
(3) XO type sex : Grasshopper
determination
(4) ABO blood grouping : Co-dominance
Answer (2)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (1, 3 & 4) are correctly
matched.
127. Select the correct match
(1) Francois Jacob and - Lac operon
Jacques Monod
(2) Alec Jeffreys - Streptococcus
pneumoniae
(3) Matthew Meselson - Pisum sativum
and F. Stahl
(4) Alfred Hershey and - TMV
Martha Chase
Answer (1)
Sol. Francois Jacob and Jacque Monod proposed
model of gene regulation known as operon
model/lac operon.
? Alec Jeffreys ? DNA fingerprinting
technique.
? Matthew Meselson and F. Stahl ? Semi-
conservative DNA replication in E. coli.
? Alfred Hershey and Martha Chase ?
Proved DNA as genetic material not
protein
128. Which of the following flowers only once in its
life-time?
(1) Papaya (2) Bamboo species
(3) Mango (4) Jackfruit
Answer (2)
Sol. Bamboo species are monocarpic i.e., flower
generally only once in its life-time after 50-
100 years.
Jackfruit, papaya and mango are polycarpic
i.e., produce flowers and fruits many times in
their life-time.
129. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Virus (2) Fungus
(3) Plant (4) Bacterium
Answer (4)
25
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
130. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) pBR 322 (2) Retrovirus
(3) X phage (4) Ti plasmid
Answer (2)
Sol. Retrovirus is commonly used as vector for
introducing a DNA fragment in human
lymphocyte.
Gene therapy : Lymphocyte from blood of
patient are grown in culture outside the body,
a functional gene is introduced by using a
retroviral vector, into these lymphocyte.
131. Select the correct match
(1) G. Mendel - Transformation
(2) Ribozyme - Nucleic acid
(3) T.H. Morgan - Transduction
(4) F
2
? Recessive parent - Dihybrid cross
Answer (2)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
132. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bioexploitation (2) Bio-infringement
(3) Biodegradation (4) Biopiracy
Answer (4)
Sol. Biopiracy is term used for or refer to the use
of bioresources by multinational companies
and other organisation without proper
authorisation from the countries and people
concerned with compensatory payment
(definition of biopiracy given in NCERT).
133. A ?new? variety of rice was patented by a
foreign company, though such varieties have
been present in India for a long time. This is
related to
(1) Basmati (2) Co-667
(3) Lerma Rojo (4) Sharbati Sonora
Answer (1)
Sol. In 1997, an American company got patent
rights on Basmati rice through the US patent
and trademark office that was actually been
derived from Indian farmer?s varieties.
The diversity of rice in India is one of the
richest in the world, 27 documented varieties
of Basmati are grown in India.
Indian basmati was crossed with semi-dwarf
varieties and claimed as an invention or a
novelty.
Sharbati Sonora and Lerma Rojo are varieties
of wheat.
134. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Genetic Engineering Appraisal Committee
(GEAC)
(2) Indian Council of Medical Research (ICMR)
(3) Research Committee on Genetic
Manipulation (RCGM)
(4) Council for Scientific and Industrial
Research (CSIR)
Answer (1)
Sol. Indian Government has setup organisation
such as GEAC (Genetic Engineering Appraisal
Committee) which will make decisions
regarding the validity of GM research and
safety of introducing GM-organism for public
services. (Direct from NCERT).
135. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Denaturation, Annealing, Extension
(2) Extension, Denaturation, Annealing
(3) Denaturation, Extension, Annealing
(4) Annealing, Extension, Denaturation
Answer (1)
Sol. This technique is used for making multiple
copies of gene (or DNA) of interest in vitro.
Each cycle has three steps
(i) Denaturation
(ii) Primer annealing
(iii) Extension of primer
26
136. Identify the major products P, Q and R in the
following sequence of reactions:
(i) O
2

P Q + R
(ii) H
8
O / A
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

P Q R
137. Which of the following compounds can form a
zwitterion?
(1) Glycine (2) Aniline
(3) Benzoic acid (4) Acetanilide
Answer (1)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
pK
,
= 9.60 pK
,
= 2.34
CH
3

CH ? CH
3

O
2

(P)
CH
3
CH
3
? CH ? CH
3

H /H O
+
2
Hydroperoxide
Rearrangement
OH
O
(R)
(Q)
3
HC?C?O?O?H
+ CH
3
? C ? CH
3

27
NEET (UG) - 2018 (Code-NN) HLAAC
OH
CH(CH
3
)
2
CH
3
? CO ? CH
3
(1)
CH
3
CH
2
? OH
(2)
(3)
(4)
Answer (1)
? ? ? ? ? ? ??
Cl
Sol
Cl Cl
CO
1, 2?H
Shift
6+ 6+ 6?
(Incipient carbocation)
Cl
Ni
CO
6?
OC
AlCl
3

CO
Now,
.
CH
3
CH
2
CH
2
? Cl + Al
CH
3
? CH ? CH
3

CH
2
CH
2
CH
3
CHO
CH
2
CH
2
CH
3
CHO COOH
CH(CH
3
)
2
OH
CH
3
CH
2
CH
2
Cl AlCl
3

, CH
3
CH(OH)CH
3
H
2
N ? CH
2
? COO
?

138. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Tetrahedral geometry and paramagnetic
(2) Square planar geometry and diamagnetic
(3) Square planar geometry and
paramagnetic
(4) Tetrahedral geometry and diamagnetic
Answer (4)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
sp -hybridisation
3
~
H
3
N ? CH
2
? COO
?
139. Iron carbonyl, Fe(CO)
5
is
(1) Dinuclear (2) Tetranuclear
(3) Trinuclear (4) Mononuclear
Answer (4)
Sol. ? Based on the number of metal atoms
present in a complex, they are classified
into mononuclear, dinuclear, trinuclear and
so on.
eg: Fe(CO)
5
: mononuclear
Co
2
(CO)
8
: dinuclear
Fe
3
(CO)
12
: trinuclear
NEET (UG) - 2018 (Code-NN) HLAAC
NO
.
NO
2

H V
~
H
(1)
V
(2)
~
(3)
H
V
~
NO
.
(4)
V H
~
NO
2
140. Match the metal ions given in Column I with
the spin magnetic moments of the ions given
in Column II and assign the correct code :
Column I Column II
a. Co
3+
i.
8
BM
b. Cr
3+
ii.
35
BM
c. Fe
3+
iii.
3
BM
d. Ni
2+
iv.
24
BM
v.
15
BM
c d
i ii
ii i
ii iii
iii iv
Answer (2)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Spin magnetic moment = 4(4 + 2) = 24 BM
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Spin magnetic moment = 3(3 + 2) = 15 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
Spin magnetic moment = 2(2 + 2) = 8 BM
141. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Linkage isomerism
(2) Geometrical isomerism
(3) Ionization isomerism
(4) Coordination isomerism
Answer (2)
Sol. In [CoCl
2
(en)
2
], Coordination number of Co
is 6 and this compound has octahedral
geometry.
? As per given option, type of isomerism is
geometrical isomerism.
142. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) MnO
4
2?
(2) CrO
4
2?

(3) MnO
4
?
(4) Cr
2
O
7
2?
Answer (1)
Sol. CrO
4
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
Cr
2
O
7
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
MnO
4
2?
= Mn
6+
= [Ar] 3d
1

Unpaired electron (n) = 1; Paramagnetic
MnO
4
?
= Mn
7+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
143. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) CH
3
? CH = CH ? CH
3

(2) HC ~ C ? C ~ CH
(3) CH
2
= CH ? CH = CH
2

(4) CH
2
= CH ? C ~ CH
Answer (4)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
144. Which of the following carbocations is
expected to be most stable?
Answer (3)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (3)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
a b
(1) iii v
(2) iv v
(3) iv i
(4) i ii
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
28
M
10 HCl + 100 mL
M
10 NaOH e. 100 mL
NEET (UG) - 2018 (Code-NN) HLAAC
145. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NR
2
> ? OR > ? F
(2) ? NH
2
< ? OR < ? F
(3) ? NH
2
> ? OR > ? F
(4) ? NR
2
< ? OR < ? F
Answer (2*)
Sol. ?I effect increases on increasing
electronegativity of atom. So, correct order of
?I effect is ?NH
2
< ? OR < ? F.
*Most appropriate Answer is option (2),
however option (4) may also be correct answer.
146. The solubility of BaSO
4
in water is 2.42 ? 10
?3
gL
?1

at 298 K. The value of its solubility product
(K
sp
) will be
(Given molar mass of BaSO
4
= 233 g mol
?1
)
(1) 1.08 ? 10
?8
mol
2
L
?2

(2) 1.08 ? 10
?10
mol
2
L
?2

(3) 1.08 ? 10
?14
mol
2
L
?2

(4) 1.08 ? 10
?12
mol
2
L
?2

Answer (2)
148. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M M
a. 60 mL 10 HCl + 40 mL 10 NaOH
M M
b. 55 mL 10 HCl + 45 mL 10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
pH of which one of them will be equal to 1?
(1) c (2) b
(3) d (4) a
Answer (1)
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
3
(mol L
?1
) Sol. Solubility of BaSO
4
, s =
2.42 10 x
233
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
= 1.04 ? 10
?5
(mol L
?1
)
10 1
=
~
BaSO (s) Ba (aq) SO (aq)
2 2 ~
~~~
~~~
+
4 4
s s
K
sp
= [Ba
2+
] [SO
42?
]= s
2
= (1.04 ? 10
?5
)
2

= 1.08 ? 10
?10
mol
2
L
?2
147. Given van der Waals constant for NH
3
, H
2
, O
2

and CO
2
are respectively 4.17, 0.244, 1.36 and
3.59, which one of the following gases is most
easily liquefied?
(1) CO
2

(2) NH
3

(3) O
2

(4) H
2

Answer (2)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
100 10
pH = ?log[H
+
] =
1
~
~ ~
~ ~
log 10
~
~ = 1.0
149. On which of the following properties does the
coagulating power of an ion depend?
(1) The sign of charge on the ion alone
(2) The magnitude of the charge on the ion
alone
(3) Both magnitude and sign of the charge on
the ion
(4) Size of the ion alone
Answer (3)
Sol. ? Coagulation of colloidal solution by using
an electrolyte depends on the charge
present (positive or negative) on colloidal
particles as well as on its size.
? Coagulating power of an electrolyte
depends on the magnitude of charge
present on effective ion of electrolyte.
29
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
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NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
15
NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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NEET (UG) - 2018 (Code-NN) HLAAC
77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
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NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
19
NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
20
NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
21
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
22
NEET (UG) - 2018 (Code-NN) HLAAC
106. Which one is wrongly matched?
(1) Unicellular organism ? Chlorella
(2) Uniflagellate gametes ? Polysiphonia
(3) Gemma cups ? Marchantia
(4) Biflagellate zoospores ? Brown algae
Answer (2)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (1, 3 & 4) are correctly
matched
107. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Herbarium (i) It is a place having a
collection of preserved
plants and animals
b. Key

(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum

(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue (iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
a b c d
(1) (iii) (iv) (i) (ii)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iv) (iii) (i)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
108. Which one of the following plants shows a
very close relationship with a species of moth,
where none of the two can complete its life
cycle without the other?
(1) Viola (2) Hydrilla
(3) Banana (4) Yucca
Answer (4)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
109. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?160?C
(2) ?120?C
(3) ?196?C
(4) ?80?C
Answer (3)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C
(Cryopreservation)
110. In which of the following forms is iron
absorbed by plants?
(1) Both ferric and ferrous
(2) Ferric
(3) Free element
(4) Ferrous
Answer (2*)
Sol. Iron is absorbed by plants in the form of ferric
ions. (According to NCERT)
*Plants absorb iron in both form i.e. Fe
++
and
Fe
+++
. (Preferably Fe
++
)
111. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Calcium (2) Magnesium
(3) Potassium (4) Sodium
Answer (3)
Sol. Potassium helps in maintaining turgidity of
cells.
112. Double fertilization is
(1) Syngamy and triple fusion
(2) Fusion of two male gametes of a pollen
tube with two different eggs
(3) Fusion of two male gametes with one egg
(4) Fusion of one male gamete with two polar
nuclei
Answer (1)
23
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
113. What is the role of NAD
+
in cellular
respiration?
(1) It is the final electron acceptor for
anaerobic respiration.
(2) It functions as an enzyme.
(3) It is a nucleotide source for ATP synthesis.
(4) It functions as an electron carrier.
Answer (4)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
114. Oxygen is not produced during photosynthesis
by
(1) Chara
(2) Green sulphur bacteria
(3) Cycas
(4) Nostoc
Answer (2)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
115. The Golgi complex participates in
(1) Activation of amino acid
(2) Fatty acid breakdown
(3) Respiration in bacteria
(4) Formation of secretory vesicles
Answer (4)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
116. Stomatal movement is not affected by
(1) CO
2
concentration
(2) Temperature
(3) O
2
concentration
(4) Light
Answer (3)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
117. Stomata in grass leaf are
(1) Barrel shaped
(2) Dumb-bell shaped
(3) Rectangular
(4) Kidney shaped
Answer (2)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
118. Which of the following is true for nucleolus?
(1) It is a site for active ribosomal RNA
synthesis
(2) Larger nucleoli are present in dividing
cells
(3) It takes part in spindle formation
(4) It is a membrane-bound structure
Answer (1)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
119. Which of the following is not a product of light
reaction of photosynthesis?
(1) Oxygen (2) ATP
(3) NADPH (4) NADH
Answer (4)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
120. The stage during which separation of the
paired homologous chromosomes begins is
(1) Zygotene (2) Pachytene
(3) Diakinesis (4) Diplotene
Answer (4)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
121. The two functional groups characteristic of
sugars are
(1) Carbonyl and hydroxyl
(2) Hydroxyl and methyl
(3) Carbonyl and phosphate
(4) Carbonyl and methyl
Answer (1)
Sol. Sugar is a common term used to denote
carbohydrate.
Carbohydrates are polyhydroxy aldehyde,
ketone or their derivatives, which means they
have carbonyl and hydroxyl groups.
122. Which among the following is not a
prokaryote?
(1) Oscillatoria
(2) Saccharomyces
(3) Nostoc
(4) Mycobacterium
Answer (2)
24
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
123. Offsets are produced by
(1) Parthenogenesis (2) Meiotic divisions
(3) Parthenocarpy (4) Mitotic divisions
Answer (4)
Sol. Offset is a vegetative part of a plant, formed
by mitosis.
? Meiotic divisions do not occur in somatic
cells.
? Parthenogenesis is the formation of
embryo from ovum or egg without
fertilisation.
? Parthenocarpy is the fruit formed without
fertilisation, (generally seedless)
124. Select the correct statement
(1) Transduction was discovered by S. Altman
(2) Franklin Stahl coined the term ?linkage?
(3) Spliceosomes take part in translation
(4) Punnett square was developed by a British
scientist
Answer (4)
Sol. Punnett square was developed by a British
geneticist, Reginald C. Punnett.
? Franklin Stahl proved semi-conservative
mode of replication.
? Transduction was discovered by Zinder
and Laderberg.
? Spliceosome formation is part of post-
transcriptional change in Eukaryotes
125. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Sporopollenin (2) Pollenkitt
(3) Oil content (4) Cellulosic intine
Answer (1)
Sol. Sporopollenin cannot be degraded by
enzyme; strong acids and alkali, therefore it is
helpful in preserving pollen as fossil.
Pollenkitt ? Help in insect pollination.
Cellulosic Intine ? Inner sporoderm layer of
pollen grain known as intine made up
cellulose & pectin.
Oil content ? No role is pollen preservation.
126. Which of the following pairs is wrongly
matched?
(1) T.H. Morgan : Linkage
(2) Starch synthesis in pea : Multiple alleles
(3) XO type sex : Grasshopper
determination
(4) ABO blood grouping : Co-dominance
Answer (2)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (1, 3 & 4) are correctly
matched.
127. Select the correct match
(1) Francois Jacob and - Lac operon
Jacques Monod
(2) Alec Jeffreys - Streptococcus
pneumoniae
(3) Matthew Meselson - Pisum sativum
and F. Stahl
(4) Alfred Hershey and - TMV
Martha Chase
Answer (1)
Sol. Francois Jacob and Jacque Monod proposed
model of gene regulation known as operon
model/lac operon.
? Alec Jeffreys ? DNA fingerprinting
technique.
? Matthew Meselson and F. Stahl ? Semi-
conservative DNA replication in E. coli.
? Alfred Hershey and Martha Chase ?
Proved DNA as genetic material not
protein
128. Which of the following flowers only once in its
life-time?
(1) Papaya (2) Bamboo species
(3) Mango (4) Jackfruit
Answer (2)
Sol. Bamboo species are monocarpic i.e., flower
generally only once in its life-time after 50-
100 years.
Jackfruit, papaya and mango are polycarpic
i.e., produce flowers and fruits many times in
their life-time.
129. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Virus (2) Fungus
(3) Plant (4) Bacterium
Answer (4)
25
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
130. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) pBR 322 (2) Retrovirus
(3) X phage (4) Ti plasmid
Answer (2)
Sol. Retrovirus is commonly used as vector for
introducing a DNA fragment in human
lymphocyte.
Gene therapy : Lymphocyte from blood of
patient are grown in culture outside the body,
a functional gene is introduced by using a
retroviral vector, into these lymphocyte.
131. Select the correct match
(1) G. Mendel - Transformation
(2) Ribozyme - Nucleic acid
(3) T.H. Morgan - Transduction
(4) F
2
? Recessive parent - Dihybrid cross
Answer (2)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
132. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bioexploitation (2) Bio-infringement
(3) Biodegradation (4) Biopiracy
Answer (4)
Sol. Biopiracy is term used for or refer to the use
of bioresources by multinational companies
and other organisation without proper
authorisation from the countries and people
concerned with compensatory payment
(definition of biopiracy given in NCERT).
133. A ?new? variety of rice was patented by a
foreign company, though such varieties have
been present in India for a long time. This is
related to
(1) Basmati (2) Co-667
(3) Lerma Rojo (4) Sharbati Sonora
Answer (1)
Sol. In 1997, an American company got patent
rights on Basmati rice through the US patent
and trademark office that was actually been
derived from Indian farmer?s varieties.
The diversity of rice in India is one of the
richest in the world, 27 documented varieties
of Basmati are grown in India.
Indian basmati was crossed with semi-dwarf
varieties and claimed as an invention or a
novelty.
Sharbati Sonora and Lerma Rojo are varieties
of wheat.
134. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Genetic Engineering Appraisal Committee
(GEAC)
(2) Indian Council of Medical Research (ICMR)
(3) Research Committee on Genetic
Manipulation (RCGM)
(4) Council for Scientific and Industrial
Research (CSIR)
Answer (1)
Sol. Indian Government has setup organisation
such as GEAC (Genetic Engineering Appraisal
Committee) which will make decisions
regarding the validity of GM research and
safety of introducing GM-organism for public
services. (Direct from NCERT).
135. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Denaturation, Annealing, Extension
(2) Extension, Denaturation, Annealing
(3) Denaturation, Extension, Annealing
(4) Annealing, Extension, Denaturation
Answer (1)
Sol. This technique is used for making multiple
copies of gene (or DNA) of interest in vitro.
Each cycle has three steps
(i) Denaturation
(ii) Primer annealing
(iii) Extension of primer
26
136. Identify the major products P, Q and R in the
following sequence of reactions:
(i) O
2

P Q + R
(ii) H
8
O / A
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

P Q R
137. Which of the following compounds can form a
zwitterion?
(1) Glycine (2) Aniline
(3) Benzoic acid (4) Acetanilide
Answer (1)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
pK
,
= 9.60 pK
,
= 2.34
CH
3

CH ? CH
3

O
2

(P)
CH
3
CH
3
? CH ? CH
3

H /H O
+
2
Hydroperoxide
Rearrangement
OH
O
(R)
(Q)
3
HC?C?O?O?H
+ CH
3
? C ? CH
3

27
NEET (UG) - 2018 (Code-NN) HLAAC
OH
CH(CH
3
)
2
CH
3
? CO ? CH
3
(1)
CH
3
CH
2
? OH
(2)
(3)
(4)
Answer (1)
? ? ? ? ? ? ??
Cl
Sol
Cl Cl
CO
1, 2?H
Shift
6+ 6+ 6?
(Incipient carbocation)
Cl
Ni
CO
6?
OC
AlCl
3

CO
Now,
.
CH
3
CH
2
CH
2
? Cl + Al
CH
3
? CH ? CH
3

CH
2
CH
2
CH
3
CHO
CH
2
CH
2
CH
3
CHO COOH
CH(CH
3
)
2
OH
CH
3
CH
2
CH
2
Cl AlCl
3

, CH
3
CH(OH)CH
3
H
2
N ? CH
2
? COO
?

138. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Tetrahedral geometry and paramagnetic
(2) Square planar geometry and diamagnetic
(3) Square planar geometry and
paramagnetic
(4) Tetrahedral geometry and diamagnetic
Answer (4)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
sp -hybridisation
3
~
H
3
N ? CH
2
? COO
?
139. Iron carbonyl, Fe(CO)
5
is
(1) Dinuclear (2) Tetranuclear
(3) Trinuclear (4) Mononuclear
Answer (4)
Sol. ? Based on the number of metal atoms
present in a complex, they are classified
into mononuclear, dinuclear, trinuclear and
so on.
eg: Fe(CO)
5
: mononuclear
Co
2
(CO)
8
: dinuclear
Fe
3
(CO)
12
: trinuclear
NEET (UG) - 2018 (Code-NN) HLAAC
NO
.
NO
2

H V
~
H
(1)
V
(2)
~
(3)
H
V
~
NO
.
(4)
V H
~
NO
2
140. Match the metal ions given in Column I with
the spin magnetic moments of the ions given
in Column II and assign the correct code :
Column I Column II
a. Co
3+
i.
8
BM
b. Cr
3+
ii.
35
BM
c. Fe
3+
iii.
3
BM
d. Ni
2+
iv.
24
BM
v.
15
BM
c d
i ii
ii i
ii iii
iii iv
Answer (2)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Spin magnetic moment = 4(4 + 2) = 24 BM
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Spin magnetic moment = 3(3 + 2) = 15 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
Spin magnetic moment = 2(2 + 2) = 8 BM
141. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Linkage isomerism
(2) Geometrical isomerism
(3) Ionization isomerism
(4) Coordination isomerism
Answer (2)
Sol. In [CoCl
2
(en)
2
], Coordination number of Co
is 6 and this compound has octahedral
geometry.
? As per given option, type of isomerism is
geometrical isomerism.
142. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) MnO
4
2?
(2) CrO
4
2?

(3) MnO
4
?
(4) Cr
2
O
7
2?
Answer (1)
Sol. CrO
4
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
Cr
2
O
7
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
MnO
4
2?
= Mn
6+
= [Ar] 3d
1

Unpaired electron (n) = 1; Paramagnetic
MnO
4
?
= Mn
7+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
143. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) CH
3
? CH = CH ? CH
3

(2) HC ~ C ? C ~ CH
(3) CH
2
= CH ? CH = CH
2

(4) CH
2
= CH ? C ~ CH
Answer (4)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
144. Which of the following carbocations is
expected to be most stable?
Answer (3)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (3)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
a b
(1) iii v
(2) iv v
(3) iv i
(4) i ii
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
28
M
10 HCl + 100 mL
M
10 NaOH e. 100 mL
NEET (UG) - 2018 (Code-NN) HLAAC
145. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NR
2
> ? OR > ? F
(2) ? NH
2
< ? OR < ? F
(3) ? NH
2
> ? OR > ? F
(4) ? NR
2
< ? OR < ? F
Answer (2*)
Sol. ?I effect increases on increasing
electronegativity of atom. So, correct order of
?I effect is ?NH
2
< ? OR < ? F.
*Most appropriate Answer is option (2),
however option (4) may also be correct answer.
146. The solubility of BaSO
4
in water is 2.42 ? 10
?3
gL
?1

at 298 K. The value of its solubility product
(K
sp
) will be
(Given molar mass of BaSO
4
= 233 g mol
?1
)
(1) 1.08 ? 10
?8
mol
2
L
?2

(2) 1.08 ? 10
?10
mol
2
L
?2

(3) 1.08 ? 10
?14
mol
2
L
?2

(4) 1.08 ? 10
?12
mol
2
L
?2

Answer (2)
148. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M M
a. 60 mL 10 HCl + 40 mL 10 NaOH
M M
b. 55 mL 10 HCl + 45 mL 10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
pH of which one of them will be equal to 1?
(1) c (2) b
(3) d (4) a
Answer (1)
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
3
(mol L
?1
) Sol. Solubility of BaSO
4
, s =
2.42 10 x
233
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
= 1.04 ? 10
?5
(mol L
?1
)
10 1
=
~
BaSO (s) Ba (aq) SO (aq)
2 2 ~
~~~
~~~
+
4 4
s s
K
sp
= [Ba
2+
] [SO
42?
]= s
2
= (1.04 ? 10
?5
)
2

= 1.08 ? 10
?10
mol
2
L
?2
147. Given van der Waals constant for NH
3
, H
2
, O
2

and CO
2
are respectively 4.17, 0.244, 1.36 and
3.59, which one of the following gases is most
easily liquefied?
(1) CO
2

(2) NH
3

(3) O
2

(4) H
2

Answer (2)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
100 10
pH = ?log[H
+
] =
1
~
~ ~
~ ~
log 10
~
~ = 1.0
149. On which of the following properties does the
coagulating power of an ion depend?
(1) The sign of charge on the ion alone
(2) The magnitude of the charge on the ion
alone
(3) Both magnitude and sign of the charge on
the ion
(4) Size of the ion alone
Answer (3)
Sol. ? Coagulation of colloidal solution by using
an electrolyte depends on the charge
present (positive or negative) on colloidal
particles as well as on its size.
? Coagulating power of an electrolyte
depends on the magnitude of charge
present on effective ion of electrolyte.
29
NEET (UG) - 2018 (Code-NN) HLAAC
S S
? ?
F
? ?
S S
F Cl
S S
F
S S
(1) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
(2) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(3) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(4) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
Answer (1)
Sol.
C
2
H
5
OH
Na
C
2
H
5
O Na
+

(A) (B)
PCl
5

C
2
H
5
Cl
(C)
C
2
H
5
O Na + + C
2
H
5
Cl
S
N
2

C
2
H
5
OC
2
H 5
(B) (C)
150. Which of the following statements is not true
for halogens?
(1) Chlorine has the highest electron-gain
enthalpy
(2) All form monobasic oxyacids
(3) All but fluorine show positive oxidation
states
(4) All are oxidizing agents
Answer (3)
Sol. Due to high electronegativity and small size,
F forms only one oxoacid, HOF known as
Fluoric (I) acid. Oxidation number of F is +1 in
HOF.
151. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
(1) Cu (2) Fe
(3) Mg (4) Zn
Answer (3)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
152. The correct order of atomic radii in group 13
elements is
(1) B < Ga < Al < In < Tl
(2) B < Al < In < Ga < Tl
(3) B < Ga < Al < Tl < In
(4) B < Al < Ga < In < Tl
Answer (1)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
153. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) Three (2) One
(3) Four (4) Two
Answer (4)
Sol. The structure of ClF
3
is
The number of lone pair of electrons on
central Cl is 2.
154.The correct order of N-compounds in its
decreasing order of oxidation states is
(1) NH
4
Cl, N
2
, NO, HNO
3

(2) HNO
3
, NO, N
2
, NH
4
Cl
(3) HNO
3
, NH
4
Cl, NO, N
2

(4) HNO
3
, NO, NH
4
Cl, N
2

Answer (2)
+ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
155. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) In
(2) Ga
(3) B
(4) Al
Answer (3)
Sol. ? 'B' has no vacant d-orbitals in its valence
shell, so it can't extend its covalency beyond
4. i.e. 'B' cannot form the ion like MF
6
3(?)
i.e.
BF
6
3(?)
.
156. The compound A on treatment with Na gives
B, and with PCl
5
gives C. B and C react
together to give diethyl ether. A, B and C are
in the order
30
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Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
2. Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.
3. Rough work is to be done on the space provided for this purpose in the Test Booklet only.
4. On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before
leaving the Room / Hall. The candidates are allowed to take away this Test Booklet with them.
5. The CODE for this Booklet is NN.
6. The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on
the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
Regulations of this examination.
11. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.
12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
14
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
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NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
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NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
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NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
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NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
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NEET (UG) - 2018 (Code-NN) HLAAC
106. Which one is wrongly matched?
(1) Unicellular organism ? Chlorella
(2) Uniflagellate gametes ? Polysiphonia
(3) Gemma cups ? Marchantia
(4) Biflagellate zoospores ? Brown algae
Answer (2)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (1, 3 & 4) are correctly
matched
107. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Herbarium (i) It is a place having a
collection of preserved
plants and animals
b. Key

(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum

(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue (iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
a b c d
(1) (iii) (iv) (i) (ii)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iv) (iii) (i)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
108. Which one of the following plants shows a
very close relationship with a species of moth,
where none of the two can complete its life
cycle without the other?
(1) Viola (2) Hydrilla
(3) Banana (4) Yucca
Answer (4)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
109. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?160?C
(2) ?120?C
(3) ?196?C
(4) ?80?C
Answer (3)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C
(Cryopreservation)
110. In which of the following forms is iron
absorbed by plants?
(1) Both ferric and ferrous
(2) Ferric
(3) Free element
(4) Ferrous
Answer (2*)
Sol. Iron is absorbed by plants in the form of ferric
ions. (According to NCERT)
*Plants absorb iron in both form i.e. Fe
++
and
Fe
+++
. (Preferably Fe
++
)
111. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Calcium (2) Magnesium
(3) Potassium (4) Sodium
Answer (3)
Sol. Potassium helps in maintaining turgidity of
cells.
112. Double fertilization is
(1) Syngamy and triple fusion
(2) Fusion of two male gametes of a pollen
tube with two different eggs
(3) Fusion of two male gametes with one egg
(4) Fusion of one male gamete with two polar
nuclei
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
113. What is the role of NAD
+
in cellular
respiration?
(1) It is the final electron acceptor for
anaerobic respiration.
(2) It functions as an enzyme.
(3) It is a nucleotide source for ATP synthesis.
(4) It functions as an electron carrier.
Answer (4)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
114. Oxygen is not produced during photosynthesis
by
(1) Chara
(2) Green sulphur bacteria
(3) Cycas
(4) Nostoc
Answer (2)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
115. The Golgi complex participates in
(1) Activation of amino acid
(2) Fatty acid breakdown
(3) Respiration in bacteria
(4) Formation of secretory vesicles
Answer (4)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
116. Stomatal movement is not affected by
(1) CO
2
concentration
(2) Temperature
(3) O
2
concentration
(4) Light
Answer (3)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
117. Stomata in grass leaf are
(1) Barrel shaped
(2) Dumb-bell shaped
(3) Rectangular
(4) Kidney shaped
Answer (2)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
118. Which of the following is true for nucleolus?
(1) It is a site for active ribosomal RNA
synthesis
(2) Larger nucleoli are present in dividing
cells
(3) It takes part in spindle formation
(4) It is a membrane-bound structure
Answer (1)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
119. Which of the following is not a product of light
reaction of photosynthesis?
(1) Oxygen (2) ATP
(3) NADPH (4) NADH
Answer (4)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
120. The stage during which separation of the
paired homologous chromosomes begins is
(1) Zygotene (2) Pachytene
(3) Diakinesis (4) Diplotene
Answer (4)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
121. The two functional groups characteristic of
sugars are
(1) Carbonyl and hydroxyl
(2) Hydroxyl and methyl
(3) Carbonyl and phosphate
(4) Carbonyl and methyl
Answer (1)
Sol. Sugar is a common term used to denote
carbohydrate.
Carbohydrates are polyhydroxy aldehyde,
ketone or their derivatives, which means they
have carbonyl and hydroxyl groups.
122. Which among the following is not a
prokaryote?
(1) Oscillatoria
(2) Saccharomyces
(3) Nostoc
(4) Mycobacterium
Answer (2)
24
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
123. Offsets are produced by
(1) Parthenogenesis (2) Meiotic divisions
(3) Parthenocarpy (4) Mitotic divisions
Answer (4)
Sol. Offset is a vegetative part of a plant, formed
by mitosis.
? Meiotic divisions do not occur in somatic
cells.
? Parthenogenesis is the formation of
embryo from ovum or egg without
fertilisation.
? Parthenocarpy is the fruit formed without
fertilisation, (generally seedless)
124. Select the correct statement
(1) Transduction was discovered by S. Altman
(2) Franklin Stahl coined the term ?linkage?
(3) Spliceosomes take part in translation
(4) Punnett square was developed by a British
scientist
Answer (4)
Sol. Punnett square was developed by a British
geneticist, Reginald C. Punnett.
? Franklin Stahl proved semi-conservative
mode of replication.
? Transduction was discovered by Zinder
and Laderberg.
? Spliceosome formation is part of post-
transcriptional change in Eukaryotes
125. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Sporopollenin (2) Pollenkitt
(3) Oil content (4) Cellulosic intine
Answer (1)
Sol. Sporopollenin cannot be degraded by
enzyme; strong acids and alkali, therefore it is
helpful in preserving pollen as fossil.
Pollenkitt ? Help in insect pollination.
Cellulosic Intine ? Inner sporoderm layer of
pollen grain known as intine made up
cellulose & pectin.
Oil content ? No role is pollen preservation.
126. Which of the following pairs is wrongly
matched?
(1) T.H. Morgan : Linkage
(2) Starch synthesis in pea : Multiple alleles
(3) XO type sex : Grasshopper
determination
(4) ABO blood grouping : Co-dominance
Answer (2)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (1, 3 & 4) are correctly
matched.
127. Select the correct match
(1) Francois Jacob and - Lac operon
Jacques Monod
(2) Alec Jeffreys - Streptococcus
pneumoniae
(3) Matthew Meselson - Pisum sativum
and F. Stahl
(4) Alfred Hershey and - TMV
Martha Chase
Answer (1)
Sol. Francois Jacob and Jacque Monod proposed
model of gene regulation known as operon
model/lac operon.
? Alec Jeffreys ? DNA fingerprinting
technique.
? Matthew Meselson and F. Stahl ? Semi-
conservative DNA replication in E. coli.
? Alfred Hershey and Martha Chase ?
Proved DNA as genetic material not
protein
128. Which of the following flowers only once in its
life-time?
(1) Papaya (2) Bamboo species
(3) Mango (4) Jackfruit
Answer (2)
Sol. Bamboo species are monocarpic i.e., flower
generally only once in its life-time after 50-
100 years.
Jackfruit, papaya and mango are polycarpic
i.e., produce flowers and fruits many times in
their life-time.
129. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Virus (2) Fungus
(3) Plant (4) Bacterium
Answer (4)
25
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
130. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) pBR 322 (2) Retrovirus
(3) X phage (4) Ti plasmid
Answer (2)
Sol. Retrovirus is commonly used as vector for
introducing a DNA fragment in human
lymphocyte.
Gene therapy : Lymphocyte from blood of
patient are grown in culture outside the body,
a functional gene is introduced by using a
retroviral vector, into these lymphocyte.
131. Select the correct match
(1) G. Mendel - Transformation
(2) Ribozyme - Nucleic acid
(3) T.H. Morgan - Transduction
(4) F
2
? Recessive parent - Dihybrid cross
Answer (2)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
132. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bioexploitation (2) Bio-infringement
(3) Biodegradation (4) Biopiracy
Answer (4)
Sol. Biopiracy is term used for or refer to the use
of bioresources by multinational companies
and other organisation without proper
authorisation from the countries and people
concerned with compensatory payment
(definition of biopiracy given in NCERT).
133. A ?new? variety of rice was patented by a
foreign company, though such varieties have
been present in India for a long time. This is
related to
(1) Basmati (2) Co-667
(3) Lerma Rojo (4) Sharbati Sonora
Answer (1)
Sol. In 1997, an American company got patent
rights on Basmati rice through the US patent
and trademark office that was actually been
derived from Indian farmer?s varieties.
The diversity of rice in India is one of the
richest in the world, 27 documented varieties
of Basmati are grown in India.
Indian basmati was crossed with semi-dwarf
varieties and claimed as an invention or a
novelty.
Sharbati Sonora and Lerma Rojo are varieties
of wheat.
134. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Genetic Engineering Appraisal Committee
(GEAC)
(2) Indian Council of Medical Research (ICMR)
(3) Research Committee on Genetic
Manipulation (RCGM)
(4) Council for Scientific and Industrial
Research (CSIR)
Answer (1)
Sol. Indian Government has setup organisation
such as GEAC (Genetic Engineering Appraisal
Committee) which will make decisions
regarding the validity of GM research and
safety of introducing GM-organism for public
services. (Direct from NCERT).
135. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Denaturation, Annealing, Extension
(2) Extension, Denaturation, Annealing
(3) Denaturation, Extension, Annealing
(4) Annealing, Extension, Denaturation
Answer (1)
Sol. This technique is used for making multiple
copies of gene (or DNA) of interest in vitro.
Each cycle has three steps
(i) Denaturation
(ii) Primer annealing
(iii) Extension of primer
26
136. Identify the major products P, Q and R in the
following sequence of reactions:
(i) O
2

P Q + R
(ii) H
8
O / A
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

P Q R
137. Which of the following compounds can form a
zwitterion?
(1) Glycine (2) Aniline
(3) Benzoic acid (4) Acetanilide
Answer (1)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
pK
,
= 9.60 pK
,
= 2.34
CH
3

CH ? CH
3

O
2

(P)
CH
3
CH
3
? CH ? CH
3

H /H O
+
2
Hydroperoxide
Rearrangement
OH
O
(R)
(Q)
3
HC?C?O?O?H
+ CH
3
? C ? CH
3

27
NEET (UG) - 2018 (Code-NN) HLAAC
OH
CH(CH
3
)
2
CH
3
? CO ? CH
3
(1)
CH
3
CH
2
? OH
(2)
(3)
(4)
Answer (1)
? ? ? ? ? ? ??
Cl
Sol
Cl Cl
CO
1, 2?H
Shift
6+ 6+ 6?
(Incipient carbocation)
Cl
Ni
CO
6?
OC
AlCl
3

CO
Now,
.
CH
3
CH
2
CH
2
? Cl + Al
CH
3
? CH ? CH
3

CH
2
CH
2
CH
3
CHO
CH
2
CH
2
CH
3
CHO COOH
CH(CH
3
)
2
OH
CH
3
CH
2
CH
2
Cl AlCl
3

, CH
3
CH(OH)CH
3
H
2
N ? CH
2
? COO
?

138. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Tetrahedral geometry and paramagnetic
(2) Square planar geometry and diamagnetic
(3) Square planar geometry and
paramagnetic
(4) Tetrahedral geometry and diamagnetic
Answer (4)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
sp -hybridisation
3
~
H
3
N ? CH
2
? COO
?
139. Iron carbonyl, Fe(CO)
5
is
(1) Dinuclear (2) Tetranuclear
(3) Trinuclear (4) Mononuclear
Answer (4)
Sol. ? Based on the number of metal atoms
present in a complex, they are classified
into mononuclear, dinuclear, trinuclear and
so on.
eg: Fe(CO)
5
: mononuclear
Co
2
(CO)
8
: dinuclear
Fe
3
(CO)
12
: trinuclear
NEET (UG) - 2018 (Code-NN) HLAAC
NO
.
NO
2

H V
~
H
(1)
V
(2)
~
(3)
H
V
~
NO
.
(4)
V H
~
NO
2
140. Match the metal ions given in Column I with
the spin magnetic moments of the ions given
in Column II and assign the correct code :
Column I Column II
a. Co
3+
i.
8
BM
b. Cr
3+
ii.
35
BM
c. Fe
3+
iii.
3
BM
d. Ni
2+
iv.
24
BM
v.
15
BM
c d
i ii
ii i
ii iii
iii iv
Answer (2)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Spin magnetic moment = 4(4 + 2) = 24 BM
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Spin magnetic moment = 3(3 + 2) = 15 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
Spin magnetic moment = 2(2 + 2) = 8 BM
141. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Linkage isomerism
(2) Geometrical isomerism
(3) Ionization isomerism
(4) Coordination isomerism
Answer (2)
Sol. In [CoCl
2
(en)
2
], Coordination number of Co
is 6 and this compound has octahedral
geometry.
? As per given option, type of isomerism is
geometrical isomerism.
142. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) MnO
4
2?
(2) CrO
4
2?

(3) MnO
4
?
(4) Cr
2
O
7
2?
Answer (1)
Sol. CrO
4
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
Cr
2
O
7
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
MnO
4
2?
= Mn
6+
= [Ar] 3d
1

Unpaired electron (n) = 1; Paramagnetic
MnO
4
?
= Mn
7+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
143. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) CH
3
? CH = CH ? CH
3

(2) HC ~ C ? C ~ CH
(3) CH
2
= CH ? CH = CH
2

(4) CH
2
= CH ? C ~ CH
Answer (4)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
144. Which of the following carbocations is
expected to be most stable?
Answer (3)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (3)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
a b
(1) iii v
(2) iv v
(3) iv i
(4) i ii
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
28
M
10 HCl + 100 mL
M
10 NaOH e. 100 mL
NEET (UG) - 2018 (Code-NN) HLAAC
145. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NR
2
> ? OR > ? F
(2) ? NH
2
< ? OR < ? F
(3) ? NH
2
> ? OR > ? F
(4) ? NR
2
< ? OR < ? F
Answer (2*)
Sol. ?I effect increases on increasing
electronegativity of atom. So, correct order of
?I effect is ?NH
2
< ? OR < ? F.
*Most appropriate Answer is option (2),
however option (4) may also be correct answer.
146. The solubility of BaSO
4
in water is 2.42 ? 10
?3
gL
?1

at 298 K. The value of its solubility product
(K
sp
) will be
(Given molar mass of BaSO
4
= 233 g mol
?1
)
(1) 1.08 ? 10
?8
mol
2
L
?2

(2) 1.08 ? 10
?10
mol
2
L
?2

(3) 1.08 ? 10
?14
mol
2
L
?2

(4) 1.08 ? 10
?12
mol
2
L
?2

Answer (2)
148. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M M
a. 60 mL 10 HCl + 40 mL 10 NaOH
M M
b. 55 mL 10 HCl + 45 mL 10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
pH of which one of them will be equal to 1?
(1) c (2) b
(3) d (4) a
Answer (1)
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
3
(mol L
?1
) Sol. Solubility of BaSO
4
, s =
2.42 10 x
233
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
= 1.04 ? 10
?5
(mol L
?1
)
10 1
=
~
BaSO (s) Ba (aq) SO (aq)
2 2 ~
~~~
~~~
+
4 4
s s
K
sp
= [Ba
2+
] [SO
42?
]= s
2
= (1.04 ? 10
?5
)
2

= 1.08 ? 10
?10
mol
2
L
?2
147. Given van der Waals constant for NH
3
, H
2
, O
2

and CO
2
are respectively 4.17, 0.244, 1.36 and
3.59, which one of the following gases is most
easily liquefied?
(1) CO
2

(2) NH
3

(3) O
2

(4) H
2

Answer (2)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
100 10
pH = ?log[H
+
] =
1
~
~ ~
~ ~
log 10
~
~ = 1.0
149. On which of the following properties does the
coagulating power of an ion depend?
(1) The sign of charge on the ion alone
(2) The magnitude of the charge on the ion
alone
(3) Both magnitude and sign of the charge on
the ion
(4) Size of the ion alone
Answer (3)
Sol. ? Coagulation of colloidal solution by using
an electrolyte depends on the charge
present (positive or negative) on colloidal
particles as well as on its size.
? Coagulating power of an electrolyte
depends on the magnitude of charge
present on effective ion of electrolyte.
29
NEET (UG) - 2018 (Code-NN) HLAAC
S S
? ?
F
? ?
S S
F Cl
S S
F
S S
(1) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
(2) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(3) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(4) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
Answer (1)
Sol.
C
2
H
5
OH
Na
C
2
H
5
O Na
+

(A) (B)
PCl
5

C
2
H
5
Cl
(C)
C
2
H
5
O Na + + C
2
H
5
Cl
S
N
2

C
2
H
5
OC
2
H 5
(B) (C)
150. Which of the following statements is not true
for halogens?
(1) Chlorine has the highest electron-gain
enthalpy
(2) All form monobasic oxyacids
(3) All but fluorine show positive oxidation
states
(4) All are oxidizing agents
Answer (3)
Sol. Due to high electronegativity and small size,
F forms only one oxoacid, HOF known as
Fluoric (I) acid. Oxidation number of F is +1 in
HOF.
151. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
(1) Cu (2) Fe
(3) Mg (4) Zn
Answer (3)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
152. The correct order of atomic radii in group 13
elements is
(1) B < Ga < Al < In < Tl
(2) B < Al < In < Ga < Tl
(3) B < Ga < Al < Tl < In
(4) B < Al < Ga < In < Tl
Answer (1)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
153. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) Three (2) One
(3) Four (4) Two
Answer (4)
Sol. The structure of ClF
3
is
The number of lone pair of electrons on
central Cl is 2.
154.The correct order of N-compounds in its
decreasing order of oxidation states is
(1) NH
4
Cl, N
2
, NO, HNO
3

(2) HNO
3
, NO, N
2
, NH
4
Cl
(3) HNO
3
, NH
4
Cl, NO, N
2

(4) HNO
3
, NO, NH
4
Cl, N
2

Answer (2)
+ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
155. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) In
(2) Ga
(3) B
(4) Al
Answer (3)
Sol. ? 'B' has no vacant d-orbitals in its valence
shell, so it can't extend its covalency beyond
4. i.e. 'B' cannot form the ion like MF
6
3(?)
i.e.
BF
6
3(?)
.
156. The compound A on treatment with Na gives
B, and with PCl
5
gives C. B and C react
together to give diethyl ether. A, B and C are
in the order
30
NEET (UG) - 2018 (Code-NN) HLAAC
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
1
,
k[A ]
0
157. Hydrocarbon (A) reacts with bromine by
substitution to form an alkyl bromide which by
Wurtz reaction is converted to gaseous
hydrocarbon containing less than four carbon
atoms. (A) is
(1) CH
4
(2) CH ~ CH
(3) CH
3
? CH
3
(4) CH
2
= CH
2

Answer (1)
Sol.
CH
4
Br
2
/h v
CH
3
Br
Na/dry ether
Wurtz reaction
CH
3
? CH
3

158. The compound C
7
H
8
undergoes the following
reactions:
3Cl
2
/ A Br
2
/Fe Zn/HCl
C
7
H
8
~~~> ~~~> ~~~~ A B C
The product 'C' is
(1) p-bromotoluene
(2) m-bromotoluene
(3) 3-bromo-2,4,6-trichlorotoluene
(4) o-bromotoluene
Answer (2)
CCl
3

3dl
2
A
(C
7
H
8
)
160. The correct difference between first and
second order reactions is that
(1) The rate of a first-order reaction does
depend on reactant concentrations; the
rate of a second-order reaction does not
depend on reactant concentrations
(2) The rate of a first-order reaction does not
depend on reactant concentrations; the
rate of a second-order reaction does
depend on reactant concentrations
(3) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(4) The half-life of a first-order reaction does
not depend on [A]
0
; the half-life of a
second-order reaction does depend on
[A]
0

Answer (4)
,
k
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
161. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BaH
2
< BeH
2
< CaH
2

(2) BeH
2
< CaH
2
< BaH
2

(3) BeH
2
< BaH
2
< CaH
2

(4) CaH
2
< BeH
2
< BaH
2

Answer (2)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
162. Consider the change in oxidation state of
Bromine corresponding to different emf values
as shown in the diagram below :
Then the species undergoing
disproportionation is
(1) HBrO (2)
~
BrO
3

(3) Br
2
(4)
~
BrO
4
(A)
CH
3
CCl
3

Sol.
Br
2
Fe
Br
(A) (B)
Zn HCl
CH
3
Br
(C)
159. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) NO
(2) N
2
O
5
(3) N
2
O
(4) NO
2

Answer (2)
Sol. Fact
Sol. ? For first order reaction,
1 /2
=
0.693
t
31
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
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NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
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NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
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NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
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NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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NEET (UG) - 2018 (Code-NN) HLAAC
77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
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NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
19
NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
20
NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
21
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
22
NEET (UG) - 2018 (Code-NN) HLAAC
106. Which one is wrongly matched?
(1) Unicellular organism ? Chlorella
(2) Uniflagellate gametes ? Polysiphonia
(3) Gemma cups ? Marchantia
(4) Biflagellate zoospores ? Brown algae
Answer (2)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (1, 3 & 4) are correctly
matched
107. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Herbarium (i) It is a place having a
collection of preserved
plants and animals
b. Key

(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum

(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue (iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
a b c d
(1) (iii) (iv) (i) (ii)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iv) (iii) (i)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
108. Which one of the following plants shows a
very close relationship with a species of moth,
where none of the two can complete its life
cycle without the other?
(1) Viola (2) Hydrilla
(3) Banana (4) Yucca
Answer (4)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
109. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?160?C
(2) ?120?C
(3) ?196?C
(4) ?80?C
Answer (3)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C
(Cryopreservation)
110. In which of the following forms is iron
absorbed by plants?
(1) Both ferric and ferrous
(2) Ferric
(3) Free element
(4) Ferrous
Answer (2*)
Sol. Iron is absorbed by plants in the form of ferric
ions. (According to NCERT)
*Plants absorb iron in both form i.e. Fe
++
and
Fe
+++
. (Preferably Fe
++
)
111. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Calcium (2) Magnesium
(3) Potassium (4) Sodium
Answer (3)
Sol. Potassium helps in maintaining turgidity of
cells.
112. Double fertilization is
(1) Syngamy and triple fusion
(2) Fusion of two male gametes of a pollen
tube with two different eggs
(3) Fusion of two male gametes with one egg
(4) Fusion of one male gamete with two polar
nuclei
Answer (1)
23
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
113. What is the role of NAD
+
in cellular
respiration?
(1) It is the final electron acceptor for
anaerobic respiration.
(2) It functions as an enzyme.
(3) It is a nucleotide source for ATP synthesis.
(4) It functions as an electron carrier.
Answer (4)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
114. Oxygen is not produced during photosynthesis
by
(1) Chara
(2) Green sulphur bacteria
(3) Cycas
(4) Nostoc
Answer (2)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
115. The Golgi complex participates in
(1) Activation of amino acid
(2) Fatty acid breakdown
(3) Respiration in bacteria
(4) Formation of secretory vesicles
Answer (4)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
116. Stomatal movement is not affected by
(1) CO
2
concentration
(2) Temperature
(3) O
2
concentration
(4) Light
Answer (3)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
117. Stomata in grass leaf are
(1) Barrel shaped
(2) Dumb-bell shaped
(3) Rectangular
(4) Kidney shaped
Answer (2)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
118. Which of the following is true for nucleolus?
(1) It is a site for active ribosomal RNA
synthesis
(2) Larger nucleoli are present in dividing
cells
(3) It takes part in spindle formation
(4) It is a membrane-bound structure
Answer (1)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
119. Which of the following is not a product of light
reaction of photosynthesis?
(1) Oxygen (2) ATP
(3) NADPH (4) NADH
Answer (4)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
120. The stage during which separation of the
paired homologous chromosomes begins is
(1) Zygotene (2) Pachytene
(3) Diakinesis (4) Diplotene
Answer (4)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
121. The two functional groups characteristic of
sugars are
(1) Carbonyl and hydroxyl
(2) Hydroxyl and methyl
(3) Carbonyl and phosphate
(4) Carbonyl and methyl
Answer (1)
Sol. Sugar is a common term used to denote
carbohydrate.
Carbohydrates are polyhydroxy aldehyde,
ketone or their derivatives, which means they
have carbonyl and hydroxyl groups.
122. Which among the following is not a
prokaryote?
(1) Oscillatoria
(2) Saccharomyces
(3) Nostoc
(4) Mycobacterium
Answer (2)
24
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
123. Offsets are produced by
(1) Parthenogenesis (2) Meiotic divisions
(3) Parthenocarpy (4) Mitotic divisions
Answer (4)
Sol. Offset is a vegetative part of a plant, formed
by mitosis.
? Meiotic divisions do not occur in somatic
cells.
? Parthenogenesis is the formation of
embryo from ovum or egg without
fertilisation.
? Parthenocarpy is the fruit formed without
fertilisation, (generally seedless)
124. Select the correct statement
(1) Transduction was discovered by S. Altman
(2) Franklin Stahl coined the term ?linkage?
(3) Spliceosomes take part in translation
(4) Punnett square was developed by a British
scientist
Answer (4)
Sol. Punnett square was developed by a British
geneticist, Reginald C. Punnett.
? Franklin Stahl proved semi-conservative
mode of replication.
? Transduction was discovered by Zinder
and Laderberg.
? Spliceosome formation is part of post-
transcriptional change in Eukaryotes
125. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Sporopollenin (2) Pollenkitt
(3) Oil content (4) Cellulosic intine
Answer (1)
Sol. Sporopollenin cannot be degraded by
enzyme; strong acids and alkali, therefore it is
helpful in preserving pollen as fossil.
Pollenkitt ? Help in insect pollination.
Cellulosic Intine ? Inner sporoderm layer of
pollen grain known as intine made up
cellulose & pectin.
Oil content ? No role is pollen preservation.
126. Which of the following pairs is wrongly
matched?
(1) T.H. Morgan : Linkage
(2) Starch synthesis in pea : Multiple alleles
(3) XO type sex : Grasshopper
determination
(4) ABO blood grouping : Co-dominance
Answer (2)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (1, 3 & 4) are correctly
matched.
127. Select the correct match
(1) Francois Jacob and - Lac operon
Jacques Monod
(2) Alec Jeffreys - Streptococcus
pneumoniae
(3) Matthew Meselson - Pisum sativum
and F. Stahl
(4) Alfred Hershey and - TMV
Martha Chase
Answer (1)
Sol. Francois Jacob and Jacque Monod proposed
model of gene regulation known as operon
model/lac operon.
? Alec Jeffreys ? DNA fingerprinting
technique.
? Matthew Meselson and F. Stahl ? Semi-
conservative DNA replication in E. coli.
? Alfred Hershey and Martha Chase ?
Proved DNA as genetic material not
protein
128. Which of the following flowers only once in its
life-time?
(1) Papaya (2) Bamboo species
(3) Mango (4) Jackfruit
Answer (2)
Sol. Bamboo species are monocarpic i.e., flower
generally only once in its life-time after 50-
100 years.
Jackfruit, papaya and mango are polycarpic
i.e., produce flowers and fruits many times in
their life-time.
129. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Virus (2) Fungus
(3) Plant (4) Bacterium
Answer (4)
25
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
130. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) pBR 322 (2) Retrovirus
(3) X phage (4) Ti plasmid
Answer (2)
Sol. Retrovirus is commonly used as vector for
introducing a DNA fragment in human
lymphocyte.
Gene therapy : Lymphocyte from blood of
patient are grown in culture outside the body,
a functional gene is introduced by using a
retroviral vector, into these lymphocyte.
131. Select the correct match
(1) G. Mendel - Transformation
(2) Ribozyme - Nucleic acid
(3) T.H. Morgan - Transduction
(4) F
2
? Recessive parent - Dihybrid cross
Answer (2)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
132. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bioexploitation (2) Bio-infringement
(3) Biodegradation (4) Biopiracy
Answer (4)
Sol. Biopiracy is term used for or refer to the use
of bioresources by multinational companies
and other organisation without proper
authorisation from the countries and people
concerned with compensatory payment
(definition of biopiracy given in NCERT).
133. A ?new? variety of rice was patented by a
foreign company, though such varieties have
been present in India for a long time. This is
related to
(1) Basmati (2) Co-667
(3) Lerma Rojo (4) Sharbati Sonora
Answer (1)
Sol. In 1997, an American company got patent
rights on Basmati rice through the US patent
and trademark office that was actually been
derived from Indian farmer?s varieties.
The diversity of rice in India is one of the
richest in the world, 27 documented varieties
of Basmati are grown in India.
Indian basmati was crossed with semi-dwarf
varieties and claimed as an invention or a
novelty.
Sharbati Sonora and Lerma Rojo are varieties
of wheat.
134. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Genetic Engineering Appraisal Committee
(GEAC)
(2) Indian Council of Medical Research (ICMR)
(3) Research Committee on Genetic
Manipulation (RCGM)
(4) Council for Scientific and Industrial
Research (CSIR)
Answer (1)
Sol. Indian Government has setup organisation
such as GEAC (Genetic Engineering Appraisal
Committee) which will make decisions
regarding the validity of GM research and
safety of introducing GM-organism for public
services. (Direct from NCERT).
135. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Denaturation, Annealing, Extension
(2) Extension, Denaturation, Annealing
(3) Denaturation, Extension, Annealing
(4) Annealing, Extension, Denaturation
Answer (1)
Sol. This technique is used for making multiple
copies of gene (or DNA) of interest in vitro.
Each cycle has three steps
(i) Denaturation
(ii) Primer annealing
(iii) Extension of primer
26
136. Identify the major products P, Q and R in the
following sequence of reactions:
(i) O
2

P Q + R
(ii) H
8
O / A
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

P Q R
137. Which of the following compounds can form a
zwitterion?
(1) Glycine (2) Aniline
(3) Benzoic acid (4) Acetanilide
Answer (1)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
pK
,
= 9.60 pK
,
= 2.34
CH
3

CH ? CH
3

O
2

(P)
CH
3
CH
3
? CH ? CH
3

H /H O
+
2
Hydroperoxide
Rearrangement
OH
O
(R)
(Q)
3
HC?C?O?O?H
+ CH
3
? C ? CH
3

27
NEET (UG) - 2018 (Code-NN) HLAAC
OH
CH(CH
3
)
2
CH
3
? CO ? CH
3
(1)
CH
3
CH
2
? OH
(2)
(3)
(4)
Answer (1)
? ? ? ? ? ? ??
Cl
Sol
Cl Cl
CO
1, 2?H
Shift
6+ 6+ 6?
(Incipient carbocation)
Cl
Ni
CO
6?
OC
AlCl
3

CO
Now,
.
CH
3
CH
2
CH
2
? Cl + Al
CH
3
? CH ? CH
3

CH
2
CH
2
CH
3
CHO
CH
2
CH
2
CH
3
CHO COOH
CH(CH
3
)
2
OH
CH
3
CH
2
CH
2
Cl AlCl
3

, CH
3
CH(OH)CH
3
H
2
N ? CH
2
? COO
?

138. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Tetrahedral geometry and paramagnetic
(2) Square planar geometry and diamagnetic
(3) Square planar geometry and
paramagnetic
(4) Tetrahedral geometry and diamagnetic
Answer (4)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
sp -hybridisation
3
~
H
3
N ? CH
2
? COO
?
139. Iron carbonyl, Fe(CO)
5
is
(1) Dinuclear (2) Tetranuclear
(3) Trinuclear (4) Mononuclear
Answer (4)
Sol. ? Based on the number of metal atoms
present in a complex, they are classified
into mononuclear, dinuclear, trinuclear and
so on.
eg: Fe(CO)
5
: mononuclear
Co
2
(CO)
8
: dinuclear
Fe
3
(CO)
12
: trinuclear
NEET (UG) - 2018 (Code-NN) HLAAC
NO
.
NO
2

H V
~
H
(1)
V
(2)
~
(3)
H
V
~
NO
.
(4)
V H
~
NO
2
140. Match the metal ions given in Column I with
the spin magnetic moments of the ions given
in Column II and assign the correct code :
Column I Column II
a. Co
3+
i.
8
BM
b. Cr
3+
ii.
35
BM
c. Fe
3+
iii.
3
BM
d. Ni
2+
iv.
24
BM
v.
15
BM
c d
i ii
ii i
ii iii
iii iv
Answer (2)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Spin magnetic moment = 4(4 + 2) = 24 BM
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Spin magnetic moment = 3(3 + 2) = 15 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
Spin magnetic moment = 2(2 + 2) = 8 BM
141. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Linkage isomerism
(2) Geometrical isomerism
(3) Ionization isomerism
(4) Coordination isomerism
Answer (2)
Sol. In [CoCl
2
(en)
2
], Coordination number of Co
is 6 and this compound has octahedral
geometry.
? As per given option, type of isomerism is
geometrical isomerism.
142. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) MnO
4
2?
(2) CrO
4
2?

(3) MnO
4
?
(4) Cr
2
O
7
2?
Answer (1)
Sol. CrO
4
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
Cr
2
O
7
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
MnO
4
2?
= Mn
6+
= [Ar] 3d
1

Unpaired electron (n) = 1; Paramagnetic
MnO
4
?
= Mn
7+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
143. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) CH
3
? CH = CH ? CH
3

(2) HC ~ C ? C ~ CH
(3) CH
2
= CH ? CH = CH
2

(4) CH
2
= CH ? C ~ CH
Answer (4)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
144. Which of the following carbocations is
expected to be most stable?
Answer (3)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (3)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
a b
(1) iii v
(2) iv v
(3) iv i
(4) i ii
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
28
M
10 HCl + 100 mL
M
10 NaOH e. 100 mL
NEET (UG) - 2018 (Code-NN) HLAAC
145. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NR
2
> ? OR > ? F
(2) ? NH
2
< ? OR < ? F
(3) ? NH
2
> ? OR > ? F
(4) ? NR
2
< ? OR < ? F
Answer (2*)
Sol. ?I effect increases on increasing
electronegativity of atom. So, correct order of
?I effect is ?NH
2
< ? OR < ? F.
*Most appropriate Answer is option (2),
however option (4) may also be correct answer.
146. The solubility of BaSO
4
in water is 2.42 ? 10
?3
gL
?1

at 298 K. The value of its solubility product
(K
sp
) will be
(Given molar mass of BaSO
4
= 233 g mol
?1
)
(1) 1.08 ? 10
?8
mol
2
L
?2

(2) 1.08 ? 10
?10
mol
2
L
?2

(3) 1.08 ? 10
?14
mol
2
L
?2

(4) 1.08 ? 10
?12
mol
2
L
?2

Answer (2)
148. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M M
a. 60 mL 10 HCl + 40 mL 10 NaOH
M M
b. 55 mL 10 HCl + 45 mL 10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
pH of which one of them will be equal to 1?
(1) c (2) b
(3) d (4) a
Answer (1)
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
3
(mol L
?1
) Sol. Solubility of BaSO
4
, s =
2.42 10 x
233
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
= 1.04 ? 10
?5
(mol L
?1
)
10 1
=
~
BaSO (s) Ba (aq) SO (aq)
2 2 ~
~~~
~~~
+
4 4
s s
K
sp
= [Ba
2+
] [SO
42?
]= s
2
= (1.04 ? 10
?5
)
2

= 1.08 ? 10
?10
mol
2
L
?2
147. Given van der Waals constant for NH
3
, H
2
, O
2

and CO
2
are respectively 4.17, 0.244, 1.36 and
3.59, which one of the following gases is most
easily liquefied?
(1) CO
2

(2) NH
3

(3) O
2

(4) H
2

Answer (2)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
100 10
pH = ?log[H
+
] =
1
~
~ ~
~ ~
log 10
~
~ = 1.0
149. On which of the following properties does the
coagulating power of an ion depend?
(1) The sign of charge on the ion alone
(2) The magnitude of the charge on the ion
alone
(3) Both magnitude and sign of the charge on
the ion
(4) Size of the ion alone
Answer (3)
Sol. ? Coagulation of colloidal solution by using
an electrolyte depends on the charge
present (positive or negative) on colloidal
particles as well as on its size.
? Coagulating power of an electrolyte
depends on the magnitude of charge
present on effective ion of electrolyte.
29
NEET (UG) - 2018 (Code-NN) HLAAC
S S
? ?
F
? ?
S S
F Cl
S S
F
S S
(1) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
(2) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(3) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(4) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
Answer (1)
Sol.
C
2
H
5
OH
Na
C
2
H
5
O Na
+

(A) (B)
PCl
5

C
2
H
5
Cl
(C)
C
2
H
5
O Na + + C
2
H
5
Cl
S
N
2

C
2
H
5
OC
2
H 5
(B) (C)
150. Which of the following statements is not true
for halogens?
(1) Chlorine has the highest electron-gain
enthalpy
(2) All form monobasic oxyacids
(3) All but fluorine show positive oxidation
states
(4) All are oxidizing agents
Answer (3)
Sol. Due to high electronegativity and small size,
F forms only one oxoacid, HOF known as
Fluoric (I) acid. Oxidation number of F is +1 in
HOF.
151. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
(1) Cu (2) Fe
(3) Mg (4) Zn
Answer (3)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
152. The correct order of atomic radii in group 13
elements is
(1) B < Ga < Al < In < Tl
(2) B < Al < In < Ga < Tl
(3) B < Ga < Al < Tl < In
(4) B < Al < Ga < In < Tl
Answer (1)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
153. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) Three (2) One
(3) Four (4) Two
Answer (4)
Sol. The structure of ClF
3
is
The number of lone pair of electrons on
central Cl is 2.
154.The correct order of N-compounds in its
decreasing order of oxidation states is
(1) NH
4
Cl, N
2
, NO, HNO
3

(2) HNO
3
, NO, N
2
, NH
4
Cl
(3) HNO
3
, NH
4
Cl, NO, N
2

(4) HNO
3
, NO, NH
4
Cl, N
2

Answer (2)
+ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
155. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) In
(2) Ga
(3) B
(4) Al
Answer (3)
Sol. ? 'B' has no vacant d-orbitals in its valence
shell, so it can't extend its covalency beyond
4. i.e. 'B' cannot form the ion like MF
6
3(?)
i.e.
BF
6
3(?)
.
156. The compound A on treatment with Na gives
B, and with PCl
5
gives C. B and C react
together to give diethyl ether. A, B and C are
in the order
30
NEET (UG) - 2018 (Code-NN) HLAAC
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
1
,
k[A ]
0
157. Hydrocarbon (A) reacts with bromine by
substitution to form an alkyl bromide which by
Wurtz reaction is converted to gaseous
hydrocarbon containing less than four carbon
atoms. (A) is
(1) CH
4
(2) CH ~ CH
(3) CH
3
? CH
3
(4) CH
2
= CH
2

Answer (1)
Sol.
CH
4
Br
2
/h v
CH
3
Br
Na/dry ether
Wurtz reaction
CH
3
? CH
3

158. The compound C
7
H
8
undergoes the following
reactions:
3Cl
2
/ A Br
2
/Fe Zn/HCl
C
7
H
8
~~~> ~~~> ~~~~ A B C
The product 'C' is
(1) p-bromotoluene
(2) m-bromotoluene
(3) 3-bromo-2,4,6-trichlorotoluene
(4) o-bromotoluene
Answer (2)
CCl
3

3dl
2
A
(C
7
H
8
)
160. The correct difference between first and
second order reactions is that
(1) The rate of a first-order reaction does
depend on reactant concentrations; the
rate of a second-order reaction does not
depend on reactant concentrations
(2) The rate of a first-order reaction does not
depend on reactant concentrations; the
rate of a second-order reaction does
depend on reactant concentrations
(3) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(4) The half-life of a first-order reaction does
not depend on [A]
0
; the half-life of a
second-order reaction does depend on
[A]
0

Answer (4)
,
k
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
161. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BaH
2
< BeH
2
< CaH
2

(2) BeH
2
< CaH
2
< BaH
2

(3) BeH
2
< BaH
2
< CaH
2

(4) CaH
2
< BeH
2
< BaH
2

Answer (2)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
162. Consider the change in oxidation state of
Bromine corresponding to different emf values
as shown in the diagram below :
Then the species undergoing
disproportionation is
(1) HBrO (2)
~
BrO
3

(3) Br
2
(4)
~
BrO
4
(A)
CH
3
CCl
3

Sol.
Br
2
Fe
Br
(A) (B)
Zn HCl
CH
3
Br
(C)
159. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) NO
(2) N
2
O
5
(3) N
2
O
(4) NO
2

Answer (2)
Sol. Fact
Sol. ? For first order reaction,
1 /2
=
0.693
t
31
NEET (UG) - 2018 (Code-NN) HLAAC
CH
3

(1) CH
3
OH and I
2

(2)
H
3
C CH
2
? OH and I
2
COONa
Sodium benzoate
Answer (1)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+5
~~* BrO , E
~
~ o
HBrO
3
BrO /HBrO
= 1.5 V
3
E for the disproportionation of HBrO,
o
cell
E
o
= E
o o
cell
~ E
~ HBrO/Br
2 BrO /HBrO
3
= 1.595 ? 1.5
= 0.095 V = + ve
163. In which case is number of molecules of
water maximum?
(1) 10
?3
mol of water
(2) 18 mL of water
(3) 0.00224 L of water vapours at 1 atm and
273 K
(4) 0.18 g of water
Answer (2)
Sol. (1) Molecules of water = mole ? N
A
= 10
?3
N
A

(2) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(4) Molecules of water = mole ? N
A
= A
18
= 10
?2
N
A

164. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intermolecular H-bonding
(2) Formation of intramolecular H-bonding
(3) More extensive association of carboxylic
acid via van der Waals force of attraction
(4) Formation of carboxylate ion
Answer (1)
Sol. Due to formation of intermolecular H-bonding
in carboxylic acid, association occurs. Hence
boiling point increases and become more
than the boiling point of aldehydes, ketones
and alcohols of comparable molecular
masses.
165. Compound A,
C
8
H
10
O,
is found to react with
NaOI (produced by reacting Y with NaOH) and
yields a yellow precipitate with characteristic
smell.
A and Y are respectively
CH ? CH
3
and I
2
(3)
OH
2 (4)
CH
2
? CH
2
? OH and I
Answer (3)
Sol. Option (3) is secondary alcohol which on
oxidation gives phenylmethyl ketone
(Acetophenone). This on reaction with I
2
and
NaOH form iodoform and sodium benzoate.
2NaOH + I -* NaOI + NaI + H
2
O
2
I
2
NaOH
166. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichlorocarbene ( : CCl
2
)
(2) Dichloromethyl cation
(
CHCl
2
)
~
(3) Dichloromethyl anion (
CHCl
2
)
?
(4) Formyl cation (
C H O
)
~
+ CHI
3

Iodoform
(Yellow PPt)
CHO
= N
A

0.00224
22.4
NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
= 10
?4

(3) Moles of water =
32
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one mark will be deducted from the total scores. The maximum marks are 720.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
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in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
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NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
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65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
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NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
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NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
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NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
22
NEET (UG) - 2018 (Code-NN) HLAAC
106. Which one is wrongly matched?
(1) Unicellular organism ? Chlorella
(2) Uniflagellate gametes ? Polysiphonia
(3) Gemma cups ? Marchantia
(4) Biflagellate zoospores ? Brown algae
Answer (2)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (1, 3 & 4) are correctly
matched
107. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Herbarium (i) It is a place having a
collection of preserved
plants and animals
b. Key

(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum

(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue (iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
a b c d
(1) (iii) (iv) (i) (ii)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iv) (iii) (i)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
108. Which one of the following plants shows a
very close relationship with a species of moth,
where none of the two can complete its life
cycle without the other?
(1) Viola (2) Hydrilla
(3) Banana (4) Yucca
Answer (4)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
109. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?160?C
(2) ?120?C
(3) ?196?C
(4) ?80?C
Answer (3)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C
(Cryopreservation)
110. In which of the following forms is iron
absorbed by plants?
(1) Both ferric and ferrous
(2) Ferric
(3) Free element
(4) Ferrous
Answer (2*)
Sol. Iron is absorbed by plants in the form of ferric
ions. (According to NCERT)
*Plants absorb iron in both form i.e. Fe
++
and
Fe
+++
. (Preferably Fe
++
)
111. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Calcium (2) Magnesium
(3) Potassium (4) Sodium
Answer (3)
Sol. Potassium helps in maintaining turgidity of
cells.
112. Double fertilization is
(1) Syngamy and triple fusion
(2) Fusion of two male gametes of a pollen
tube with two different eggs
(3) Fusion of two male gametes with one egg
(4) Fusion of one male gamete with two polar
nuclei
Answer (1)
23
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
113. What is the role of NAD
+
in cellular
respiration?
(1) It is the final electron acceptor for
anaerobic respiration.
(2) It functions as an enzyme.
(3) It is a nucleotide source for ATP synthesis.
(4) It functions as an electron carrier.
Answer (4)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
114. Oxygen is not produced during photosynthesis
by
(1) Chara
(2) Green sulphur bacteria
(3) Cycas
(4) Nostoc
Answer (2)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
115. The Golgi complex participates in
(1) Activation of amino acid
(2) Fatty acid breakdown
(3) Respiration in bacteria
(4) Formation of secretory vesicles
Answer (4)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
116. Stomatal movement is not affected by
(1) CO
2
concentration
(2) Temperature
(3) O
2
concentration
(4) Light
Answer (3)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
117. Stomata in grass leaf are
(1) Barrel shaped
(2) Dumb-bell shaped
(3) Rectangular
(4) Kidney shaped
Answer (2)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
118. Which of the following is true for nucleolus?
(1) It is a site for active ribosomal RNA
synthesis
(2) Larger nucleoli are present in dividing
cells
(3) It takes part in spindle formation
(4) It is a membrane-bound structure
Answer (1)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
119. Which of the following is not a product of light
reaction of photosynthesis?
(1) Oxygen (2) ATP
(3) NADPH (4) NADH
Answer (4)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
120. The stage during which separation of the
paired homologous chromosomes begins is
(1) Zygotene (2) Pachytene
(3) Diakinesis (4) Diplotene
Answer (4)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
121. The two functional groups characteristic of
sugars are
(1) Carbonyl and hydroxyl
(2) Hydroxyl and methyl
(3) Carbonyl and phosphate
(4) Carbonyl and methyl
Answer (1)
Sol. Sugar is a common term used to denote
carbohydrate.
Carbohydrates are polyhydroxy aldehyde,
ketone or their derivatives, which means they
have carbonyl and hydroxyl groups.
122. Which among the following is not a
prokaryote?
(1) Oscillatoria
(2) Saccharomyces
(3) Nostoc
(4) Mycobacterium
Answer (2)
24
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
123. Offsets are produced by
(1) Parthenogenesis (2) Meiotic divisions
(3) Parthenocarpy (4) Mitotic divisions
Answer (4)
Sol. Offset is a vegetative part of a plant, formed
by mitosis.
? Meiotic divisions do not occur in somatic
cells.
? Parthenogenesis is the formation of
embryo from ovum or egg without
fertilisation.
? Parthenocarpy is the fruit formed without
fertilisation, (generally seedless)
124. Select the correct statement
(1) Transduction was discovered by S. Altman
(2) Franklin Stahl coined the term ?linkage?
(3) Spliceosomes take part in translation
(4) Punnett square was developed by a British
scientist
Answer (4)
Sol. Punnett square was developed by a British
geneticist, Reginald C. Punnett.
? Franklin Stahl proved semi-conservative
mode of replication.
? Transduction was discovered by Zinder
and Laderberg.
? Spliceosome formation is part of post-
transcriptional change in Eukaryotes
125. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Sporopollenin (2) Pollenkitt
(3) Oil content (4) Cellulosic intine
Answer (1)
Sol. Sporopollenin cannot be degraded by
enzyme; strong acids and alkali, therefore it is
helpful in preserving pollen as fossil.
Pollenkitt ? Help in insect pollination.
Cellulosic Intine ? Inner sporoderm layer of
pollen grain known as intine made up
cellulose & pectin.
Oil content ? No role is pollen preservation.
126. Which of the following pairs is wrongly
matched?
(1) T.H. Morgan : Linkage
(2) Starch synthesis in pea : Multiple alleles
(3) XO type sex : Grasshopper
determination
(4) ABO blood grouping : Co-dominance
Answer (2)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (1, 3 & 4) are correctly
matched.
127. Select the correct match
(1) Francois Jacob and - Lac operon
Jacques Monod
(2) Alec Jeffreys - Streptococcus
pneumoniae
(3) Matthew Meselson - Pisum sativum
and F. Stahl
(4) Alfred Hershey and - TMV
Martha Chase
Answer (1)
Sol. Francois Jacob and Jacque Monod proposed
model of gene regulation known as operon
model/lac operon.
? Alec Jeffreys ? DNA fingerprinting
technique.
? Matthew Meselson and F. Stahl ? Semi-
conservative DNA replication in E. coli.
? Alfred Hershey and Martha Chase ?
Proved DNA as genetic material not
protein
128. Which of the following flowers only once in its
life-time?
(1) Papaya (2) Bamboo species
(3) Mango (4) Jackfruit
Answer (2)
Sol. Bamboo species are monocarpic i.e., flower
generally only once in its life-time after 50-
100 years.
Jackfruit, papaya and mango are polycarpic
i.e., produce flowers and fruits many times in
their life-time.
129. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Virus (2) Fungus
(3) Plant (4) Bacterium
Answer (4)
25
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
130. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) pBR 322 (2) Retrovirus
(3) X phage (4) Ti plasmid
Answer (2)
Sol. Retrovirus is commonly used as vector for
introducing a DNA fragment in human
lymphocyte.
Gene therapy : Lymphocyte from blood of
patient are grown in culture outside the body,
a functional gene is introduced by using a
retroviral vector, into these lymphocyte.
131. Select the correct match
(1) G. Mendel - Transformation
(2) Ribozyme - Nucleic acid
(3) T.H. Morgan - Transduction
(4) F
2
? Recessive parent - Dihybrid cross
Answer (2)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
132. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bioexploitation (2) Bio-infringement
(3) Biodegradation (4) Biopiracy
Answer (4)
Sol. Biopiracy is term used for or refer to the use
of bioresources by multinational companies
and other organisation without proper
authorisation from the countries and people
concerned with compensatory payment
(definition of biopiracy given in NCERT).
133. A ?new? variety of rice was patented by a
foreign company, though such varieties have
been present in India for a long time. This is
related to
(1) Basmati (2) Co-667
(3) Lerma Rojo (4) Sharbati Sonora
Answer (1)
Sol. In 1997, an American company got patent
rights on Basmati rice through the US patent
and trademark office that was actually been
derived from Indian farmer?s varieties.
The diversity of rice in India is one of the
richest in the world, 27 documented varieties
of Basmati are grown in India.
Indian basmati was crossed with semi-dwarf
varieties and claimed as an invention or a
novelty.
Sharbati Sonora and Lerma Rojo are varieties
of wheat.
134. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Genetic Engineering Appraisal Committee
(GEAC)
(2) Indian Council of Medical Research (ICMR)
(3) Research Committee on Genetic
Manipulation (RCGM)
(4) Council for Scientific and Industrial
Research (CSIR)
Answer (1)
Sol. Indian Government has setup organisation
such as GEAC (Genetic Engineering Appraisal
Committee) which will make decisions
regarding the validity of GM research and
safety of introducing GM-organism for public
services. (Direct from NCERT).
135. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Denaturation, Annealing, Extension
(2) Extension, Denaturation, Annealing
(3) Denaturation, Extension, Annealing
(4) Annealing, Extension, Denaturation
Answer (1)
Sol. This technique is used for making multiple
copies of gene (or DNA) of interest in vitro.
Each cycle has three steps
(i) Denaturation
(ii) Primer annealing
(iii) Extension of primer
26
136. Identify the major products P, Q and R in the
following sequence of reactions:
(i) O
2

P Q + R
(ii) H
8
O / A
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

P Q R
137. Which of the following compounds can form a
zwitterion?
(1) Glycine (2) Aniline
(3) Benzoic acid (4) Acetanilide
Answer (1)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
pK
,
= 9.60 pK
,
= 2.34
CH
3

CH ? CH
3

O
2

(P)
CH
3
CH
3
? CH ? CH
3

H /H O
+
2
Hydroperoxide
Rearrangement
OH
O
(R)
(Q)
3
HC?C?O?O?H
+ CH
3
? C ? CH
3

27
NEET (UG) - 2018 (Code-NN) HLAAC
OH
CH(CH
3
)
2
CH
3
? CO ? CH
3
(1)
CH
3
CH
2
? OH
(2)
(3)
(4)
Answer (1)
? ? ? ? ? ? ??
Cl
Sol
Cl Cl
CO
1, 2?H
Shift
6+ 6+ 6?
(Incipient carbocation)
Cl
Ni
CO
6?
OC
AlCl
3

CO
Now,
.
CH
3
CH
2
CH
2
? Cl + Al
CH
3
? CH ? CH
3

CH
2
CH
2
CH
3
CHO
CH
2
CH
2
CH
3
CHO COOH
CH(CH
3
)
2
OH
CH
3
CH
2
CH
2
Cl AlCl
3

, CH
3
CH(OH)CH
3
H
2
N ? CH
2
? COO
?

138. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Tetrahedral geometry and paramagnetic
(2) Square planar geometry and diamagnetic
(3) Square planar geometry and
paramagnetic
(4) Tetrahedral geometry and diamagnetic
Answer (4)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
sp -hybridisation
3
~
H
3
N ? CH
2
? COO
?
139. Iron carbonyl, Fe(CO)
5
is
(1) Dinuclear (2) Tetranuclear
(3) Trinuclear (4) Mononuclear
Answer (4)
Sol. ? Based on the number of metal atoms
present in a complex, they are classified
into mononuclear, dinuclear, trinuclear and
so on.
eg: Fe(CO)
5
: mononuclear
Co
2
(CO)
8
: dinuclear
Fe
3
(CO)
12
: trinuclear
NEET (UG) - 2018 (Code-NN) HLAAC
NO
.
NO
2

H V
~
H
(1)
V
(2)
~
(3)
H
V
~
NO
.
(4)
V H
~
NO
2
140. Match the metal ions given in Column I with
the spin magnetic moments of the ions given
in Column II and assign the correct code :
Column I Column II
a. Co
3+
i.
8
BM
b. Cr
3+
ii.
35
BM
c. Fe
3+
iii.
3
BM
d. Ni
2+
iv.
24
BM
v.
15
BM
c d
i ii
ii i
ii iii
iii iv
Answer (2)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Spin magnetic moment = 4(4 + 2) = 24 BM
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Spin magnetic moment = 3(3 + 2) = 15 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
Spin magnetic moment = 2(2 + 2) = 8 BM
141. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Linkage isomerism
(2) Geometrical isomerism
(3) Ionization isomerism
(4) Coordination isomerism
Answer (2)
Sol. In [CoCl
2
(en)
2
], Coordination number of Co
is 6 and this compound has octahedral
geometry.
? As per given option, type of isomerism is
geometrical isomerism.
142. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) MnO
4
2?
(2) CrO
4
2?

(3) MnO
4
?
(4) Cr
2
O
7
2?
Answer (1)
Sol. CrO
4
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
Cr
2
O
7
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
MnO
4
2?
= Mn
6+
= [Ar] 3d
1

Unpaired electron (n) = 1; Paramagnetic
MnO
4
?
= Mn
7+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
143. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) CH
3
? CH = CH ? CH
3

(2) HC ~ C ? C ~ CH
(3) CH
2
= CH ? CH = CH
2

(4) CH
2
= CH ? C ~ CH
Answer (4)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
144. Which of the following carbocations is
expected to be most stable?
Answer (3)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (3)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
a b
(1) iii v
(2) iv v
(3) iv i
(4) i ii
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
28
M
10 HCl + 100 mL
M
10 NaOH e. 100 mL
NEET (UG) - 2018 (Code-NN) HLAAC
145. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NR
2
> ? OR > ? F
(2) ? NH
2
< ? OR < ? F
(3) ? NH
2
> ? OR > ? F
(4) ? NR
2
< ? OR < ? F
Answer (2*)
Sol. ?I effect increases on increasing
electronegativity of atom. So, correct order of
?I effect is ?NH
2
< ? OR < ? F.
*Most appropriate Answer is option (2),
however option (4) may also be correct answer.
146. The solubility of BaSO
4
in water is 2.42 ? 10
?3
gL
?1

at 298 K. The value of its solubility product
(K
sp
) will be
(Given molar mass of BaSO
4
= 233 g mol
?1
)
(1) 1.08 ? 10
?8
mol
2
L
?2

(2) 1.08 ? 10
?10
mol
2
L
?2

(3) 1.08 ? 10
?14
mol
2
L
?2

(4) 1.08 ? 10
?12
mol
2
L
?2

Answer (2)
148. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M M
a. 60 mL 10 HCl + 40 mL 10 NaOH
M M
b. 55 mL 10 HCl + 45 mL 10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
pH of which one of them will be equal to 1?
(1) c (2) b
(3) d (4) a
Answer (1)
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
3
(mol L
?1
) Sol. Solubility of BaSO
4
, s =
2.42 10 x
233
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
= 1.04 ? 10
?5
(mol L
?1
)
10 1
=
~
BaSO (s) Ba (aq) SO (aq)
2 2 ~
~~~
~~~
+
4 4
s s
K
sp
= [Ba
2+
] [SO
42?
]= s
2
= (1.04 ? 10
?5
)
2

= 1.08 ? 10
?10
mol
2
L
?2
147. Given van der Waals constant for NH
3
, H
2
, O
2

and CO
2
are respectively 4.17, 0.244, 1.36 and
3.59, which one of the following gases is most
easily liquefied?
(1) CO
2

(2) NH
3

(3) O
2

(4) H
2

Answer (2)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
100 10
pH = ?log[H
+
] =
1
~
~ ~
~ ~
log 10
~
~ = 1.0
149. On which of the following properties does the
coagulating power of an ion depend?
(1) The sign of charge on the ion alone
(2) The magnitude of the charge on the ion
alone
(3) Both magnitude and sign of the charge on
the ion
(4) Size of the ion alone
Answer (3)
Sol. ? Coagulation of colloidal solution by using
an electrolyte depends on the charge
present (positive or negative) on colloidal
particles as well as on its size.
? Coagulating power of an electrolyte
depends on the magnitude of charge
present on effective ion of electrolyte.
29
NEET (UG) - 2018 (Code-NN) HLAAC
S S
? ?
F
? ?
S S
F Cl
S S
F
S S
(1) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
(2) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(3) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(4) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
Answer (1)
Sol.
C
2
H
5
OH
Na
C
2
H
5
O Na
+

(A) (B)
PCl
5

C
2
H
5
Cl
(C)
C
2
H
5
O Na + + C
2
H
5
Cl
S
N
2

C
2
H
5
OC
2
H 5
(B) (C)
150. Which of the following statements is not true
for halogens?
(1) Chlorine has the highest electron-gain
enthalpy
(2) All form monobasic oxyacids
(3) All but fluorine show positive oxidation
states
(4) All are oxidizing agents
Answer (3)
Sol. Due to high electronegativity and small size,
F forms only one oxoacid, HOF known as
Fluoric (I) acid. Oxidation number of F is +1 in
HOF.
151. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
(1) Cu (2) Fe
(3) Mg (4) Zn
Answer (3)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
152. The correct order of atomic radii in group 13
elements is
(1) B < Ga < Al < In < Tl
(2) B < Al < In < Ga < Tl
(3) B < Ga < Al < Tl < In
(4) B < Al < Ga < In < Tl
Answer (1)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
153. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) Three (2) One
(3) Four (4) Two
Answer (4)
Sol. The structure of ClF
3
is
The number of lone pair of electrons on
central Cl is 2.
154.The correct order of N-compounds in its
decreasing order of oxidation states is
(1) NH
4
Cl, N
2
, NO, HNO
3

(2) HNO
3
, NO, N
2
, NH
4
Cl
(3) HNO
3
, NH
4
Cl, NO, N
2

(4) HNO
3
, NO, NH
4
Cl, N
2

Answer (2)
+ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
155. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) In
(2) Ga
(3) B
(4) Al
Answer (3)
Sol. ? 'B' has no vacant d-orbitals in its valence
shell, so it can't extend its covalency beyond
4. i.e. 'B' cannot form the ion like MF
6
3(?)
i.e.
BF
6
3(?)
.
156. The compound A on treatment with Na gives
B, and with PCl
5
gives C. B and C react
together to give diethyl ether. A, B and C are
in the order
30
NEET (UG) - 2018 (Code-NN) HLAAC
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
1
,
k[A ]
0
157. Hydrocarbon (A) reacts with bromine by
substitution to form an alkyl bromide which by
Wurtz reaction is converted to gaseous
hydrocarbon containing less than four carbon
atoms. (A) is
(1) CH
4
(2) CH ~ CH
(3) CH
3
? CH
3
(4) CH
2
= CH
2

Answer (1)
Sol.
CH
4
Br
2
/h v
CH
3
Br
Na/dry ether
Wurtz reaction
CH
3
? CH
3

158. The compound C
7
H
8
undergoes the following
reactions:
3Cl
2
/ A Br
2
/Fe Zn/HCl
C
7
H
8
~~~> ~~~> ~~~~ A B C
The product 'C' is
(1) p-bromotoluene
(2) m-bromotoluene
(3) 3-bromo-2,4,6-trichlorotoluene
(4) o-bromotoluene
Answer (2)
CCl
3

3dl
2
A
(C
7
H
8
)
160. The correct difference between first and
second order reactions is that
(1) The rate of a first-order reaction does
depend on reactant concentrations; the
rate of a second-order reaction does not
depend on reactant concentrations
(2) The rate of a first-order reaction does not
depend on reactant concentrations; the
rate of a second-order reaction does
depend on reactant concentrations
(3) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(4) The half-life of a first-order reaction does
not depend on [A]
0
; the half-life of a
second-order reaction does depend on
[A]
0

Answer (4)
,
k
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
161. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BaH
2
< BeH
2
< CaH
2

(2) BeH
2
< CaH
2
< BaH
2

(3) BeH
2
< BaH
2
< CaH
2

(4) CaH
2
< BeH
2
< BaH
2

Answer (2)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
162. Consider the change in oxidation state of
Bromine corresponding to different emf values
as shown in the diagram below :
Then the species undergoing
disproportionation is
(1) HBrO (2)
~
BrO
3

(3) Br
2
(4)
~
BrO
4
(A)
CH
3
CCl
3

Sol.
Br
2
Fe
Br
(A) (B)
Zn HCl
CH
3
Br
(C)
159. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) NO
(2) N
2
O
5
(3) N
2
O
(4) NO
2

Answer (2)
Sol. Fact
Sol. ? For first order reaction,
1 /2
=
0.693
t
31
NEET (UG) - 2018 (Code-NN) HLAAC
CH
3

(1) CH
3
OH and I
2

(2)
H
3
C CH
2
? OH and I
2
COONa
Sodium benzoate
Answer (1)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+5
~~* BrO , E
~
~ o
HBrO
3
BrO /HBrO
= 1.5 V
3
E for the disproportionation of HBrO,
o
cell
E
o
= E
o o
cell
~ E
~ HBrO/Br
2 BrO /HBrO
3
= 1.595 ? 1.5
= 0.095 V = + ve
163. In which case is number of molecules of
water maximum?
(1) 10
?3
mol of water
(2) 18 mL of water
(3) 0.00224 L of water vapours at 1 atm and
273 K
(4) 0.18 g of water
Answer (2)
Sol. (1) Molecules of water = mole ? N
A
= 10
?3
N
A

(2) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(4) Molecules of water = mole ? N
A
= A
18
= 10
?2
N
A

164. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intermolecular H-bonding
(2) Formation of intramolecular H-bonding
(3) More extensive association of carboxylic
acid via van der Waals force of attraction
(4) Formation of carboxylate ion
Answer (1)
Sol. Due to formation of intermolecular H-bonding
in carboxylic acid, association occurs. Hence
boiling point increases and become more
than the boiling point of aldehydes, ketones
and alcohols of comparable molecular
masses.
165. Compound A,
C
8
H
10
O,
is found to react with
NaOI (produced by reacting Y with NaOH) and
yields a yellow precipitate with characteristic
smell.
A and Y are respectively
CH ? CH
3
and I
2
(3)
OH
2 (4)
CH
2
? CH
2
? OH and I
Answer (3)
Sol. Option (3) is secondary alcohol which on
oxidation gives phenylmethyl ketone
(Acetophenone). This on reaction with I
2
and
NaOH form iodoform and sodium benzoate.
2NaOH + I -* NaOI + NaI + H
2
O
2
I
2
NaOH
166. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichlorocarbene ( : CCl
2
)
(2) Dichloromethyl cation
(
CHCl
2
)
~
(3) Dichloromethyl anion (
CHCl
2
)
?
(4) Formyl cation (
C H O
)
~
+ CHI
3

Iodoform
(Yellow PPt)
CHO
= N
A

0.00224
22.4
NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
= 10
?4

(3) Moles of water =
32
?
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
NEET (UG) - 2018 (Code-NN) HLAAC
Answer (1)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
CCl ~~~ : CCl + Cl
3 2
Electrophile
167. The bond dissociation energies of X
2
, Y
2
and
XY are in the ratio of 1 : 0.5 : 1. AH for the
formation of XY is ?200 kJ mol
?1
. The bond
dissociation energy of X
2
will be
(1) 400 kJ mol
?1

(2) 200 kJ mol
?1

(3) 800 kJ mol
?1

(4) 100 kJ mol
?1

Answer (3)
Sol. The reaction for A
f
H?(XY)
1
+
1
X (g) Y (g) XY(g) ~~~
2 2
2 2
Bond energies of X
2
, Y
2
and XY are X, X 2 , X
respectively
A =
?
+
?
_ = _
? ? X X
H X 200
? ?
2 4
On solving, we get
? - + = - X X 200
2 4
= X = 800 kJ/mole
168. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Remains unchanged
(2) Is halved
(3) Is tripled
(4) Is doubled
Answer (4)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
169. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Forces of attraction between the gas
molecules
(2) Density of the gas molecules
(3) Electric field present between the gas
molecules
(4) Volume of the gas molecules
Answer (1)
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
170. Which one of the following conditions will
favour maximum formation of the product in
the reaction,
A
2
(g) + B
2
(g) ~~=
~~~ X
2
(g) A
r
H = -X kJ?
(1) High temperature and low pressure
(2) Low temperature and high pressure
(3) High temperature and high pressure
(4) Low temperature and low pressure
Answer (2)
Sol.
A
2
(g) + B
2
(g)
=~~
~~=
X
2
(g); A H = -x kJ
On increasing pressure equilibrium shifts in a
direction where pressure decreases i.e.
forward direction.
On decreasing temperature, equilibrium shifts
in exothermic direction i.e., forward direction.
So, high pressure and low temperature
favours maximum formation of product.
171. For the redox reaction
~ 2 ~ ~ ~
+ Mn CO
2
H
2
O + ~~~ + +
2
MnO C O H
4 2 4
The correct coefficients of the reactants for
the balanced equation are
MnO
4
~
C
2
O
4
H
+ 2~
(1) 5 16 2
(2) 16 5 2
(3) 2 16 5
(4) 2 5 16
~
t
1/2
~
Sol. In real gas equation,
?
? ?
2
?
+ - =
an
P (V nb) nRT
2 ?
V
?
33
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1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
10. The candidates are governed by all Rules and Regulations of the examination with regard to their
conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and
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12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
14
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
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NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
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NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
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NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
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NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
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NEET (UG) - 2018 (Code-NN) HLAAC
106. Which one is wrongly matched?
(1) Unicellular organism ? Chlorella
(2) Uniflagellate gametes ? Polysiphonia
(3) Gemma cups ? Marchantia
(4) Biflagellate zoospores ? Brown algae
Answer (2)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (1, 3 & 4) are correctly
matched
107. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Herbarium (i) It is a place having a
collection of preserved
plants and animals
b. Key

(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum

(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue (iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
a b c d
(1) (iii) (iv) (i) (ii)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iv) (iii) (i)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
108. Which one of the following plants shows a
very close relationship with a species of moth,
where none of the two can complete its life
cycle without the other?
(1) Viola (2) Hydrilla
(3) Banana (4) Yucca
Answer (4)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
109. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?160?C
(2) ?120?C
(3) ?196?C
(4) ?80?C
Answer (3)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C
(Cryopreservation)
110. In which of the following forms is iron
absorbed by plants?
(1) Both ferric and ferrous
(2) Ferric
(3) Free element
(4) Ferrous
Answer (2*)
Sol. Iron is absorbed by plants in the form of ferric
ions. (According to NCERT)
*Plants absorb iron in both form i.e. Fe
++
and
Fe
+++
. (Preferably Fe
++
)
111. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Calcium (2) Magnesium
(3) Potassium (4) Sodium
Answer (3)
Sol. Potassium helps in maintaining turgidity of
cells.
112. Double fertilization is
(1) Syngamy and triple fusion
(2) Fusion of two male gametes of a pollen
tube with two different eggs
(3) Fusion of two male gametes with one egg
(4) Fusion of one male gamete with two polar
nuclei
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
113. What is the role of NAD
+
in cellular
respiration?
(1) It is the final electron acceptor for
anaerobic respiration.
(2) It functions as an enzyme.
(3) It is a nucleotide source for ATP synthesis.
(4) It functions as an electron carrier.
Answer (4)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
114. Oxygen is not produced during photosynthesis
by
(1) Chara
(2) Green sulphur bacteria
(3) Cycas
(4) Nostoc
Answer (2)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
115. The Golgi complex participates in
(1) Activation of amino acid
(2) Fatty acid breakdown
(3) Respiration in bacteria
(4) Formation of secretory vesicles
Answer (4)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
116. Stomatal movement is not affected by
(1) CO
2
concentration
(2) Temperature
(3) O
2
concentration
(4) Light
Answer (3)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
117. Stomata in grass leaf are
(1) Barrel shaped
(2) Dumb-bell shaped
(3) Rectangular
(4) Kidney shaped
Answer (2)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
118. Which of the following is true for nucleolus?
(1) It is a site for active ribosomal RNA
synthesis
(2) Larger nucleoli are present in dividing
cells
(3) It takes part in spindle formation
(4) It is a membrane-bound structure
Answer (1)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
119. Which of the following is not a product of light
reaction of photosynthesis?
(1) Oxygen (2) ATP
(3) NADPH (4) NADH
Answer (4)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
120. The stage during which separation of the
paired homologous chromosomes begins is
(1) Zygotene (2) Pachytene
(3) Diakinesis (4) Diplotene
Answer (4)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
121. The two functional groups characteristic of
sugars are
(1) Carbonyl and hydroxyl
(2) Hydroxyl and methyl
(3) Carbonyl and phosphate
(4) Carbonyl and methyl
Answer (1)
Sol. Sugar is a common term used to denote
carbohydrate.
Carbohydrates are polyhydroxy aldehyde,
ketone or their derivatives, which means they
have carbonyl and hydroxyl groups.
122. Which among the following is not a
prokaryote?
(1) Oscillatoria
(2) Saccharomyces
(3) Nostoc
(4) Mycobacterium
Answer (2)
24
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
123. Offsets are produced by
(1) Parthenogenesis (2) Meiotic divisions
(3) Parthenocarpy (4) Mitotic divisions
Answer (4)
Sol. Offset is a vegetative part of a plant, formed
by mitosis.
? Meiotic divisions do not occur in somatic
cells.
? Parthenogenesis is the formation of
embryo from ovum or egg without
fertilisation.
? Parthenocarpy is the fruit formed without
fertilisation, (generally seedless)
124. Select the correct statement
(1) Transduction was discovered by S. Altman
(2) Franklin Stahl coined the term ?linkage?
(3) Spliceosomes take part in translation
(4) Punnett square was developed by a British
scientist
Answer (4)
Sol. Punnett square was developed by a British
geneticist, Reginald C. Punnett.
? Franklin Stahl proved semi-conservative
mode of replication.
? Transduction was discovered by Zinder
and Laderberg.
? Spliceosome formation is part of post-
transcriptional change in Eukaryotes
125. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Sporopollenin (2) Pollenkitt
(3) Oil content (4) Cellulosic intine
Answer (1)
Sol. Sporopollenin cannot be degraded by
enzyme; strong acids and alkali, therefore it is
helpful in preserving pollen as fossil.
Pollenkitt ? Help in insect pollination.
Cellulosic Intine ? Inner sporoderm layer of
pollen grain known as intine made up
cellulose & pectin.
Oil content ? No role is pollen preservation.
126. Which of the following pairs is wrongly
matched?
(1) T.H. Morgan : Linkage
(2) Starch synthesis in pea : Multiple alleles
(3) XO type sex : Grasshopper
determination
(4) ABO blood grouping : Co-dominance
Answer (2)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (1, 3 & 4) are correctly
matched.
127. Select the correct match
(1) Francois Jacob and - Lac operon
Jacques Monod
(2) Alec Jeffreys - Streptococcus
pneumoniae
(3) Matthew Meselson - Pisum sativum
and F. Stahl
(4) Alfred Hershey and - TMV
Martha Chase
Answer (1)
Sol. Francois Jacob and Jacque Monod proposed
model of gene regulation known as operon
model/lac operon.
? Alec Jeffreys ? DNA fingerprinting
technique.
? Matthew Meselson and F. Stahl ? Semi-
conservative DNA replication in E. coli.
? Alfred Hershey and Martha Chase ?
Proved DNA as genetic material not
protein
128. Which of the following flowers only once in its
life-time?
(1) Papaya (2) Bamboo species
(3) Mango (4) Jackfruit
Answer (2)
Sol. Bamboo species are monocarpic i.e., flower
generally only once in its life-time after 50-
100 years.
Jackfruit, papaya and mango are polycarpic
i.e., produce flowers and fruits many times in
their life-time.
129. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Virus (2) Fungus
(3) Plant (4) Bacterium
Answer (4)
25
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
130. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) pBR 322 (2) Retrovirus
(3) X phage (4) Ti plasmid
Answer (2)
Sol. Retrovirus is commonly used as vector for
introducing a DNA fragment in human
lymphocyte.
Gene therapy : Lymphocyte from blood of
patient are grown in culture outside the body,
a functional gene is introduced by using a
retroviral vector, into these lymphocyte.
131. Select the correct match
(1) G. Mendel - Transformation
(2) Ribozyme - Nucleic acid
(3) T.H. Morgan - Transduction
(4) F
2
? Recessive parent - Dihybrid cross
Answer (2)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
132. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bioexploitation (2) Bio-infringement
(3) Biodegradation (4) Biopiracy
Answer (4)
Sol. Biopiracy is term used for or refer to the use
of bioresources by multinational companies
and other organisation without proper
authorisation from the countries and people
concerned with compensatory payment
(definition of biopiracy given in NCERT).
133. A ?new? variety of rice was patented by a
foreign company, though such varieties have
been present in India for a long time. This is
related to
(1) Basmati (2) Co-667
(3) Lerma Rojo (4) Sharbati Sonora
Answer (1)
Sol. In 1997, an American company got patent
rights on Basmati rice through the US patent
and trademark office that was actually been
derived from Indian farmer?s varieties.
The diversity of rice in India is one of the
richest in the world, 27 documented varieties
of Basmati are grown in India.
Indian basmati was crossed with semi-dwarf
varieties and claimed as an invention or a
novelty.
Sharbati Sonora and Lerma Rojo are varieties
of wheat.
134. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Genetic Engineering Appraisal Committee
(GEAC)
(2) Indian Council of Medical Research (ICMR)
(3) Research Committee on Genetic
Manipulation (RCGM)
(4) Council for Scientific and Industrial
Research (CSIR)
Answer (1)
Sol. Indian Government has setup organisation
such as GEAC (Genetic Engineering Appraisal
Committee) which will make decisions
regarding the validity of GM research and
safety of introducing GM-organism for public
services. (Direct from NCERT).
135. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Denaturation, Annealing, Extension
(2) Extension, Denaturation, Annealing
(3) Denaturation, Extension, Annealing
(4) Annealing, Extension, Denaturation
Answer (1)
Sol. This technique is used for making multiple
copies of gene (or DNA) of interest in vitro.
Each cycle has three steps
(i) Denaturation
(ii) Primer annealing
(iii) Extension of primer
26
136. Identify the major products P, Q and R in the
following sequence of reactions:
(i) O
2

P Q + R
(ii) H
8
O / A
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

P Q R
137. Which of the following compounds can form a
zwitterion?
(1) Glycine (2) Aniline
(3) Benzoic acid (4) Acetanilide
Answer (1)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
pK
,
= 9.60 pK
,
= 2.34
CH
3

CH ? CH
3

O
2

(P)
CH
3
CH
3
? CH ? CH
3

H /H O
+
2
Hydroperoxide
Rearrangement
OH
O
(R)
(Q)
3
HC?C?O?O?H
+ CH
3
? C ? CH
3

27
NEET (UG) - 2018 (Code-NN) HLAAC
OH
CH(CH
3
)
2
CH
3
? CO ? CH
3
(1)
CH
3
CH
2
? OH
(2)
(3)
(4)
Answer (1)
? ? ? ? ? ? ??
Cl
Sol
Cl Cl
CO
1, 2?H
Shift
6+ 6+ 6?
(Incipient carbocation)
Cl
Ni
CO
6?
OC
AlCl
3

CO
Now,
.
CH
3
CH
2
CH
2
? Cl + Al
CH
3
? CH ? CH
3

CH
2
CH
2
CH
3
CHO
CH
2
CH
2
CH
3
CHO COOH
CH(CH
3
)
2
OH
CH
3
CH
2
CH
2
Cl AlCl
3

, CH
3
CH(OH)CH
3
H
2
N ? CH
2
? COO
?

138. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Tetrahedral geometry and paramagnetic
(2) Square planar geometry and diamagnetic
(3) Square planar geometry and
paramagnetic
(4) Tetrahedral geometry and diamagnetic
Answer (4)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
sp -hybridisation
3
~
H
3
N ? CH
2
? COO
?
139. Iron carbonyl, Fe(CO)
5
is
(1) Dinuclear (2) Tetranuclear
(3) Trinuclear (4) Mononuclear
Answer (4)
Sol. ? Based on the number of metal atoms
present in a complex, they are classified
into mononuclear, dinuclear, trinuclear and
so on.
eg: Fe(CO)
5
: mononuclear
Co
2
(CO)
8
: dinuclear
Fe
3
(CO)
12
: trinuclear
NEET (UG) - 2018 (Code-NN) HLAAC
NO
.
NO
2

H V
~
H
(1)
V
(2)
~
(3)
H
V
~
NO
.
(4)
V H
~
NO
2
140. Match the metal ions given in Column I with
the spin magnetic moments of the ions given
in Column II and assign the correct code :
Column I Column II
a. Co
3+
i.
8
BM
b. Cr
3+
ii.
35
BM
c. Fe
3+
iii.
3
BM
d. Ni
2+
iv.
24
BM
v.
15
BM
c d
i ii
ii i
ii iii
iii iv
Answer (2)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Spin magnetic moment = 4(4 + 2) = 24 BM
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Spin magnetic moment = 3(3 + 2) = 15 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
Spin magnetic moment = 2(2 + 2) = 8 BM
141. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Linkage isomerism
(2) Geometrical isomerism
(3) Ionization isomerism
(4) Coordination isomerism
Answer (2)
Sol. In [CoCl
2
(en)
2
], Coordination number of Co
is 6 and this compound has octahedral
geometry.
? As per given option, type of isomerism is
geometrical isomerism.
142. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) MnO
4
2?
(2) CrO
4
2?

(3) MnO
4
?
(4) Cr
2
O
7
2?
Answer (1)
Sol. CrO
4
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
Cr
2
O
7
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
MnO
4
2?
= Mn
6+
= [Ar] 3d
1

Unpaired electron (n) = 1; Paramagnetic
MnO
4
?
= Mn
7+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
143. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) CH
3
? CH = CH ? CH
3

(2) HC ~ C ? C ~ CH
(3) CH
2
= CH ? CH = CH
2

(4) CH
2
= CH ? C ~ CH
Answer (4)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
144. Which of the following carbocations is
expected to be most stable?
Answer (3)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (3)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
a b
(1) iii v
(2) iv v
(3) iv i
(4) i ii
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
28
M
10 HCl + 100 mL
M
10 NaOH e. 100 mL
NEET (UG) - 2018 (Code-NN) HLAAC
145. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NR
2
> ? OR > ? F
(2) ? NH
2
< ? OR < ? F
(3) ? NH
2
> ? OR > ? F
(4) ? NR
2
< ? OR < ? F
Answer (2*)
Sol. ?I effect increases on increasing
electronegativity of atom. So, correct order of
?I effect is ?NH
2
< ? OR < ? F.
*Most appropriate Answer is option (2),
however option (4) may also be correct answer.
146. The solubility of BaSO
4
in water is 2.42 ? 10
?3
gL
?1

at 298 K. The value of its solubility product
(K
sp
) will be
(Given molar mass of BaSO
4
= 233 g mol
?1
)
(1) 1.08 ? 10
?8
mol
2
L
?2

(2) 1.08 ? 10
?10
mol
2
L
?2

(3) 1.08 ? 10
?14
mol
2
L
?2

(4) 1.08 ? 10
?12
mol
2
L
?2

Answer (2)
148. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M M
a. 60 mL 10 HCl + 40 mL 10 NaOH
M M
b. 55 mL 10 HCl + 45 mL 10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
pH of which one of them will be equal to 1?
(1) c (2) b
(3) d (4) a
Answer (1)
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
3
(mol L
?1
) Sol. Solubility of BaSO
4
, s =
2.42 10 x
233
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
= 1.04 ? 10
?5
(mol L
?1
)
10 1
=
~
BaSO (s) Ba (aq) SO (aq)
2 2 ~
~~~
~~~
+
4 4
s s
K
sp
= [Ba
2+
] [SO
42?
]= s
2
= (1.04 ? 10
?5
)
2

= 1.08 ? 10
?10
mol
2
L
?2
147. Given van der Waals constant for NH
3
, H
2
, O
2

and CO
2
are respectively 4.17, 0.244, 1.36 and
3.59, which one of the following gases is most
easily liquefied?
(1) CO
2

(2) NH
3

(3) O
2

(4) H
2

Answer (2)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
100 10
pH = ?log[H
+
] =
1
~
~ ~
~ ~
log 10
~
~ = 1.0
149. On which of the following properties does the
coagulating power of an ion depend?
(1) The sign of charge on the ion alone
(2) The magnitude of the charge on the ion
alone
(3) Both magnitude and sign of the charge on
the ion
(4) Size of the ion alone
Answer (3)
Sol. ? Coagulation of colloidal solution by using
an electrolyte depends on the charge
present (positive or negative) on colloidal
particles as well as on its size.
? Coagulating power of an electrolyte
depends on the magnitude of charge
present on effective ion of electrolyte.
29
NEET (UG) - 2018 (Code-NN) HLAAC
S S
? ?
F
? ?
S S
F Cl
S S
F
S S
(1) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
(2) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(3) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(4) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
Answer (1)
Sol.
C
2
H
5
OH
Na
C
2
H
5
O Na
+

(A) (B)
PCl
5

C
2
H
5
Cl
(C)
C
2
H
5
O Na + + C
2
H
5
Cl
S
N
2

C
2
H
5
OC
2
H 5
(B) (C)
150. Which of the following statements is not true
for halogens?
(1) Chlorine has the highest electron-gain
enthalpy
(2) All form monobasic oxyacids
(3) All but fluorine show positive oxidation
states
(4) All are oxidizing agents
Answer (3)
Sol. Due to high electronegativity and small size,
F forms only one oxoacid, HOF known as
Fluoric (I) acid. Oxidation number of F is +1 in
HOF.
151. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
(1) Cu (2) Fe
(3) Mg (4) Zn
Answer (3)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
152. The correct order of atomic radii in group 13
elements is
(1) B < Ga < Al < In < Tl
(2) B < Al < In < Ga < Tl
(3) B < Ga < Al < Tl < In
(4) B < Al < Ga < In < Tl
Answer (1)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
153. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) Three (2) One
(3) Four (4) Two
Answer (4)
Sol. The structure of ClF
3
is
The number of lone pair of electrons on
central Cl is 2.
154.The correct order of N-compounds in its
decreasing order of oxidation states is
(1) NH
4
Cl, N
2
, NO, HNO
3

(2) HNO
3
, NO, N
2
, NH
4
Cl
(3) HNO
3
, NH
4
Cl, NO, N
2

(4) HNO
3
, NO, NH
4
Cl, N
2

Answer (2)
+ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
155. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) In
(2) Ga
(3) B
(4) Al
Answer (3)
Sol. ? 'B' has no vacant d-orbitals in its valence
shell, so it can't extend its covalency beyond
4. i.e. 'B' cannot form the ion like MF
6
3(?)
i.e.
BF
6
3(?)
.
156. The compound A on treatment with Na gives
B, and with PCl
5
gives C. B and C react
together to give diethyl ether. A, B and C are
in the order
30
NEET (UG) - 2018 (Code-NN) HLAAC
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
1
,
k[A ]
0
157. Hydrocarbon (A) reacts with bromine by
substitution to form an alkyl bromide which by
Wurtz reaction is converted to gaseous
hydrocarbon containing less than four carbon
atoms. (A) is
(1) CH
4
(2) CH ~ CH
(3) CH
3
? CH
3
(4) CH
2
= CH
2

Answer (1)
Sol.
CH
4
Br
2
/h v
CH
3
Br
Na/dry ether
Wurtz reaction
CH
3
? CH
3

158. The compound C
7
H
8
undergoes the following
reactions:
3Cl
2
/ A Br
2
/Fe Zn/HCl
C
7
H
8
~~~> ~~~> ~~~~ A B C
The product 'C' is
(1) p-bromotoluene
(2) m-bromotoluene
(3) 3-bromo-2,4,6-trichlorotoluene
(4) o-bromotoluene
Answer (2)
CCl
3

3dl
2
A
(C
7
H
8
)
160. The correct difference between first and
second order reactions is that
(1) The rate of a first-order reaction does
depend on reactant concentrations; the
rate of a second-order reaction does not
depend on reactant concentrations
(2) The rate of a first-order reaction does not
depend on reactant concentrations; the
rate of a second-order reaction does
depend on reactant concentrations
(3) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(4) The half-life of a first-order reaction does
not depend on [A]
0
; the half-life of a
second-order reaction does depend on
[A]
0

Answer (4)
,
k
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
161. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BaH
2
< BeH
2
< CaH
2

(2) BeH
2
< CaH
2
< BaH
2

(3) BeH
2
< BaH
2
< CaH
2

(4) CaH
2
< BeH
2
< BaH
2

Answer (2)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
162. Consider the change in oxidation state of
Bromine corresponding to different emf values
as shown in the diagram below :
Then the species undergoing
disproportionation is
(1) HBrO (2)
~
BrO
3

(3) Br
2
(4)
~
BrO
4
(A)
CH
3
CCl
3

Sol.
Br
2
Fe
Br
(A) (B)
Zn HCl
CH
3
Br
(C)
159. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) NO
(2) N
2
O
5
(3) N
2
O
(4) NO
2

Answer (2)
Sol. Fact
Sol. ? For first order reaction,
1 /2
=
0.693
t
31
NEET (UG) - 2018 (Code-NN) HLAAC
CH
3

(1) CH
3
OH and I
2

(2)
H
3
C CH
2
? OH and I
2
COONa
Sodium benzoate
Answer (1)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+5
~~* BrO , E
~
~ o
HBrO
3
BrO /HBrO
= 1.5 V
3
E for the disproportionation of HBrO,
o
cell
E
o
= E
o o
cell
~ E
~ HBrO/Br
2 BrO /HBrO
3
= 1.595 ? 1.5
= 0.095 V = + ve
163. In which case is number of molecules of
water maximum?
(1) 10
?3
mol of water
(2) 18 mL of water
(3) 0.00224 L of water vapours at 1 atm and
273 K
(4) 0.18 g of water
Answer (2)
Sol. (1) Molecules of water = mole ? N
A
= 10
?3
N
A

(2) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(4) Molecules of water = mole ? N
A
= A
18
= 10
?2
N
A

164. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intermolecular H-bonding
(2) Formation of intramolecular H-bonding
(3) More extensive association of carboxylic
acid via van der Waals force of attraction
(4) Formation of carboxylate ion
Answer (1)
Sol. Due to formation of intermolecular H-bonding
in carboxylic acid, association occurs. Hence
boiling point increases and become more
than the boiling point of aldehydes, ketones
and alcohols of comparable molecular
masses.
165. Compound A,
C
8
H
10
O,
is found to react with
NaOI (produced by reacting Y with NaOH) and
yields a yellow precipitate with characteristic
smell.
A and Y are respectively
CH ? CH
3
and I
2
(3)
OH
2 (4)
CH
2
? CH
2
? OH and I
Answer (3)
Sol. Option (3) is secondary alcohol which on
oxidation gives phenylmethyl ketone
(Acetophenone). This on reaction with I
2
and
NaOH form iodoform and sodium benzoate.
2NaOH + I -* NaOI + NaI + H
2
O
2
I
2
NaOH
166. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichlorocarbene ( : CCl
2
)
(2) Dichloromethyl cation
(
CHCl
2
)
~
(3) Dichloromethyl anion (
CHCl
2
)
?
(4) Formyl cation (
C H O
)
~
+ CHI
3

Iodoform
(Yellow PPt)
CHO
= N
A

0.00224
22.4
NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
= 10
?4

(3) Moles of water =
32
?
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
NEET (UG) - 2018 (Code-NN) HLAAC
Answer (1)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
CCl ~~~ : CCl + Cl
3 2
Electrophile
167. The bond dissociation energies of X
2
, Y
2
and
XY are in the ratio of 1 : 0.5 : 1. AH for the
formation of XY is ?200 kJ mol
?1
. The bond
dissociation energy of X
2
will be
(1) 400 kJ mol
?1

(2) 200 kJ mol
?1

(3) 800 kJ mol
?1

(4) 100 kJ mol
?1

Answer (3)
Sol. The reaction for A
f
H?(XY)
1
+
1
X (g) Y (g) XY(g) ~~~
2 2
2 2
Bond energies of X
2
, Y
2
and XY are X, X 2 , X
respectively
A =
?
+
?
_ = _
? ? X X
H X 200
? ?
2 4
On solving, we get
? - + = - X X 200
2 4
= X = 800 kJ/mole
168. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Remains unchanged
(2) Is halved
(3) Is tripled
(4) Is doubled
Answer (4)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
169. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Forces of attraction between the gas
molecules
(2) Density of the gas molecules
(3) Electric field present between the gas
molecules
(4) Volume of the gas molecules
Answer (1)
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
170. Which one of the following conditions will
favour maximum formation of the product in
the reaction,
A
2
(g) + B
2
(g) ~~=
~~~ X
2
(g) A
r
H = -X kJ?
(1) High temperature and low pressure
(2) Low temperature and high pressure
(3) High temperature and high pressure
(4) Low temperature and low pressure
Answer (2)
Sol.
A
2
(g) + B
2
(g)
=~~
~~=
X
2
(g); A H = -x kJ
On increasing pressure equilibrium shifts in a
direction where pressure decreases i.e.
forward direction.
On decreasing temperature, equilibrium shifts
in exothermic direction i.e., forward direction.
So, high pressure and low temperature
favours maximum formation of product.
171. For the redox reaction
~ 2 ~ ~ ~
+ Mn CO
2
H
2
O + ~~~ + +
2
MnO C O H
4 2 4
The correct coefficients of the reactants for
the balanced equation are
MnO
4
~
C
2
O
4
H
+ 2~
(1) 5 16 2
(2) 16 5 2
(3) 2 16 5
(4) 2 5 16
~
t
1/2
~
Sol. In real gas equation,
?
? ?
2
?
+ - =
an
P (V nb) nRT
2 ?
V
?
33
NEET (UG) - 2018 (Code-NN) HLAAC
Answer (4)
Reduction
+7 +3 +4
Sol.
MnO
4
?
+ C
2
O
4
2?
+ H
+
Mn
2+
+ CO
2
+ H
2
O
Oxidation
n-factor of MnO
~
4
~ 5
n-factor of
2
C
2
O
4
~ ~ 2
Ratio of n-factors of MnO
~
4
and
2
C
2
O
4
~
is 5 : 2
So, molar ratio in balanced reaction is 2 : 5
~ The balanced equation is
2MnO 5C O 16H 2Mn 10CO
2
8H
2
O
~ 2 ~
2
+ +
4
+ + ?
2 4
172. Regarding cross-linked or network polymers,
which of the following statements is
incorrect ?
(1) They contain strong covalents bonds in
their polymer chains.
(2) They contain covalent bonds between
various linear polymer chains.
(3) Examples are bakelite and melamine.
(4) They are formed from bi- and tri-functional
monomers.
Answer (1)
Sol. Cross linked or network polymers are formed
from bi-functional and tri-functional monomers
and contain strong covalent bonds between
various linear polymer chains, e.g. bakelite,
melamine etc. Option (1) is not related to
cross-linking.
173. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) In acidic (strong) medium aniline is
present as anilinium ion.
(2) Inspite of substituents nitro group always
goes to only m-position.
(3) In absence of substituents nitro group
always goes to m-position.
(4) In electrophilic substitution reactions
amino group is meta directive.
Answer (1)
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
174. Which of the following oxides is most acidic in
nature?
(1) CaO (2) MgO
(3) BaO (4) BeO
Answer (4)
Sol. BeO < MgO < CaO < BaO
Basic character increases.
So, the most acidic should be BeO. In fact,
BeO is amphoteric oxide while other given
oxides are basic.
175. The difference between amylose and
amylopectin is
(1) Amylose is made up of glucose and
galactose
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(3) Amylopectin have 1 - 4 a-linkage and
1 ? 6 3-linkage
(4) Amylose have 1 - 4 a-linkage and 1 - 6
3-linkage
Answer (2)
Sol. Amylose and Amylopectin are polymers of a-
D-glucose, so 13-link is not possible. Amylose is
linear with 1 - 4 a-linkage whereas
Amylopectin is branched and has both 1 - 4
and 1 - 6 a-linkages.
176. A mixture of 2.3 g formic acid and 4.5 g oxalic
acid is treated with conc. H
2
SO
4
. The evolved
gaseous mixture is passed through KOH
pellets. Weight (in g) of the remaining product
at STP will be
(1) 4.4 (2) 1.4
(3) 2.8 (4) 3.0
NH
2
NH
3
Sol.
H
34
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4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
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7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
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in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
14
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
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NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
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NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
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NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
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NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
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NEET (UG) - 2018 (Code-NN) HLAAC
106. Which one is wrongly matched?
(1) Unicellular organism ? Chlorella
(2) Uniflagellate gametes ? Polysiphonia
(3) Gemma cups ? Marchantia
(4) Biflagellate zoospores ? Brown algae
Answer (2)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (1, 3 & 4) are correctly
matched
107. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Herbarium (i) It is a place having a
collection of preserved
plants and animals
b. Key

(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum

(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue (iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
a b c d
(1) (iii) (iv) (i) (ii)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iv) (iii) (i)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
108. Which one of the following plants shows a
very close relationship with a species of moth,
where none of the two can complete its life
cycle without the other?
(1) Viola (2) Hydrilla
(3) Banana (4) Yucca
Answer (4)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
109. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?160?C
(2) ?120?C
(3) ?196?C
(4) ?80?C
Answer (3)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C
(Cryopreservation)
110. In which of the following forms is iron
absorbed by plants?
(1) Both ferric and ferrous
(2) Ferric
(3) Free element
(4) Ferrous
Answer (2*)
Sol. Iron is absorbed by plants in the form of ferric
ions. (According to NCERT)
*Plants absorb iron in both form i.e. Fe
++
and
Fe
+++
. (Preferably Fe
++
)
111. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Calcium (2) Magnesium
(3) Potassium (4) Sodium
Answer (3)
Sol. Potassium helps in maintaining turgidity of
cells.
112. Double fertilization is
(1) Syngamy and triple fusion
(2) Fusion of two male gametes of a pollen
tube with two different eggs
(3) Fusion of two male gametes with one egg
(4) Fusion of one male gamete with two polar
nuclei
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
113. What is the role of NAD
+
in cellular
respiration?
(1) It is the final electron acceptor for
anaerobic respiration.
(2) It functions as an enzyme.
(3) It is a nucleotide source for ATP synthesis.
(4) It functions as an electron carrier.
Answer (4)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
114. Oxygen is not produced during photosynthesis
by
(1) Chara
(2) Green sulphur bacteria
(3) Cycas
(4) Nostoc
Answer (2)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
115. The Golgi complex participates in
(1) Activation of amino acid
(2) Fatty acid breakdown
(3) Respiration in bacteria
(4) Formation of secretory vesicles
Answer (4)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
116. Stomatal movement is not affected by
(1) CO
2
concentration
(2) Temperature
(3) O
2
concentration
(4) Light
Answer (3)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
117. Stomata in grass leaf are
(1) Barrel shaped
(2) Dumb-bell shaped
(3) Rectangular
(4) Kidney shaped
Answer (2)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
118. Which of the following is true for nucleolus?
(1) It is a site for active ribosomal RNA
synthesis
(2) Larger nucleoli are present in dividing
cells
(3) It takes part in spindle formation
(4) It is a membrane-bound structure
Answer (1)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
119. Which of the following is not a product of light
reaction of photosynthesis?
(1) Oxygen (2) ATP
(3) NADPH (4) NADH
Answer (4)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
120. The stage during which separation of the
paired homologous chromosomes begins is
(1) Zygotene (2) Pachytene
(3) Diakinesis (4) Diplotene
Answer (4)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
121. The two functional groups characteristic of
sugars are
(1) Carbonyl and hydroxyl
(2) Hydroxyl and methyl
(3) Carbonyl and phosphate
(4) Carbonyl and methyl
Answer (1)
Sol. Sugar is a common term used to denote
carbohydrate.
Carbohydrates are polyhydroxy aldehyde,
ketone or their derivatives, which means they
have carbonyl and hydroxyl groups.
122. Which among the following is not a
prokaryote?
(1) Oscillatoria
(2) Saccharomyces
(3) Nostoc
(4) Mycobacterium
Answer (2)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
123. Offsets are produced by
(1) Parthenogenesis (2) Meiotic divisions
(3) Parthenocarpy (4) Mitotic divisions
Answer (4)
Sol. Offset is a vegetative part of a plant, formed
by mitosis.
? Meiotic divisions do not occur in somatic
cells.
? Parthenogenesis is the formation of
embryo from ovum or egg without
fertilisation.
? Parthenocarpy is the fruit formed without
fertilisation, (generally seedless)
124. Select the correct statement
(1) Transduction was discovered by S. Altman
(2) Franklin Stahl coined the term ?linkage?
(3) Spliceosomes take part in translation
(4) Punnett square was developed by a British
scientist
Answer (4)
Sol. Punnett square was developed by a British
geneticist, Reginald C. Punnett.
? Franklin Stahl proved semi-conservative
mode of replication.
? Transduction was discovered by Zinder
and Laderberg.
? Spliceosome formation is part of post-
transcriptional change in Eukaryotes
125. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Sporopollenin (2) Pollenkitt
(3) Oil content (4) Cellulosic intine
Answer (1)
Sol. Sporopollenin cannot be degraded by
enzyme; strong acids and alkali, therefore it is
helpful in preserving pollen as fossil.
Pollenkitt ? Help in insect pollination.
Cellulosic Intine ? Inner sporoderm layer of
pollen grain known as intine made up
cellulose & pectin.
Oil content ? No role is pollen preservation.
126. Which of the following pairs is wrongly
matched?
(1) T.H. Morgan : Linkage
(2) Starch synthesis in pea : Multiple alleles
(3) XO type sex : Grasshopper
determination
(4) ABO blood grouping : Co-dominance
Answer (2)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (1, 3 & 4) are correctly
matched.
127. Select the correct match
(1) Francois Jacob and - Lac operon
Jacques Monod
(2) Alec Jeffreys - Streptococcus
pneumoniae
(3) Matthew Meselson - Pisum sativum
and F. Stahl
(4) Alfred Hershey and - TMV
Martha Chase
Answer (1)
Sol. Francois Jacob and Jacque Monod proposed
model of gene regulation known as operon
model/lac operon.
? Alec Jeffreys ? DNA fingerprinting
technique.
? Matthew Meselson and F. Stahl ? Semi-
conservative DNA replication in E. coli.
? Alfred Hershey and Martha Chase ?
Proved DNA as genetic material not
protein
128. Which of the following flowers only once in its
life-time?
(1) Papaya (2) Bamboo species
(3) Mango (4) Jackfruit
Answer (2)
Sol. Bamboo species are monocarpic i.e., flower
generally only once in its life-time after 50-
100 years.
Jackfruit, papaya and mango are polycarpic
i.e., produce flowers and fruits many times in
their life-time.
129. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Virus (2) Fungus
(3) Plant (4) Bacterium
Answer (4)
25
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
130. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) pBR 322 (2) Retrovirus
(3) X phage (4) Ti plasmid
Answer (2)
Sol. Retrovirus is commonly used as vector for
introducing a DNA fragment in human
lymphocyte.
Gene therapy : Lymphocyte from blood of
patient are grown in culture outside the body,
a functional gene is introduced by using a
retroviral vector, into these lymphocyte.
131. Select the correct match
(1) G. Mendel - Transformation
(2) Ribozyme - Nucleic acid
(3) T.H. Morgan - Transduction
(4) F
2
? Recessive parent - Dihybrid cross
Answer (2)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
132. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bioexploitation (2) Bio-infringement
(3) Biodegradation (4) Biopiracy
Answer (4)
Sol. Biopiracy is term used for or refer to the use
of bioresources by multinational companies
and other organisation without proper
authorisation from the countries and people
concerned with compensatory payment
(definition of biopiracy given in NCERT).
133. A ?new? variety of rice was patented by a
foreign company, though such varieties have
been present in India for a long time. This is
related to
(1) Basmati (2) Co-667
(3) Lerma Rojo (4) Sharbati Sonora
Answer (1)
Sol. In 1997, an American company got patent
rights on Basmati rice through the US patent
and trademark office that was actually been
derived from Indian farmer?s varieties.
The diversity of rice in India is one of the
richest in the world, 27 documented varieties
of Basmati are grown in India.
Indian basmati was crossed with semi-dwarf
varieties and claimed as an invention or a
novelty.
Sharbati Sonora and Lerma Rojo are varieties
of wheat.
134. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Genetic Engineering Appraisal Committee
(GEAC)
(2) Indian Council of Medical Research (ICMR)
(3) Research Committee on Genetic
Manipulation (RCGM)
(4) Council for Scientific and Industrial
Research (CSIR)
Answer (1)
Sol. Indian Government has setup organisation
such as GEAC (Genetic Engineering Appraisal
Committee) which will make decisions
regarding the validity of GM research and
safety of introducing GM-organism for public
services. (Direct from NCERT).
135. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Denaturation, Annealing, Extension
(2) Extension, Denaturation, Annealing
(3) Denaturation, Extension, Annealing
(4) Annealing, Extension, Denaturation
Answer (1)
Sol. This technique is used for making multiple
copies of gene (or DNA) of interest in vitro.
Each cycle has three steps
(i) Denaturation
(ii) Primer annealing
(iii) Extension of primer
26
136. Identify the major products P, Q and R in the
following sequence of reactions:
(i) O
2

P Q + R
(ii) H
8
O / A
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

P Q R
137. Which of the following compounds can form a
zwitterion?
(1) Glycine (2) Aniline
(3) Benzoic acid (4) Acetanilide
Answer (1)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
pK
,
= 9.60 pK
,
= 2.34
CH
3

CH ? CH
3

O
2

(P)
CH
3
CH
3
? CH ? CH
3

H /H O
+
2
Hydroperoxide
Rearrangement
OH
O
(R)
(Q)
3
HC?C?O?O?H
+ CH
3
? C ? CH
3

27
NEET (UG) - 2018 (Code-NN) HLAAC
OH
CH(CH
3
)
2
CH
3
? CO ? CH
3
(1)
CH
3
CH
2
? OH
(2)
(3)
(4)
Answer (1)
? ? ? ? ? ? ??
Cl
Sol
Cl Cl
CO
1, 2?H
Shift
6+ 6+ 6?
(Incipient carbocation)
Cl
Ni
CO
6?
OC
AlCl
3

CO
Now,
.
CH
3
CH
2
CH
2
? Cl + Al
CH
3
? CH ? CH
3

CH
2
CH
2
CH
3
CHO
CH
2
CH
2
CH
3
CHO COOH
CH(CH
3
)
2
OH
CH
3
CH
2
CH
2
Cl AlCl
3

, CH
3
CH(OH)CH
3
H
2
N ? CH
2
? COO
?

138. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Tetrahedral geometry and paramagnetic
(2) Square planar geometry and diamagnetic
(3) Square planar geometry and
paramagnetic
(4) Tetrahedral geometry and diamagnetic
Answer (4)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
sp -hybridisation
3
~
H
3
N ? CH
2
? COO
?
139. Iron carbonyl, Fe(CO)
5
is
(1) Dinuclear (2) Tetranuclear
(3) Trinuclear (4) Mononuclear
Answer (4)
Sol. ? Based on the number of metal atoms
present in a complex, they are classified
into mononuclear, dinuclear, trinuclear and
so on.
eg: Fe(CO)
5
: mononuclear
Co
2
(CO)
8
: dinuclear
Fe
3
(CO)
12
: trinuclear
NEET (UG) - 2018 (Code-NN) HLAAC
NO
.
NO
2

H V
~
H
(1)
V
(2)
~
(3)
H
V
~
NO
.
(4)
V H
~
NO
2
140. Match the metal ions given in Column I with
the spin magnetic moments of the ions given
in Column II and assign the correct code :
Column I Column II
a. Co
3+
i.
8
BM
b. Cr
3+
ii.
35
BM
c. Fe
3+
iii.
3
BM
d. Ni
2+
iv.
24
BM
v.
15
BM
c d
i ii
ii i
ii iii
iii iv
Answer (2)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Spin magnetic moment = 4(4 + 2) = 24 BM
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Spin magnetic moment = 3(3 + 2) = 15 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
Spin magnetic moment = 2(2 + 2) = 8 BM
141. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Linkage isomerism
(2) Geometrical isomerism
(3) Ionization isomerism
(4) Coordination isomerism
Answer (2)
Sol. In [CoCl
2
(en)
2
], Coordination number of Co
is 6 and this compound has octahedral
geometry.
? As per given option, type of isomerism is
geometrical isomerism.
142. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) MnO
4
2?
(2) CrO
4
2?

(3) MnO
4
?
(4) Cr
2
O
7
2?
Answer (1)
Sol. CrO
4
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
Cr
2
O
7
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
MnO
4
2?
= Mn
6+
= [Ar] 3d
1

Unpaired electron (n) = 1; Paramagnetic
MnO
4
?
= Mn
7+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
143. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) CH
3
? CH = CH ? CH
3

(2) HC ~ C ? C ~ CH
(3) CH
2
= CH ? CH = CH
2

(4) CH
2
= CH ? C ~ CH
Answer (4)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
144. Which of the following carbocations is
expected to be most stable?
Answer (3)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (3)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
a b
(1) iii v
(2) iv v
(3) iv i
(4) i ii
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
28
M
10 HCl + 100 mL
M
10 NaOH e. 100 mL
NEET (UG) - 2018 (Code-NN) HLAAC
145. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NR
2
> ? OR > ? F
(2) ? NH
2
< ? OR < ? F
(3) ? NH
2
> ? OR > ? F
(4) ? NR
2
< ? OR < ? F
Answer (2*)
Sol. ?I effect increases on increasing
electronegativity of atom. So, correct order of
?I effect is ?NH
2
< ? OR < ? F.
*Most appropriate Answer is option (2),
however option (4) may also be correct answer.
146. The solubility of BaSO
4
in water is 2.42 ? 10
?3
gL
?1

at 298 K. The value of its solubility product
(K
sp
) will be
(Given molar mass of BaSO
4
= 233 g mol
?1
)
(1) 1.08 ? 10
?8
mol
2
L
?2

(2) 1.08 ? 10
?10
mol
2
L
?2

(3) 1.08 ? 10
?14
mol
2
L
?2

(4) 1.08 ? 10
?12
mol
2
L
?2

Answer (2)
148. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M M
a. 60 mL 10 HCl + 40 mL 10 NaOH
M M
b. 55 mL 10 HCl + 45 mL 10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
pH of which one of them will be equal to 1?
(1) c (2) b
(3) d (4) a
Answer (1)
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
3
(mol L
?1
) Sol. Solubility of BaSO
4
, s =
2.42 10 x
233
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
= 1.04 ? 10
?5
(mol L
?1
)
10 1
=
~
BaSO (s) Ba (aq) SO (aq)
2 2 ~
~~~
~~~
+
4 4
s s
K
sp
= [Ba
2+
] [SO
42?
]= s
2
= (1.04 ? 10
?5
)
2

= 1.08 ? 10
?10
mol
2
L
?2
147. Given van der Waals constant for NH
3
, H
2
, O
2

and CO
2
are respectively 4.17, 0.244, 1.36 and
3.59, which one of the following gases is most
easily liquefied?
(1) CO
2

(2) NH
3

(3) O
2

(4) H
2

Answer (2)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
100 10
pH = ?log[H
+
] =
1
~
~ ~
~ ~
log 10
~
~ = 1.0
149. On which of the following properties does the
coagulating power of an ion depend?
(1) The sign of charge on the ion alone
(2) The magnitude of the charge on the ion
alone
(3) Both magnitude and sign of the charge on
the ion
(4) Size of the ion alone
Answer (3)
Sol. ? Coagulation of colloidal solution by using
an electrolyte depends on the charge
present (positive or negative) on colloidal
particles as well as on its size.
? Coagulating power of an electrolyte
depends on the magnitude of charge
present on effective ion of electrolyte.
29
NEET (UG) - 2018 (Code-NN) HLAAC
S S
? ?
F
? ?
S S
F Cl
S S
F
S S
(1) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
(2) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(3) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(4) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
Answer (1)
Sol.
C
2
H
5
OH
Na
C
2
H
5
O Na
+

(A) (B)
PCl
5

C
2
H
5
Cl
(C)
C
2
H
5
O Na + + C
2
H
5
Cl
S
N
2

C
2
H
5
OC
2
H 5
(B) (C)
150. Which of the following statements is not true
for halogens?
(1) Chlorine has the highest electron-gain
enthalpy
(2) All form monobasic oxyacids
(3) All but fluorine show positive oxidation
states
(4) All are oxidizing agents
Answer (3)
Sol. Due to high electronegativity and small size,
F forms only one oxoacid, HOF known as
Fluoric (I) acid. Oxidation number of F is +1 in
HOF.
151. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
(1) Cu (2) Fe
(3) Mg (4) Zn
Answer (3)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
152. The correct order of atomic radii in group 13
elements is
(1) B < Ga < Al < In < Tl
(2) B < Al < In < Ga < Tl
(3) B < Ga < Al < Tl < In
(4) B < Al < Ga < In < Tl
Answer (1)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
153. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) Three (2) One
(3) Four (4) Two
Answer (4)
Sol. The structure of ClF
3
is
The number of lone pair of electrons on
central Cl is 2.
154.The correct order of N-compounds in its
decreasing order of oxidation states is
(1) NH
4
Cl, N
2
, NO, HNO
3

(2) HNO
3
, NO, N
2
, NH
4
Cl
(3) HNO
3
, NH
4
Cl, NO, N
2

(4) HNO
3
, NO, NH
4
Cl, N
2

Answer (2)
+ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
155. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) In
(2) Ga
(3) B
(4) Al
Answer (3)
Sol. ? 'B' has no vacant d-orbitals in its valence
shell, so it can't extend its covalency beyond
4. i.e. 'B' cannot form the ion like MF
6
3(?)
i.e.
BF
6
3(?)
.
156. The compound A on treatment with Na gives
B, and with PCl
5
gives C. B and C react
together to give diethyl ether. A, B and C are
in the order
30
NEET (UG) - 2018 (Code-NN) HLAAC
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
1
,
k[A ]
0
157. Hydrocarbon (A) reacts with bromine by
substitution to form an alkyl bromide which by
Wurtz reaction is converted to gaseous
hydrocarbon containing less than four carbon
atoms. (A) is
(1) CH
4
(2) CH ~ CH
(3) CH
3
? CH
3
(4) CH
2
= CH
2

Answer (1)
Sol.
CH
4
Br
2
/h v
CH
3
Br
Na/dry ether
Wurtz reaction
CH
3
? CH
3

158. The compound C
7
H
8
undergoes the following
reactions:
3Cl
2
/ A Br
2
/Fe Zn/HCl
C
7
H
8
~~~> ~~~> ~~~~ A B C
The product 'C' is
(1) p-bromotoluene
(2) m-bromotoluene
(3) 3-bromo-2,4,6-trichlorotoluene
(4) o-bromotoluene
Answer (2)
CCl
3

3dl
2
A
(C
7
H
8
)
160. The correct difference between first and
second order reactions is that
(1) The rate of a first-order reaction does
depend on reactant concentrations; the
rate of a second-order reaction does not
depend on reactant concentrations
(2) The rate of a first-order reaction does not
depend on reactant concentrations; the
rate of a second-order reaction does
depend on reactant concentrations
(3) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(4) The half-life of a first-order reaction does
not depend on [A]
0
; the half-life of a
second-order reaction does depend on
[A]
0

Answer (4)
,
k
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
161. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BaH
2
< BeH
2
< CaH
2

(2) BeH
2
< CaH
2
< BaH
2

(3) BeH
2
< BaH
2
< CaH
2

(4) CaH
2
< BeH
2
< BaH
2

Answer (2)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
162. Consider the change in oxidation state of
Bromine corresponding to different emf values
as shown in the diagram below :
Then the species undergoing
disproportionation is
(1) HBrO (2)
~
BrO
3

(3) Br
2
(4)
~
BrO
4
(A)
CH
3
CCl
3

Sol.
Br
2
Fe
Br
(A) (B)
Zn HCl
CH
3
Br
(C)
159. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) NO
(2) N
2
O
5
(3) N
2
O
(4) NO
2

Answer (2)
Sol. Fact
Sol. ? For first order reaction,
1 /2
=
0.693
t
31
NEET (UG) - 2018 (Code-NN) HLAAC
CH
3

(1) CH
3
OH and I
2

(2)
H
3
C CH
2
? OH and I
2
COONa
Sodium benzoate
Answer (1)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+5
~~* BrO , E
~
~ o
HBrO
3
BrO /HBrO
= 1.5 V
3
E for the disproportionation of HBrO,
o
cell
E
o
= E
o o
cell
~ E
~ HBrO/Br
2 BrO /HBrO
3
= 1.595 ? 1.5
= 0.095 V = + ve
163. In which case is number of molecules of
water maximum?
(1) 10
?3
mol of water
(2) 18 mL of water
(3) 0.00224 L of water vapours at 1 atm and
273 K
(4) 0.18 g of water
Answer (2)
Sol. (1) Molecules of water = mole ? N
A
= 10
?3
N
A

(2) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(4) Molecules of water = mole ? N
A
= A
18
= 10
?2
N
A

164. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intermolecular H-bonding
(2) Formation of intramolecular H-bonding
(3) More extensive association of carboxylic
acid via van der Waals force of attraction
(4) Formation of carboxylate ion
Answer (1)
Sol. Due to formation of intermolecular H-bonding
in carboxylic acid, association occurs. Hence
boiling point increases and become more
than the boiling point of aldehydes, ketones
and alcohols of comparable molecular
masses.
165. Compound A,
C
8
H
10
O,
is found to react with
NaOI (produced by reacting Y with NaOH) and
yields a yellow precipitate with characteristic
smell.
A and Y are respectively
CH ? CH
3
and I
2
(3)
OH
2 (4)
CH
2
? CH
2
? OH and I
Answer (3)
Sol. Option (3) is secondary alcohol which on
oxidation gives phenylmethyl ketone
(Acetophenone). This on reaction with I
2
and
NaOH form iodoform and sodium benzoate.
2NaOH + I -* NaOI + NaI + H
2
O
2
I
2
NaOH
166. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichlorocarbene ( : CCl
2
)
(2) Dichloromethyl cation
(
CHCl
2
)
~
(3) Dichloromethyl anion (
CHCl
2
)
?
(4) Formyl cation (
C H O
)
~
+ CHI
3

Iodoform
(Yellow PPt)
CHO
= N
A

0.00224
22.4
NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
= 10
?4

(3) Moles of water =
32
?
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
NEET (UG) - 2018 (Code-NN) HLAAC
Answer (1)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
CCl ~~~ : CCl + Cl
3 2
Electrophile
167. The bond dissociation energies of X
2
, Y
2
and
XY are in the ratio of 1 : 0.5 : 1. AH for the
formation of XY is ?200 kJ mol
?1
. The bond
dissociation energy of X
2
will be
(1) 400 kJ mol
?1

(2) 200 kJ mol
?1

(3) 800 kJ mol
?1

(4) 100 kJ mol
?1

Answer (3)
Sol. The reaction for A
f
H?(XY)
1
+
1
X (g) Y (g) XY(g) ~~~
2 2
2 2
Bond energies of X
2
, Y
2
and XY are X, X 2 , X
respectively
A =
?
+
?
_ = _
? ? X X
H X 200
? ?
2 4
On solving, we get
? - + = - X X 200
2 4
= X = 800 kJ/mole
168. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Remains unchanged
(2) Is halved
(3) Is tripled
(4) Is doubled
Answer (4)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
169. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Forces of attraction between the gas
molecules
(2) Density of the gas molecules
(3) Electric field present between the gas
molecules
(4) Volume of the gas molecules
Answer (1)
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
170. Which one of the following conditions will
favour maximum formation of the product in
the reaction,
A
2
(g) + B
2
(g) ~~=
~~~ X
2
(g) A
r
H = -X kJ?
(1) High temperature and low pressure
(2) Low temperature and high pressure
(3) High temperature and high pressure
(4) Low temperature and low pressure
Answer (2)
Sol.
A
2
(g) + B
2
(g)
=~~
~~=
X
2
(g); A H = -x kJ
On increasing pressure equilibrium shifts in a
direction where pressure decreases i.e.
forward direction.
On decreasing temperature, equilibrium shifts
in exothermic direction i.e., forward direction.
So, high pressure and low temperature
favours maximum formation of product.
171. For the redox reaction
~ 2 ~ ~ ~
+ Mn CO
2
H
2
O + ~~~ + +
2
MnO C O H
4 2 4
The correct coefficients of the reactants for
the balanced equation are
MnO
4
~
C
2
O
4
H
+ 2~
(1) 5 16 2
(2) 16 5 2
(3) 2 16 5
(4) 2 5 16
~
t
1/2
~
Sol. In real gas equation,
?
? ?
2
?
+ - =
an
P (V nb) nRT
2 ?
V
?
33
NEET (UG) - 2018 (Code-NN) HLAAC
Answer (4)
Reduction
+7 +3 +4
Sol.
MnO
4
?
+ C
2
O
4
2?
+ H
+
Mn
2+
+ CO
2
+ H
2
O
Oxidation
n-factor of MnO
~
4
~ 5
n-factor of
2
C
2
O
4
~ ~ 2
Ratio of n-factors of MnO
~
4
and
2
C
2
O
4
~
is 5 : 2
So, molar ratio in balanced reaction is 2 : 5
~ The balanced equation is
2MnO 5C O 16H 2Mn 10CO
2
8H
2
O
~ 2 ~
2
+ +
4
+ + ?
2 4
172. Regarding cross-linked or network polymers,
which of the following statements is
incorrect ?
(1) They contain strong covalents bonds in
their polymer chains.
(2) They contain covalent bonds between
various linear polymer chains.
(3) Examples are bakelite and melamine.
(4) They are formed from bi- and tri-functional
monomers.
Answer (1)
Sol. Cross linked or network polymers are formed
from bi-functional and tri-functional monomers
and contain strong covalent bonds between
various linear polymer chains, e.g. bakelite,
melamine etc. Option (1) is not related to
cross-linking.
173. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) In acidic (strong) medium aniline is
present as anilinium ion.
(2) Inspite of substituents nitro group always
goes to only m-position.
(3) In absence of substituents nitro group
always goes to m-position.
(4) In electrophilic substitution reactions
amino group is meta directive.
Answer (1)
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
174. Which of the following oxides is most acidic in
nature?
(1) CaO (2) MgO
(3) BaO (4) BeO
Answer (4)
Sol. BeO < MgO < CaO < BaO
Basic character increases.
So, the most acidic should be BeO. In fact,
BeO is amphoteric oxide while other given
oxides are basic.
175. The difference between amylose and
amylopectin is
(1) Amylose is made up of glucose and
galactose
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(3) Amylopectin have 1 - 4 a-linkage and
1 ? 6 3-linkage
(4) Amylose have 1 - 4 a-linkage and 1 - 6
3-linkage
Answer (2)
Sol. Amylose and Amylopectin are polymers of a-
D-glucose, so 13-link is not possible. Amylose is
linear with 1 - 4 a-linkage whereas
Amylopectin is branched and has both 1 - 4
and 1 - 6 a-linkages.
176. A mixture of 2.3 g formic acid and 4.5 g oxalic
acid is treated with conc. H
2
SO
4
. The evolved
gaseous mixture is passed through KOH
pellets. Weight (in g) of the remaining product
at STP will be
(1) 4.4 (2) 1.4
(3) 2.8 (4) 3.0
NH
2
NH
3
Sol.
H
34
H
2
O(l)
O(l)
mol
Sol.
HCOOH
Conc.H
2
SO
4
CO(g) + H
2

? ? 1
1

2.3g or
? mol ?
? 20 ? 20
2
x 28 = 2.8g
20
1s
2
2s
2
2p
3

OR
1s
2
2s
2
2p
3

~ Option (3) violates Hund's Rule.
NEET (UG) - 2018 (Code-NN) HLAAC
Answer (3)
COOH
? 1 ?
4.5g or
?
mol
?

? 20 ?
- =
3
BO =
10 4
2
NO and CN
Which one ofthese will have the highestbond
(4)
= (7t2py)2,(7t*2px)1
1
COOH
Conc.H 2SO4
CO(g) + CO 2 (g) +
20
mol
1
20
mol
CN :
(a2s)2,(a*2s)2,
(a1s)
2
, (a* 1s)
2
, (72p
x
)
2
(72p
y
)
2
, (a2p
z
)
1

BO = ~ ~
9 4
2 2.5
178. Considerthe following
1s
2
2s
2
2p 2p 2p
1
z

(4) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
Answer (3)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
: (
a*1
s
)2, (a2s)2, (a* 2s)2 ,
CN
+
(a1s)
2
, (72p
x
)
2

= (i2p
y
)
2

order?
(1) CN
(2) NO
(3)
Answer (4)
Sol. NO :
BO = 8
-
4
2
2
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So th e remaining gas is CO.
So, weight of remaining gaseous product CO
is
species :
CN+,CN?,
CN+
CN?
(a1s)2,(a*1s)2,(c2s)2,(c*2s)2,(c2pz)2,
(72px)2 = (~*2py)0
BO = 10 _5 25
2
CN? : (a1s)2,(a*1s)2,
(a2s)2,(a*2s)2,
(72px)2
= (72py)2,(a2pz)2
=
177. Which one is a wrong statement?
(1) The value of m for d
z
2 is zero
(2) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(3) The electronic configuration of N atom is
35
179. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 2s
2
2p
3
,
the simplest formula for this compound is
(1)
(2) Mg
2
X
3

(3) Mg
2
X
(4)
Answer (1)
Sol. Element (X) electronic configuration
1s
2
Mg
3
X
2
MgX
2
1s
2
2s
2
2p
3

So, valency of X will be 3.
Valency of Mg is 2.
Formula of compound formed by Mg and X will
be Mg
3
X
2
.
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Important Instructions :
1. The test is of 3 hours duration and this Test Booklet contains 180 questions. Each question carries
4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response,
one mark will be deducted from the total scores. The maximum marks are 720.
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the Answer Sheet. Do not write your Roll No. anywhere else except in the specified space in the
Test Booklet/Answer Sheet.
7. Each candidate must show on demand his/her Admission Card to the Invigilator.
8. No candidate, without special permission of the Superintendent or Invigilator, would leave his/her
seat.
9. Use of Electronic/Manual Calculator is prohibited.
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12. The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet
in the Attendance Sheet.
DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
NN
HLAAC
1
NEET (UG) - 2018 (Code-NN) HLAAC
Now,
u
2
= ?20
1 1 1
= +
f v u
2 2
1 1 1
~ ?
?15 v 20
2
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
3. An em wave is propagating in a medium with
1 1 1
= +
f v u
1
1
= +
1 1
v ?15 40
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
=
1 1 1
? ~ ?
15 v 40
1
1. The magnetic potential energy stored in a
certain inductor is 25 mJ, when the current in
the inductor is 60 mA. This inductor is of
inductance
(1) 13.89 H (2) 0.138 H
(3) 1.389 H (4) 138.88 H
Answer (1)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
~
a velocity V Vi
?
= . The instantaneous
25 2 10 10 x x x
6 ?3
L ~
3600
500
~
36
= 13.89 H
2. An object is placed at a distance of 40 cm
from a concave mirror of focal length 15 cm.
If the object is displaced through a distance
of 20 cm towards the mirror, the
displacement of the image will be
(1) 36 cm towards the mirror
(2) 30 cm away from the mirror
(3) 30 cm towards the mirror
(4) 36 cm away from the mirror
Answer (4)
Sol.
f = 15 cm
O 40 cm
oscillating electric field of this em wave is
along +y axis. Then the direction of oscillating
magnetic field of the em wave will be along
(1) ?x direction
(2) ?z direction
(3) ?y direction
(4) +z direction
Answer (4)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
~
Direction of propagation is along +z direction.
4. The refractive index of the material of a
prism is 2 and the angle of the prism is 30?.
One of the two refracting surfaces of the
prism is made a mirror inwards, by silver
coating. A beam of monochromatic light
entering the prism from the other face will
retrace its path (after reflection from the
silvered surface) if its angle of incidence on
the prism is
(1) Zero
(2) 60?
(3) 30?
(4) 45?
Answer (4)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
So, 6 6k
?
=
2
(20 0) -
C
4 10
3
x
I
C
= 5 * 10
?3
= 5 mA
V
i
= V
BE
+ I
B
R
B
V
i
= 0 + I
B
R
B
20 = I
B
* 500 * 10
3

20
I = = ~ 40 A
B
500 10
3
x
I 25 10
125
~
"
C
= R _ _
~ 6
3
I
40 10 x
b
6. In a p-n junction diode, change in temperature
due to heating
(1) Affects the overall V - I characteristics of
p-n junction
(2) Affects only reverse resistance
(3) Does not affect resistance of p-n junction
(4) Affects only forward resistance
Answer (1)
Sol. Due to heating, number of electron-hole pairs
will increase, so overall resistance of diode
will change.
Due to which forward biasing and reversed
biasing both are changed.
7. In the combination of the following gates the
output Y can be written in terms of inputs A
and B as
A
B
Y
(1) A + B
(2) A ? B
(3) A ? B + A ? B
(4) A ? B + A ? B
Answer (4)
sin i ~
1
i.e. i = 45?
2
5. In the circuit shown in the figure, the input
voltage V
i
is 20 V, V
BE
= 0 and V
CE
= 0. The
values of I
B
, I
C
and 13 are given by
20 V
R
C
4 kQ
C
V
i
500 kQ
R
B
B
E
(1) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
(2) I
B
= 40 JLA, I
C
= 10 mA, 3 = 250
(3) I
B
= 20 ~A, I
C
= 5 mA, 13 = 250
(4) I
B
= 25 ~A, I
C
= 5 mA, 13 = 200
Answer (1)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
20 V
I
A ? B
A
Sol. A
B
V
B
A
A ? B
B
V = (A . B + A . B)
3
R
B

V
i
V
b
500 kQ I
b
R
0
= 4 kQ
I
C
NEET (UG) - 2018 (Code-NN) HLAAC
i
30?
M
60?
p = 2
30?
Applying Snell's law at M,
sin i 2
~
sin30 1 ?
1
= sin i 2 = x
2
NEET (UG) - 2018 (Code-NN) HLAAC
A, 3l
F
F'
3A, l
8. The power radiated by a black body is P and
it radiates maximum energy at wavelength, X
0
.
If the temperature of the black body is now
changed so that it radiates maximum energy
at wavelength X
0

3
4
the power radiated by it ,
becomes nP. The value of n is
(1)
81
(2)
3
256 4
(3)
256
(4)
4
81 3
Answer (3)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ = A
max
2
T
2
~ X =
3 ~
0
' T T
0
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
9. Two wires are made of the same material and
have the same volume. The first wire has
cross-sectional area A and the second wire
has cross-sectional area 3A. If the length of
the first wire is increased by Al on applying a
force F, how much force is needed to stretch
the socond wire by the same amount?
(1) F (2) 9 F
(3) 4 F (4) 6 F
Answer (2)
Sol. Wire 1 :
Wire 2 :
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F' A l
_ Y
3A l
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
From equation (i) & (ii),
A =
~ ~
~
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
= F' = 9F
10. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 ? 10
5
Nm
?2
) requires
54 cal of heat energy to convert to steam at
100?C. If the volume of the steam produced is
167.1 cc, the change in internal energy of the
sample, is
(1) 84.5 J (2) 104.3 J
(3) 42.2 J (4) 208.7 J
Answer (4)
Sol. AQ = AU + AW
= 54 ? 4.18 = AU + 1.013 ? 10
5
(167.1 ? 10
?6
? 0)
= AU = 208.7 J
11. A small sphere of radius 'r' falls from rest in a
viscous liquid. As a result, heat is produced
due to viscous force. The rate of production of
heat when the sphere attains its terminal
velocity, is proportional to
(1) r
4
(2) r
3
(3) r
5
(4) r
2
Answer (3)
2
Sol. Power = lrll . = lrll 6 rV
T
V
T
6 rV
T
V
T
r ?
2
~ Power r oc
5

12. When the light of frequency 2?
0
(where v
0
is
threshold frequency), is incident on a metal
plate, the maximum velocity of electrons
emitted is v
1
. When the frequency of the
incident radiation is increased to 5 v
0
, the
maximum velocity of electrons emitted from
the same plate is v
2
. The ratio of v
1
to v
2
is
(1) 2 : 1
(2) 1 : 2
(3) 4 : 1
(4) 1 : 4
Answer (2)
4 5
h
mV
0

...(i) x
0
=
E
0
a -
e E
0

Sol. = +
2
1
E W mv
0
2
v = v +
1
2
h(2 ) h mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
2
h(5 ) h v = v ~ 1 mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
1 v
_
2
2
4 v
~
v 1
1
v 2
2
13. For a radioactive material, half-life is 10
minutes. If initially there are 600 number of
nuclei, the time taken (in minutes) for the
disintegration of 450 nuclei is
(1) 15 (2) 20
(3) 30 (4) 10
Answer (2)
Sol. Number of nuclei remaining = 600 ? 450 = 150
150 1
_
~ ~
( ~
600 2
~
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 x 10
= 20 minute
14. An electron of mass m with an initial velocity
~
E ?E
0
?

i =
(E
0
= constant > 0) at t = 0. If X
0
is its
de-Broglie wavelength initially, then its de-
Broglie wavelength at time t is
(1) A
0
x
0
~ ~
(2) ~
+
eE
0
1
t ~
~
mV
0 ~
(3) X
0
t
~ ~
0
(4) ~
+
eE
1 t
0 ~
~
mV
0 ~
Answer (2)
Sol. Initial de-Broglie wavelength
Acceleration of electron
m
Velocity after time ?t?
~ ~
0
=
~
+
eE
V V t
0 ~
~ m ~
h h
mV m V
eE
t
I
+
~
I ~
0
0
~ m ~
h
~ ~
mV
0

Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
15. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : ?2 (2) 1 : 1
(3) 2 : ?1 (4) 1 : ?1
Answer (4)
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
16. A tuning fork is used to produce resonance in
a glass tube. The length of the air column in
this tube can be adjusted by a variable piston.
At room temperature of 27?C two successive
resonances are produced at 20 cm and 73 cm
of column length. If the frequency of the
tuning fork is 320 Hz, the velocity of sound in
air at 27?C is
(1) 300 m/s (2) 330 m/s
(3) 350 m/s (4) 339 m/s
Answer (4)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 x 320 [73 ? 20] x 10
?2

= 339.2 ms
?1

= 339 m/s
V
0
F
So, X = =
~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
...(ii)
NEET (UG) - 2018 (Code-NN) HLAAC
N 1
_
~
('
N
0
2
n
~
~
~
t
t
1/2
t
t
1/2
V V
0
?

i = (V
0
> 0) enters an electric field
-*
x
0
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
S
2hm
eE
17. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Inversely proportional to the distance
between the plates
(2) Independent of the distance between the
plates
(3) Proportional to the square root of the
distance between the plates
(4) Linearly proportional to the distance
between the plates
Answer (2)
Sol. For isolated capacitor Q = Constant
F
plate
0
F is Independent of the distance between
plates.
18. A pendulum is hung from the roof of a
sufficiently high building and is moving freely
to and fro like a simple harmonic oscillator.
The acceleration of the bob of the pendulum
is 20 m/s
2
at a distance of 5 m from the mean
position. The time period of oscillation is
(1) 1 s (2) 2ir s
(3) 2 s (4) ir s
Answer (4)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
19. An electron falls from rest through a vertical
distance h in a uniform and vertically upward
directed electric field E. The direction of
electric field is now reversed, keeping its
magnitude the same. A proton is allowed to
fall from rest in it through the same vertical
distance h. The time of fall of the electron, in
comparison to the time of fall of the proton is
(1) Equal (2) Smaller
(3) 10 times greater (4) 5 times greater
Answer (2)
Sol. =
1 eE
2
h t
2 m
~ = t
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
20. The kinetic energies of a planet in an
elliptical orbit about the Sun, at positions A, B
and C are K
A
, K
B
and K
C
, respectively. AC is
the major axis and SB is perpendicular to AC
at the position of the Sun S as shown in the
figure. Then
(1) K
B
> K
A
> K
C

(2) K
A
< K
B
< K
C

(3) K
B
< K
A
< K
C

(4) K
A
> K
B
> K
C

Answer (4)
Sol. B
perihelion
A
V
A
Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

21. A solid sphere is in rolling motion. In rolling
motion a body possesses translational kinetic
energy (K
t
) as well as rotational kinetic
energy (K
r
) simultaneously. The ratio
K
t
: (K
t
+ K
r
) for the sphere is
(1) 2 :
(3) 10
Answer (4)
Sol. =
2
1
K mv
t
2
K K mv I
t r ~~ ~
+ ~ + 0) =
2 2 2 2 5
1 1 ( ~~ ~
2
mv
2
+
~
2
1 1 2 v
~ ~~ ~
mr 2

r
2
7
= my
2
10
So, K K 7
t
+
r
K
~
5
t
Q
2
~
2AE
5 (2) 7 : 10
: 7 (4) 5 : 7
S
C
V
C

C
aphelion
6
0.5 9.8 x
= 11.32 A
0.25 3 x
NEET (UG) - 2018 (Code-NN) HLAAC
22. If the mass of the Sun were ten times smaller
and the universal gravitational constant were
ten times larger in magnitude, which of the
following is not correct?
(1) ?g? on the Earth will not change
(2) Raindrops will fall faster
(3) Time period of a simple pendulum on the
Earth would decrease
(4) Walking on the ground would become
more difficult
Answer (1)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (1) is wrong option.
23. A solid sphere is rotating freely about its
symmetry axis in free space. The radius of the
sphere is increased keeping its mass same.
Which of the following physical quantities
would remain constant for the sphere?
(1) Angular momentum
(2) Angular velocity
(3) Rotational kinetic energy
(4) Moment of inertia
Answer (1)
Sol. ~
ex
= 0
So
,
=
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
24. A metallic rod of mass per unit length
0.5 kg m
?1
is lying horizontally on a smooth
inclined plane which makes an angle of 30?
with the horizontal. The rod is not allowed to
slide down by flowing a current through it
when a magnetic field of induction 0.25 T is
acting on it in the vertical direction. The
current flowing in the rod to keep it stationary
is
(1) 11.32 A (2) 7.14 A
(3) 14.76 A (4) 5.98 A
Answer (1)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
=
mg
I tan30?
B
30?
25. An inductor 20 mH, a capacitor 100 jtF and a
resistor 50 Q are connected in series across
a source of emf, V = 10 sin 314 t. The power
loss in the circuit is
(1) 1.13 W (2) 0.79 W
(3) 2.74 W (4) 0.43 W
Answer (2)
2
( ~ V
RMS
Sol. P =
~ ~ R av
~ ~
Z
2
= +
~
w -
~
= Q
~ 1 ~
Z R
2
L 56
~
w
)
C
2
~ ~
10
~
~ ~
P =
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
26. A thin diamagnetic rod is placed vertically
between the poles of an electromagnet. When
the current in the electromagnet is switched
on, then the diamagnetic rod is pushed up, out
of the horizontal magnetic field. Hence the
rod gains gravitational potential energy. The
work required to do this comes from
(1) The induced electric field due to the
changing magnetic field
(2) The current source
(3) The lattice structure of the material of the
rod
(4) The magnetic field
Answer (2)
Sol. Energy of current source will be converted
into potential energy of the rod.
27. Current sensitivity of a moving coil
galvanometer is 5 div/mA and its voltage
sensitivity (angular deflection per unit voltage
applied) is 20 div/V. The resistance of the
galvanometer is
(1) 500 Q (2) 40 Q
(3) 250 Q (4) 25 Q
Answer (3)
30?
l lB
7
NEET (UG) - 2018 (Code-NN) HLAAC
NBA
~ V
S

CR
G
~
2l'
3v v
~
4 l 2l '
= '
l
4 2 l l
~ ~
3 2 3 x
Sol. Current sensitivity
NBA
~
C
Voltage sensitivity
So, resistance of galvanometer
5 1 ~ 5000
= =
I
S
R ~ ~ ~ 250 G

x ~
3
V 20 10 20
S
28. The volume (V) of a monatomic gas varies
with its temperature (T), as shown in the
graph. The ratio of work done by the gas, to
the heat absorbed by it, when it undergoes a
change from state A to state B, is
(1) 2 (2) 2
7 5
(3) 1 (4) 2
3 3
Answer (2)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
dW nRdT 2
Required ratio
= = =
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
29. The fundamental frequency in an open organ
pipe is equal to the third harmonic of a closed
organ pipe. If the length of the closed organ
pipe is 20 cm, the length of the open organ
pipe is
(1) 16 cm
(2) 13.2 cm
(3) 12.5 cm
(4) 8 cm
Answer (2)
Sol. For closed organ pipe, third harmonic
~
3v
4l
For open organ pipe, fundamental frequency
v
Given,
2 20
13.33 cm
x
= =
3
30. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
(1) 12.5%
(2) 26.8%
(3) 6.25%
(4) 20%
Answer (2)
Sol. Efficiency of ideal heat engine,
~ ~
~ ~
~
~
~
T
1 T
2
~ ~ 1
T
2
: Sink temperature
T
1
: Source temperature
ii =
~
_
~
x
~ ~ T
% 1
2
100
~ ~
T
1
~
i
-
I
X
( 273 ~
1
~ 373_I
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
31. At what temperature will the rms speed of
oxygen molecules become just sufficient for
escaping from the Earth's atmosphere?
(Given :
Mass of oxygen molecule (m) = 2.76 * 10
?26
kg
Boltzmann's constant k
B
= 1.38 * 10
?23
JK
?1
)
(1) 1.254 * 10
4
K (2) 2.508 * 10
4
K
(3) 5.016 * 10
4
K (4) 8.360 * 10
4
K
Answer (4)
I
S
100
8
NEET (UG) - 2018 (Code-NN) HLAAC
3k T
B
So, = 11200 m/s
m
O
2
h
B
A
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
On solving,
T = 8.360 ? 10 K
4
32. Unpolarised light is incident from air on a
plane surface of a material of refractive index
'j'. At a particular angle of incidence 'i', it is
found that the reflected and refracted rays are
perpendicular to each other. Which of the
following options is correct for this situation?
(1) i tan
_
~ ~
1
1
=
~ ~
~
~
~
(2) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(3) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(4) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
Answer (4)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
~
Also, tan i = p. (Brewster angle)
33. In Young's double slit experiment the
separation d between the slits is 2 mm, the
wavelength X of the light used is 5896 ? and
distance D between the screen and slits is
100 cm. It is found that the angular width of
the fringes is 0.20?. To increase the fringe
angular width to 0.21? (with same X and D) the
separation between the slits needs to be
changed to
(1) 1.7 mm (2) 1.8 mm
(3) 2.1 mm (4) 1.9 mm
Answer (4)
x
Sol. Angular width d
~
0.20 ? =
...(i)
2 mm
...(ii)
d
0.20 d
Dividing we get,
~ d = 1.9 mm
34. An astronomical refracting telescope will
have large angular magnification and high
angular resolution, when it has an objective
lens of
(1) Small focal length and small diameter
(2) Small focal length and large diameter
(3) Large focal length and large diameter
(4) Large focal length and small diameter
Answer (3)
Sol. For telescope, angular magnification = f
0
f
E
So, focal length of objective lens should be
large.
D
Angular resolution =
X 1.22
should be large.
So, objective should have large focal length
(f
0
) and large diameter D.
35. A body initially at rest and sliding along a
frictionless track from a height h (as shown in
the figure) just completes a vertical circle of
diameter AB = D. The height h is equal to
5 D
(2) 3 D (1)
4 2
7
(3) (4) D
5
Answer (1)
--
0.21 0 _ ~
~
0.21 2 mm
9
h
v
L
2
~
A =
1
W I ~
2
2
A
W I for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
=
=
=
W
C
> W
B
> W
A

2
:
5
4 :
1
:
2
5 :
1
10
h
NEET (UG) - 2018 (Code-NN) HLAAC
B
A
As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

1
0 mgh mv 0
2
+ = +
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
36. Three objects, A : (a solid sphere), B : (a thin
circular disk) and C : (a circular ring), each
have the same mass M and radius R. They all
spin with the same angular speed w about
their own symmetry axes. The amounts of
work (W) required to bring them to rest, would
satisfy the relation
(1) W
A
> W
C
> W
B

(2) W
C
> W
B
> W
A

(3) W
B
> W
A
> W
C

(4) W
A
> W
B
> W
C

Answer (2)
Sol. Work done required to bring them rest
~W = AKE
37. Which one of the following statements is
incorrect?
(1) Coefficient of sliding friction has
dimensions of length.
(2) Rolling friction is smaller than sliding
friction.
(3) Frictional force opposes the relative
motion.
(4) Limiting value of static friction is directly
proportional to normal reaction.
Answer (1)
Sol. Coefficient of sliding friction has no
dimension.
f = ~
s
N
~ ~
s
~
38. A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block is
v, then the value of coefficient of restitution
(e) will be
(1) 0.4 (2) 0.5
(3) 0.8 (4) 0.25
Answer (4)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
v ~ ~
v
4
v
e =
Relative velocity of separation 4
=
Relative velocity of approach v
e = =
1
0.25
4
39. A carbon resistor of (47 ? 4.7) kQ is to be
marked with rings of different colours for its
identification. The colour code sequence will
be
(1) Green ? Orange ? Violet ? Gold
(2) Violet ? Yellow ? Orange ? Silver
(3) Yellow ? Green ? Violet ? Gold
(4) Yellow ? Violet ? Orange ? Silver
Answer (4)
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
Sol.
v
L

f
N
10
(1)
O
I
(3)
O
I
n
O
NEET (UG) - 2018 (Code-NN) HLAAC
40. A set of 'n' equal resistors, of value 'R' each,
are connected in series to a battery of emf 'E'
and internal resistance 'R'. The current drawn
is I. Now, the 'n' resistors are connected in
parallel to the same battery. Then the current
drawn from battery becomes 10 I. The value of
'n' is
(1) 9 (2) 10
(3) 20 (4) 11
Answer (2)
Sol. = I
nR R +
10 I =
R R
E
+
n
Dividing (ii) by (i),
(n 1)R +
~ ~
~
+
~
1
1 R
~ )
n
After solving the equation, n = 10
41. A battery consists of a variable number 'n' of
identical cells (having internal resistance 'r'
each) which are connected in series. The
terminals of the battery are short-circuited
and the current I is measured. Which of the
graphs shows the correct relationship
between I and n?
I
n
I
n
Answer (2)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
42. A toy car with charge q moves on a
frictionless horizontal plane surface under
~
the influence of a uniform electric field E .
~
Due to the force q E , its velocity increases
from 0 to 6 m/s in one second duration. At
that instant the direction of the field is
reversed. The car continues to move for two
more seconds under the influence of this field.
The average velocity and the average speed
of the toy car between 0 to 3 seconds are
respectively
(1) 1.5 m/s, 3 m/s
(2) 2 m/s, 4 m/s
(3) 1 m/s, 3.5 m/s
(4) 1 m/s, 3 m/s
Answer (4)
a
t = 1 ?a
t = 2
B
v = 0
C
t = 3
v = ?6 ms
?1

-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
S 6(1)
1
2
= ~
2
1
= 3 m ...(i)
For t = 1 s to t = 2 s,
= _ x =
1
S 6.1 6(1) 3 m
2
...(ii)
2
2
For t = 2 s to t = 3 s,
S 0 6(1) 3 m
3
2
= - x = -
1
2
...(iii)
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
Total distance travelled = 9 m
E
...(i)
...(ii)
10 ~
I
(2)
O
(4)
O
n
n
Sol. t = 0
A
v = 0
v = 6 ms
?1

?a
-2
- 1
-1
Average speed =
9
= 3 ms
3
11
NEET (UG) - 2018 (Code-NN) HLAAC
C B
A
m
a
0
~
45. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(5) -4i
?
-
j
-
8k
?
Answer (1)
N
N sin b
Sol.
V
0
0
ma
(pseudo)
g
(2) =
a cosec 0
(3) a = g cos 0
g
(4) =
a sin G
Answer (1)
Sol.
N cos O
(2) - 8i
?
- 4 ? j - 7k
?
(1) - 7i
?
- 4 ? j - 8k
?
(3) - 7i
?
- 8j
?
- 4k
?
F
mg
0
a
In non-inertial frame,
N sin O = ma ...(i)
N cos o = mg ...(ii)
A
r - r
0
P
r
0
r
tan ~ ~
a
O
x
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j 3k)
?
(2i ? 2j 2k)
?
- = + - - - -
? ? ?
= 0i + 2j - k
? ? ?
i j k
? ? ?
0 2 1 -
4 5 6 -
= - - - 7i 4 j 8k ~ ~
43. A block of mass m is placed on a smooth
inclined wedge ABC of inclination 0 as shown
in the figure. The wedge is given an
acceleration 'a' towards the right. The
relation between a and 0 for the block to
remain stationary on the wedge is
(1) a = g tan o
(1) 0.529 cm
(2) 0.521 cm
(3) 0.053 cm
(4) 0.525 cm
Answer (1)
Sol. Diameter of the ball
= MSR + CSR x (Least count) ? Zero error
= 0.5 cm + 25 x 0.001 ? (?0.004)
= 0.5 + 0.025 + 0.004
= 0.529 cm
g
a = g tan 0
44. A student measured the diameter of a small
steel ball using a screw gauge of least count
0.001 cm. The main scale reading is 5 mm and
zero of circular scale division coincides with
25 divisions above the reference level. If
screw gauge has a zero error of ?0.004 cm,
the correct diameter of the ball is
12
NEET (UG) - 2018 (Code-NN) HLAAC
46. Which of the following hormones can play a
significant role in osteoporosis?
(1) Parathyroid hormone and Prolactin
(2) Aldosterone and Prolactin
(3) Estrogen and Parathyroid hormone
(4) Progesterone and Aldosterone
Answer (3)
Sol. Estrogen promotes the activity of osteoblast
and inhibits osteoclast. In an ageing female
osteoporosis occurs due to deficiency of
estrogen. Parathormone promotes
mobilisation of calcium from bone into blood.
Excessive activity of parathormone causes
demineralisation leading to osteoporosis.
47. Which of the following is an amino acid
derived hormone?
(1) Estriol
(2) Epinephrine
(3) Estradiol
(4) Ecdysone
Answer (2)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
48. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
(2) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(3) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(4) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
Answer (4)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
49. The transparent lens in the human eye is held
in its place by
(1) smooth muscles attached to the ciliary
body
(2) ligaments attached to the ciliary body
(3) smooth muscles attached to the iris
(4) ligaments attached to the iris
Answer (2)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
50. The amnion of mammalian embryo is derived
from
(1) ectoderm and endoderm
(2) ectoderm and mesoderm
(3) mesoderm and trophoblast
(4) endoderm and mesoderm
Answer (2)
Sol. The extraembryonic or foetal membranes are
amnion, chorion, allantois and Yolk sac.
Amnion is formed from mesoderm on outer
side and ectoderm on inner side.
Chorion is formed from trophoectoderm and
mesoderm whereas allantois and Yolk sac
membrane have mesoderm on outerside and
endoderm in inner side.
51. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, progestogens, estrogens,
glucocorticoids
(2) hCG, hPL, progestogens, prolactin
(3) hCG, hPL, progestogens, estrogens
(4) hCG, hPL, estrogens, relaxin, oxytocin
Answer (3)
Sol. Placenta releases human chorionic
gonadotropic hormone (hCG) which
stimulates the Corpus luteum during
pregnancy to release estrogen and
progesterone and also rescues corpus
luteum from regression. Human placental
lactogen (hPL) is involved in growth of body of
mother and breast. Progesterone maintains
pregnancy, keeps the uterus silent by
increasing uterine threshold to contractile
stimuli.
13
NEET (UG) - 2018 (Code-NN) HLAAC
52. The contraceptive ?SAHELI?
(1) is a post-coital contraceptive.
(2) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(3) is an IUD.
(4) increases the concentration of estrogen
and prevents ovulation in females.
Answer (2)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
53. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
(2) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(3) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
Answer (1)
Sol. Spermiogenesis is transformation of
spermatids into spermatozoa whereas
spermiation is the release of the sperms from
sertoli cells into the lumen of seminiferous
tubule.
54. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Both sons and daughters
(2) Only daughters
(3) Only grandchildren
(4) Only sons
Answer (1)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
55. According to Hugo de Vries, the mechanism
of evolution is
(1) Minor mutations
(2) Multiple step mutations
(3) Phenotypic variations
(4) Saltation
Answer (4)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
56. All of the following are part of an operon
except
(1) a promoter (2) an operator
(3) an enhancer (4) structural genes
Answer (3)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
57. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) UCCAUAGCGUA
(2) AGGUAUCGCAU
(3) ACCUAUGCGAU
(4) UGGTUTCGCAT
Answer (2)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
58. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Proliferative Phase i. Breakdown of
endometrial
lining
b. Secretory Phase ii. Follicular Phase
c. Menstruation iii. Luteal Phase
a b c
(1) iii i ii
(2) iii ii i
(3) ii iii i
(4) i iii ii
Answer (3)
14
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. During proliferative phase, the follicles start
developing, hence, called follicular phase.
Secretory phase is also called as luteal phase
mainly controlled by progesterone secreted
by corpus luteum. Estrogen further thickens
the endometrium maintained by
progesterone.
Menstruation occurs due to decline in
progesterone level and involves breakdown of
overgrown endometrial lining.
59. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Amoebiasis
(2) Elephantiasis
(3) Ringworm disease
(4) Ascariasis
Answer (2)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
60. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Eye of octopus, bat and man
(2) Forelimbs of man, bat and cheetah
(3) Brain of bat, man and cheetah
(4) Heart of bat, man and cheetah
Answer (1)
Sol. Divergent evolution occurs in the same
structure, example - forelimbs, heart, brain of
vertebrates which have developed along
different directions due to adaptation to
different needs whereas eye of octopus, bat
and man are examples of analogous organs
showing convergent evolution.
61. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Adaptive radiation
(2) Homology
(3) Convergent evolution
(4) Analogy
Answer (2)
Sol. In different vertebrates, bones of forelimbs
are similar but their forelimbs are adapted in
different way as per their adaptation, hence
example of homology.
62. Which of the following is not an autoimmune
disease?
(1) Vitiligo
(2) Psoriasis
(3) Alzheimer's disease
(4) Rheumatoid arthritis
Answer (3)
Sol. Rheumatoid arthritis is an autoimmune
disorder in which antibodies are produced
against the synovial membrane and cartilage.
Vitiligo causes white patches on skin also
characterised as autoimmune disorder.
Psoriasis is a skin disease that causes itchy
or sore patches of thick red skin and is also
autoimmune whereas Alzheimer's disease is
due to deficiency of neurotransmitter
acetylcholine.
63. Which of the following characteristics
represent ?Inheritance of blood groups? in
humans?
a. Dominance
b. Co-dominance
c. Multiple allele
d. Incomplete dominance
e. Polygenic inheritance
(1) a, c and e (2) b, c and e
(3) b, d and e (4) a, b and c
Answer (4)
Sol. ? I
A
I
O
, I
B
I
O
- Dominant?recessive
relationship
? I
A
I
B
- Codominance
? I
A
, I
B
& I
O
- 3-different allelic
forms of a gene
(multiple allelism)
64. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin E
(2) Vitamin D
(3) Vitamin B
12
(4) Vitamin A
Answer (3)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
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NEET (UG) - 2018 (Code-NN) HLAAC
65. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column-I Column-II
a. Eutrophication i. UV-B radiation
b. Sanitary landfill ii. Deforestation
c. Snow blindness iii. Nutrient
enrichment
d. Jhum cultivation iv. Waste disposal
a b c d
(1) i ii iv iii
(2) ii i iii iv
(3) iii iv i ii
(4) i iii iv ii
Answer (3)
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
66. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Amensalism
(2) Commensalism
(3) Parasitism
(4) Mutualism
Answer (1)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicilium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicilium or the
organism which produces it.
67. All of the following are included in ? ex-situ
conservation? except
(1) Seed banks
(2) Wildlife safari parks
(3) Botanical gardens
(4) Sacred groves
Answer (4)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
68. In a growing population of a country,
(1) pre-reproductive individuals are less than
the reproductive individuals.
(2) pre-reproductive individuals are more
than the reproductive individuals.
(3) reproductive and pre-reproductive
individuals are equal in number.
(4) reproductive individuals are less than the
post-reproductive individuals.
Answer (2)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
69. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Leaves (2) Flowers
(3) Roots (4) Latex
Answer (4)
Sol. ?Smack? also called as brown sugar/Heroin is
formed by acetylation of morphine. It is
obtained from the latex of unripe capsule of
Poppy plant.
70. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Parietal cells (2) Chief cells
(3) Goblet cells (4) Mucous cells
Answer (1)
Sol. Parietal or oxyntic cell is a source of HCl and
intrinsic factor. HCl converts iron present in
diet from ferric to ferrous form so that it can
be absorbed easily and used during
erythropoiesis.
Intrinsic factor is essential for the absorption
of vitamin B
12
and its deficiency causes
pernicious anaemia.
71. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Fibrinogen (i) Osmotic balance
b. Globulin (ii) Blood clotting
c. Albumin (iii) Defence
mechanism
a b c
(1) (ii) (iii) (i)
(2) (iii) (ii) (i)
(3) (i) (iii) (ii)
(4) (i) (ii) (iii)
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Fibrinogen forms fibrin strands during
coagulation. These strands forms a network
and the meshes of which are occupied by
blood cells, this structure finally forms a clot.
Antibodies are derived from y-Globulin fraction
of plasma proteins which means globulins are
involved in defence mechanisms.
Albumin is a plasma protein mainly
responsible for BCOP.
72. Which of the following is an occupational
respiratory disorder?
(1) Emphysema
(2) Anthracis
(3) Botulism
(4) Silicosis
Answer (4)
Sol. Silicosis is due to excess inhalation of silica
dust in the workers involved grinding or stone
breaking industries.
Long exposure can give rise to inflammation
leading to fibrosis and thus causing serious
lung damage.
Anthrax is a serious infectious disease caused
by Bacillus anthracis. It commonly affects
domestic and wild animals. Emphysema is a
chronic disorder in which alveolar walls are
damaged due to which respiratory surface is
decreased.
Botulism is a form of food poisoning caused
by Clostridium botulinum.
73. Calcium is important in skeletal muscle
contraction because it
(1) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
(2) Binds to troponin to remove the masking
of active sites on actin for myosin.
(3) Detaches the myosin head from the actin
filament.
(4) Activates the myosin ATPase by binding to
it.
Answer (2)
Sol. ? Signal for contraction increase Ca
++
level
many folds in the sarcoplasm.
? Ca
++
now binds with sub-unit of troponin
(troponin "C") which is masking the active
site on actin filament and displaces the
sub-unit of troponin.
? Once the active site is exposed, head of
the myosin attaches and initiate
contraction by sliding the actin over
myosin.
74. Nissl bodies are mainly composed of
(1) Free ribosomes and RER
(2) Proteins and lipids
(3) Nucleic acids and SER
(4) DNA and RNA
Answer (1)
Sol. Nissl granules are present in the cyton and
even extend into the dendrite but absent in
axon and rest of the neuron.
Nissl granules are in fact composed of free
ribosomes and RER. They are responsible for
protein synthesis.
75. Which of these statements is incorrect?
(1) Oxidative phosphorylation takes place in
outer mitochondrial membrane
(2) Enzymes of TCA cycle are present in
mitochondrial matrix
(3) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(4) Glycolysis occurs in cytosol
Answer (1)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
76. Select the incorrect match :
(1) Polytene ? Oocytes of
chromosomes amphibians
(2) Lampbrush ? Diplotene bivalents
chromosomes
(3) Submetacentric ? L-shaped
chromosomes chromosomes
(4) Allosomes ? Sex chromosomes
Answer (1)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
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77. Which of the following terms describe human
dentition?
(1) Pleurodont, Diphyodont, Heterodont
(2) Thecodont, Diphyodont, Homodont
(3) Pleurodont, Monophyodont, Homodont
(4) Thecodont, Diphyodont, Heterodont
Answer (4)
Sol. In humans, dentition is
? Thecodont : Teeth are present in the
sockets of the jaw bone called alveoli.
? Diphyodont : Teeth erupts twice,
temporary milk or deciduous teeth are
replaced by a set of permanent or adult
teeth.
? Heterodont dentition : Dentition consists
of different types of teeth namely incisors,
canine, premolars and molars.
78. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Phospholipid synthesis
(2) Protein folding
(3) Cleavage of signal peptide
(4) Protein glycosylation
Answer (1)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
79. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Nucleosome (2) Polysome
(3) Plastidome (4) Polyhedral bodies
Answer (2)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
80. Ciliates differ from all other protozoans in
(1) having two types of nuclei
(2) using flagella for locomotion
(3) using pseudopodia for capturing prey
(4) having a contractile vacuole for removing
excess water
Answer (1)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
81. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Osteichthyes
(2) Amphibia
(3) Aves
(4) Reptilia
Answer (3)
Sol. The digestive tract of Aves has additional
chambers in their digestive system as crop
and Gizzard.
Crop is concerned with storage of food grains.
Gizzard is a masticatory organ in birds used to
crush food grain.
82. Which one of these animals is not a
homeotherm?
(1) Psittacula
(2) Macropus
(3) Camelus
(4) Chelone
Answer (4)
Sol. Homeotherm are animals that maintain
constant body temperature, irrespective of
surrounding temperature.
Birds and mammals are homeotherm.
Chelone (Turtle) belongs to class reptilia
which is Poikilotherm or cold blood.
83. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of anal cerci
(2) Presence of a boat shaped sternum on the
9
th
abdominal segment
(3) Forewings with darker tegmina
(4) Presence of caudal styles
Answer (4)
Sol. Males bear a pair of short, thread like anal
styles which are absent in females.
Anal/caudal styles arise from 9
th
abdominal
segment in male cockroach.
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NEET (UG) - 2018 (Code-NN) HLAAC
84. Which of the following animals does not
undergo metamorphosis?
(1) Starfish (2) Earthworm
(3) Moth (4) Tunicate
Answer (2)
Sol. Metamorphosis refers to transformation of
larva into adult.
Animal that perform metamorphosis are said
to have indirect development.
In earthworm development is direct which
means no larval stage and hence no
metamorphosis.
85. Which of the following organisms are known
as chief producers in the oceans?
(1) Euglenoids (2) Dinoflagellates
(3) Cyanobacteria (4) Diatoms
Answer (4)
Sol. Diatoms are chief producers of the ocean.
86. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Decreased respiratory surface;
Inflammation of bronchioles
(2) Inflammation of bronchioles; Decreased
respiratory surface
(3) Increased respiratory surface;
Inflammation of bronchioles
(4) Increased number of bronchioles;
Increased respiratory surface
Answer (2)
Sol. Asthma is a difficulty in breathing causing
wheezing due to inflammation of bronchi and
bronchioles. Emphysema is a chronic disorder
in which alveolar walls are damaged due to
which respiratory surface is decreased.
87. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Tricuspid valve i. Between left atrium
and left ventricle
b. Bicuspid valve ii. Between right
ventricle and
pulmonary artery
c. Semilunar valve iii. Between right
atrium and right
ventricle
a b c
(1) ii i iii
(2) iii i ii
(3) i ii iii
(4) i iii ii
Answer (2)
Sol. Tricuspid valves are AV valve present
between right atrium and right ventricle.
Bicuspid valves are AV valve present between
left atrium and left ventricle. Semilunar valves
are present at the openings of aortic and
pulmonary aorta.
88. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Tidal volume i. 2500 ? 3000 mL
b. Inspiratory Reserve ii. 1100 ? 1200 mL
volume
c. Expiratory Reserve iii. 500 ? 550 mL
volume
d. Residual volume iv. 1000 ? 1100 mL
a b c d
(1) iv iii ii i
(2) iii ii i iv
(3) i iv ii iii
(4) iii i iv ii
Answer (4)
Sol. Tidal volume is volume of air inspired or
expired during normal respiration. It is
approximately 500 mL. Inspiratory reserve
volume is additional volume of air a person
can inspire by a forceful inspiration. It is
around 2500 ? 3000 mL. Expiratory reserve
volume is additional volume of air a person
can be expired by a forceful expiration. This
averages 1000 ? 1100 mL.
Residual volume is volume of air remaining in
lungs even after forceful expiration. This
averages 1100 ? 1200 mL.
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NEET (UG) - 2018 (Code-NN) HLAAC
89. Match the items given in Column I with those
in Column II and select the correct option
given below :
Column I Column II
a. Glycosuria

i. Accumulation of
uric acid in joints
b. Gout ii. Mass of crystallised
salts within the
kidney
c. Renal calculi

iii. Inflammation in
glomeruli
d. Glomerular iv. Presence of in
nephritis glucose urine
d
iii
i
iv
iv
Answer (1)
Sol. Glycosuria denotes presence of glucose in the
urine. This is observed when blood glucose
level rises above 180 mg/100 ml of blood, this
is called renal threshold value for glucose.
Gout is due to deposition of uric acid crystals
in the joint.
Renal calculi are precipitates of calcium
phosphate produced in the pelvis of the
kidney.
Glomerular nephritis is the inflammatory
condition of glomerulus characterised by
proteinuria and haematuria.
90. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
(Function) (Part of Excretory
system)
a. Ultrafiltration i. Henle's loop
b. Concentration ii. Ureter
of urine
c. Transport of iii. Urinary bladder
urine
d. Storage of iv. Malpighian
urine corpuscle
v. Proximal
convoluted tubule
a b c d
(1) v iv i iii
(2) iv v ii iii
(3) v iv i ii
(4) iv i ii iii
Answer (4)
Sol. Ultrafiltration refers to filtration of very fine
particles having molecular weight less than
68,000 daltons through malpighian corpuscle.
Concentration of urine refers to water
absorption from glomerular filtrate as a result
of hyperosmolarity in the medulla created by
counter-current mechanism in Henle's loop.
Urine is carried from kidney to bladder
through ureter.
Urinary bladder is concerned with storage of
urine.
91. Secondary xylem and phloem in dicot stem
are produced by
(1) Axillary meristems
(2) Apical meristems
(3) Phellogen
(4) Vascular cambium
Answer (4)
Sol. ? Vascular cambium is partially secondary
? Form secondary xylem towards its inside
and secondary phloem towards outsides.
? 4 ? 10 times more secondary xylem is
produced than secondary phloem.
92. Pneumatophores occur in
(1) Submerged hydrophytes
(2) Halophytes
(3) Carnivorous plants
(4) Free-floating hydrophytes
Answer (2)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
a b c
(1) iv i ii
(2) iii ii iv
(3) ii iii i
(4) i ii iii
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NEET (UG) - 2018 (Code-NN) HLAAC
93. Plants having little or no secondary growth
are
(1) Cycads
(2) Grasses
(3) Conifers
(4) Deciduous angiosperms
Answer (2)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
94. Select the wrong statement :
(1) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
(2) Cell wall is present in members of Fungi
and Plantae
(3) Pseudopodia are locomotory and feeding
structures in Sporozoans
(4) Mushrooms belong to Basidiomycetes
Answer (3)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
95. Casparian strips occur in
(1) Endodermis
(2) Epidermis
(3) Cortex
(4) Pericycle
Answer (1)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Sweet potato is a modified
(1) Rhizome
(2) Stem
(3) Tap root
(4) Adventitious root
Answer (4)
Sol. Sweet potato is a modified adventitious root
for storage of food
? Rhizomes are underground modified stem
? Tap root is primary root directly elongated
from the redicle
97. Which of the following statements is correct?
(1) Stems are usually unbranched in both
Cycas and Cedrus
(2) Ovules are not enclosed by ovary wall in
gymnosperms
(3) Horsetails are gymnosperms
(4) Selaginella is heterosporous, while
Salvinia is homosporous
Answer (2)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
98. What type of ecological pyramid would be
obtained with the following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g
(1) Upright pyramid of biomass
(2) Inverted pyramid of biomass
(3) Upright pyramid of numbers
(4) Pyramid of energy
Answer (2)
Sol. ? The given data depicts the inverted
pyramid of biomass, usually found in
aquatic ecosystem.
? Pyramid of energy is always upright
? Upright pyramid of biomass and numbers
are not possible, as the data depicts
primary producer is less than primary
consumer and this is less than secondary
consumers.
99. World Ozone Day is celebrated on
(1) 22
nd
April
(2) 5
th
June
(3) 16
th
September
(4) 21
st
April
Answer (3)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. World Ozone day is celebrated on 16
th

September.
5
th
June - World Environment Day
21
st
April - National Yellow Bat Day
22
nd
April - National Earth Day
100. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Oxygen (2) Carbon
(3) Fe (4) Cl
Answer (4)
Sol. UV rays act on CFCs, releasing Cl atoms,
chlorine reacts with ozone in sequential
method converting into oxygen
Carbon, oxygen and Fe are not related to
ozone layer depletion
101. Natality refers to
(1) Number of individuals entering a habitat
(2) Death rate
(3) Number of individuals leaving the habitat
(4) Birth rate
Answer (4)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual ? Immigration
entering a habitat is
? Number of individual ? Emigration
leaving the habital
102. Niche is
(1) the functional role played by the organism
where it lives
(2) all the biological factors in the organism's
environment
(3) the range of temperature that the
organism needs to live
(4) the physical space where an organism
lives
Answer (1)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
103. Which of the following is a secondary
pollutant?
(1) O
3

(2) CO
(3) SO
2

(4) CO
2

Answer (1)
Sol. O
3
(ozone) is a secondary pollutant. These are
formed by the reaction of primary pollutant.
CO ? Quantitative pollutant
CO
2
? Primary pollutant
SO
2
? Primary pollutant
104. Winged pollen grains are present in
(1) Pinus
(2) Mustard
(3) Mango
(4) Cycas
Answer (1)
Sol. In Pinus, winged pollen grains are present. It
is extended outer exine on two lateral sides to
form the wings of pollen. It is the
characteristic feature, only in Pinus.
Pollen grains of Mustard, Cycas & Mango are
not winged shaped.
105. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Saccharomyces
(2) Neurospora
(3) Agaricus
(4) Alternaria
Answer (3)
Sol. ? In Agaricus (a genus of basidiomycetes),
basidiospores or meiospores are
produced exogenously.
? Neurospora (a genus of ascomycetes)
produces ascospores as meiospores but
endogenously inside the ascus.)
? Alternaria (a genus of deuteromycetes)
does not produce sexual spores.
? Saccharomyces (Unicellular
ascomycetes) produces ascospores,
endogenously.
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NEET (UG) - 2018 (Code-NN) HLAAC
106. Which one is wrongly matched?
(1) Unicellular organism ? Chlorella
(2) Uniflagellate gametes ? Polysiphonia
(3) Gemma cups ? Marchantia
(4) Biflagellate zoospores ? Brown algae
Answer (2)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (1, 3 & 4) are correctly
matched
107. Match the items given in Column I with those
in Column II and select the correct option
given below:
Column I Column II
a. Herbarium (i) It is a place having a
collection of preserved
plants and animals
b. Key

(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum

(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue (iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
a b c d
(1) (iii) (iv) (i) (ii)
(2) (i) (iv) (iii) (ii)
(3) (ii) (iv) (iii) (i)
(4) (iii) (ii) (i) (iv)
Answer (1)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
108. Which one of the following plants shows a
very close relationship with a species of moth,
where none of the two can complete its life
cycle without the other?
(1) Viola (2) Hydrilla
(3) Banana (4) Yucca
Answer (4)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
109. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?160?C
(2) ?120?C
(3) ?196?C
(4) ?80?C
Answer (3)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C
(Cryopreservation)
110. In which of the following forms is iron
absorbed by plants?
(1) Both ferric and ferrous
(2) Ferric
(3) Free element
(4) Ferrous
Answer (2*)
Sol. Iron is absorbed by plants in the form of ferric
ions. (According to NCERT)
*Plants absorb iron in both form i.e. Fe
++
and
Fe
+++
. (Preferably Fe
++
)
111. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Calcium (2) Magnesium
(3) Potassium (4) Sodium
Answer (3)
Sol. Potassium helps in maintaining turgidity of
cells.
112. Double fertilization is
(1) Syngamy and triple fusion
(2) Fusion of two male gametes of a pollen
tube with two different eggs
(3) Fusion of two male gametes with one egg
(4) Fusion of one male gamete with two polar
nuclei
Answer (1)
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NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
113. What is the role of NAD
+
in cellular
respiration?
(1) It is the final electron acceptor for
anaerobic respiration.
(2) It functions as an enzyme.
(3) It is a nucleotide source for ATP synthesis.
(4) It functions as an electron carrier.
Answer (4)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
114. Oxygen is not produced during photosynthesis
by
(1) Chara
(2) Green sulphur bacteria
(3) Cycas
(4) Nostoc
Answer (2)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
115. The Golgi complex participates in
(1) Activation of amino acid
(2) Fatty acid breakdown
(3) Respiration in bacteria
(4) Formation of secretory vesicles
Answer (4)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
116. Stomatal movement is not affected by
(1) CO
2
concentration
(2) Temperature
(3) O
2
concentration
(4) Light
Answer (3)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
117. Stomata in grass leaf are
(1) Barrel shaped
(2) Dumb-bell shaped
(3) Rectangular
(4) Kidney shaped
Answer (2)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
118. Which of the following is true for nucleolus?
(1) It is a site for active ribosomal RNA
synthesis
(2) Larger nucleoli are present in dividing
cells
(3) It takes part in spindle formation
(4) It is a membrane-bound structure
Answer (1)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
119. Which of the following is not a product of light
reaction of photosynthesis?
(1) Oxygen (2) ATP
(3) NADPH (4) NADH
Answer (4)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
120. The stage during which separation of the
paired homologous chromosomes begins is
(1) Zygotene (2) Pachytene
(3) Diakinesis (4) Diplotene
Answer (4)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
121. The two functional groups characteristic of
sugars are
(1) Carbonyl and hydroxyl
(2) Hydroxyl and methyl
(3) Carbonyl and phosphate
(4) Carbonyl and methyl
Answer (1)
Sol. Sugar is a common term used to denote
carbohydrate.
Carbohydrates are polyhydroxy aldehyde,
ketone or their derivatives, which means they
have carbonyl and hydroxyl groups.
122. Which among the following is not a
prokaryote?
(1) Oscillatoria
(2) Saccharomyces
(3) Nostoc
(4) Mycobacterium
Answer (2)
24
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
123. Offsets are produced by
(1) Parthenogenesis (2) Meiotic divisions
(3) Parthenocarpy (4) Mitotic divisions
Answer (4)
Sol. Offset is a vegetative part of a plant, formed
by mitosis.
? Meiotic divisions do not occur in somatic
cells.
? Parthenogenesis is the formation of
embryo from ovum or egg without
fertilisation.
? Parthenocarpy is the fruit formed without
fertilisation, (generally seedless)
124. Select the correct statement
(1) Transduction was discovered by S. Altman
(2) Franklin Stahl coined the term ?linkage?
(3) Spliceosomes take part in translation
(4) Punnett square was developed by a British
scientist
Answer (4)
Sol. Punnett square was developed by a British
geneticist, Reginald C. Punnett.
? Franklin Stahl proved semi-conservative
mode of replication.
? Transduction was discovered by Zinder
and Laderberg.
? Spliceosome formation is part of post-
transcriptional change in Eukaryotes
125. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Sporopollenin (2) Pollenkitt
(3) Oil content (4) Cellulosic intine
Answer (1)
Sol. Sporopollenin cannot be degraded by
enzyme; strong acids and alkali, therefore it is
helpful in preserving pollen as fossil.
Pollenkitt ? Help in insect pollination.
Cellulosic Intine ? Inner sporoderm layer of
pollen grain known as intine made up
cellulose & pectin.
Oil content ? No role is pollen preservation.
126. Which of the following pairs is wrongly
matched?
(1) T.H. Morgan : Linkage
(2) Starch synthesis in pea : Multiple alleles
(3) XO type sex : Grasshopper
determination
(4) ABO blood grouping : Co-dominance
Answer (2)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (1, 3 & 4) are correctly
matched.
127. Select the correct match
(1) Francois Jacob and - Lac operon
Jacques Monod
(2) Alec Jeffreys - Streptococcus
pneumoniae
(3) Matthew Meselson - Pisum sativum
and F. Stahl
(4) Alfred Hershey and - TMV
Martha Chase
Answer (1)
Sol. Francois Jacob and Jacque Monod proposed
model of gene regulation known as operon
model/lac operon.
? Alec Jeffreys ? DNA fingerprinting
technique.
? Matthew Meselson and F. Stahl ? Semi-
conservative DNA replication in E. coli.
? Alfred Hershey and Martha Chase ?
Proved DNA as genetic material not
protein
128. Which of the following flowers only once in its
life-time?
(1) Papaya (2) Bamboo species
(3) Mango (4) Jackfruit
Answer (2)
Sol. Bamboo species are monocarpic i.e., flower
generally only once in its life-time after 50-
100 years.
Jackfruit, papaya and mango are polycarpic
i.e., produce flowers and fruits many times in
their life-time.
129. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Virus (2) Fungus
(3) Plant (4) Bacterium
Answer (4)
25
NEET (UG) - 2018 (Code-NN) HLAAC
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
130. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) pBR 322 (2) Retrovirus
(3) X phage (4) Ti plasmid
Answer (2)
Sol. Retrovirus is commonly used as vector for
introducing a DNA fragment in human
lymphocyte.
Gene therapy : Lymphocyte from blood of
patient are grown in culture outside the body,
a functional gene is introduced by using a
retroviral vector, into these lymphocyte.
131. Select the correct match
(1) G. Mendel - Transformation
(2) Ribozyme - Nucleic acid
(3) T.H. Morgan - Transduction
(4) F
2
? Recessive parent - Dihybrid cross
Answer (2)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
132. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bioexploitation (2) Bio-infringement
(3) Biodegradation (4) Biopiracy
Answer (4)
Sol. Biopiracy is term used for or refer to the use
of bioresources by multinational companies
and other organisation without proper
authorisation from the countries and people
concerned with compensatory payment
(definition of biopiracy given in NCERT).
133. A ?new? variety of rice was patented by a
foreign company, though such varieties have
been present in India for a long time. This is
related to
(1) Basmati (2) Co-667
(3) Lerma Rojo (4) Sharbati Sonora
Answer (1)
Sol. In 1997, an American company got patent
rights on Basmati rice through the US patent
and trademark office that was actually been
derived from Indian farmer?s varieties.
The diversity of rice in India is one of the
richest in the world, 27 documented varieties
of Basmati are grown in India.
Indian basmati was crossed with semi-dwarf
varieties and claimed as an invention or a
novelty.
Sharbati Sonora and Lerma Rojo are varieties
of wheat.
134. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Genetic Engineering Appraisal Committee
(GEAC)
(2) Indian Council of Medical Research (ICMR)
(3) Research Committee on Genetic
Manipulation (RCGM)
(4) Council for Scientific and Industrial
Research (CSIR)
Answer (1)
Sol. Indian Government has setup organisation
such as GEAC (Genetic Engineering Appraisal
Committee) which will make decisions
regarding the validity of GM research and
safety of introducing GM-organism for public
services. (Direct from NCERT).
135. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Denaturation, Annealing, Extension
(2) Extension, Denaturation, Annealing
(3) Denaturation, Extension, Annealing
(4) Annealing, Extension, Denaturation
Answer (1)
Sol. This technique is used for making multiple
copies of gene (or DNA) of interest in vitro.
Each cycle has three steps
(i) Denaturation
(ii) Primer annealing
(iii) Extension of primer
26
136. Identify the major products P, Q and R in the
following sequence of reactions:
(i) O
2

P Q + R
(ii) H
8
O / A
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

P Q R
137. Which of the following compounds can form a
zwitterion?
(1) Glycine (2) Aniline
(3) Benzoic acid (4) Acetanilide
Answer (1)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
pK
,
= 9.60 pK
,
= 2.34
CH
3

CH ? CH
3

O
2

(P)
CH
3
CH
3
? CH ? CH
3

H /H O
+
2
Hydroperoxide
Rearrangement
OH
O
(R)
(Q)
3
HC?C?O?O?H
+ CH
3
? C ? CH
3

27
NEET (UG) - 2018 (Code-NN) HLAAC
OH
CH(CH
3
)
2
CH
3
? CO ? CH
3
(1)
CH
3
CH
2
? OH
(2)
(3)
(4)
Answer (1)
? ? ? ? ? ? ??
Cl
Sol
Cl Cl
CO
1, 2?H
Shift
6+ 6+ 6?
(Incipient carbocation)
Cl
Ni
CO
6?
OC
AlCl
3

CO
Now,
.
CH
3
CH
2
CH
2
? Cl + Al
CH
3
? CH ? CH
3

CH
2
CH
2
CH
3
CHO
CH
2
CH
2
CH
3
CHO COOH
CH(CH
3
)
2
OH
CH
3
CH
2
CH
2
Cl AlCl
3

, CH
3
CH(OH)CH
3
H
2
N ? CH
2
? COO
?

138. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Tetrahedral geometry and paramagnetic
(2) Square planar geometry and diamagnetic
(3) Square planar geometry and
paramagnetic
(4) Tetrahedral geometry and diamagnetic
Answer (4)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
sp -hybridisation
3
~
H
3
N ? CH
2
? COO
?
139. Iron carbonyl, Fe(CO)
5
is
(1) Dinuclear (2) Tetranuclear
(3) Trinuclear (4) Mononuclear
Answer (4)
Sol. ? Based on the number of metal atoms
present in a complex, they are classified
into mononuclear, dinuclear, trinuclear and
so on.
eg: Fe(CO)
5
: mononuclear
Co
2
(CO)
8
: dinuclear
Fe
3
(CO)
12
: trinuclear
NEET (UG) - 2018 (Code-NN) HLAAC
NO
.
NO
2

H V
~
H
(1)
V
(2)
~
(3)
H
V
~
NO
.
(4)
V H
~
NO
2
140. Match the metal ions given in Column I with
the spin magnetic moments of the ions given
in Column II and assign the correct code :
Column I Column II
a. Co
3+
i.
8
BM
b. Cr
3+
ii.
35
BM
c. Fe
3+
iii.
3
BM
d. Ni
2+
iv.
24
BM
v.
15
BM
c d
i ii
ii i
ii iii
iii iv
Answer (2)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Spin magnetic moment = 4(4 + 2) = 24 BM
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Spin magnetic moment = 3(3 + 2) = 15 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
Spin magnetic moment = 2(2 + 2) = 8 BM
141. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Linkage isomerism
(2) Geometrical isomerism
(3) Ionization isomerism
(4) Coordination isomerism
Answer (2)
Sol. In [CoCl
2
(en)
2
], Coordination number of Co
is 6 and this compound has octahedral
geometry.
? As per given option, type of isomerism is
geometrical isomerism.
142. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) MnO
4
2?
(2) CrO
4
2?

(3) MnO
4
?
(4) Cr
2
O
7
2?
Answer (1)
Sol. CrO
4
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
Cr
2
O
7
2?
= Cr
6+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
MnO
4
2?
= Mn
6+
= [Ar] 3d
1

Unpaired electron (n) = 1; Paramagnetic
MnO
4
?
= Mn
7+
= [Ar]
Unpaired electron (n) = 0; Diamagnetic
143. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) CH
3
? CH = CH ? CH
3

(2) HC ~ C ? C ~ CH
(3) CH
2
= CH ? CH = CH
2

(4) CH
2
= CH ? C ~ CH
Answer (4)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
144. Which of the following carbocations is
expected to be most stable?
Answer (3)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (3)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
a b
(1) iii v
(2) iv v
(3) iv i
(4) i ii
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
28
M
10 HCl + 100 mL
M
10 NaOH e. 100 mL
NEET (UG) - 2018 (Code-NN) HLAAC
145. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NR
2
> ? OR > ? F
(2) ? NH
2
< ? OR < ? F
(3) ? NH
2
> ? OR > ? F
(4) ? NR
2
< ? OR < ? F
Answer (2*)
Sol. ?I effect increases on increasing
electronegativity of atom. So, correct order of
?I effect is ?NH
2
< ? OR < ? F.
*Most appropriate Answer is option (2),
however option (4) may also be correct answer.
146. The solubility of BaSO
4
in water is 2.42 ? 10
?3
gL
?1

at 298 K. The value of its solubility product
(K
sp
) will be
(Given molar mass of BaSO
4
= 233 g mol
?1
)
(1) 1.08 ? 10
?8
mol
2
L
?2

(2) 1.08 ? 10
?10
mol
2
L
?2

(3) 1.08 ? 10
?14
mol
2
L
?2

(4) 1.08 ? 10
?12
mol
2
L
?2

Answer (2)
148. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M M
a. 60 mL 10 HCl + 40 mL 10 NaOH
M M
b. 55 mL 10 HCl + 45 mL 10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
pH of which one of them will be equal to 1?
(1) c (2) b
(3) d (4) a
Answer (1)
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
3
(mol L
?1
) Sol. Solubility of BaSO
4
, s =
2.42 10 x
233
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
= 1.04 ? 10
?5
(mol L
?1
)
10 1
=
~
BaSO (s) Ba (aq) SO (aq)
2 2 ~
~~~
~~~
+
4 4
s s
K
sp
= [Ba
2+
] [SO
42?
]= s
2
= (1.04 ? 10
?5
)
2

= 1.08 ? 10
?10
mol
2
L
?2
147. Given van der Waals constant for NH
3
, H
2
, O
2

and CO
2
are respectively 4.17, 0.244, 1.36 and
3.59, which one of the following gases is most
easily liquefied?
(1) CO
2

(2) NH
3

(3) O
2

(4) H
2

Answer (2)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
100 10
pH = ?log[H
+
] =
1
~
~ ~
~ ~
log 10
~
~ = 1.0
149. On which of the following properties does the
coagulating power of an ion depend?
(1) The sign of charge on the ion alone
(2) The magnitude of the charge on the ion
alone
(3) Both magnitude and sign of the charge on
the ion
(4) Size of the ion alone
Answer (3)
Sol. ? Coagulation of colloidal solution by using
an electrolyte depends on the charge
present (positive or negative) on colloidal
particles as well as on its size.
? Coagulating power of an electrolyte
depends on the magnitude of charge
present on effective ion of electrolyte.
29
NEET (UG) - 2018 (Code-NN) HLAAC
S S
? ?
F
? ?
S S
F Cl
S S
F
S S
(1) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
(2) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(3) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(4) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
Answer (1)
Sol.
C
2
H
5
OH
Na
C
2
H
5
O Na
+

(A) (B)
PCl
5

C
2
H
5
Cl
(C)
C
2
H
5
O Na + + C
2
H
5
Cl
S
N
2

C
2
H
5
OC
2
H 5
(B) (C)
150. Which of the following statements is not true
for halogens?
(1) Chlorine has the highest electron-gain
enthalpy
(2) All form monobasic oxyacids
(3) All but fluorine show positive oxidation
states
(4) All are oxidizing agents
Answer (3)
Sol. Due to high electronegativity and small size,
F forms only one oxoacid, HOF known as
Fluoric (I) acid. Oxidation number of F is +1 in
HOF.
151. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
(1) Cu (2) Fe
(3) Mg (4) Zn
Answer (3)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
152. The correct order of atomic radii in group 13
elements is
(1) B < Ga < Al < In < Tl
(2) B < Al < In < Ga < Tl
(3) B < Ga < Al < Tl < In
(4) B < Al < Ga < In < Tl
Answer (1)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
153. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) Three (2) One
(3) Four (4) Two
Answer (4)
Sol. The structure of ClF
3
is
The number of lone pair of electrons on
central Cl is 2.
154.The correct order of N-compounds in its
decreasing order of oxidation states is
(1) NH
4
Cl, N
2
, NO, HNO
3

(2) HNO
3
, NO, N
2
, NH
4
Cl
(3) HNO
3
, NH
4
Cl, NO, N
2

(4) HNO
3
, NO, NH
4
Cl, N
2

Answer (2)
+ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
155. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) In
(2) Ga
(3) B
(4) Al
Answer (3)
Sol. ? 'B' has no vacant d-orbitals in its valence
shell, so it can't extend its covalency beyond
4. i.e. 'B' cannot form the ion like MF
6
3(?)
i.e.
BF
6
3(?)
.
156. The compound A on treatment with Na gives
B, and with PCl
5
gives C. B and C react
together to give diethyl ether. A, B and C are
in the order
30
NEET (UG) - 2018 (Code-NN) HLAAC
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
1
,
k[A ]
0
157. Hydrocarbon (A) reacts with bromine by
substitution to form an alkyl bromide which by
Wurtz reaction is converted to gaseous
hydrocarbon containing less than four carbon
atoms. (A) is
(1) CH
4
(2) CH ~ CH
(3) CH
3
? CH
3
(4) CH
2
= CH
2

Answer (1)
Sol.
CH
4
Br
2
/h v
CH
3
Br
Na/dry ether
Wurtz reaction
CH
3
? CH
3

158. The compound C
7
H
8
undergoes the following
reactions:
3Cl
2
/ A Br
2
/Fe Zn/HCl
C
7
H
8
~~~> ~~~> ~~~~ A B C
The product 'C' is
(1) p-bromotoluene
(2) m-bromotoluene
(3) 3-bromo-2,4,6-trichlorotoluene
(4) o-bromotoluene
Answer (2)
CCl
3

3dl
2
A
(C
7
H
8
)
160. The correct difference between first and
second order reactions is that
(1) The rate of a first-order reaction does
depend on reactant concentrations; the
rate of a second-order reaction does not
depend on reactant concentrations
(2) The rate of a first-order reaction does not
depend on reactant concentrations; the
rate of a second-order reaction does
depend on reactant concentrations
(3) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(4) The half-life of a first-order reaction does
not depend on [A]
0
; the half-life of a
second-order reaction does depend on
[A]
0

Answer (4)
,
k
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
161. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BaH
2
< BeH
2
< CaH
2

(2) BeH
2
< CaH
2
< BaH
2

(3) BeH
2
< BaH
2
< CaH
2

(4) CaH
2
< BeH
2
< BaH
2

Answer (2)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
162. Consider the change in oxidation state of
Bromine corresponding to different emf values
as shown in the diagram below :
Then the species undergoing
disproportionation is
(1) HBrO (2)
~
BrO
3

(3) Br
2
(4)
~
BrO
4
(A)
CH
3
CCl
3

Sol.
Br
2
Fe
Br
(A) (B)
Zn HCl
CH
3
Br
(C)
159. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) NO
(2) N
2
O
5
(3) N
2
O
(4) NO
2

Answer (2)
Sol. Fact
Sol. ? For first order reaction,
1 /2
=
0.693
t
31
NEET (UG) - 2018 (Code-NN) HLAAC
CH
3

(1) CH
3
OH and I
2

(2)
H
3
C CH
2
? OH and I
2
COONa
Sodium benzoate
Answer (1)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+5
~~* BrO , E
~
~ o
HBrO
3
BrO /HBrO
= 1.5 V
3
E for the disproportionation of HBrO,
o
cell
E
o
= E
o o
cell
~ E
~ HBrO/Br
2 BrO /HBrO
3
= 1.595 ? 1.5
= 0.095 V = + ve
163. In which case is number of molecules of
water maximum?
(1) 10
?3
mol of water
(2) 18 mL of water
(3) 0.00224 L of water vapours at 1 atm and
273 K
(4) 0.18 g of water
Answer (2)
Sol. (1) Molecules of water = mole ? N
A
= 10
?3
N
A

(2) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(4) Molecules of water = mole ? N
A
= A
18
= 10
?2
N
A

164. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intermolecular H-bonding
(2) Formation of intramolecular H-bonding
(3) More extensive association of carboxylic
acid via van der Waals force of attraction
(4) Formation of carboxylate ion
Answer (1)
Sol. Due to formation of intermolecular H-bonding
in carboxylic acid, association occurs. Hence
boiling point increases and become more
than the boiling point of aldehydes, ketones
and alcohols of comparable molecular
masses.
165. Compound A,
C
8
H
10
O,
is found to react with
NaOI (produced by reacting Y with NaOH) and
yields a yellow precipitate with characteristic
smell.
A and Y are respectively
CH ? CH
3
and I
2
(3)
OH
2 (4)
CH
2
? CH
2
? OH and I
Answer (3)
Sol. Option (3) is secondary alcohol which on
oxidation gives phenylmethyl ketone
(Acetophenone). This on reaction with I
2
and
NaOH form iodoform and sodium benzoate.
2NaOH + I -* NaOI + NaI + H
2
O
2
I
2
NaOH
166. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichlorocarbene ( : CCl
2
)
(2) Dichloromethyl cation
(
CHCl
2
)
~
(3) Dichloromethyl anion (
CHCl
2
)
?
(4) Formyl cation (
C H O
)
~
+ CHI
3

Iodoform
(Yellow PPt)
CHO
= N
A

0.00224
22.4
NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
= 10
?4

(3) Moles of water =
32
?
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
NEET (UG) - 2018 (Code-NN) HLAAC
Answer (1)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
CCl ~~~ : CCl + Cl
3 2
Electrophile
167. The bond dissociation energies of X
2
, Y
2
and
XY are in the ratio of 1 : 0.5 : 1. AH for the
formation of XY is ?200 kJ mol
?1
. The bond
dissociation energy of X
2
will be
(1) 400 kJ mol
?1

(2) 200 kJ mol
?1

(3) 800 kJ mol
?1

(4) 100 kJ mol
?1

Answer (3)
Sol. The reaction for A
f
H?(XY)
1
+
1
X (g) Y (g) XY(g) ~~~
2 2
2 2
Bond energies of X
2
, Y
2
and XY are X, X 2 , X
respectively
A =
?
+
?
_ = _
? ? X X
H X 200
? ?
2 4
On solving, we get
? - + = - X X 200
2 4
= X = 800 kJ/mole
168. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Remains unchanged
(2) Is halved
(3) Is tripled
(4) Is doubled
Answer (4)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
169. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Forces of attraction between the gas
molecules
(2) Density of the gas molecules
(3) Electric field present between the gas
molecules
(4) Volume of the gas molecules
Answer (1)
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
170. Which one of the following conditions will
favour maximum formation of the product in
the reaction,
A
2
(g) + B
2
(g) ~~=
~~~ X
2
(g) A
r
H = -X kJ?
(1) High temperature and low pressure
(2) Low temperature and high pressure
(3) High temperature and high pressure
(4) Low temperature and low pressure
Answer (2)
Sol.
A
2
(g) + B
2
(g)
=~~
~~=
X
2
(g); A H = -x kJ
On increasing pressure equilibrium shifts in a
direction where pressure decreases i.e.
forward direction.
On decreasing temperature, equilibrium shifts
in exothermic direction i.e., forward direction.
So, high pressure and low temperature
favours maximum formation of product.
171. For the redox reaction
~ 2 ~ ~ ~
+ Mn CO
2
H
2
O + ~~~ + +
2
MnO C O H
4 2 4
The correct coefficients of the reactants for
the balanced equation are
MnO
4
~
C
2
O
4
H
+ 2~
(1) 5 16 2
(2) 16 5 2
(3) 2 16 5
(4) 2 5 16
~
t
1/2
~
Sol. In real gas equation,
?
? ?
2
?
+ - =
an
P (V nb) nRT
2 ?
V
?
33
NEET (UG) - 2018 (Code-NN) HLAAC
Answer (4)
Reduction
+7 +3 +4
Sol.
MnO
4
?
+ C
2
O
4
2?
+ H
+
Mn
2+
+ CO
2
+ H
2
O
Oxidation
n-factor of MnO
~
4
~ 5
n-factor of
2
C
2
O
4
~ ~ 2
Ratio of n-factors of MnO
~
4
and
2
C
2
O
4
~
is 5 : 2
So, molar ratio in balanced reaction is 2 : 5
~ The balanced equation is
2MnO 5C O 16H 2Mn 10CO
2
8H
2
O
~ 2 ~
2
+ +
4
+ + ?
2 4
172. Regarding cross-linked or network polymers,
which of the following statements is
incorrect ?
(1) They contain strong covalents bonds in
their polymer chains.
(2) They contain covalent bonds between
various linear polymer chains.
(3) Examples are bakelite and melamine.
(4) They are formed from bi- and tri-functional
monomers.
Answer (1)
Sol. Cross linked or network polymers are formed
from bi-functional and tri-functional monomers
and contain strong covalent bonds between
various linear polymer chains, e.g. bakelite,
melamine etc. Option (1) is not related to
cross-linking.
173. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) In acidic (strong) medium aniline is
present as anilinium ion.
(2) Inspite of substituents nitro group always
goes to only m-position.
(3) In absence of substituents nitro group
always goes to m-position.
(4) In electrophilic substitution reactions
amino group is meta directive.
Answer (1)
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
174. Which of the following oxides is most acidic in
nature?
(1) CaO (2) MgO
(3) BaO (4) BeO
Answer (4)
Sol. BeO < MgO < CaO < BaO
Basic character increases.
So, the most acidic should be BeO. In fact,
BeO is amphoteric oxide while other given
oxides are basic.
175. The difference between amylose and
amylopectin is
(1) Amylose is made up of glucose and
galactose
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(3) Amylopectin have 1 - 4 a-linkage and
1 ? 6 3-linkage
(4) Amylose have 1 - 4 a-linkage and 1 - 6
3-linkage
Answer (2)
Sol. Amylose and Amylopectin are polymers of a-
D-glucose, so 13-link is not possible. Amylose is
linear with 1 - 4 a-linkage whereas
Amylopectin is branched and has both 1 - 4
and 1 - 6 a-linkages.
176. A mixture of 2.3 g formic acid and 4.5 g oxalic
acid is treated with conc. H
2
SO
4
. The evolved
gaseous mixture is passed through KOH
pellets. Weight (in g) of the remaining product
at STP will be
(1) 4.4 (2) 1.4
(3) 2.8 (4) 3.0
NH
2
NH
3
Sol.
H
34
H
2
O(l)
O(l)
mol
Sol.
HCOOH
Conc.H
2
SO
4
CO(g) + H
2

? ? 1
1

2.3g or
? mol ?
? 20 ? 20
2
x 28 = 2.8g
20
1s
2
2s
2
2p
3

OR
1s
2
2s
2
2p
3

~ Option (3) violates Hund's Rule.
NEET (UG) - 2018 (Code-NN) HLAAC
Answer (3)
COOH
? 1 ?
4.5g or
?
mol
?

? 20 ?
- =
3
BO =
10 4
2
NO and CN
Which one ofthese will have the highestbond
(4)
= (7t2py)2,(7t*2px)1
1
COOH
Conc.H 2SO4
CO(g) + CO 2 (g) +
20
mol
1
20
mol
CN :
(a2s)2,(a*2s)2,
(a1s)
2
, (a* 1s)
2
, (72p
x
)
2
(72p
y
)
2
, (a2p
z
)
1

BO = ~ ~
9 4
2 2.5
178. Considerthe following
1s
2
2s
2
2p 2p 2p
1
z

(4) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
Answer (3)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
: (
a*1
s
)2, (a2s)2, (a* 2s)2 ,
CN
+
(a1s)
2
, (72p
x
)
2

= (i2p
y
)
2

order?
(1) CN
(2) NO
(3)
Answer (4)
Sol. NO :
BO = 8
-
4
2
2
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So th e remaining gas is CO.
So, weight of remaining gaseous product CO
is
species :
CN+,CN?,
CN+
CN?
(a1s)2,(a*1s)2,(c2s)2,(c*2s)2,(c2pz)2,
(72px)2 = (~*2py)0
BO = 10 _5 25
2
CN? : (a1s)2,(a*1s)2,
(a2s)2,(a*2s)2,
(72px)2
= (72py)2,(a2pz)2
=
177. Which one is a wrong statement?
(1) The value of m for d
z
2 is zero
(2) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(3) The electronic configuration of N atom is
35
179. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 2s
2
2p
3
,
the simplest formula for this compound is
(1)
(2) Mg
2
X
3

(3) Mg
2
X
(4)
Answer (1)
Sol. Element (X) electronic configuration
1s
2
Mg
3
X
2
MgX
2
1s
2
2s
2
2p
3

So, valency of X will be 3.
Valency of Mg is 2.
Formula of compound formed by Mg and X will
be Mg
3
X
2
.
Answer (3)
Sol. For BCC lattice : Z = 2,
= a
For FCC lattice : Z = 4, a = 2 2 r
4r
3
1
(2)
2
3 3
(4)
4 2
(1)
(3)
2
3
4 3
3 2
NEET (UG) - 20 18 (Code-NN) HLAAC
180. Iron exhibits bcc structure at room
temperature. Above 900?C, it transforms to
fcc structure. The ratio of density of iron at
room temperature to that at 900?C (assuming
molar mass and atomic radii of iron remains
constant with temperature) is
? ? ZM
? 3 ?
d ? ?
N a
25 C o
A
BCC
d ? ? ZM
900 C 0
? ?
? 3 ?
N a
A
3
? ? ? ?
=
?
2 2 2 r 3 3
?
~
?
? ?
?
4
4r
4 2
? ?
? ?
3
~ ~
FCC
U U U
36
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