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

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

Important Instructions :
1. The test is of 3 hours duration and 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 EE.
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
EE
CHLAA
1
FirstRanker.com - FirstRanker's Choice
Important Instructions :
1. The test is of 3 hours duration and 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 EE.
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
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
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Important Instructions :
1. The test is of 3 hours duration and 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 EE.
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
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
FirstRanker.com - FirstRanker's Choice
Important Instructions :
1. The test is of 3 hours duration and 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 EE.
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
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
FirstRanker.com - FirstRanker's Choice
Important Instructions :
1. The test is of 3 hours duration and 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 EE.
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
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
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Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
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Important Instructions :
1. The test is of 3 hours duration and 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 EE.
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
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
FirstRanker.com - FirstRanker's Choice
Important Instructions :
1. The test is of 3 hours duration and 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 EE.
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
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
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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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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
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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
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
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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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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
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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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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15
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Important Instructions :
1. The test is of 3 hours duration and 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.
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
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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
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
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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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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
NEET (UG) - 2018 (Code-EE) CHLAA
91. Which of the following statements is correct?
(1) Ovules are not enclosed by ovary wall in
gymnosperms
(2) Horsetails are gymnosperms
(3) Selaginella is heterosporous, while
Salvinia is homosporous
(4) Stems are usually unbranched in both
Cycas and Cedrus
Answer (1)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
92. Pneumatophores occur in
(1) Halophytes
(2) Carnivorous plants
(3) Free-floating hydrophytes
(4) Submerged hydrophytes
Answer (1)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
93. Sweet potato is a modified
(1) Stem
(2) Tap root
(3) Adventitious root
(4) Rhizome
Answer (3)
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
94. Plants having little or no secondary growth
are
(1) Grasses
(2) Conifers
(3) Deciduous angiosperms
(4) Cycads
Answer (1)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
95. Casparian strips occur in
(1) Epidermis
(2) Cortex
(3) Pericycle
(4) Endodermi
Answer (4)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Secondary xylem and phloem in dicot stem
are produced by
(1) Apical meristems
(2) Phellogen
(3) Vascular cambium
(4) Axillary meristems
Answer (3)
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.
97. Select the wrong statement :
(1) Cell wall is present in members of Fungi
and Plantae
(2) Pseudopodia are locomotory and feeding
structures in Sporozoans
(3) Mushrooms belong to Basidiomycetes
(4) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
Answer (2)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
98. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Fungus (2) Plant
(3) Bacterium (4) Virus
Answer (3)
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
23
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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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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
NEET (UG) - 2018 (Code-EE) CHLAA
91. Which of the following statements is correct?
(1) Ovules are not enclosed by ovary wall in
gymnosperms
(2) Horsetails are gymnosperms
(3) Selaginella is heterosporous, while
Salvinia is homosporous
(4) Stems are usually unbranched in both
Cycas and Cedrus
Answer (1)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
92. Pneumatophores occur in
(1) Halophytes
(2) Carnivorous plants
(3) Free-floating hydrophytes
(4) Submerged hydrophytes
Answer (1)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
93. Sweet potato is a modified
(1) Stem
(2) Tap root
(3) Adventitious root
(4) Rhizome
Answer (3)
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
94. Plants having little or no secondary growth
are
(1) Grasses
(2) Conifers
(3) Deciduous angiosperms
(4) Cycads
Answer (1)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
95. Casparian strips occur in
(1) Epidermis
(2) Cortex
(3) Pericycle
(4) Endodermi
Answer (4)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Secondary xylem and phloem in dicot stem
are produced by
(1) Apical meristems
(2) Phellogen
(3) Vascular cambium
(4) Axillary meristems
Answer (3)
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.
97. Select the wrong statement :
(1) Cell wall is present in members of Fungi
and Plantae
(2) Pseudopodia are locomotory and feeding
structures in Sporozoans
(3) Mushrooms belong to Basidiomycetes
(4) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
Answer (2)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
98. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Fungus (2) Plant
(3) Bacterium (4) Virus
Answer (3)
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
23
NEET (UG) - 2018 (Code-EE) CHLAAA
99. Select the correct match
(1) Alec Jeffreys

- Streptococcus
pneumoniae
(2) Matthew Meselson - Pisum sativum
and F. Stahl
(3) Alfred Hershey and - TMV
Martha Chase
(4) Francois Jacob and - Lac operon
Jacques Monod
Answer (4)
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
100. Select the correct statement
(1) Franklin Stahl coined the term ?linkage?
(2) Spliceosomes take part in translation
(3) Punnett square was developed by a British
scientist
(4) Transduction was discovered by S. Altman
Answer (3)
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
101. Which of the following pairs is wrongly
matched?
(1) Starch synthesis in pea : Multiple alleles
(2) XO type sex : Grasshopper
determination
(3) ABO blood grouping : Co-dominance
(4) T.H. Morgan : Linkage
Answer (1)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (2, 3 & 4) are correctly
matched.
102. Offsets are produced by
(1) Meiotic divisions
(2) Parthenocarpy
(3) Mitotic divisions
(4) Parthenogenesis
Answer (3)
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)
103. Which of the following flowers only once in its
life-time?
(1) Bamboo species
(2) Mango
(3) Jackfruit
(4) Papaya
Answer (1)
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.
104. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Pollenkitt (2) Oil content
(3) Cellulosic intine (4) Sporopollenin
Answer (4)
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.
24
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Important Instructions :
1. The test is of 3 hours duration and 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.
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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
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
NEET (UG) - 2018 (Code-EE) CHLAA
91. Which of the following statements is correct?
(1) Ovules are not enclosed by ovary wall in
gymnosperms
(2) Horsetails are gymnosperms
(3) Selaginella is heterosporous, while
Salvinia is homosporous
(4) Stems are usually unbranched in both
Cycas and Cedrus
Answer (1)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
92. Pneumatophores occur in
(1) Halophytes
(2) Carnivorous plants
(3) Free-floating hydrophytes
(4) Submerged hydrophytes
Answer (1)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
93. Sweet potato is a modified
(1) Stem
(2) Tap root
(3) Adventitious root
(4) Rhizome
Answer (3)
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
94. Plants having little or no secondary growth
are
(1) Grasses
(2) Conifers
(3) Deciduous angiosperms
(4) Cycads
Answer (1)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
95. Casparian strips occur in
(1) Epidermis
(2) Cortex
(3) Pericycle
(4) Endodermi
Answer (4)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Secondary xylem and phloem in dicot stem
are produced by
(1) Apical meristems
(2) Phellogen
(3) Vascular cambium
(4) Axillary meristems
Answer (3)
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.
97. Select the wrong statement :
(1) Cell wall is present in members of Fungi
and Plantae
(2) Pseudopodia are locomotory and feeding
structures in Sporozoans
(3) Mushrooms belong to Basidiomycetes
(4) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
Answer (2)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
98. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Fungus (2) Plant
(3) Bacterium (4) Virus
Answer (3)
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
23
NEET (UG) - 2018 (Code-EE) CHLAAA
99. Select the correct match
(1) Alec Jeffreys

- Streptococcus
pneumoniae
(2) Matthew Meselson - Pisum sativum
and F. Stahl
(3) Alfred Hershey and - TMV
Martha Chase
(4) Francois Jacob and - Lac operon
Jacques Monod
Answer (4)
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
100. Select the correct statement
(1) Franklin Stahl coined the term ?linkage?
(2) Spliceosomes take part in translation
(3) Punnett square was developed by a British
scientist
(4) Transduction was discovered by S. Altman
Answer (3)
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
101. Which of the following pairs is wrongly
matched?
(1) Starch synthesis in pea : Multiple alleles
(2) XO type sex : Grasshopper
determination
(3) ABO blood grouping : Co-dominance
(4) T.H. Morgan : Linkage
Answer (1)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (2, 3 & 4) are correctly
matched.
102. Offsets are produced by
(1) Meiotic divisions
(2) Parthenocarpy
(3) Mitotic divisions
(4) Parthenogenesis
Answer (3)
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)
103. Which of the following flowers only once in its
life-time?
(1) Bamboo species
(2) Mango
(3) Jackfruit
(4) Papaya
Answer (1)
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.
104. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Pollenkitt (2) Oil content
(3) Cellulosic intine (4) Sporopollenin
Answer (4)
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.
24
NEET (UG) - 2018 (Code-EE) CHLAA
105. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) Retrovirus
(2) X phage
(3) Ti plasmid
(4) pBR 322
Answer (1)
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.
106. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Extension, Denaturation, Annealing
(2) Denaturation, Extension, Annealing
(3) Annealing, Extension, Denaturation
(4) Denaturation, Annealing, Extension
Answer (4)
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
107. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Indian Council of Medical Research (ICMR)
(2) Research Committee on Genetic
Manipulation (RCGM)
(3) Council for Scientific and Industrial
Research (CSIR)
(4) Genetic Engineering Appraisal Committee
(GEAC)
Answer (4)
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).
108. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bio-infringement
(2) Biodegradation
(3) Biopiracy
(4) Bioexploitation
Answer (3)
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).
109. 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) Co-667
(2) Lerma Rojo
(3) Sharbati Sonora
(4) Basmati
Answer (4)
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.
110. Select the correct match
(1) Ribozyme - Nucleic acid
(2) T.H. Morgan - Transduction
(3) F
2
? Recessive parent - Dihybrid cross
(4) G. Mendel - Transformation
Answer (1)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
25
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Important Instructions :
1. The test is of 3 hours duration and 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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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
NEET (UG) - 2018 (Code-EE) CHLAA
91. Which of the following statements is correct?
(1) Ovules are not enclosed by ovary wall in
gymnosperms
(2) Horsetails are gymnosperms
(3) Selaginella is heterosporous, while
Salvinia is homosporous
(4) Stems are usually unbranched in both
Cycas and Cedrus
Answer (1)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
92. Pneumatophores occur in
(1) Halophytes
(2) Carnivorous plants
(3) Free-floating hydrophytes
(4) Submerged hydrophytes
Answer (1)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
93. Sweet potato is a modified
(1) Stem
(2) Tap root
(3) Adventitious root
(4) Rhizome
Answer (3)
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
94. Plants having little or no secondary growth
are
(1) Grasses
(2) Conifers
(3) Deciduous angiosperms
(4) Cycads
Answer (1)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
95. Casparian strips occur in
(1) Epidermis
(2) Cortex
(3) Pericycle
(4) Endodermi
Answer (4)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Secondary xylem and phloem in dicot stem
are produced by
(1) Apical meristems
(2) Phellogen
(3) Vascular cambium
(4) Axillary meristems
Answer (3)
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.
97. Select the wrong statement :
(1) Cell wall is present in members of Fungi
and Plantae
(2) Pseudopodia are locomotory and feeding
structures in Sporozoans
(3) Mushrooms belong to Basidiomycetes
(4) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
Answer (2)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
98. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Fungus (2) Plant
(3) Bacterium (4) Virus
Answer (3)
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
23
NEET (UG) - 2018 (Code-EE) CHLAAA
99. Select the correct match
(1) Alec Jeffreys

- Streptococcus
pneumoniae
(2) Matthew Meselson - Pisum sativum
and F. Stahl
(3) Alfred Hershey and - TMV
Martha Chase
(4) Francois Jacob and - Lac operon
Jacques Monod
Answer (4)
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
100. Select the correct statement
(1) Franklin Stahl coined the term ?linkage?
(2) Spliceosomes take part in translation
(3) Punnett square was developed by a British
scientist
(4) Transduction was discovered by S. Altman
Answer (3)
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
101. Which of the following pairs is wrongly
matched?
(1) Starch synthesis in pea : Multiple alleles
(2) XO type sex : Grasshopper
determination
(3) ABO blood grouping : Co-dominance
(4) T.H. Morgan : Linkage
Answer (1)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (2, 3 & 4) are correctly
matched.
102. Offsets are produced by
(1) Meiotic divisions
(2) Parthenocarpy
(3) Mitotic divisions
(4) Parthenogenesis
Answer (3)
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)
103. Which of the following flowers only once in its
life-time?
(1) Bamboo species
(2) Mango
(3) Jackfruit
(4) Papaya
Answer (1)
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.
104. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Pollenkitt (2) Oil content
(3) Cellulosic intine (4) Sporopollenin
Answer (4)
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.
24
NEET (UG) - 2018 (Code-EE) CHLAA
105. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) Retrovirus
(2) X phage
(3) Ti plasmid
(4) pBR 322
Answer (1)
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.
106. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Extension, Denaturation, Annealing
(2) Denaturation, Extension, Annealing
(3) Annealing, Extension, Denaturation
(4) Denaturation, Annealing, Extension
Answer (4)
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
107. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Indian Council of Medical Research (ICMR)
(2) Research Committee on Genetic
Manipulation (RCGM)
(3) Council for Scientific and Industrial
Research (CSIR)
(4) Genetic Engineering Appraisal Committee
(GEAC)
Answer (4)
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).
108. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bio-infringement
(2) Biodegradation
(3) Biopiracy
(4) Bioexploitation
Answer (3)
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).
109. 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) Co-667
(2) Lerma Rojo
(3) Sharbati Sonora
(4) Basmati
Answer (4)
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.
110. Select the correct match
(1) Ribozyme - Nucleic acid
(2) T.H. Morgan - Transduction
(3) F
2
? Recessive parent - Dihybrid cross
(4) G. Mendel - Transformation
Answer (1)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
25
NEET (UG) - 2018 (Code-EE) CHLAAA
111. Niche is
(1) all the biological factors in the organism's
environment
(2) the range of temperature that the
organism needs to live
(3) the physical space where an organism
lives
(4) the functional role played by the organism
where it lives
Answer (4)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
112. Which of the following is a secondary
pollutant?
(1) CO (2) SO
2

(3) CO
2
(4) O
3

Answer (4)
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
113. World Ozone Day is celebrated on
(1) 5
th
June (2) 16
th
September
(3) 21
st
April (4) 22
nd
April
Answer (2)
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
114. Natality refers to
(1) Death rate
(2) Number of individuals leaving the habitat
(3) Birth rate
(4) Number of individuals entering a habitat
Answer (3)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual
entering a habitat is
? Immigration
? Number of individual
leaving the habital
? Emigration
115. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Carbon (2) Fe
(3) Cl (4) Oxygen
Answer (3)
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
116. 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) Inverted pyramid of biomass
(2) Upright pyramid of numbers
(3) Pyramid of energy
(4) Upright pyramid of biomass
Answer (1)
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.
117. The Golgi complex participates in
(1) Fatty acid breakdown
(2) Respiration in bacteria
(3) Formation of secretory vesicles
(4) Activation of amino acid
Answer (3)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
118. Which of the following is not a product of light
reaction of photosynthesis?
(1) ATP (2) NADPH
(3) NADH (4) Oxygen
Answer (3)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
26
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
NEET (UG) - 2018 (Code-EE) CHLAA
91. Which of the following statements is correct?
(1) Ovules are not enclosed by ovary wall in
gymnosperms
(2) Horsetails are gymnosperms
(3) Selaginella is heterosporous, while
Salvinia is homosporous
(4) Stems are usually unbranched in both
Cycas and Cedrus
Answer (1)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
92. Pneumatophores occur in
(1) Halophytes
(2) Carnivorous plants
(3) Free-floating hydrophytes
(4) Submerged hydrophytes
Answer (1)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
93. Sweet potato is a modified
(1) Stem
(2) Tap root
(3) Adventitious root
(4) Rhizome
Answer (3)
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
94. Plants having little or no secondary growth
are
(1) Grasses
(2) Conifers
(3) Deciduous angiosperms
(4) Cycads
Answer (1)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
95. Casparian strips occur in
(1) Epidermis
(2) Cortex
(3) Pericycle
(4) Endodermi
Answer (4)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Secondary xylem and phloem in dicot stem
are produced by
(1) Apical meristems
(2) Phellogen
(3) Vascular cambium
(4) Axillary meristems
Answer (3)
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.
97. Select the wrong statement :
(1) Cell wall is present in members of Fungi
and Plantae
(2) Pseudopodia are locomotory and feeding
structures in Sporozoans
(3) Mushrooms belong to Basidiomycetes
(4) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
Answer (2)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
98. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Fungus (2) Plant
(3) Bacterium (4) Virus
Answer (3)
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
23
NEET (UG) - 2018 (Code-EE) CHLAAA
99. Select the correct match
(1) Alec Jeffreys

- Streptococcus
pneumoniae
(2) Matthew Meselson - Pisum sativum
and F. Stahl
(3) Alfred Hershey and - TMV
Martha Chase
(4) Francois Jacob and - Lac operon
Jacques Monod
Answer (4)
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
100. Select the correct statement
(1) Franklin Stahl coined the term ?linkage?
(2) Spliceosomes take part in translation
(3) Punnett square was developed by a British
scientist
(4) Transduction was discovered by S. Altman
Answer (3)
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
101. Which of the following pairs is wrongly
matched?
(1) Starch synthesis in pea : Multiple alleles
(2) XO type sex : Grasshopper
determination
(3) ABO blood grouping : Co-dominance
(4) T.H. Morgan : Linkage
Answer (1)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (2, 3 & 4) are correctly
matched.
102. Offsets are produced by
(1) Meiotic divisions
(2) Parthenocarpy
(3) Mitotic divisions
(4) Parthenogenesis
Answer (3)
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)
103. Which of the following flowers only once in its
life-time?
(1) Bamboo species
(2) Mango
(3) Jackfruit
(4) Papaya
Answer (1)
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.
104. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Pollenkitt (2) Oil content
(3) Cellulosic intine (4) Sporopollenin
Answer (4)
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.
24
NEET (UG) - 2018 (Code-EE) CHLAA
105. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) Retrovirus
(2) X phage
(3) Ti plasmid
(4) pBR 322
Answer (1)
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.
106. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Extension, Denaturation, Annealing
(2) Denaturation, Extension, Annealing
(3) Annealing, Extension, Denaturation
(4) Denaturation, Annealing, Extension
Answer (4)
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
107. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Indian Council of Medical Research (ICMR)
(2) Research Committee on Genetic
Manipulation (RCGM)
(3) Council for Scientific and Industrial
Research (CSIR)
(4) Genetic Engineering Appraisal Committee
(GEAC)
Answer (4)
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).
108. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bio-infringement
(2) Biodegradation
(3) Biopiracy
(4) Bioexploitation
Answer (3)
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).
109. 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) Co-667
(2) Lerma Rojo
(3) Sharbati Sonora
(4) Basmati
Answer (4)
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.
110. Select the correct match
(1) Ribozyme - Nucleic acid
(2) T.H. Morgan - Transduction
(3) F
2
? Recessive parent - Dihybrid cross
(4) G. Mendel - Transformation
Answer (1)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
25
NEET (UG) - 2018 (Code-EE) CHLAAA
111. Niche is
(1) all the biological factors in the organism's
environment
(2) the range of temperature that the
organism needs to live
(3) the physical space where an organism
lives
(4) the functional role played by the organism
where it lives
Answer (4)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
112. Which of the following is a secondary
pollutant?
(1) CO (2) SO
2

(3) CO
2
(4) O
3

Answer (4)
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
113. World Ozone Day is celebrated on
(1) 5
th
June (2) 16
th
September
(3) 21
st
April (4) 22
nd
April
Answer (2)
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
114. Natality refers to
(1) Death rate
(2) Number of individuals leaving the habitat
(3) Birth rate
(4) Number of individuals entering a habitat
Answer (3)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual
entering a habitat is
? Immigration
? Number of individual
leaving the habital
? Emigration
115. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Carbon (2) Fe
(3) Cl (4) Oxygen
Answer (3)
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
116. 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) Inverted pyramid of biomass
(2) Upright pyramid of numbers
(3) Pyramid of energy
(4) Upright pyramid of biomass
Answer (1)
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.
117. The Golgi complex participates in
(1) Fatty acid breakdown
(2) Respiration in bacteria
(3) Formation of secretory vesicles
(4) Activation of amino acid
Answer (3)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
118. Which of the following is not a product of light
reaction of photosynthesis?
(1) ATP (2) NADPH
(3) NADH (4) Oxygen
Answer (3)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
26
NEET (UG) - 2018 (Code-EE) CHLAA
119. Which among the following is not a
prokaryote?
(1) Saccharomyces (2) Nostoc
(3) Mycobacterium (4) Oscillatoria
Answer (1)
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
120. Stomatal movement is not affected by
(1) Temperature (2) O
2
concentration
(3) Light (4) CO
2
concentration
Answer (2)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
121. Which of the following is true for nucleolus?
(1) Larger nucleoli are present in dividing
cells
(2) It takes part in spindle formation
(3) It is a membrane-bound structure
(4) It is a site for active ribosomal RNA
synthesis
Answer (4)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
122. The stage during which separation of the
paired homologous chromosomes begins is
(1) Pachytene (2) Diakinesis
(3) Diplotene (4) Zygotene
Answer (3)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
123. The two functional groups characteristic of
sugars are
(1) Hydroxyl and methyl
(2) Carbonyl and phosphate
(3) Carbonyl and methyl
(4) Carbonyl and hydroxyl
Answer (4)
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.
124. Stomata in grass leaf are
(1) Dumb-bell shaped
(2) Rectangular
(3) Kidney shaped
(4) Barrel shaped
Answer (1)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
125. 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) Hydrilla (2) Banana
(3) Yucca (4) Viola
Answer (3)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
126. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?120?C (2) ?196?C
(3) ?80?C (4) ?160?C
Answer (2)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C (Cryopreservation)
127. Double fertilization is
(1) Fusion of two male gametes of a pollen
tube with two different eggs
(2) Fusion of two male gametes with one egg
(3) Fusion of one male gamete with two polar
nuclei
(4) Syngamy and triple fusion
Answer (4)
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
128. Oxygen is not produced during photosynthesis
by
(1) Green sulphur bacteria
(2) Cycas
(3) Nostoc
(4) Chara
Answer (1)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
27
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1. The test is of 3 hours duration and Test Booklet contains 180 questions. Each question carries
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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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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
NEET (UG) - 2018 (Code-EE) CHLAA
91. Which of the following statements is correct?
(1) Ovules are not enclosed by ovary wall in
gymnosperms
(2) Horsetails are gymnosperms
(3) Selaginella is heterosporous, while
Salvinia is homosporous
(4) Stems are usually unbranched in both
Cycas and Cedrus
Answer (1)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
92. Pneumatophores occur in
(1) Halophytes
(2) Carnivorous plants
(3) Free-floating hydrophytes
(4) Submerged hydrophytes
Answer (1)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
93. Sweet potato is a modified
(1) Stem
(2) Tap root
(3) Adventitious root
(4) Rhizome
Answer (3)
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
94. Plants having little or no secondary growth
are
(1) Grasses
(2) Conifers
(3) Deciduous angiosperms
(4) Cycads
Answer (1)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
95. Casparian strips occur in
(1) Epidermis
(2) Cortex
(3) Pericycle
(4) Endodermi
Answer (4)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Secondary xylem and phloem in dicot stem
are produced by
(1) Apical meristems
(2) Phellogen
(3) Vascular cambium
(4) Axillary meristems
Answer (3)
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.
97. Select the wrong statement :
(1) Cell wall is present in members of Fungi
and Plantae
(2) Pseudopodia are locomotory and feeding
structures in Sporozoans
(3) Mushrooms belong to Basidiomycetes
(4) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
Answer (2)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
98. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Fungus (2) Plant
(3) Bacterium (4) Virus
Answer (3)
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
23
NEET (UG) - 2018 (Code-EE) CHLAAA
99. Select the correct match
(1) Alec Jeffreys

- Streptococcus
pneumoniae
(2) Matthew Meselson - Pisum sativum
and F. Stahl
(3) Alfred Hershey and - TMV
Martha Chase
(4) Francois Jacob and - Lac operon
Jacques Monod
Answer (4)
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
100. Select the correct statement
(1) Franklin Stahl coined the term ?linkage?
(2) Spliceosomes take part in translation
(3) Punnett square was developed by a British
scientist
(4) Transduction was discovered by S. Altman
Answer (3)
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
101. Which of the following pairs is wrongly
matched?
(1) Starch synthesis in pea : Multiple alleles
(2) XO type sex : Grasshopper
determination
(3) ABO blood grouping : Co-dominance
(4) T.H. Morgan : Linkage
Answer (1)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (2, 3 & 4) are correctly
matched.
102. Offsets are produced by
(1) Meiotic divisions
(2) Parthenocarpy
(3) Mitotic divisions
(4) Parthenogenesis
Answer (3)
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)
103. Which of the following flowers only once in its
life-time?
(1) Bamboo species
(2) Mango
(3) Jackfruit
(4) Papaya
Answer (1)
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.
104. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Pollenkitt (2) Oil content
(3) Cellulosic intine (4) Sporopollenin
Answer (4)
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.
24
NEET (UG) - 2018 (Code-EE) CHLAA
105. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) Retrovirus
(2) X phage
(3) Ti plasmid
(4) pBR 322
Answer (1)
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.
106. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Extension, Denaturation, Annealing
(2) Denaturation, Extension, Annealing
(3) Annealing, Extension, Denaturation
(4) Denaturation, Annealing, Extension
Answer (4)
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
107. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Indian Council of Medical Research (ICMR)
(2) Research Committee on Genetic
Manipulation (RCGM)
(3) Council for Scientific and Industrial
Research (CSIR)
(4) Genetic Engineering Appraisal Committee
(GEAC)
Answer (4)
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).
108. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bio-infringement
(2) Biodegradation
(3) Biopiracy
(4) Bioexploitation
Answer (3)
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).
109. 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) Co-667
(2) Lerma Rojo
(3) Sharbati Sonora
(4) Basmati
Answer (4)
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.
110. Select the correct match
(1) Ribozyme - Nucleic acid
(2) T.H. Morgan - Transduction
(3) F
2
? Recessive parent - Dihybrid cross
(4) G. Mendel - Transformation
Answer (1)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
25
NEET (UG) - 2018 (Code-EE) CHLAAA
111. Niche is
(1) all the biological factors in the organism's
environment
(2) the range of temperature that the
organism needs to live
(3) the physical space where an organism
lives
(4) the functional role played by the organism
where it lives
Answer (4)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
112. Which of the following is a secondary
pollutant?
(1) CO (2) SO
2

(3) CO
2
(4) O
3

Answer (4)
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
113. World Ozone Day is celebrated on
(1) 5
th
June (2) 16
th
September
(3) 21
st
April (4) 22
nd
April
Answer (2)
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
114. Natality refers to
(1) Death rate
(2) Number of individuals leaving the habitat
(3) Birth rate
(4) Number of individuals entering a habitat
Answer (3)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual
entering a habitat is
? Immigration
? Number of individual
leaving the habital
? Emigration
115. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Carbon (2) Fe
(3) Cl (4) Oxygen
Answer (3)
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
116. 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) Inverted pyramid of biomass
(2) Upright pyramid of numbers
(3) Pyramid of energy
(4) Upright pyramid of biomass
Answer (1)
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.
117. The Golgi complex participates in
(1) Fatty acid breakdown
(2) Respiration in bacteria
(3) Formation of secretory vesicles
(4) Activation of amino acid
Answer (3)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
118. Which of the following is not a product of light
reaction of photosynthesis?
(1) ATP (2) NADPH
(3) NADH (4) Oxygen
Answer (3)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
26
NEET (UG) - 2018 (Code-EE) CHLAA
119. Which among the following is not a
prokaryote?
(1) Saccharomyces (2) Nostoc
(3) Mycobacterium (4) Oscillatoria
Answer (1)
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
120. Stomatal movement is not affected by
(1) Temperature (2) O
2
concentration
(3) Light (4) CO
2
concentration
Answer (2)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
121. Which of the following is true for nucleolus?
(1) Larger nucleoli are present in dividing
cells
(2) It takes part in spindle formation
(3) It is a membrane-bound structure
(4) It is a site for active ribosomal RNA
synthesis
Answer (4)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
122. The stage during which separation of the
paired homologous chromosomes begins is
(1) Pachytene (2) Diakinesis
(3) Diplotene (4) Zygotene
Answer (3)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
123. The two functional groups characteristic of
sugars are
(1) Hydroxyl and methyl
(2) Carbonyl and phosphate
(3) Carbonyl and methyl
(4) Carbonyl and hydroxyl
Answer (4)
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.
124. Stomata in grass leaf are
(1) Dumb-bell shaped
(2) Rectangular
(3) Kidney shaped
(4) Barrel shaped
Answer (1)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
125. 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) Hydrilla (2) Banana
(3) Yucca (4) Viola
Answer (3)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
126. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?120?C (2) ?196?C
(3) ?80?C (4) ?160?C
Answer (2)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C (Cryopreservation)
127. Double fertilization is
(1) Fusion of two male gametes of a pollen
tube with two different eggs
(2) Fusion of two male gametes with one egg
(3) Fusion of one male gamete with two polar
nuclei
(4) Syngamy and triple fusion
Answer (4)
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
128. Oxygen is not produced during photosynthesis
by
(1) Green sulphur bacteria
(2) Cycas
(3) Nostoc
(4) Chara
Answer (1)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
27
NEET (UG) - 2018 (Code-EE) CHLAAA
129. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Magnesium (2) Potassium
(3) Sodium (4) Calcium
Answer (2)
Sol. Potassium helps in maintaining turgidity of
cells.
130. What is the role of NAD
+
in cellular
respiration?
(1) It functions as an enzyme.
(2) It is a nucleotide source for ATP synthesis.
(3) It functions as an electron carrier.
(4) It is the final electron acceptor for
anaerobic respiration.
Answer (3)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
131. In which of the following forms is iron
absorbed by plants?
(1) Ferric
(2) Free element
(3) Ferrous
(4) Both ferric and ferrous
Answer (1*)
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
++
)
132. Winged pollen grains are present in
(1) Mustard (2) Mango
(3) Cycas (4) Pinus
Answer (4)
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.
133. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Neurospora
(2) Agaricus
(3) Alternaria
(4) Saccharomyces
Answer (2)
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.
134. Which one is wrongly matched?
(1) Uniflagellate gametes ? Polysiphonia
(2) Gemma cups ? Marchantia
(3) Biflagellate zoospores ? Brown algae
(4) Unicellular organism ? Chlorella
Answer (1)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (2, 3 & 4) are correctly
matched
135. 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.
28
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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
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
NEET (UG) - 2018 (Code-EE) CHLAA
91. Which of the following statements is correct?
(1) Ovules are not enclosed by ovary wall in
gymnosperms
(2) Horsetails are gymnosperms
(3) Selaginella is heterosporous, while
Salvinia is homosporous
(4) Stems are usually unbranched in both
Cycas and Cedrus
Answer (1)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
92. Pneumatophores occur in
(1) Halophytes
(2) Carnivorous plants
(3) Free-floating hydrophytes
(4) Submerged hydrophytes
Answer (1)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
93. Sweet potato is a modified
(1) Stem
(2) Tap root
(3) Adventitious root
(4) Rhizome
Answer (3)
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
94. Plants having little or no secondary growth
are
(1) Grasses
(2) Conifers
(3) Deciduous angiosperms
(4) Cycads
Answer (1)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
95. Casparian strips occur in
(1) Epidermis
(2) Cortex
(3) Pericycle
(4) Endodermi
Answer (4)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Secondary xylem and phloem in dicot stem
are produced by
(1) Apical meristems
(2) Phellogen
(3) Vascular cambium
(4) Axillary meristems
Answer (3)
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.
97. Select the wrong statement :
(1) Cell wall is present in members of Fungi
and Plantae
(2) Pseudopodia are locomotory and feeding
structures in Sporozoans
(3) Mushrooms belong to Basidiomycetes
(4) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
Answer (2)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
98. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Fungus (2) Plant
(3) Bacterium (4) Virus
Answer (3)
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
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NEET (UG) - 2018 (Code-EE) CHLAAA
99. Select the correct match
(1) Alec Jeffreys

- Streptococcus
pneumoniae
(2) Matthew Meselson - Pisum sativum
and F. Stahl
(3) Alfred Hershey and - TMV
Martha Chase
(4) Francois Jacob and - Lac operon
Jacques Monod
Answer (4)
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
100. Select the correct statement
(1) Franklin Stahl coined the term ?linkage?
(2) Spliceosomes take part in translation
(3) Punnett square was developed by a British
scientist
(4) Transduction was discovered by S. Altman
Answer (3)
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
101. Which of the following pairs is wrongly
matched?
(1) Starch synthesis in pea : Multiple alleles
(2) XO type sex : Grasshopper
determination
(3) ABO blood grouping : Co-dominance
(4) T.H. Morgan : Linkage
Answer (1)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (2, 3 & 4) are correctly
matched.
102. Offsets are produced by
(1) Meiotic divisions
(2) Parthenocarpy
(3) Mitotic divisions
(4) Parthenogenesis
Answer (3)
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)
103. Which of the following flowers only once in its
life-time?
(1) Bamboo species
(2) Mango
(3) Jackfruit
(4) Papaya
Answer (1)
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.
104. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Pollenkitt (2) Oil content
(3) Cellulosic intine (4) Sporopollenin
Answer (4)
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.
24
NEET (UG) - 2018 (Code-EE) CHLAA
105. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) Retrovirus
(2) X phage
(3) Ti plasmid
(4) pBR 322
Answer (1)
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.
106. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Extension, Denaturation, Annealing
(2) Denaturation, Extension, Annealing
(3) Annealing, Extension, Denaturation
(4) Denaturation, Annealing, Extension
Answer (4)
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
107. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Indian Council of Medical Research (ICMR)
(2) Research Committee on Genetic
Manipulation (RCGM)
(3) Council for Scientific and Industrial
Research (CSIR)
(4) Genetic Engineering Appraisal Committee
(GEAC)
Answer (4)
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).
108. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bio-infringement
(2) Biodegradation
(3) Biopiracy
(4) Bioexploitation
Answer (3)
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).
109. 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) Co-667
(2) Lerma Rojo
(3) Sharbati Sonora
(4) Basmati
Answer (4)
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.
110. Select the correct match
(1) Ribozyme - Nucleic acid
(2) T.H. Morgan - Transduction
(3) F
2
? Recessive parent - Dihybrid cross
(4) G. Mendel - Transformation
Answer (1)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
25
NEET (UG) - 2018 (Code-EE) CHLAAA
111. Niche is
(1) all the biological factors in the organism's
environment
(2) the range of temperature that the
organism needs to live
(3) the physical space where an organism
lives
(4) the functional role played by the organism
where it lives
Answer (4)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
112. Which of the following is a secondary
pollutant?
(1) CO (2) SO
2

(3) CO
2
(4) O
3

Answer (4)
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
113. World Ozone Day is celebrated on
(1) 5
th
June (2) 16
th
September
(3) 21
st
April (4) 22
nd
April
Answer (2)
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
114. Natality refers to
(1) Death rate
(2) Number of individuals leaving the habitat
(3) Birth rate
(4) Number of individuals entering a habitat
Answer (3)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual
entering a habitat is
? Immigration
? Number of individual
leaving the habital
? Emigration
115. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Carbon (2) Fe
(3) Cl (4) Oxygen
Answer (3)
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
116. 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) Inverted pyramid of biomass
(2) Upright pyramid of numbers
(3) Pyramid of energy
(4) Upright pyramid of biomass
Answer (1)
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.
117. The Golgi complex participates in
(1) Fatty acid breakdown
(2) Respiration in bacteria
(3) Formation of secretory vesicles
(4) Activation of amino acid
Answer (3)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
118. Which of the following is not a product of light
reaction of photosynthesis?
(1) ATP (2) NADPH
(3) NADH (4) Oxygen
Answer (3)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
26
NEET (UG) - 2018 (Code-EE) CHLAA
119. Which among the following is not a
prokaryote?
(1) Saccharomyces (2) Nostoc
(3) Mycobacterium (4) Oscillatoria
Answer (1)
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
120. Stomatal movement is not affected by
(1) Temperature (2) O
2
concentration
(3) Light (4) CO
2
concentration
Answer (2)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
121. Which of the following is true for nucleolus?
(1) Larger nucleoli are present in dividing
cells
(2) It takes part in spindle formation
(3) It is a membrane-bound structure
(4) It is a site for active ribosomal RNA
synthesis
Answer (4)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
122. The stage during which separation of the
paired homologous chromosomes begins is
(1) Pachytene (2) Diakinesis
(3) Diplotene (4) Zygotene
Answer (3)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
123. The two functional groups characteristic of
sugars are
(1) Hydroxyl and methyl
(2) Carbonyl and phosphate
(3) Carbonyl and methyl
(4) Carbonyl and hydroxyl
Answer (4)
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.
124. Stomata in grass leaf are
(1) Dumb-bell shaped
(2) Rectangular
(3) Kidney shaped
(4) Barrel shaped
Answer (1)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
125. 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) Hydrilla (2) Banana
(3) Yucca (4) Viola
Answer (3)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
126. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?120?C (2) ?196?C
(3) ?80?C (4) ?160?C
Answer (2)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C (Cryopreservation)
127. Double fertilization is
(1) Fusion of two male gametes of a pollen
tube with two different eggs
(2) Fusion of two male gametes with one egg
(3) Fusion of one male gamete with two polar
nuclei
(4) Syngamy and triple fusion
Answer (4)
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
128. Oxygen is not produced during photosynthesis
by
(1) Green sulphur bacteria
(2) Cycas
(3) Nostoc
(4) Chara
Answer (1)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
27
NEET (UG) - 2018 (Code-EE) CHLAAA
129. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Magnesium (2) Potassium
(3) Sodium (4) Calcium
Answer (2)
Sol. Potassium helps in maintaining turgidity of
cells.
130. What is the role of NAD
+
in cellular
respiration?
(1) It functions as an enzyme.
(2) It is a nucleotide source for ATP synthesis.
(3) It functions as an electron carrier.
(4) It is the final electron acceptor for
anaerobic respiration.
Answer (3)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
131. In which of the following forms is iron
absorbed by plants?
(1) Ferric
(2) Free element
(3) Ferrous
(4) Both ferric and ferrous
Answer (1*)
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
++
)
132. Winged pollen grains are present in
(1) Mustard (2) Mango
(3) Cycas (4) Pinus
Answer (4)
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.
133. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Neurospora
(2) Agaricus
(3) Alternaria
(4) Saccharomyces
Answer (2)
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.
134. Which one is wrongly matched?
(1) Uniflagellate gametes ? Polysiphonia
(2) Gemma cups ? Marchantia
(3) Biflagellate zoospores ? Brown algae
(4) Unicellular organism ? Chlorella
Answer (1)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (2, 3 & 4) are correctly
matched
135. 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.
28
NEET (UG) - 2018 (Code-EE) CHLAA
a b
(1) (i) (iv)
(2) (ii) (iv)
(3) (iii) (ii)
(4) (iii) (iv)
c d
(iii) (ii)
(iii) (i)
(i) (iv)
(i) (ii)
Answer (4)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
136. Which of the following is an amino acid
derived hormone?
(1) Epinephrine (2) Estradiol
(3) Ecdysone (4) Estriol
Answer (1)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
137. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(2) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(3) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
(4) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
Answer (3)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
138. The transparent lens in the human eye is held
in its place by
(1) ligaments attached to the ciliary body
(2) smooth muscles attached to the iris
(3) ligaments attached to the iris
(4) smooth muscles attached to the ciliary
body
Answer (1)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
139. Which of the following hormones can play a
significant role in osteoporosis?
(1) Aldosterone and Prolactin
(2) Estrogen and Parathyroid hormone
(3) Progesterone and Aldosterone
(4) Parathyroid hormone and Prolactin
Answer (2)
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.
140. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Forelimbs of man, bat and cheetah
(2) Brain of bat, man and cheetah
(3) Heart of bat, man and cheetah
(4) Eye of octopus, bat and man
Answer (4)
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.
141. Which of the following is not an autoimmune
disease?
(1) Psoriasis
(2) Alzheimer's disease
(3) Rheumatoid arthritis
(4) Vitiligo
Answer (2)
29
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
NEET (UG) - 2018 (Code-EE) CHLAA
91. Which of the following statements is correct?
(1) Ovules are not enclosed by ovary wall in
gymnosperms
(2) Horsetails are gymnosperms
(3) Selaginella is heterosporous, while
Salvinia is homosporous
(4) Stems are usually unbranched in both
Cycas and Cedrus
Answer (1)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
92. Pneumatophores occur in
(1) Halophytes
(2) Carnivorous plants
(3) Free-floating hydrophytes
(4) Submerged hydrophytes
Answer (1)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
93. Sweet potato is a modified
(1) Stem
(2) Tap root
(3) Adventitious root
(4) Rhizome
Answer (3)
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
94. Plants having little or no secondary growth
are
(1) Grasses
(2) Conifers
(3) Deciduous angiosperms
(4) Cycads
Answer (1)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
95. Casparian strips occur in
(1) Epidermis
(2) Cortex
(3) Pericycle
(4) Endodermi
Answer (4)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Secondary xylem and phloem in dicot stem
are produced by
(1) Apical meristems
(2) Phellogen
(3) Vascular cambium
(4) Axillary meristems
Answer (3)
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.
97. Select the wrong statement :
(1) Cell wall is present in members of Fungi
and Plantae
(2) Pseudopodia are locomotory and feeding
structures in Sporozoans
(3) Mushrooms belong to Basidiomycetes
(4) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
Answer (2)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
98. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Fungus (2) Plant
(3) Bacterium (4) Virus
Answer (3)
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
23
NEET (UG) - 2018 (Code-EE) CHLAAA
99. Select the correct match
(1) Alec Jeffreys

- Streptococcus
pneumoniae
(2) Matthew Meselson - Pisum sativum
and F. Stahl
(3) Alfred Hershey and - TMV
Martha Chase
(4) Francois Jacob and - Lac operon
Jacques Monod
Answer (4)
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
100. Select the correct statement
(1) Franklin Stahl coined the term ?linkage?
(2) Spliceosomes take part in translation
(3) Punnett square was developed by a British
scientist
(4) Transduction was discovered by S. Altman
Answer (3)
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
101. Which of the following pairs is wrongly
matched?
(1) Starch synthesis in pea : Multiple alleles
(2) XO type sex : Grasshopper
determination
(3) ABO blood grouping : Co-dominance
(4) T.H. Morgan : Linkage
Answer (1)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (2, 3 & 4) are correctly
matched.
102. Offsets are produced by
(1) Meiotic divisions
(2) Parthenocarpy
(3) Mitotic divisions
(4) Parthenogenesis
Answer (3)
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)
103. Which of the following flowers only once in its
life-time?
(1) Bamboo species
(2) Mango
(3) Jackfruit
(4) Papaya
Answer (1)
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.
104. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Pollenkitt (2) Oil content
(3) Cellulosic intine (4) Sporopollenin
Answer (4)
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.
24
NEET (UG) - 2018 (Code-EE) CHLAA
105. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) Retrovirus
(2) X phage
(3) Ti plasmid
(4) pBR 322
Answer (1)
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.
106. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Extension, Denaturation, Annealing
(2) Denaturation, Extension, Annealing
(3) Annealing, Extension, Denaturation
(4) Denaturation, Annealing, Extension
Answer (4)
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
107. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Indian Council of Medical Research (ICMR)
(2) Research Committee on Genetic
Manipulation (RCGM)
(3) Council for Scientific and Industrial
Research (CSIR)
(4) Genetic Engineering Appraisal Committee
(GEAC)
Answer (4)
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).
108. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bio-infringement
(2) Biodegradation
(3) Biopiracy
(4) Bioexploitation
Answer (3)
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).
109. 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) Co-667
(2) Lerma Rojo
(3) Sharbati Sonora
(4) Basmati
Answer (4)
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.
110. Select the correct match
(1) Ribozyme - Nucleic acid
(2) T.H. Morgan - Transduction
(3) F
2
? Recessive parent - Dihybrid cross
(4) G. Mendel - Transformation
Answer (1)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
25
NEET (UG) - 2018 (Code-EE) CHLAAA
111. Niche is
(1) all the biological factors in the organism's
environment
(2) the range of temperature that the
organism needs to live
(3) the physical space where an organism
lives
(4) the functional role played by the organism
where it lives
Answer (4)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
112. Which of the following is a secondary
pollutant?
(1) CO (2) SO
2

(3) CO
2
(4) O
3

Answer (4)
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
113. World Ozone Day is celebrated on
(1) 5
th
June (2) 16
th
September
(3) 21
st
April (4) 22
nd
April
Answer (2)
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
114. Natality refers to
(1) Death rate
(2) Number of individuals leaving the habitat
(3) Birth rate
(4) Number of individuals entering a habitat
Answer (3)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual
entering a habitat is
? Immigration
? Number of individual
leaving the habital
? Emigration
115. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Carbon (2) Fe
(3) Cl (4) Oxygen
Answer (3)
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
116. 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) Inverted pyramid of biomass
(2) Upright pyramid of numbers
(3) Pyramid of energy
(4) Upright pyramid of biomass
Answer (1)
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.
117. The Golgi complex participates in
(1) Fatty acid breakdown
(2) Respiration in bacteria
(3) Formation of secretory vesicles
(4) Activation of amino acid
Answer (3)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
118. Which of the following is not a product of light
reaction of photosynthesis?
(1) ATP (2) NADPH
(3) NADH (4) Oxygen
Answer (3)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
26
NEET (UG) - 2018 (Code-EE) CHLAA
119. Which among the following is not a
prokaryote?
(1) Saccharomyces (2) Nostoc
(3) Mycobacterium (4) Oscillatoria
Answer (1)
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
120. Stomatal movement is not affected by
(1) Temperature (2) O
2
concentration
(3) Light (4) CO
2
concentration
Answer (2)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
121. Which of the following is true for nucleolus?
(1) Larger nucleoli are present in dividing
cells
(2) It takes part in spindle formation
(3) It is a membrane-bound structure
(4) It is a site for active ribosomal RNA
synthesis
Answer (4)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
122. The stage during which separation of the
paired homologous chromosomes begins is
(1) Pachytene (2) Diakinesis
(3) Diplotene (4) Zygotene
Answer (3)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
123. The two functional groups characteristic of
sugars are
(1) Hydroxyl and methyl
(2) Carbonyl and phosphate
(3) Carbonyl and methyl
(4) Carbonyl and hydroxyl
Answer (4)
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.
124. Stomata in grass leaf are
(1) Dumb-bell shaped
(2) Rectangular
(3) Kidney shaped
(4) Barrel shaped
Answer (1)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
125. 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) Hydrilla (2) Banana
(3) Yucca (4) Viola
Answer (3)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
126. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?120?C (2) ?196?C
(3) ?80?C (4) ?160?C
Answer (2)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C (Cryopreservation)
127. Double fertilization is
(1) Fusion of two male gametes of a pollen
tube with two different eggs
(2) Fusion of two male gametes with one egg
(3) Fusion of one male gamete with two polar
nuclei
(4) Syngamy and triple fusion
Answer (4)
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
128. Oxygen is not produced during photosynthesis
by
(1) Green sulphur bacteria
(2) Cycas
(3) Nostoc
(4) Chara
Answer (1)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
27
NEET (UG) - 2018 (Code-EE) CHLAAA
129. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Magnesium (2) Potassium
(3) Sodium (4) Calcium
Answer (2)
Sol. Potassium helps in maintaining turgidity of
cells.
130. What is the role of NAD
+
in cellular
respiration?
(1) It functions as an enzyme.
(2) It is a nucleotide source for ATP synthesis.
(3) It functions as an electron carrier.
(4) It is the final electron acceptor for
anaerobic respiration.
Answer (3)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
131. In which of the following forms is iron
absorbed by plants?
(1) Ferric
(2) Free element
(3) Ferrous
(4) Both ferric and ferrous
Answer (1*)
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
++
)
132. Winged pollen grains are present in
(1) Mustard (2) Mango
(3) Cycas (4) Pinus
Answer (4)
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.
133. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Neurospora
(2) Agaricus
(3) Alternaria
(4) Saccharomyces
Answer (2)
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.
134. Which one is wrongly matched?
(1) Uniflagellate gametes ? Polysiphonia
(2) Gemma cups ? Marchantia
(3) Biflagellate zoospores ? Brown algae
(4) Unicellular organism ? Chlorella
Answer (1)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (2, 3 & 4) are correctly
matched
135. 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.
28
NEET (UG) - 2018 (Code-EE) CHLAA
a b
(1) (i) (iv)
(2) (ii) (iv)
(3) (iii) (ii)
(4) (iii) (iv)
c d
(iii) (ii)
(iii) (i)
(i) (iv)
(i) (ii)
Answer (4)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
136. Which of the following is an amino acid
derived hormone?
(1) Epinephrine (2) Estradiol
(3) Ecdysone (4) Estriol
Answer (1)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
137. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(2) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(3) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
(4) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
Answer (3)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
138. The transparent lens in the human eye is held
in its place by
(1) ligaments attached to the ciliary body
(2) smooth muscles attached to the iris
(3) ligaments attached to the iris
(4) smooth muscles attached to the ciliary
body
Answer (1)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
139. Which of the following hormones can play a
significant role in osteoporosis?
(1) Aldosterone and Prolactin
(2) Estrogen and Parathyroid hormone
(3) Progesterone and Aldosterone
(4) Parathyroid hormone and Prolactin
Answer (2)
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.
140. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Forelimbs of man, bat and cheetah
(2) Brain of bat, man and cheetah
(3) Heart of bat, man and cheetah
(4) Eye of octopus, bat and man
Answer (4)
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.
141. Which of the following is not an autoimmune
disease?
(1) Psoriasis
(2) Alzheimer's disease
(3) Rheumatoid arthritis
(4) Vitiligo
Answer (2)
29
NEET (UG) - 2018 (Code-EE) CHLAAA
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.
142. 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) b, c and e (2) b, d and e
(3) a, b and c (4) a, c and e
Answer (3)
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)
143. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Elephantiasis (2) Ringworm disease
(3) Ascariasis (4) Amoebiasis
Answer (1)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
144. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Homology
(2) Convergent evolution
(3) Analogy
(4) Adaptive radiation
Answer (1)
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.
145. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin D (2) Vitamin B
12
(3) Vitamin A (4) Vitamin E
Answer (2)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
146. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Commensalism (2) Parasitism
(3) Mutualism (4) Amensalism
Answer (4)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicillium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicillium or the
organism which produces it.
147. All of the following are included in ? ex-situ
conservation? except
(1) Wildlife safari parks
(2) Botanical gardens
(3) Sacred groves
(4) Seed banks
Answer (3)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
148. 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) ii i iii iv
(2) iii iv i ii
(3) i iii iv ii
(4) i ii iv iii
Answer (2)
30
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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
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
NEET (UG) - 2018 (Code-EE) CHLAA
91. Which of the following statements is correct?
(1) Ovules are not enclosed by ovary wall in
gymnosperms
(2) Horsetails are gymnosperms
(3) Selaginella is heterosporous, while
Salvinia is homosporous
(4) Stems are usually unbranched in both
Cycas and Cedrus
Answer (1)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
92. Pneumatophores occur in
(1) Halophytes
(2) Carnivorous plants
(3) Free-floating hydrophytes
(4) Submerged hydrophytes
Answer (1)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
93. Sweet potato is a modified
(1) Stem
(2) Tap root
(3) Adventitious root
(4) Rhizome
Answer (3)
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
94. Plants having little or no secondary growth
are
(1) Grasses
(2) Conifers
(3) Deciduous angiosperms
(4) Cycads
Answer (1)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
95. Casparian strips occur in
(1) Epidermis
(2) Cortex
(3) Pericycle
(4) Endodermi
Answer (4)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Secondary xylem and phloem in dicot stem
are produced by
(1) Apical meristems
(2) Phellogen
(3) Vascular cambium
(4) Axillary meristems
Answer (3)
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.
97. Select the wrong statement :
(1) Cell wall is present in members of Fungi
and Plantae
(2) Pseudopodia are locomotory and feeding
structures in Sporozoans
(3) Mushrooms belong to Basidiomycetes
(4) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
Answer (2)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
98. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Fungus (2) Plant
(3) Bacterium (4) Virus
Answer (3)
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
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NEET (UG) - 2018 (Code-EE) CHLAAA
99. Select the correct match
(1) Alec Jeffreys

- Streptococcus
pneumoniae
(2) Matthew Meselson - Pisum sativum
and F. Stahl
(3) Alfred Hershey and - TMV
Martha Chase
(4) Francois Jacob and - Lac operon
Jacques Monod
Answer (4)
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
100. Select the correct statement
(1) Franklin Stahl coined the term ?linkage?
(2) Spliceosomes take part in translation
(3) Punnett square was developed by a British
scientist
(4) Transduction was discovered by S. Altman
Answer (3)
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
101. Which of the following pairs is wrongly
matched?
(1) Starch synthesis in pea : Multiple alleles
(2) XO type sex : Grasshopper
determination
(3) ABO blood grouping : Co-dominance
(4) T.H. Morgan : Linkage
Answer (1)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (2, 3 & 4) are correctly
matched.
102. Offsets are produced by
(1) Meiotic divisions
(2) Parthenocarpy
(3) Mitotic divisions
(4) Parthenogenesis
Answer (3)
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)
103. Which of the following flowers only once in its
life-time?
(1) Bamboo species
(2) Mango
(3) Jackfruit
(4) Papaya
Answer (1)
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.
104. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Pollenkitt (2) Oil content
(3) Cellulosic intine (4) Sporopollenin
Answer (4)
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.
24
NEET (UG) - 2018 (Code-EE) CHLAA
105. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) Retrovirus
(2) X phage
(3) Ti plasmid
(4) pBR 322
Answer (1)
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.
106. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Extension, Denaturation, Annealing
(2) Denaturation, Extension, Annealing
(3) Annealing, Extension, Denaturation
(4) Denaturation, Annealing, Extension
Answer (4)
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
107. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Indian Council of Medical Research (ICMR)
(2) Research Committee on Genetic
Manipulation (RCGM)
(3) Council for Scientific and Industrial
Research (CSIR)
(4) Genetic Engineering Appraisal Committee
(GEAC)
Answer (4)
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).
108. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bio-infringement
(2) Biodegradation
(3) Biopiracy
(4) Bioexploitation
Answer (3)
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).
109. 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) Co-667
(2) Lerma Rojo
(3) Sharbati Sonora
(4) Basmati
Answer (4)
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.
110. Select the correct match
(1) Ribozyme - Nucleic acid
(2) T.H. Morgan - Transduction
(3) F
2
? Recessive parent - Dihybrid cross
(4) G. Mendel - Transformation
Answer (1)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
25
NEET (UG) - 2018 (Code-EE) CHLAAA
111. Niche is
(1) all the biological factors in the organism's
environment
(2) the range of temperature that the
organism needs to live
(3) the physical space where an organism
lives
(4) the functional role played by the organism
where it lives
Answer (4)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
112. Which of the following is a secondary
pollutant?
(1) CO (2) SO
2

(3) CO
2
(4) O
3

Answer (4)
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
113. World Ozone Day is celebrated on
(1) 5
th
June (2) 16
th
September
(3) 21
st
April (4) 22
nd
April
Answer (2)
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
114. Natality refers to
(1) Death rate
(2) Number of individuals leaving the habitat
(3) Birth rate
(4) Number of individuals entering a habitat
Answer (3)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual
entering a habitat is
? Immigration
? Number of individual
leaving the habital
? Emigration
115. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Carbon (2) Fe
(3) Cl (4) Oxygen
Answer (3)
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
116. 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) Inverted pyramid of biomass
(2) Upright pyramid of numbers
(3) Pyramid of energy
(4) Upright pyramid of biomass
Answer (1)
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.
117. The Golgi complex participates in
(1) Fatty acid breakdown
(2) Respiration in bacteria
(3) Formation of secretory vesicles
(4) Activation of amino acid
Answer (3)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
118. Which of the following is not a product of light
reaction of photosynthesis?
(1) ATP (2) NADPH
(3) NADH (4) Oxygen
Answer (3)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
26
NEET (UG) - 2018 (Code-EE) CHLAA
119. Which among the following is not a
prokaryote?
(1) Saccharomyces (2) Nostoc
(3) Mycobacterium (4) Oscillatoria
Answer (1)
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
120. Stomatal movement is not affected by
(1) Temperature (2) O
2
concentration
(3) Light (4) CO
2
concentration
Answer (2)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
121. Which of the following is true for nucleolus?
(1) Larger nucleoli are present in dividing
cells
(2) It takes part in spindle formation
(3) It is a membrane-bound structure
(4) It is a site for active ribosomal RNA
synthesis
Answer (4)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
122. The stage during which separation of the
paired homologous chromosomes begins is
(1) Pachytene (2) Diakinesis
(3) Diplotene (4) Zygotene
Answer (3)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
123. The two functional groups characteristic of
sugars are
(1) Hydroxyl and methyl
(2) Carbonyl and phosphate
(3) Carbonyl and methyl
(4) Carbonyl and hydroxyl
Answer (4)
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.
124. Stomata in grass leaf are
(1) Dumb-bell shaped
(2) Rectangular
(3) Kidney shaped
(4) Barrel shaped
Answer (1)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
125. 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) Hydrilla (2) Banana
(3) Yucca (4) Viola
Answer (3)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
126. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?120?C (2) ?196?C
(3) ?80?C (4) ?160?C
Answer (2)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C (Cryopreservation)
127. Double fertilization is
(1) Fusion of two male gametes of a pollen
tube with two different eggs
(2) Fusion of two male gametes with one egg
(3) Fusion of one male gamete with two polar
nuclei
(4) Syngamy and triple fusion
Answer (4)
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
128. Oxygen is not produced during photosynthesis
by
(1) Green sulphur bacteria
(2) Cycas
(3) Nostoc
(4) Chara
Answer (1)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
27
NEET (UG) - 2018 (Code-EE) CHLAAA
129. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Magnesium (2) Potassium
(3) Sodium (4) Calcium
Answer (2)
Sol. Potassium helps in maintaining turgidity of
cells.
130. What is the role of NAD
+
in cellular
respiration?
(1) It functions as an enzyme.
(2) It is a nucleotide source for ATP synthesis.
(3) It functions as an electron carrier.
(4) It is the final electron acceptor for
anaerobic respiration.
Answer (3)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
131. In which of the following forms is iron
absorbed by plants?
(1) Ferric
(2) Free element
(3) Ferrous
(4) Both ferric and ferrous
Answer (1*)
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
++
)
132. Winged pollen grains are present in
(1) Mustard (2) Mango
(3) Cycas (4) Pinus
Answer (4)
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.
133. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Neurospora
(2) Agaricus
(3) Alternaria
(4) Saccharomyces
Answer (2)
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.
134. Which one is wrongly matched?
(1) Uniflagellate gametes ? Polysiphonia
(2) Gemma cups ? Marchantia
(3) Biflagellate zoospores ? Brown algae
(4) Unicellular organism ? Chlorella
Answer (1)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (2, 3 & 4) are correctly
matched
135. 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.
28
NEET (UG) - 2018 (Code-EE) CHLAA
a b
(1) (i) (iv)
(2) (ii) (iv)
(3) (iii) (ii)
(4) (iii) (iv)
c d
(iii) (ii)
(iii) (i)
(i) (iv)
(i) (ii)
Answer (4)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
136. Which of the following is an amino acid
derived hormone?
(1) Epinephrine (2) Estradiol
(3) Ecdysone (4) Estriol
Answer (1)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
137. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(2) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(3) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
(4) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
Answer (3)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
138. The transparent lens in the human eye is held
in its place by
(1) ligaments attached to the ciliary body
(2) smooth muscles attached to the iris
(3) ligaments attached to the iris
(4) smooth muscles attached to the ciliary
body
Answer (1)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
139. Which of the following hormones can play a
significant role in osteoporosis?
(1) Aldosterone and Prolactin
(2) Estrogen and Parathyroid hormone
(3) Progesterone and Aldosterone
(4) Parathyroid hormone and Prolactin
Answer (2)
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.
140. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Forelimbs of man, bat and cheetah
(2) Brain of bat, man and cheetah
(3) Heart of bat, man and cheetah
(4) Eye of octopus, bat and man
Answer (4)
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.
141. Which of the following is not an autoimmune
disease?
(1) Psoriasis
(2) Alzheimer's disease
(3) Rheumatoid arthritis
(4) Vitiligo
Answer (2)
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NEET (UG) - 2018 (Code-EE) CHLAAA
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.
142. 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) b, c and e (2) b, d and e
(3) a, b and c (4) a, c and e
Answer (3)
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)
143. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Elephantiasis (2) Ringworm disease
(3) Ascariasis (4) Amoebiasis
Answer (1)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
144. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Homology
(2) Convergent evolution
(3) Analogy
(4) Adaptive radiation
Answer (1)
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.
145. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin D (2) Vitamin B
12
(3) Vitamin A (4) Vitamin E
Answer (2)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
146. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Commensalism (2) Parasitism
(3) Mutualism (4) Amensalism
Answer (4)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicillium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicillium or the
organism which produces it.
147. All of the following are included in ? ex-situ
conservation? except
(1) Wildlife safari parks
(2) Botanical gardens
(3) Sacred groves
(4) Seed banks
Answer (3)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
148. 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) ii i iii iv
(2) iii iv i ii
(3) i iii iv ii
(4) i ii iv iii
Answer (2)
30
NEET (UG) - 2018 (Code-EE) CHLAA
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
149. In a growing population of a country,
(1) pre-reproductive individuals are more
than the reproductive individuals.
(2) reproductive and pre-reproductive
individuals are equal in number.
(3) reproductive individuals are less than the
post-reproductive individuals.
(4) pre-reproductive individuals are less than
the reproductive individuals.
Answer (1)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
150. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Flowers (2) Roots
(3) Latex (4) Leaves
Answer (3)
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.
151. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, hPL, progestogens, prolactin
(2) hCG, hPL, progestogens, estrogens
(3) hCG, hPL, estrogens, relaxin, oxytocin
(4) hCG, progestogens, estrogens,
glucocorticoids
Answer (2)
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.
152. The contraceptive ?SAHELI?
(1) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(2) is an IUD.
(3) increases the concentration of estrogen
and prevents ovulation in females.
(4) is a post-coital contraceptive.
Answer (1)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
153. The amnion of mammalian embryo is derived
from
(1) ectoderm and mesoderm
(2) mesoderm and trophoblast
(3) endoderm and mesoderm
(4) ectoderm and endoderm
Answer (1)
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.
154. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(2) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(3) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
Answer (4)
31
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
NEET (UG) - 2018 (Code-EE) CHLAA
91. Which of the following statements is correct?
(1) Ovules are not enclosed by ovary wall in
gymnosperms
(2) Horsetails are gymnosperms
(3) Selaginella is heterosporous, while
Salvinia is homosporous
(4) Stems are usually unbranched in both
Cycas and Cedrus
Answer (1)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
92. Pneumatophores occur in
(1) Halophytes
(2) Carnivorous plants
(3) Free-floating hydrophytes
(4) Submerged hydrophytes
Answer (1)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
93. Sweet potato is a modified
(1) Stem
(2) Tap root
(3) Adventitious root
(4) Rhizome
Answer (3)
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
94. Plants having little or no secondary growth
are
(1) Grasses
(2) Conifers
(3) Deciduous angiosperms
(4) Cycads
Answer (1)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
95. Casparian strips occur in
(1) Epidermis
(2) Cortex
(3) Pericycle
(4) Endodermi
Answer (4)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Secondary xylem and phloem in dicot stem
are produced by
(1) Apical meristems
(2) Phellogen
(3) Vascular cambium
(4) Axillary meristems
Answer (3)
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.
97. Select the wrong statement :
(1) Cell wall is present in members of Fungi
and Plantae
(2) Pseudopodia are locomotory and feeding
structures in Sporozoans
(3) Mushrooms belong to Basidiomycetes
(4) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
Answer (2)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
98. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Fungus (2) Plant
(3) Bacterium (4) Virus
Answer (3)
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
23
NEET (UG) - 2018 (Code-EE) CHLAAA
99. Select the correct match
(1) Alec Jeffreys

- Streptococcus
pneumoniae
(2) Matthew Meselson - Pisum sativum
and F. Stahl
(3) Alfred Hershey and - TMV
Martha Chase
(4) Francois Jacob and - Lac operon
Jacques Monod
Answer (4)
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
100. Select the correct statement
(1) Franklin Stahl coined the term ?linkage?
(2) Spliceosomes take part in translation
(3) Punnett square was developed by a British
scientist
(4) Transduction was discovered by S. Altman
Answer (3)
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
101. Which of the following pairs is wrongly
matched?
(1) Starch synthesis in pea : Multiple alleles
(2) XO type sex : Grasshopper
determination
(3) ABO blood grouping : Co-dominance
(4) T.H. Morgan : Linkage
Answer (1)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (2, 3 & 4) are correctly
matched.
102. Offsets are produced by
(1) Meiotic divisions
(2) Parthenocarpy
(3) Mitotic divisions
(4) Parthenogenesis
Answer (3)
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)
103. Which of the following flowers only once in its
life-time?
(1) Bamboo species
(2) Mango
(3) Jackfruit
(4) Papaya
Answer (1)
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.
104. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Pollenkitt (2) Oil content
(3) Cellulosic intine (4) Sporopollenin
Answer (4)
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.
24
NEET (UG) - 2018 (Code-EE) CHLAA
105. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) Retrovirus
(2) X phage
(3) Ti plasmid
(4) pBR 322
Answer (1)
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.
106. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Extension, Denaturation, Annealing
(2) Denaturation, Extension, Annealing
(3) Annealing, Extension, Denaturation
(4) Denaturation, Annealing, Extension
Answer (4)
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
107. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Indian Council of Medical Research (ICMR)
(2) Research Committee on Genetic
Manipulation (RCGM)
(3) Council for Scientific and Industrial
Research (CSIR)
(4) Genetic Engineering Appraisal Committee
(GEAC)
Answer (4)
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).
108. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bio-infringement
(2) Biodegradation
(3) Biopiracy
(4) Bioexploitation
Answer (3)
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).
109. 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) Co-667
(2) Lerma Rojo
(3) Sharbati Sonora
(4) Basmati
Answer (4)
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.
110. Select the correct match
(1) Ribozyme - Nucleic acid
(2) T.H. Morgan - Transduction
(3) F
2
? Recessive parent - Dihybrid cross
(4) G. Mendel - Transformation
Answer (1)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
25
NEET (UG) - 2018 (Code-EE) CHLAAA
111. Niche is
(1) all the biological factors in the organism's
environment
(2) the range of temperature that the
organism needs to live
(3) the physical space where an organism
lives
(4) the functional role played by the organism
where it lives
Answer (4)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
112. Which of the following is a secondary
pollutant?
(1) CO (2) SO
2

(3) CO
2
(4) O
3

Answer (4)
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
113. World Ozone Day is celebrated on
(1) 5
th
June (2) 16
th
September
(3) 21
st
April (4) 22
nd
April
Answer (2)
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
114. Natality refers to
(1) Death rate
(2) Number of individuals leaving the habitat
(3) Birth rate
(4) Number of individuals entering a habitat
Answer (3)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual
entering a habitat is
? Immigration
? Number of individual
leaving the habital
? Emigration
115. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Carbon (2) Fe
(3) Cl (4) Oxygen
Answer (3)
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
116. 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) Inverted pyramid of biomass
(2) Upright pyramid of numbers
(3) Pyramid of energy
(4) Upright pyramid of biomass
Answer (1)
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.
117. The Golgi complex participates in
(1) Fatty acid breakdown
(2) Respiration in bacteria
(3) Formation of secretory vesicles
(4) Activation of amino acid
Answer (3)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
118. Which of the following is not a product of light
reaction of photosynthesis?
(1) ATP (2) NADPH
(3) NADH (4) Oxygen
Answer (3)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
26
NEET (UG) - 2018 (Code-EE) CHLAA
119. Which among the following is not a
prokaryote?
(1) Saccharomyces (2) Nostoc
(3) Mycobacterium (4) Oscillatoria
Answer (1)
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
120. Stomatal movement is not affected by
(1) Temperature (2) O
2
concentration
(3) Light (4) CO
2
concentration
Answer (2)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
121. Which of the following is true for nucleolus?
(1) Larger nucleoli are present in dividing
cells
(2) It takes part in spindle formation
(3) It is a membrane-bound structure
(4) It is a site for active ribosomal RNA
synthesis
Answer (4)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
122. The stage during which separation of the
paired homologous chromosomes begins is
(1) Pachytene (2) Diakinesis
(3) Diplotene (4) Zygotene
Answer (3)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
123. The two functional groups characteristic of
sugars are
(1) Hydroxyl and methyl
(2) Carbonyl and phosphate
(3) Carbonyl and methyl
(4) Carbonyl and hydroxyl
Answer (4)
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.
124. Stomata in grass leaf are
(1) Dumb-bell shaped
(2) Rectangular
(3) Kidney shaped
(4) Barrel shaped
Answer (1)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
125. 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) Hydrilla (2) Banana
(3) Yucca (4) Viola
Answer (3)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
126. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?120?C (2) ?196?C
(3) ?80?C (4) ?160?C
Answer (2)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C (Cryopreservation)
127. Double fertilization is
(1) Fusion of two male gametes of a pollen
tube with two different eggs
(2) Fusion of two male gametes with one egg
(3) Fusion of one male gamete with two polar
nuclei
(4) Syngamy and triple fusion
Answer (4)
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
128. Oxygen is not produced during photosynthesis
by
(1) Green sulphur bacteria
(2) Cycas
(3) Nostoc
(4) Chara
Answer (1)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
27
NEET (UG) - 2018 (Code-EE) CHLAAA
129. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Magnesium (2) Potassium
(3) Sodium (4) Calcium
Answer (2)
Sol. Potassium helps in maintaining turgidity of
cells.
130. What is the role of NAD
+
in cellular
respiration?
(1) It functions as an enzyme.
(2) It is a nucleotide source for ATP synthesis.
(3) It functions as an electron carrier.
(4) It is the final electron acceptor for
anaerobic respiration.
Answer (3)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
131. In which of the following forms is iron
absorbed by plants?
(1) Ferric
(2) Free element
(3) Ferrous
(4) Both ferric and ferrous
Answer (1*)
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
++
)
132. Winged pollen grains are present in
(1) Mustard (2) Mango
(3) Cycas (4) Pinus
Answer (4)
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.
133. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Neurospora
(2) Agaricus
(3) Alternaria
(4) Saccharomyces
Answer (2)
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.
134. Which one is wrongly matched?
(1) Uniflagellate gametes ? Polysiphonia
(2) Gemma cups ? Marchantia
(3) Biflagellate zoospores ? Brown algae
(4) Unicellular organism ? Chlorella
Answer (1)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (2, 3 & 4) are correctly
matched
135. 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.
28
NEET (UG) - 2018 (Code-EE) CHLAA
a b
(1) (i) (iv)
(2) (ii) (iv)
(3) (iii) (ii)
(4) (iii) (iv)
c d
(iii) (ii)
(iii) (i)
(i) (iv)
(i) (ii)
Answer (4)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
136. Which of the following is an amino acid
derived hormone?
(1) Epinephrine (2) Estradiol
(3) Ecdysone (4) Estriol
Answer (1)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
137. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(2) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(3) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
(4) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
Answer (3)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
138. The transparent lens in the human eye is held
in its place by
(1) ligaments attached to the ciliary body
(2) smooth muscles attached to the iris
(3) ligaments attached to the iris
(4) smooth muscles attached to the ciliary
body
Answer (1)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
139. Which of the following hormones can play a
significant role in osteoporosis?
(1) Aldosterone and Prolactin
(2) Estrogen and Parathyroid hormone
(3) Progesterone and Aldosterone
(4) Parathyroid hormone and Prolactin
Answer (2)
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.
140. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Forelimbs of man, bat and cheetah
(2) Brain of bat, man and cheetah
(3) Heart of bat, man and cheetah
(4) Eye of octopus, bat and man
Answer (4)
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.
141. Which of the following is not an autoimmune
disease?
(1) Psoriasis
(2) Alzheimer's disease
(3) Rheumatoid arthritis
(4) Vitiligo
Answer (2)
29
NEET (UG) - 2018 (Code-EE) CHLAAA
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.
142. 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) b, c and e (2) b, d and e
(3) a, b and c (4) a, c and e
Answer (3)
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)
143. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Elephantiasis (2) Ringworm disease
(3) Ascariasis (4) Amoebiasis
Answer (1)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
144. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Homology
(2) Convergent evolution
(3) Analogy
(4) Adaptive radiation
Answer (1)
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.
145. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin D (2) Vitamin B
12
(3) Vitamin A (4) Vitamin E
Answer (2)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
146. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Commensalism (2) Parasitism
(3) Mutualism (4) Amensalism
Answer (4)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicillium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicillium or the
organism which produces it.
147. All of the following are included in ? ex-situ
conservation? except
(1) Wildlife safari parks
(2) Botanical gardens
(3) Sacred groves
(4) Seed banks
Answer (3)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
148. 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) ii i iii iv
(2) iii iv i ii
(3) i iii iv ii
(4) i ii iv iii
Answer (2)
30
NEET (UG) - 2018 (Code-EE) CHLAA
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
149. In a growing population of a country,
(1) pre-reproductive individuals are more
than the reproductive individuals.
(2) reproductive and pre-reproductive
individuals are equal in number.
(3) reproductive individuals are less than the
post-reproductive individuals.
(4) pre-reproductive individuals are less than
the reproductive individuals.
Answer (1)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
150. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Flowers (2) Roots
(3) Latex (4) Leaves
Answer (3)
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.
151. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, hPL, progestogens, prolactin
(2) hCG, hPL, progestogens, estrogens
(3) hCG, hPL, estrogens, relaxin, oxytocin
(4) hCG, progestogens, estrogens,
glucocorticoids
Answer (2)
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.
152. The contraceptive ?SAHELI?
(1) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(2) is an IUD.
(3) increases the concentration of estrogen
and prevents ovulation in females.
(4) is a post-coital contraceptive.
Answer (1)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
153. The amnion of mammalian embryo is derived
from
(1) ectoderm and mesoderm
(2) mesoderm and trophoblast
(3) endoderm and mesoderm
(4) ectoderm and endoderm
Answer (1)
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.
154. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(2) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(3) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
Answer (4)
31
NEET (UG) - 2018 (Code-EE) CHLAAA
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.
155. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Inflammation of bronchioles; Decreased
respiratory surface
(2) Increased respiratory surface;
Inflammation of bronchioles
(3) Increased number of bronchioles;
Increased respiratory surface
(4) Decreased respiratory surface;
Inflammation of bronchioles
Answer (1)
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.
156. 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
c
ii
iii
ii
iii
Answer (1)
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.
157. 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) iii ii i iv
(2) i iv ii iii
(3) iii i iv ii
(4) iv iii ii i
Answer (3)
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.
158. 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
a b
(1) iii i
(2) i ii
(3) i iii
(4) ii i
32
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
NEET (UG) - 2018 (Code-EE) CHLAA
91. Which of the following statements is correct?
(1) Ovules are not enclosed by ovary wall in
gymnosperms
(2) Horsetails are gymnosperms
(3) Selaginella is heterosporous, while
Salvinia is homosporous
(4) Stems are usually unbranched in both
Cycas and Cedrus
Answer (1)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
92. Pneumatophores occur in
(1) Halophytes
(2) Carnivorous plants
(3) Free-floating hydrophytes
(4) Submerged hydrophytes
Answer (1)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
93. Sweet potato is a modified
(1) Stem
(2) Tap root
(3) Adventitious root
(4) Rhizome
Answer (3)
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
94. Plants having little or no secondary growth
are
(1) Grasses
(2) Conifers
(3) Deciduous angiosperms
(4) Cycads
Answer (1)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
95. Casparian strips occur in
(1) Epidermis
(2) Cortex
(3) Pericycle
(4) Endodermi
Answer (4)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Secondary xylem and phloem in dicot stem
are produced by
(1) Apical meristems
(2) Phellogen
(3) Vascular cambium
(4) Axillary meristems
Answer (3)
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.
97. Select the wrong statement :
(1) Cell wall is present in members of Fungi
and Plantae
(2) Pseudopodia are locomotory and feeding
structures in Sporozoans
(3) Mushrooms belong to Basidiomycetes
(4) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
Answer (2)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
98. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Fungus (2) Plant
(3) Bacterium (4) Virus
Answer (3)
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
23
NEET (UG) - 2018 (Code-EE) CHLAAA
99. Select the correct match
(1) Alec Jeffreys

- Streptococcus
pneumoniae
(2) Matthew Meselson - Pisum sativum
and F. Stahl
(3) Alfred Hershey and - TMV
Martha Chase
(4) Francois Jacob and - Lac operon
Jacques Monod
Answer (4)
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
100. Select the correct statement
(1) Franklin Stahl coined the term ?linkage?
(2) Spliceosomes take part in translation
(3) Punnett square was developed by a British
scientist
(4) Transduction was discovered by S. Altman
Answer (3)
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
101. Which of the following pairs is wrongly
matched?
(1) Starch synthesis in pea : Multiple alleles
(2) XO type sex : Grasshopper
determination
(3) ABO blood grouping : Co-dominance
(4) T.H. Morgan : Linkage
Answer (1)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (2, 3 & 4) are correctly
matched.
102. Offsets are produced by
(1) Meiotic divisions
(2) Parthenocarpy
(3) Mitotic divisions
(4) Parthenogenesis
Answer (3)
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)
103. Which of the following flowers only once in its
life-time?
(1) Bamboo species
(2) Mango
(3) Jackfruit
(4) Papaya
Answer (1)
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.
104. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Pollenkitt (2) Oil content
(3) Cellulosic intine (4) Sporopollenin
Answer (4)
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.
24
NEET (UG) - 2018 (Code-EE) CHLAA
105. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) Retrovirus
(2) X phage
(3) Ti plasmid
(4) pBR 322
Answer (1)
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.
106. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Extension, Denaturation, Annealing
(2) Denaturation, Extension, Annealing
(3) Annealing, Extension, Denaturation
(4) Denaturation, Annealing, Extension
Answer (4)
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
107. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Indian Council of Medical Research (ICMR)
(2) Research Committee on Genetic
Manipulation (RCGM)
(3) Council for Scientific and Industrial
Research (CSIR)
(4) Genetic Engineering Appraisal Committee
(GEAC)
Answer (4)
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).
108. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bio-infringement
(2) Biodegradation
(3) Biopiracy
(4) Bioexploitation
Answer (3)
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).
109. 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) Co-667
(2) Lerma Rojo
(3) Sharbati Sonora
(4) Basmati
Answer (4)
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.
110. Select the correct match
(1) Ribozyme - Nucleic acid
(2) T.H. Morgan - Transduction
(3) F
2
? Recessive parent - Dihybrid cross
(4) G. Mendel - Transformation
Answer (1)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
25
NEET (UG) - 2018 (Code-EE) CHLAAA
111. Niche is
(1) all the biological factors in the organism's
environment
(2) the range of temperature that the
organism needs to live
(3) the physical space where an organism
lives
(4) the functional role played by the organism
where it lives
Answer (4)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
112. Which of the following is a secondary
pollutant?
(1) CO (2) SO
2

(3) CO
2
(4) O
3

Answer (4)
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
113. World Ozone Day is celebrated on
(1) 5
th
June (2) 16
th
September
(3) 21
st
April (4) 22
nd
April
Answer (2)
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
114. Natality refers to
(1) Death rate
(2) Number of individuals leaving the habitat
(3) Birth rate
(4) Number of individuals entering a habitat
Answer (3)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual
entering a habitat is
? Immigration
? Number of individual
leaving the habital
? Emigration
115. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Carbon (2) Fe
(3) Cl (4) Oxygen
Answer (3)
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
116. 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) Inverted pyramid of biomass
(2) Upright pyramid of numbers
(3) Pyramid of energy
(4) Upright pyramid of biomass
Answer (1)
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.
117. The Golgi complex participates in
(1) Fatty acid breakdown
(2) Respiration in bacteria
(3) Formation of secretory vesicles
(4) Activation of amino acid
Answer (3)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
118. Which of the following is not a product of light
reaction of photosynthesis?
(1) ATP (2) NADPH
(3) NADH (4) Oxygen
Answer (3)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
26
NEET (UG) - 2018 (Code-EE) CHLAA
119. Which among the following is not a
prokaryote?
(1) Saccharomyces (2) Nostoc
(3) Mycobacterium (4) Oscillatoria
Answer (1)
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
120. Stomatal movement is not affected by
(1) Temperature (2) O
2
concentration
(3) Light (4) CO
2
concentration
Answer (2)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
121. Which of the following is true for nucleolus?
(1) Larger nucleoli are present in dividing
cells
(2) It takes part in spindle formation
(3) It is a membrane-bound structure
(4) It is a site for active ribosomal RNA
synthesis
Answer (4)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
122. The stage during which separation of the
paired homologous chromosomes begins is
(1) Pachytene (2) Diakinesis
(3) Diplotene (4) Zygotene
Answer (3)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
123. The two functional groups characteristic of
sugars are
(1) Hydroxyl and methyl
(2) Carbonyl and phosphate
(3) Carbonyl and methyl
(4) Carbonyl and hydroxyl
Answer (4)
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.
124. Stomata in grass leaf are
(1) Dumb-bell shaped
(2) Rectangular
(3) Kidney shaped
(4) Barrel shaped
Answer (1)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
125. 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) Hydrilla (2) Banana
(3) Yucca (4) Viola
Answer (3)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
126. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?120?C (2) ?196?C
(3) ?80?C (4) ?160?C
Answer (2)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C (Cryopreservation)
127. Double fertilization is
(1) Fusion of two male gametes of a pollen
tube with two different eggs
(2) Fusion of two male gametes with one egg
(3) Fusion of one male gamete with two polar
nuclei
(4) Syngamy and triple fusion
Answer (4)
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
128. Oxygen is not produced during photosynthesis
by
(1) Green sulphur bacteria
(2) Cycas
(3) Nostoc
(4) Chara
Answer (1)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
27
NEET (UG) - 2018 (Code-EE) CHLAAA
129. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Magnesium (2) Potassium
(3) Sodium (4) Calcium
Answer (2)
Sol. Potassium helps in maintaining turgidity of
cells.
130. What is the role of NAD
+
in cellular
respiration?
(1) It functions as an enzyme.
(2) It is a nucleotide source for ATP synthesis.
(3) It functions as an electron carrier.
(4) It is the final electron acceptor for
anaerobic respiration.
Answer (3)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
131. In which of the following forms is iron
absorbed by plants?
(1) Ferric
(2) Free element
(3) Ferrous
(4) Both ferric and ferrous
Answer (1*)
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
++
)
132. Winged pollen grains are present in
(1) Mustard (2) Mango
(3) Cycas (4) Pinus
Answer (4)
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.
133. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Neurospora
(2) Agaricus
(3) Alternaria
(4) Saccharomyces
Answer (2)
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.
134. Which one is wrongly matched?
(1) Uniflagellate gametes ? Polysiphonia
(2) Gemma cups ? Marchantia
(3) Biflagellate zoospores ? Brown algae
(4) Unicellular organism ? Chlorella
Answer (1)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (2, 3 & 4) are correctly
matched
135. 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.
28
NEET (UG) - 2018 (Code-EE) CHLAA
a b
(1) (i) (iv)
(2) (ii) (iv)
(3) (iii) (ii)
(4) (iii) (iv)
c d
(iii) (ii)
(iii) (i)
(i) (iv)
(i) (ii)
Answer (4)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
136. Which of the following is an amino acid
derived hormone?
(1) Epinephrine (2) Estradiol
(3) Ecdysone (4) Estriol
Answer (1)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
137. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(2) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(3) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
(4) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
Answer (3)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
138. The transparent lens in the human eye is held
in its place by
(1) ligaments attached to the ciliary body
(2) smooth muscles attached to the iris
(3) ligaments attached to the iris
(4) smooth muscles attached to the ciliary
body
Answer (1)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
139. Which of the following hormones can play a
significant role in osteoporosis?
(1) Aldosterone and Prolactin
(2) Estrogen and Parathyroid hormone
(3) Progesterone and Aldosterone
(4) Parathyroid hormone and Prolactin
Answer (2)
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.
140. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Forelimbs of man, bat and cheetah
(2) Brain of bat, man and cheetah
(3) Heart of bat, man and cheetah
(4) Eye of octopus, bat and man
Answer (4)
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.
141. Which of the following is not an autoimmune
disease?
(1) Psoriasis
(2) Alzheimer's disease
(3) Rheumatoid arthritis
(4) Vitiligo
Answer (2)
29
NEET (UG) - 2018 (Code-EE) CHLAAA
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.
142. 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) b, c and e (2) b, d and e
(3) a, b and c (4) a, c and e
Answer (3)
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)
143. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Elephantiasis (2) Ringworm disease
(3) Ascariasis (4) Amoebiasis
Answer (1)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
144. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Homology
(2) Convergent evolution
(3) Analogy
(4) Adaptive radiation
Answer (1)
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.
145. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin D (2) Vitamin B
12
(3) Vitamin A (4) Vitamin E
Answer (2)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
146. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Commensalism (2) Parasitism
(3) Mutualism (4) Amensalism
Answer (4)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicillium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicillium or the
organism which produces it.
147. All of the following are included in ? ex-situ
conservation? except
(1) Wildlife safari parks
(2) Botanical gardens
(3) Sacred groves
(4) Seed banks
Answer (3)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
148. 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) ii i iii iv
(2) iii iv i ii
(3) i iii iv ii
(4) i ii iv iii
Answer (2)
30
NEET (UG) - 2018 (Code-EE) CHLAA
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
149. In a growing population of a country,
(1) pre-reproductive individuals are more
than the reproductive individuals.
(2) reproductive and pre-reproductive
individuals are equal in number.
(3) reproductive individuals are less than the
post-reproductive individuals.
(4) pre-reproductive individuals are less than
the reproductive individuals.
Answer (1)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
150. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Flowers (2) Roots
(3) Latex (4) Leaves
Answer (3)
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.
151. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, hPL, progestogens, prolactin
(2) hCG, hPL, progestogens, estrogens
(3) hCG, hPL, estrogens, relaxin, oxytocin
(4) hCG, progestogens, estrogens,
glucocorticoids
Answer (2)
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.
152. The contraceptive ?SAHELI?
(1) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(2) is an IUD.
(3) increases the concentration of estrogen
and prevents ovulation in females.
(4) is a post-coital contraceptive.
Answer (1)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
153. The amnion of mammalian embryo is derived
from
(1) ectoderm and mesoderm
(2) mesoderm and trophoblast
(3) endoderm and mesoderm
(4) ectoderm and endoderm
Answer (1)
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.
154. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(2) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(3) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
Answer (4)
31
NEET (UG) - 2018 (Code-EE) CHLAAA
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.
155. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Inflammation of bronchioles; Decreased
respiratory surface
(2) Increased respiratory surface;
Inflammation of bronchioles
(3) Increased number of bronchioles;
Increased respiratory surface
(4) Decreased respiratory surface;
Inflammation of bronchioles
Answer (1)
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.
156. 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
c
ii
iii
ii
iii
Answer (1)
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.
157. 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) iii ii i iv
(2) i iv ii iii
(3) iii i iv ii
(4) iv iii ii i
Answer (3)
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.
158. 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
a b
(1) iii i
(2) i ii
(3) i iii
(4) ii i
32
NEET (UG) - 2018 (Code-EE) CHLAA
c d
iv i
i iv
iii iv
ii iii
a b
(1) iii ii
(2) ii iii
(3) i ii
(4) iv i
Answer (4)
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.
159. 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) iv v ii iii
(2) v iv i ii
(3) iv i ii iii
(4) v iv i iii
Answer (3)
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.
160. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Protein folding
(2) Cleavage of signal peptide
(3) Protein glycosylation
(4) Phospholipid synthesis
Answer (4)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
161. Which of these statements is incorrect?
(1) Enzymes of TCA cycle are present in
mitochondrial matrix
(2) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(3) Glycolysis occurs in cytosol
(4) Oxidative phosphorylation takes place in
outer mitochondrial membrane
Answer (4)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
162. Nissl bodies are mainly composed of
(1) Proteins and lipids
(2) Nucleic acids and SER
(3) DNA and RNA
(4) Free ribosomes and RER
Answer (4)
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.
163. Which of the following terms describe human
dentition?
(1) Thecodont, Diphyodont, Homodont
(2) Pleurodont, Monophyodont, Homodont
(3) Thecodont, Diphyodont, Heterodont
(4) Pleurodont, Diphyodont, Heterodont
Answer (3)
33
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
NEET (UG) - 2018 (Code-EE) CHLAA
91. Which of the following statements is correct?
(1) Ovules are not enclosed by ovary wall in
gymnosperms
(2) Horsetails are gymnosperms
(3) Selaginella is heterosporous, while
Salvinia is homosporous
(4) Stems are usually unbranched in both
Cycas and Cedrus
Answer (1)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
92. Pneumatophores occur in
(1) Halophytes
(2) Carnivorous plants
(3) Free-floating hydrophytes
(4) Submerged hydrophytes
Answer (1)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
93. Sweet potato is a modified
(1) Stem
(2) Tap root
(3) Adventitious root
(4) Rhizome
Answer (3)
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
94. Plants having little or no secondary growth
are
(1) Grasses
(2) Conifers
(3) Deciduous angiosperms
(4) Cycads
Answer (1)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
95. Casparian strips occur in
(1) Epidermis
(2) Cortex
(3) Pericycle
(4) Endodermi
Answer (4)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Secondary xylem and phloem in dicot stem
are produced by
(1) Apical meristems
(2) Phellogen
(3) Vascular cambium
(4) Axillary meristems
Answer (3)
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.
97. Select the wrong statement :
(1) Cell wall is present in members of Fungi
and Plantae
(2) Pseudopodia are locomotory and feeding
structures in Sporozoans
(3) Mushrooms belong to Basidiomycetes
(4) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
Answer (2)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
98. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Fungus (2) Plant
(3) Bacterium (4) Virus
Answer (3)
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
23
NEET (UG) - 2018 (Code-EE) CHLAAA
99. Select the correct match
(1) Alec Jeffreys

- Streptococcus
pneumoniae
(2) Matthew Meselson - Pisum sativum
and F. Stahl
(3) Alfred Hershey and - TMV
Martha Chase
(4) Francois Jacob and - Lac operon
Jacques Monod
Answer (4)
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
100. Select the correct statement
(1) Franklin Stahl coined the term ?linkage?
(2) Spliceosomes take part in translation
(3) Punnett square was developed by a British
scientist
(4) Transduction was discovered by S. Altman
Answer (3)
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
101. Which of the following pairs is wrongly
matched?
(1) Starch synthesis in pea : Multiple alleles
(2) XO type sex : Grasshopper
determination
(3) ABO blood grouping : Co-dominance
(4) T.H. Morgan : Linkage
Answer (1)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (2, 3 & 4) are correctly
matched.
102. Offsets are produced by
(1) Meiotic divisions
(2) Parthenocarpy
(3) Mitotic divisions
(4) Parthenogenesis
Answer (3)
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)
103. Which of the following flowers only once in its
life-time?
(1) Bamboo species
(2) Mango
(3) Jackfruit
(4) Papaya
Answer (1)
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.
104. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Pollenkitt (2) Oil content
(3) Cellulosic intine (4) Sporopollenin
Answer (4)
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.
24
NEET (UG) - 2018 (Code-EE) CHLAA
105. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) Retrovirus
(2) X phage
(3) Ti plasmid
(4) pBR 322
Answer (1)
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.
106. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Extension, Denaturation, Annealing
(2) Denaturation, Extension, Annealing
(3) Annealing, Extension, Denaturation
(4) Denaturation, Annealing, Extension
Answer (4)
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
107. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Indian Council of Medical Research (ICMR)
(2) Research Committee on Genetic
Manipulation (RCGM)
(3) Council for Scientific and Industrial
Research (CSIR)
(4) Genetic Engineering Appraisal Committee
(GEAC)
Answer (4)
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).
108. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bio-infringement
(2) Biodegradation
(3) Biopiracy
(4) Bioexploitation
Answer (3)
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).
109. 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) Co-667
(2) Lerma Rojo
(3) Sharbati Sonora
(4) Basmati
Answer (4)
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.
110. Select the correct match
(1) Ribozyme - Nucleic acid
(2) T.H. Morgan - Transduction
(3) F
2
? Recessive parent - Dihybrid cross
(4) G. Mendel - Transformation
Answer (1)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
25
NEET (UG) - 2018 (Code-EE) CHLAAA
111. Niche is
(1) all the biological factors in the organism's
environment
(2) the range of temperature that the
organism needs to live
(3) the physical space where an organism
lives
(4) the functional role played by the organism
where it lives
Answer (4)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
112. Which of the following is a secondary
pollutant?
(1) CO (2) SO
2

(3) CO
2
(4) O
3

Answer (4)
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
113. World Ozone Day is celebrated on
(1) 5
th
June (2) 16
th
September
(3) 21
st
April (4) 22
nd
April
Answer (2)
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
114. Natality refers to
(1) Death rate
(2) Number of individuals leaving the habitat
(3) Birth rate
(4) Number of individuals entering a habitat
Answer (3)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual
entering a habitat is
? Immigration
? Number of individual
leaving the habital
? Emigration
115. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Carbon (2) Fe
(3) Cl (4) Oxygen
Answer (3)
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
116. 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) Inverted pyramid of biomass
(2) Upright pyramid of numbers
(3) Pyramid of energy
(4) Upright pyramid of biomass
Answer (1)
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.
117. The Golgi complex participates in
(1) Fatty acid breakdown
(2) Respiration in bacteria
(3) Formation of secretory vesicles
(4) Activation of amino acid
Answer (3)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
118. Which of the following is not a product of light
reaction of photosynthesis?
(1) ATP (2) NADPH
(3) NADH (4) Oxygen
Answer (3)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
26
NEET (UG) - 2018 (Code-EE) CHLAA
119. Which among the following is not a
prokaryote?
(1) Saccharomyces (2) Nostoc
(3) Mycobacterium (4) Oscillatoria
Answer (1)
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
120. Stomatal movement is not affected by
(1) Temperature (2) O
2
concentration
(3) Light (4) CO
2
concentration
Answer (2)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
121. Which of the following is true for nucleolus?
(1) Larger nucleoli are present in dividing
cells
(2) It takes part in spindle formation
(3) It is a membrane-bound structure
(4) It is a site for active ribosomal RNA
synthesis
Answer (4)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
122. The stage during which separation of the
paired homologous chromosomes begins is
(1) Pachytene (2) Diakinesis
(3) Diplotene (4) Zygotene
Answer (3)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
123. The two functional groups characteristic of
sugars are
(1) Hydroxyl and methyl
(2) Carbonyl and phosphate
(3) Carbonyl and methyl
(4) Carbonyl and hydroxyl
Answer (4)
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.
124. Stomata in grass leaf are
(1) Dumb-bell shaped
(2) Rectangular
(3) Kidney shaped
(4) Barrel shaped
Answer (1)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
125. 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) Hydrilla (2) Banana
(3) Yucca (4) Viola
Answer (3)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
126. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?120?C (2) ?196?C
(3) ?80?C (4) ?160?C
Answer (2)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C (Cryopreservation)
127. Double fertilization is
(1) Fusion of two male gametes of a pollen
tube with two different eggs
(2) Fusion of two male gametes with one egg
(3) Fusion of one male gamete with two polar
nuclei
(4) Syngamy and triple fusion
Answer (4)
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
128. Oxygen is not produced during photosynthesis
by
(1) Green sulphur bacteria
(2) Cycas
(3) Nostoc
(4) Chara
Answer (1)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
27
NEET (UG) - 2018 (Code-EE) CHLAAA
129. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Magnesium (2) Potassium
(3) Sodium (4) Calcium
Answer (2)
Sol. Potassium helps in maintaining turgidity of
cells.
130. What is the role of NAD
+
in cellular
respiration?
(1) It functions as an enzyme.
(2) It is a nucleotide source for ATP synthesis.
(3) It functions as an electron carrier.
(4) It is the final electron acceptor for
anaerobic respiration.
Answer (3)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
131. In which of the following forms is iron
absorbed by plants?
(1) Ferric
(2) Free element
(3) Ferrous
(4) Both ferric and ferrous
Answer (1*)
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
++
)
132. Winged pollen grains are present in
(1) Mustard (2) Mango
(3) Cycas (4) Pinus
Answer (4)
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.
133. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Neurospora
(2) Agaricus
(3) Alternaria
(4) Saccharomyces
Answer (2)
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.
134. Which one is wrongly matched?
(1) Uniflagellate gametes ? Polysiphonia
(2) Gemma cups ? Marchantia
(3) Biflagellate zoospores ? Brown algae
(4) Unicellular organism ? Chlorella
Answer (1)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (2, 3 & 4) are correctly
matched
135. 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.
28
NEET (UG) - 2018 (Code-EE) CHLAA
a b
(1) (i) (iv)
(2) (ii) (iv)
(3) (iii) (ii)
(4) (iii) (iv)
c d
(iii) (ii)
(iii) (i)
(i) (iv)
(i) (ii)
Answer (4)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
136. Which of the following is an amino acid
derived hormone?
(1) Epinephrine (2) Estradiol
(3) Ecdysone (4) Estriol
Answer (1)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
137. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(2) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(3) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
(4) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
Answer (3)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
138. The transparent lens in the human eye is held
in its place by
(1) ligaments attached to the ciliary body
(2) smooth muscles attached to the iris
(3) ligaments attached to the iris
(4) smooth muscles attached to the ciliary
body
Answer (1)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
139. Which of the following hormones can play a
significant role in osteoporosis?
(1) Aldosterone and Prolactin
(2) Estrogen and Parathyroid hormone
(3) Progesterone and Aldosterone
(4) Parathyroid hormone and Prolactin
Answer (2)
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.
140. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Forelimbs of man, bat and cheetah
(2) Brain of bat, man and cheetah
(3) Heart of bat, man and cheetah
(4) Eye of octopus, bat and man
Answer (4)
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.
141. Which of the following is not an autoimmune
disease?
(1) Psoriasis
(2) Alzheimer's disease
(3) Rheumatoid arthritis
(4) Vitiligo
Answer (2)
29
NEET (UG) - 2018 (Code-EE) CHLAAA
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.
142. 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) b, c and e (2) b, d and e
(3) a, b and c (4) a, c and e
Answer (3)
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)
143. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Elephantiasis (2) Ringworm disease
(3) Ascariasis (4) Amoebiasis
Answer (1)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
144. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Homology
(2) Convergent evolution
(3) Analogy
(4) Adaptive radiation
Answer (1)
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.
145. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin D (2) Vitamin B
12
(3) Vitamin A (4) Vitamin E
Answer (2)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
146. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Commensalism (2) Parasitism
(3) Mutualism (4) Amensalism
Answer (4)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicillium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicillium or the
organism which produces it.
147. All of the following are included in ? ex-situ
conservation? except
(1) Wildlife safari parks
(2) Botanical gardens
(3) Sacred groves
(4) Seed banks
Answer (3)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
148. 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) ii i iii iv
(2) iii iv i ii
(3) i iii iv ii
(4) i ii iv iii
Answer (2)
30
NEET (UG) - 2018 (Code-EE) CHLAA
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
149. In a growing population of a country,
(1) pre-reproductive individuals are more
than the reproductive individuals.
(2) reproductive and pre-reproductive
individuals are equal in number.
(3) reproductive individuals are less than the
post-reproductive individuals.
(4) pre-reproductive individuals are less than
the reproductive individuals.
Answer (1)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
150. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Flowers (2) Roots
(3) Latex (4) Leaves
Answer (3)
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.
151. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, hPL, progestogens, prolactin
(2) hCG, hPL, progestogens, estrogens
(3) hCG, hPL, estrogens, relaxin, oxytocin
(4) hCG, progestogens, estrogens,
glucocorticoids
Answer (2)
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.
152. The contraceptive ?SAHELI?
(1) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(2) is an IUD.
(3) increases the concentration of estrogen
and prevents ovulation in females.
(4) is a post-coital contraceptive.
Answer (1)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
153. The amnion of mammalian embryo is derived
from
(1) ectoderm and mesoderm
(2) mesoderm and trophoblast
(3) endoderm and mesoderm
(4) ectoderm and endoderm
Answer (1)
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.
154. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(2) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(3) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
Answer (4)
31
NEET (UG) - 2018 (Code-EE) CHLAAA
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.
155. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Inflammation of bronchioles; Decreased
respiratory surface
(2) Increased respiratory surface;
Inflammation of bronchioles
(3) Increased number of bronchioles;
Increased respiratory surface
(4) Decreased respiratory surface;
Inflammation of bronchioles
Answer (1)
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.
156. 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
c
ii
iii
ii
iii
Answer (1)
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.
157. 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) iii ii i iv
(2) i iv ii iii
(3) iii i iv ii
(4) iv iii ii i
Answer (3)
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.
158. 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
a b
(1) iii i
(2) i ii
(3) i iii
(4) ii i
32
NEET (UG) - 2018 (Code-EE) CHLAA
c d
iv i
i iv
iii iv
ii iii
a b
(1) iii ii
(2) ii iii
(3) i ii
(4) iv i
Answer (4)
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.
159. 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) iv v ii iii
(2) v iv i ii
(3) iv i ii iii
(4) v iv i iii
Answer (3)
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.
160. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Protein folding
(2) Cleavage of signal peptide
(3) Protein glycosylation
(4) Phospholipid synthesis
Answer (4)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
161. Which of these statements is incorrect?
(1) Enzymes of TCA cycle are present in
mitochondrial matrix
(2) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(3) Glycolysis occurs in cytosol
(4) Oxidative phosphorylation takes place in
outer mitochondrial membrane
Answer (4)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
162. Nissl bodies are mainly composed of
(1) Proteins and lipids
(2) Nucleic acids and SER
(3) DNA and RNA
(4) Free ribosomes and RER
Answer (4)
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.
163. Which of the following terms describe human
dentition?
(1) Thecodont, Diphyodont, Homodont
(2) Pleurodont, Monophyodont, Homodont
(3) Thecodont, Diphyodont, Heterodont
(4) Pleurodont, Diphyodont, Heterodont
Answer (3)
33
NEET (UG) - 2018 (Code-EE) CHLAAA
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.
164. Select the incorrect match :
(1) Lampbrush ? Diplotene bivalents
chromosomes
(2) Submetacentric ? L-shaped
chromosomes chromosomes
(3) Allosomes ? Sex chromosomes
(4) Polytene ? Oocytes of
chromosomes amphibians
Answer (4)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
165. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Polysome
(2) Plastidome
(3) Polyhedral bodies
(4) Nucleosome
Answer (1)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
166. According to Hugo de Vries, the mechanism
of evolution is
(1) Multiple step mutations
(2) Phenotypic variations
(3) Saltation
(4) Minor mutations
Answer (3)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
167. 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 ii i
(2) ii iii i
(3) i iii ii
(4) iii i ii
Answer (2)
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.
168. All of the following are part of an operon
except
(1) an operator
(2) an enhancer
(3) structural genes
(4) a promoter
Answer (2)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
169. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) AGGUAUCGCAU
(2) ACCUAUGCGAU
(3) UGGTUTCGCAT
(4) UCCAUAGCGUA
Answer (1)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
34
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DATE : 06/05/2018
Test Booklet Code


Time : 3 hrs. Max. Marks : 720
Answers & Solutions
for
NEET (UG) - 2018
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
NEET (UG) - 2018 (Code-EE) CHLAA
91. Which of the following statements is correct?
(1) Ovules are not enclosed by ovary wall in
gymnosperms
(2) Horsetails are gymnosperms
(3) Selaginella is heterosporous, while
Salvinia is homosporous
(4) Stems are usually unbranched in both
Cycas and Cedrus
Answer (1)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
92. Pneumatophores occur in
(1) Halophytes
(2) Carnivorous plants
(3) Free-floating hydrophytes
(4) Submerged hydrophytes
Answer (1)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
93. Sweet potato is a modified
(1) Stem
(2) Tap root
(3) Adventitious root
(4) Rhizome
Answer (3)
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
94. Plants having little or no secondary growth
are
(1) Grasses
(2) Conifers
(3) Deciduous angiosperms
(4) Cycads
Answer (1)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
95. Casparian strips occur in
(1) Epidermis
(2) Cortex
(3) Pericycle
(4) Endodermi
Answer (4)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Secondary xylem and phloem in dicot stem
are produced by
(1) Apical meristems
(2) Phellogen
(3) Vascular cambium
(4) Axillary meristems
Answer (3)
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.
97. Select the wrong statement :
(1) Cell wall is present in members of Fungi
and Plantae
(2) Pseudopodia are locomotory and feeding
structures in Sporozoans
(3) Mushrooms belong to Basidiomycetes
(4) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
Answer (2)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
98. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Fungus (2) Plant
(3) Bacterium (4) Virus
Answer (3)
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
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NEET (UG) - 2018 (Code-EE) CHLAAA
99. Select the correct match
(1) Alec Jeffreys

- Streptococcus
pneumoniae
(2) Matthew Meselson - Pisum sativum
and F. Stahl
(3) Alfred Hershey and - TMV
Martha Chase
(4) Francois Jacob and - Lac operon
Jacques Monod
Answer (4)
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
100. Select the correct statement
(1) Franklin Stahl coined the term ?linkage?
(2) Spliceosomes take part in translation
(3) Punnett square was developed by a British
scientist
(4) Transduction was discovered by S. Altman
Answer (3)
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
101. Which of the following pairs is wrongly
matched?
(1) Starch synthesis in pea : Multiple alleles
(2) XO type sex : Grasshopper
determination
(3) ABO blood grouping : Co-dominance
(4) T.H. Morgan : Linkage
Answer (1)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (2, 3 & 4) are correctly
matched.
102. Offsets are produced by
(1) Meiotic divisions
(2) Parthenocarpy
(3) Mitotic divisions
(4) Parthenogenesis
Answer (3)
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)
103. Which of the following flowers only once in its
life-time?
(1) Bamboo species
(2) Mango
(3) Jackfruit
(4) Papaya
Answer (1)
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.
104. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Pollenkitt (2) Oil content
(3) Cellulosic intine (4) Sporopollenin
Answer (4)
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.
24
NEET (UG) - 2018 (Code-EE) CHLAA
105. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) Retrovirus
(2) X phage
(3) Ti plasmid
(4) pBR 322
Answer (1)
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.
106. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Extension, Denaturation, Annealing
(2) Denaturation, Extension, Annealing
(3) Annealing, Extension, Denaturation
(4) Denaturation, Annealing, Extension
Answer (4)
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
107. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Indian Council of Medical Research (ICMR)
(2) Research Committee on Genetic
Manipulation (RCGM)
(3) Council for Scientific and Industrial
Research (CSIR)
(4) Genetic Engineering Appraisal Committee
(GEAC)
Answer (4)
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).
108. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bio-infringement
(2) Biodegradation
(3) Biopiracy
(4) Bioexploitation
Answer (3)
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).
109. 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) Co-667
(2) Lerma Rojo
(3) Sharbati Sonora
(4) Basmati
Answer (4)
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.
110. Select the correct match
(1) Ribozyme - Nucleic acid
(2) T.H. Morgan - Transduction
(3) F
2
? Recessive parent - Dihybrid cross
(4) G. Mendel - Transformation
Answer (1)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
25
NEET (UG) - 2018 (Code-EE) CHLAAA
111. Niche is
(1) all the biological factors in the organism's
environment
(2) the range of temperature that the
organism needs to live
(3) the physical space where an organism
lives
(4) the functional role played by the organism
where it lives
Answer (4)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
112. Which of the following is a secondary
pollutant?
(1) CO (2) SO
2

(3) CO
2
(4) O
3

Answer (4)
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
113. World Ozone Day is celebrated on
(1) 5
th
June (2) 16
th
September
(3) 21
st
April (4) 22
nd
April
Answer (2)
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
114. Natality refers to
(1) Death rate
(2) Number of individuals leaving the habitat
(3) Birth rate
(4) Number of individuals entering a habitat
Answer (3)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual
entering a habitat is
? Immigration
? Number of individual
leaving the habital
? Emigration
115. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Carbon (2) Fe
(3) Cl (4) Oxygen
Answer (3)
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
116. 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) Inverted pyramid of biomass
(2) Upright pyramid of numbers
(3) Pyramid of energy
(4) Upright pyramid of biomass
Answer (1)
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.
117. The Golgi complex participates in
(1) Fatty acid breakdown
(2) Respiration in bacteria
(3) Formation of secretory vesicles
(4) Activation of amino acid
Answer (3)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
118. Which of the following is not a product of light
reaction of photosynthesis?
(1) ATP (2) NADPH
(3) NADH (4) Oxygen
Answer (3)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
26
NEET (UG) - 2018 (Code-EE) CHLAA
119. Which among the following is not a
prokaryote?
(1) Saccharomyces (2) Nostoc
(3) Mycobacterium (4) Oscillatoria
Answer (1)
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
120. Stomatal movement is not affected by
(1) Temperature (2) O
2
concentration
(3) Light (4) CO
2
concentration
Answer (2)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
121. Which of the following is true for nucleolus?
(1) Larger nucleoli are present in dividing
cells
(2) It takes part in spindle formation
(3) It is a membrane-bound structure
(4) It is a site for active ribosomal RNA
synthesis
Answer (4)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
122. The stage during which separation of the
paired homologous chromosomes begins is
(1) Pachytene (2) Diakinesis
(3) Diplotene (4) Zygotene
Answer (3)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
123. The two functional groups characteristic of
sugars are
(1) Hydroxyl and methyl
(2) Carbonyl and phosphate
(3) Carbonyl and methyl
(4) Carbonyl and hydroxyl
Answer (4)
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.
124. Stomata in grass leaf are
(1) Dumb-bell shaped
(2) Rectangular
(3) Kidney shaped
(4) Barrel shaped
Answer (1)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
125. 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) Hydrilla (2) Banana
(3) Yucca (4) Viola
Answer (3)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
126. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?120?C (2) ?196?C
(3) ?80?C (4) ?160?C
Answer (2)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C (Cryopreservation)
127. Double fertilization is
(1) Fusion of two male gametes of a pollen
tube with two different eggs
(2) Fusion of two male gametes with one egg
(3) Fusion of one male gamete with two polar
nuclei
(4) Syngamy and triple fusion
Answer (4)
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
128. Oxygen is not produced during photosynthesis
by
(1) Green sulphur bacteria
(2) Cycas
(3) Nostoc
(4) Chara
Answer (1)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
27
NEET (UG) - 2018 (Code-EE) CHLAAA
129. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Magnesium (2) Potassium
(3) Sodium (4) Calcium
Answer (2)
Sol. Potassium helps in maintaining turgidity of
cells.
130. What is the role of NAD
+
in cellular
respiration?
(1) It functions as an enzyme.
(2) It is a nucleotide source for ATP synthesis.
(3) It functions as an electron carrier.
(4) It is the final electron acceptor for
anaerobic respiration.
Answer (3)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
131. In which of the following forms is iron
absorbed by plants?
(1) Ferric
(2) Free element
(3) Ferrous
(4) Both ferric and ferrous
Answer (1*)
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
++
)
132. Winged pollen grains are present in
(1) Mustard (2) Mango
(3) Cycas (4) Pinus
Answer (4)
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.
133. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Neurospora
(2) Agaricus
(3) Alternaria
(4) Saccharomyces
Answer (2)
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.
134. Which one is wrongly matched?
(1) Uniflagellate gametes ? Polysiphonia
(2) Gemma cups ? Marchantia
(3) Biflagellate zoospores ? Brown algae
(4) Unicellular organism ? Chlorella
Answer (1)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (2, 3 & 4) are correctly
matched
135. 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.
28
NEET (UG) - 2018 (Code-EE) CHLAA
a b
(1) (i) (iv)
(2) (ii) (iv)
(3) (iii) (ii)
(4) (iii) (iv)
c d
(iii) (ii)
(iii) (i)
(i) (iv)
(i) (ii)
Answer (4)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
136. Which of the following is an amino acid
derived hormone?
(1) Epinephrine (2) Estradiol
(3) Ecdysone (4) Estriol
Answer (1)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
137. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(2) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(3) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
(4) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
Answer (3)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
138. The transparent lens in the human eye is held
in its place by
(1) ligaments attached to the ciliary body
(2) smooth muscles attached to the iris
(3) ligaments attached to the iris
(4) smooth muscles attached to the ciliary
body
Answer (1)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
139. Which of the following hormones can play a
significant role in osteoporosis?
(1) Aldosterone and Prolactin
(2) Estrogen and Parathyroid hormone
(3) Progesterone and Aldosterone
(4) Parathyroid hormone and Prolactin
Answer (2)
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.
140. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Forelimbs of man, bat and cheetah
(2) Brain of bat, man and cheetah
(3) Heart of bat, man and cheetah
(4) Eye of octopus, bat and man
Answer (4)
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.
141. Which of the following is not an autoimmune
disease?
(1) Psoriasis
(2) Alzheimer's disease
(3) Rheumatoid arthritis
(4) Vitiligo
Answer (2)
29
NEET (UG) - 2018 (Code-EE) CHLAAA
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.
142. 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) b, c and e (2) b, d and e
(3) a, b and c (4) a, c and e
Answer (3)
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)
143. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Elephantiasis (2) Ringworm disease
(3) Ascariasis (4) Amoebiasis
Answer (1)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
144. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Homology
(2) Convergent evolution
(3) Analogy
(4) Adaptive radiation
Answer (1)
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.
145. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin D (2) Vitamin B
12
(3) Vitamin A (4) Vitamin E
Answer (2)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
146. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Commensalism (2) Parasitism
(3) Mutualism (4) Amensalism
Answer (4)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicillium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicillium or the
organism which produces it.
147. All of the following are included in ? ex-situ
conservation? except
(1) Wildlife safari parks
(2) Botanical gardens
(3) Sacred groves
(4) Seed banks
Answer (3)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
148. 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) ii i iii iv
(2) iii iv i ii
(3) i iii iv ii
(4) i ii iv iii
Answer (2)
30
NEET (UG) - 2018 (Code-EE) CHLAA
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
149. In a growing population of a country,
(1) pre-reproductive individuals are more
than the reproductive individuals.
(2) reproductive and pre-reproductive
individuals are equal in number.
(3) reproductive individuals are less than the
post-reproductive individuals.
(4) pre-reproductive individuals are less than
the reproductive individuals.
Answer (1)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
150. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Flowers (2) Roots
(3) Latex (4) Leaves
Answer (3)
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.
151. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, hPL, progestogens, prolactin
(2) hCG, hPL, progestogens, estrogens
(3) hCG, hPL, estrogens, relaxin, oxytocin
(4) hCG, progestogens, estrogens,
glucocorticoids
Answer (2)
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.
152. The contraceptive ?SAHELI?
(1) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(2) is an IUD.
(3) increases the concentration of estrogen
and prevents ovulation in females.
(4) is a post-coital contraceptive.
Answer (1)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
153. The amnion of mammalian embryo is derived
from
(1) ectoderm and mesoderm
(2) mesoderm and trophoblast
(3) endoderm and mesoderm
(4) ectoderm and endoderm
Answer (1)
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.
154. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(2) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(3) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
Answer (4)
31
NEET (UG) - 2018 (Code-EE) CHLAAA
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.
155. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Inflammation of bronchioles; Decreased
respiratory surface
(2) Increased respiratory surface;
Inflammation of bronchioles
(3) Increased number of bronchioles;
Increased respiratory surface
(4) Decreased respiratory surface;
Inflammation of bronchioles
Answer (1)
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.
156. 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
c
ii
iii
ii
iii
Answer (1)
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.
157. 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) iii ii i iv
(2) i iv ii iii
(3) iii i iv ii
(4) iv iii ii i
Answer (3)
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.
158. 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
a b
(1) iii i
(2) i ii
(3) i iii
(4) ii i
32
NEET (UG) - 2018 (Code-EE) CHLAA
c d
iv i
i iv
iii iv
ii iii
a b
(1) iii ii
(2) ii iii
(3) i ii
(4) iv i
Answer (4)
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.
159. 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) iv v ii iii
(2) v iv i ii
(3) iv i ii iii
(4) v iv i iii
Answer (3)
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.
160. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Protein folding
(2) Cleavage of signal peptide
(3) Protein glycosylation
(4) Phospholipid synthesis
Answer (4)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
161. Which of these statements is incorrect?
(1) Enzymes of TCA cycle are present in
mitochondrial matrix
(2) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(3) Glycolysis occurs in cytosol
(4) Oxidative phosphorylation takes place in
outer mitochondrial membrane
Answer (4)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
162. Nissl bodies are mainly composed of
(1) Proteins and lipids
(2) Nucleic acids and SER
(3) DNA and RNA
(4) Free ribosomes and RER
Answer (4)
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.
163. Which of the following terms describe human
dentition?
(1) Thecodont, Diphyodont, Homodont
(2) Pleurodont, Monophyodont, Homodont
(3) Thecodont, Diphyodont, Heterodont
(4) Pleurodont, Diphyodont, Heterodont
Answer (3)
33
NEET (UG) - 2018 (Code-EE) CHLAAA
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.
164. Select the incorrect match :
(1) Lampbrush ? Diplotene bivalents
chromosomes
(2) Submetacentric ? L-shaped
chromosomes chromosomes
(3) Allosomes ? Sex chromosomes
(4) Polytene ? Oocytes of
chromosomes amphibians
Answer (4)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
165. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Polysome
(2) Plastidome
(3) Polyhedral bodies
(4) Nucleosome
Answer (1)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
166. According to Hugo de Vries, the mechanism
of evolution is
(1) Multiple step mutations
(2) Phenotypic variations
(3) Saltation
(4) Minor mutations
Answer (3)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
167. 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 ii i
(2) ii iii i
(3) i iii ii
(4) iii i ii
Answer (2)
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.
168. All of the following are part of an operon
except
(1) an operator
(2) an enhancer
(3) structural genes
(4) a promoter
Answer (2)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
169. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) AGGUAUCGCAU
(2) ACCUAUGCGAU
(3) UGGTUTCGCAT
(4) UCCAUAGCGUA
Answer (1)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
34
NEET (UG) - 2018 (Code-EE) CHLAA
170. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Only daughters
(2) Only grandchildren
(3) Only sons
(4) Both sons and daughters
Answer (4)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
171. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Chief cells (2) Goblet cells
(3) Mucous cells (4) Parietal cells
Answer (4)
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.
172. 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) (iii) (ii) (i)
(2) (i) (iii) (ii)
(3) (i) (ii) (iii)
(4) (ii) (iii) (i)
Answer (4)
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.
173. Which of the following is an occupational
respiratory disorder?
(1) Anthracis (2) Botulism
(3) Silicosis (4) Emphysema
Answer (3)
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.
174. Calcium is important in skeletal muscle
contraction because it
(1) Binds to troponin to remove the masking
of active sites on actin for myosin.
(2) Detaches the myosin head from the actin
filament.
(3) Activates the myosin ATPase by binding to
it.
(4) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
Answer (1)
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.
35
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1. The test is of 3 hours duration and 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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seat.
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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
EE
CHLAA
1
NEET (UG) - 2018 (Code-EE) CHLAA
3. The efficiency of an ideal heat engine working
between the freezing point and boiling point of
water, is
1. 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) 26.8%
(2) 6.25%
(3) 20%
(4) 12.5%
Answer (1)
(1) 2.508 ? 10
4
K (2) 5.016 ? 10
4
K
(3) 8.360 ? 10
4
K (4) 1.254 ? 10
4
K
Answer (3)
Sol. V
escape
= 11200 m/s
Say at temperature T it attains V
escape
3k T
B
So, = 11200 m/s
m
O
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
On solving,
~
i
-
I
X
( 273 ~
1
~ 373_I
100
T = 8.360 ? 10 K
4
2. 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) 1
5 3
2
4. 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) 13.2 cm
(2) 12.5 cm
(3) 8 cm
(4) 16 cm
Answer (1)
Sol. For closed organ pipe, third harmonic
(3)
=
~ ~
x =
( ~
100 100 26.8%
~ ~ 373
2
(4)
3 7
3v
~
4l
For open organ pipe, fundamental frequency
Answer (1)
Sol. Given process is isobaric
dQ = n C
,
dT
=
~ ~
~ ~ 5
dQ n R dT
~ ~ 2
dW = P dV = n RdT
=
2
20 13.33 cm
x
=
Required ratio
= =
dW nRdT 2
=
dQ ~ ~ 5 5
n R dT
~ ~
~ ~ 2
3
2
v
~
2l'
Given,
3v v
4 l 2l '
~ ~
l
4 2 l l
~ ~
3 2 3 x
~
NEET (UG) - 2018 (Code-EE) CHLAA
NBA
~
NBA
CR
G
C
Voltage sensitivity
I
S
V
S
~
20 V
R
0
= 4 kQ
I
C
R
B
500 kQ
V
i
I
b
V
b
~
mg
I tan30 ?
B
0.5 9.8 x
0.25 3 x
= 11.32 A
5. 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) 7.14 A (2) 14.76 A
(3) 5.98 A (4) 11.32 A
Answer (4)
Sol. For equilibrium,
B
mg sin30 ? = I l Bcos30 ?
l lB
30?
6. 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) 0.79 W (2) 2.74 W
(3) 0.43 W (4) 1.13 W
Answer (1)
2
~ ~ V
RMS
Sol. 1' =
~ J 1 av
~ ~
Z
Z
2
~ ~
10
~
~ ~
P ~
~ ~
x = 50 0.79 W
av
~ ~
2 56
~ ~
7. 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) 40 Q (2) 250 Q
(3) 25 Q (4) 500 Q
Answer (2)
Sol. Current sensitivity
So, resistance of galvanometer
I 5 1 x
- ~
S
R =
5000
G
~ 3
V 20 10 x 20
S
8. 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 current source
(2) The lattice structure of the material of the
rod
(3) The magnetic field
(4) The induced electric field due to the
changing magnetic field
Answer (1)
Sol. Energy of current source will be converted
into potential energy of the rod.
9. 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
R
B

B
E
(1) I
B
= 40 JIA, I
C
= 10 mA, 13 = 250
(2) I
B
= 20 jiA, I
C
= 5 mA, 13 = 250
(3) I
B
= 25 jiA, I
C
= 5 mA, 13 = 200
(4) I
B
= 40 ~A, I
C
= 5 mA, 13 = 125
Answer (4)
Sol. V
BE
= 0
V
CE
= 0
V
b
= 0
30?
2
= +
~
w _
~
= Q
~ 1 ~
R
2
L 56
~
~
~
C
= Q 250
V
i

500 kQ
3
NEET (UG) - 2018 (Code-EE) CHLAA
x
...(i)
2 mm
0.20 0
I
C
=
4 10
3
(20 0)
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
3
~ ~ ~
I 25 10
I
40 10
_ 6
C
b
x
x
= 125
10. 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 + A ? B
(3) A ? B + A ? B
(4) A + B
Answer (3)
12. 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) Reflected light is polarised with its
electric vector parallel to the plane of
incidence
(2) i sin
_
~ ~
1
1
=
~ ~
~
~
~
(3) Reflected light is polarised with its
electric vector perpendicular to the plane
of incidence
(4) i tan
_
" ~
1
1
=
I ~
~
~
~
Answer (3)
Sol. When reflected light rays and refracted rays
are perpendicular, reflected light is polarised
with electric field vector perpendicular to the
plane of incidence.
Sol. A
B
A
A ? B
p.
B
A
A ? B
B
V = (A . B + A . B)
11. In a p-n junction diode, change in temperature
due to heating
(1) Affects only reverse resistance
(2) Does not affect resistance of p-n junction
(3) Affects only forward resistance
(4) Affects the overall V - I characteristics of
p-n junction
Answer (4)
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.
Also, tan i = p. (Brewster angle)
13. 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) 1.8 mm (2) 2.1 mm
(3) 1.9 mm (4) 1.7 mm
Answer (3)
Sol. Angular width d
~ x
V
4
NEET (UG) - 2018 (Code-EE) CHLAA
E
0
V
0
F
So, X = =
mV m V
eE
t
I
+
~
~
0
0
~ m ~
h h
h
mV
0

...(i)
n
~
~~
0.21 o ? x
...(ii)
d
Acceleration of electron

0.20 d
Dividing we get,

~


0.21 2 mm
a -
e E
0

~ d = 1.9 mm
14. 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 large diameter
(2) Large focal length and large diameter
(3) Large focal length and small diameter
(4) Small focal length and small diameter
Answer (2)
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.
15. 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) ~ ~
~
~
eE
0
1
t ~
~
mV
0 ~
(2) X
0
t
~ ~
(3) ~
+
eE
0
0 ~
1 t
~
mV
0 ~
(4) ?
0
Answer (1)
Sol. Initial de-Broglie wavelength
x
0
=
m
Velocity after time ?t?
~ eE ~
0
V V =
~
+ t
0 ~
~ m ~
h
~ ~
x
0

~ ~ ~ ~
0
~
+
eE
~ ~
+
eE
0
1 t 1
t ~
[
mV
] [ mV 0 0 ~
Divide (ii) by (i),
~
0
x =
~ ~
0
~
eE
1
t ~
[
mV
0 ~
16. 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) 20 (2) 30
(3) 10 (4) 15
Answer (1)
Sol. Number of nuclei remaining = 600 ? 450 = 150
t
150 1
600 2
_
~ ~
( ~
~ ~
t
1/2
( ( ~ 1
2
1
~ ~
-
~ ~
~ ~ ~
2 2
t = 2t
1/2
= 2 ? 10
= 20 minute
17. The ratio of kinetic energy to the total energy
of an electron in a Bohr orbit of the hydrogen
atom, is
(1) 1 : 1 (2) 2 : ?1
(3) 1 : ?1 (4) 1 : ?2
Answer (3)
V V
0
?

i = (V
0
> 0) enters an electric field
x
0
...(ii)
mV
0

N 1
_
~
('
N 2
0
t
t
1/2
5
NEET (UG) - 2018 (Code-EE) CHLAA
2
2
4 v
1 1 1
? _ ?
15 v 40
1
1 1 1
v ?15 40
1
= = +
i
30?
M
60?
p = 2
30?
f = 15 cm
O 40 cm
1 1 1
= +
f v u
2 2
sin i 2
~
sin30 1 ?
Sol. KE = ?(total energy)
So, Kinetic energy : total energy = 1 : ?1
18. 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) 1 : 2 (2) 4 : 1
(3) 1 : 4 (4) 2 : 1
Answer (1)
Sol. = +
2
1
E W mv
0
2
2
h(2 ) h v = v ~ 1 mv
0 0 1
2
1
2
h v = mv ...(i)
0 1
2
v = v ~
1
2
h(5 ) h mv
0 0 2
2
1
2
4h v = mv ...(ii)
0 2
2
Divide (i) by (ii),
2
1
v 1
1
v 2
2
19. 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) 30 cm away from the mirror
(2) 30 cm towards the mirror
(3) 36 cm away from the mirror
(4) 36 cm towards the mirror
Answer (3)
Sol.
1 1 1
= +
f v u
1
v
1
= ?24 cm
When object is displaced by 20 cm towards
mirror.
Now,
u
2
= ?20
1 1 1
_ ?
v 20 15
2
v
2
= ?60 cm
So, image shifts away from mirror by
= 60 ? 24 = 36 cm.
20. 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) 60? (2) 30?
(3) 45? (4) Zero
Answer (3)
Sol. For retracing its path, light ray should be
normally incident on silvered face.
Applying Snell's law at M,
1 v
_
~
1 1 1
~ ?
?15 v 20
2
6
NEET (UG) - 2018 (Code-EE) CHLAA
=
sin i 2 = x
2
1
1
~ i.e. i = 45? sin i
2
I I
(2)
O
(1)
O
n n
a velocity
V Vi
?
=
. The instantaneous
~
(3)
O
I
n
21. An em wave is propagating in a medium with
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) ?z direction
(2) ?y direction
(3) +z direction
(4) ?x direction
Answer (3)
-, -, -,
Sol. E B V x =
-*
(Ej)
?
(B) Vi
?
x =
So, 6 6k
?
~
=
Direction of propagation is along +z direction.
22. 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) 0.138 H (2) 1.389 H
(3) 138.88 H (4) 13.89 H
Answer (4)
Sol. Energy stored in inductor
U Ll =
1

2
2
x = x x x
1
?3 ?3 2
25 10 L (60 10 )
2
25 2 10 10 x x x
6 ?3
~
3600
500
~
36
= 13.89 H
23. 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
(4)
O
Answer (1)
E E
Sol. I ~ ~
n
nr r
So, I is independent of n and I is constant.
~
O
24. 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) 10 (2) 20
(3) 11 (4) 9
Answer (1)
Sol. = I
nR R +
...(ii)
fl
Dividing (ii) by (i),
(n 1)R +
~
~ ~
~
~
~
1
1 R
~ )
n
After solving the equation, n = 10
25. 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) Violet ? Yellow ? Orange ? Silver
(2) Yellow ? Green ? Violet ? Gold
(3) Yellow ? Violet ? Orange ? Silver
(4) Green ? Orange ? Violet ? Gold
Answer (3)
L
n
I
n
E
...(i)
E
10 I =
R R +
10
7
NEET (UG) - 2018 (Code-EE) CHLAA
2
Q
~ F
plate
0
2A E
Sol. (47 ? 4.7) kQ = 47 ? 10
3
? 10%
~ Yellow ? Violet ? Orange ? Silver
26. 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) 330 m/s (2) 350 m/s
(3) 339 m/s (4) 300 m/s
Answer (3)
Sol. v = 2 (v) [L
2
? L
1
]
= 2 ? 320 [73 ? 20] ? 10
?2

= 339.2 ms
?1

= 339 m/s
27. The electrostatic force between the metal
plates of an isolated parallel plate capacitor
C having a charge Q and area A, is
(1) Independent of the distance between the
plates
(2) Proportional to the square root of the
distance between the plates
(3) Linearly proportional to the distance
between the plates
(4) Inversely proportional to the distance
between the plates
Answer (1)
Sol. For isolated capacitor Q = Constant
F is Independent of the distance between
plates.
28. 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) Smaller
(2) 10 times greater
(3) 5 times greater
(4) Equal
Answer (1)
Sol. =
1 eE
2
h t
2 m
~ = t
eE
~ t oc m as ?e? is same for electron and
proton.
? Electron has smaller mass so it will take
smaller time.
29. 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) 2ir s (2) 2 s
(3) ir s (4) 1 s
Answer (3)
Sol. |a| = w
2
y
= 20 = ~
2
(5)
= w = 2 rad/s
= = = it
2 2 ir ir
T s
(1) 2
30. 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)
3
(2)
256
4 81
(3)
4
(4)
81
3 256
Answer (2)
Sol. We know,
X
max
T = constant (Wien's law)
So,
max
1
T
1
~ ~ ~
max
2
T
2
~ X =
3
~ ' T T
4
~ ' =
4
T T
3
(
'
~ ( ~
4 4
P T
=
1 I
=
1 I
=
4 256
So,
2
P T ) ) 3 81
1
2hm
8
? ? ?
i j k
0 2 1 -
4 5 6 -
~ ~
9
= - - - 7i 4 j 8k
V
T
r OC
2

~ Power r oc
5

NEET (UG) - 2018 (Code-EE) CHLAA
31. 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) 9 F
(2) 4 F
(3) 6 F
(4) F
Answer (1)
Sol. Wire 1 :

F

A, 3l
Wire 2 :
F'
3A, l
For wire 1,
~ ~
~ ~
~ ~ F
l 3l ...(i)
~ )
AY
For wire 2,
F ' A l
_ Y
3A l
From equation (i) & (ii),
A =
~ ~
=
~ ~
~ ~ ~ ' ~ F F
l 3l l
L ) )
AY 3AY
33. A sample of 0.1 g of water at 100?C and
normal pressure (1.013 x 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) 104.3 J
(2) 42.2 J
(3) 208.7 J
(4) 84.5 J
Answer (3)
Sol. AQ = AU + AW
= 54 x 4.18 = AU + 1.013 x 10
5
(167.1 x 10
?6
? 0)
= AU = 208.7 J
~
34. The moment of the force, = + - F 4i
? 5 ?
j 6k ? at
(2, 0, ?3), about the point (2, ?2, ?2), is given
by
(1) - 8i
?
- 4 ? j - 7k
?
(2) - 7i
?
- 8j
?
- 4k
?
(3) -4i
?
-
j
-
8k
?
(4) - 7i
?
- 4 ? j - 8k
?
' ' F
~ A =
~ ~
l l ...(ii)
~ ~
3AY
= F' = 9F
32. 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
3
(2) r
5
(3) r
2
(4) r
4
Answer (2)
Sol. Power = lrl . = lrl
T
6 rV
T
V
T
6 rV
2
Answer (4)
Sol.
O
~ ~ ~ ~
~ ~ ~ ~ (r r
0
)
F ...(i)
~ ~
r r (2i ? 0j ? 3k)
?
(2i ? 2j ? 2k)
?
- = + - - - -
V
r
0
? ?
= 0i + 2j - k
A
? ? ?
?
r - r
0
r
P
F
x
NEET (UG) - 2018 (Code-EE) CHLAA
B
v = 0
-
Acceleration a =
6 0
= 6 ms
1
For t = 0 to t = 1 s,
-2
Answer (4)
Sol.
N cos O
N sin b
a
N
0
0
ma
(pseudo)
mg
0
Total displacement S = S
1
+ S
2
+ S
3
= 3 m
Average velocity =
3
= 1 ms
3
-1
35. 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
(1) 0.521 cm
(2) 0.053 cm
(3) 0.525 cm
(4) 0.529 cm
Answer (4)
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
36. A block of mass m is placed on a smooth
inclined wedge ABC of inclination e 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
A
m
a
N sin O = ma ...(i)
N cos O = mg ...(ii)
tan ~ ~
a
g
a = g tan 0
37. 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) 2 m/s, 4 m/s
(2) 1 m/s, 3.5 m/s
(3) 1 m/s, 3 m/s
(4) 1.5 m/s, 3 m/s
Answer (3)
a
t = 1 ?a
t = 2
v = 6 ms
?1

C
t = 3
v = ?6 ms
?1

Sol. t = 0
A
v = 0
?a
0
C
g
(1) = a (2) a = g cos o
cosec 0
g
(3) = a (4) a = g tan o
sin O
= ~
2
1
S 6(1)
2
= 3 m ...(i)
B
In non-inertial frame,
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,
= - x = -
1
S 0 6(1) 3 m
2
3
2
...(iii)
10
NEET (UG) - 2018 (Code-EE) CHLAA
Total distance travelled = 9 m
Average speed =
9
= 3 ms
3
-1
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) 0.5
(2) 0.8
(3) 0.25
(4) 0.4
Answer (3)
Sol. According to law of conservation of linear
momentum,
mv + 4m x 0 = 4mv ' + 0
V ~ ~
V
4
v
-
Relative velocity of separation 4
-
Relative velocity of approach v
e = =
1
0.25
4
39. 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
h
A
(1) 3D
2
(2)
D
7
5
(3) D
Sol.
h
v
L

As track is frictionless, so total mechanical
energy will remain constant
T.M.E
I
=T.M.E
F

0 mgh mv 0 + =
1
2
+
L
2
v
L
2
2g
For completing the vertical circle, v
L
~ 5gR
h = = =
5gR 5 5
R D
2g 2 4
40. 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
C
> W
B
> W
A

(2) W
B
> W
A
> W
C

(3) W
A
> W
B
> W
C

(4) W
A
> W
C
> W
B

Answer (1)
Sol. Work done required to bring them rest
~W = AKE
1
A = W I ~
2
2
AW I
for same (0
1
2 2
=
2
2
W : W : W MR : MR : MR
A B C
5 2
2 1
=
: : 1
5 2
e
B
B
A
h ~
= 4 : 5 : 10
= W
C
> W
B
> W
A
5 D
(4)
4
Answer (4)
11
NEET (UG) - 2018 (Code-EE) CHLAA
B
A
S
41. Which one of the following statements is
incorrect?
(1) Rolling friction is smaller than sliding
friction.
(2) Frictional force opposes the relative
motion.
(3) Limiting value of static friction is directly
proportional to normal reaction.
(4) Coefficient of sliding friction has
dimensions of length.
Answer (4)
Sol. Coefficient of sliding friction has no
dimension.
f = JL
s
N
f
~ ~
s
~
N
42. 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) 7 : 10 (2) 10 : 7
(3) 5 : 7 (4) 2 : 5
Answer (3)
Sol. =
2
1
K mv
t
2
1 ( '1 ' v
~ ~
1
~ ~ ~
1
2 2
mv
2
~
~
1 2 mr
2
K K mv I
t r ~~ ~
2 2 2 2 5
~ )~ )
r
7
= my
2
10
So, K K 7
t
+
r
K
=
5
t
43. 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 velocity
(2) Rotational kinetic energy
(3) Moment of inertia
(4) Angular momentum
Answer (4)
Sol.
ex
= 0
So
,
~
dL 0
dt
i.e. L = constant
So angular momentum remains constant.
44. 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) Raindrops will fall faster
(2) Time period of a simple pendulum on the
Earth would decrease
(3) Walking on the ground would become
more difficult
(4) ?g? on the Earth will not change
Answer (4)
Sol. If Universal Gravitational constant becomes
ten times, then G ' = 10 G
So, acceleration due to gravity increases.
i.e. (4) is wrong option.
45. 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
A
< K
B
< K
C

(2) K
B
< K
A
< K
C

(3) K
A
> K
B
> K
C

(4) K
B
> K
A
> K
C

Answer (3)
Sol. B
perihelion
A
S
V
A

Point A is perihelion and C is aphelion.
So, V
A
> V
B
> V
C

So, K
A
> K
B
> K
C

2
C
V
C

C
aphelion
12
M
b. 55 mL 10 HCl + 45 mL
3
2.42 10 x
(mol L
?1
)
233
10 1
=
~
100 10
NEET (UG) - 2018 (Code-EE) CHLAA
46. 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) NH
3

(2) O
2

(3) H
2

(4) CO
2

Answer (1)
Sol. ? van der waal constant ?a?, signifies
intermolecular forces of attraction.
? Higher is the value of ?a?, easier will be
the liquefaction of gas.
47. Following solutions were prepared by mixing
different volumes of NaOH and HCl of
different concentrations :
M
a. 60 mL
M
10 NaOH
48. 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
?10
mol
2
L
?2

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

(3) 1.08 * 10
?12
mol
2
L
?2

(4) 1.08 * 10
?8
mol
2
L
?2

Answer (1)
Sol. Solubility of BaSO
4
, s =
= 1.04 * 10
?5
(mol L
?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
49. On which of the following properties does the
coagulating power of an ion depend?
(1) The magnitude of the charge on the ion
alone
HCl + 40 mL
10
M
10 NaOH
c. 75 mL
M
5 HCl + 25 mL
M
5 NaOH
M
10 NaOH
(2) Both magnitude and sign of the charge on
the ion
d. 100 mL
M
HCl + 100 mL
10
pH of which one of them will be equal to 1?
(1) b
(2) d
(3) a
(4) c
Answer (4)
(3) Size of the ion alone
(4) The sign of charge on the ion alone
Answer (2)
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.
50. Considering Ellingham diagram, which of the
following metals can be used to reduce
alumina?
Sol. ? Meq of HCl =
1
75 ~ x = 15 1
5
? Meq of NaOH =
1
25 ~ x = 5 1
5
? Meq of HCl in resulting solution = 10
? Molarity of [H
+
] in resulting mixture
pH = ?log[H
+
] =
_
~ ~
~ ~ 1
log ~
~ = 1.0
10
(1) Fe
(2) Mg
(3) Zn
(4) Cu
13
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (2)
Sol. The metal which is more reactive than 'Al'
can reduce alumina i.e. 'Mg' should be the
correct option.
51. The correct order of atomic radii in group 13
elements is
(1) B < Al < In < Ga < Tl
(2) B < Ga < Al < Tl < In
(3) B < Al < Ga < In < Tl
(4) B < Ga < Al < In < Tl
Answer (4)
Sol.
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
52. In the structure of ClF
3
, the number of lone
pairs of electrons on central atom ?Cl? is
(1) One
(2) Four
(3) Two
(4) Three
Answer (3)
Sol. The structure of ClF
3
is
S S
F
S S
Cl
S S
F
S S
The number of lone pair of electrons on
central Cl is 2.
53. The correct order of N-compounds in its
decreasing order of oxidation states is
(1) HNO
3
, NO, N
2
, NH
4
Cl
(2) HNO
3
, NH
4
Cl, NO, N
2

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

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

Answer (1)
~ 5 +2 0 ?3
Sol.
HNO , N O, N , NH
4
Cl
3
2
Hence, the correct option is (1).
54. Which of the following statements is not true
for halogens?
(1) All form monobasic oxyacids
(2) All but fluorine show positive oxidation
states
(3) All are oxidizing agents
(4) Chlorine has the highest electron-gain
enthalpy
Answer (2)
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.
55. Which one of the following elements is unable
to form MF
6
3?
ion?
(1) Ga
(2) B
(3) Al
(4) In
Answer (2)
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(?)
.
Hence, the correct option is (2).
56. Regarding cross-linked or network polymers,
which of the following statements is
incorrect?
(1) They contain covalent bonds between
various linear polymer chains.
(2) Examples are bakelite and melamine.
(3) They are formed from bi- and tri-functional
monomers.
(4) They contain strong covalent bonds in
their polymer chains.
Answer (4)
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 (4) is not related to
cross-linking.
So option (4) should be the correct option.
? ?
? ?
F
14
NEET (UG) - 2018 (Code-EE) CHLAA
57. Nitration of aniline in strong acidic medium
also gives m-nitroaniline because
(1) Inspite of substituents nitro group always
goes to only m-position.
(2) In absence of substituents nitro group
always goes to m-position.
(3) In electrophilic substitution reactions
amino group is meta directive.
(4) In acidic (strong) medium aniline is
present as anilinium ion.
Answer (4)
NH
3
H
Anilinium ion
?NH
,
is m-directing, hence besides para
(51%) and ortho (2%), meta product (47%) is
also formed in significant yield.
58. Which of the following oxides is most acidic in
nature?
(1) MgO (2) BaO
(3) BeO (4) CaO
Answer (3)
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.
59. The difference between amylose and
amylopectin is
(1) Amylopectin have 1 - 4 a-linkage and
1 - 6 a-linkage
(2) Amylopectin have 1 - 4 a-linkage and
1 - 6 13-linkage
(3) Amylose have 1 - 4 a-linkage and 1 - 6
13-linkage
(4) Amylose is made up of glucose and
galactose
Answer (1)
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.
So option (1) should be the correct option.
60. 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) 1.4 (2) 2.8
(3) 3.0 (4) 4.4
Answer (2)
Sol.
~~~~~~
Conc.H
2
SO
4
HCOOH CO(g) H O(l) +
2
~ ~ 1 1
2.3g or
~
mol mol
~ 20
20
COOH
Conc.H
2
SO
4
CO(g) + CO
2
(g) + H
2
O(l)
1
COOH
~ 1 ~
4.5g or
l
mol ~
~ 20 ~
Gaseous mixture formed is CO and CO
2
when
it is passed through KOH, only CO
2
is
absorbed. So the remaining gas is CO.
So, weight of remaining gaseous product CO
is
2 28 2.8g x =
So, the correct option is (2)
61. In the reaction
+
OH O Na
?
+ CHCl
3
+ NaOH
The electrophile involved is
(1) Dichloromethyl cation
( CHCl
2
)
~
(2) Dichloromethyl anion ( CHCl
2
)
~
(3) Formyl cation (
C HO
)
~
(4) Dichlorocarbene ( : CCl
2
)
NH
2
Sol.
20
mol
20
1
mol
20
CHO
15 16
Iodoform
(Yellow PPt) Sodium benzoate
NaOH
COONa
Answer (4)
Sol. It is Reimer-Tiemann reaction. The electrophile
formed is :CCl
2
(Dichlorocarbene) according to
the following reaction
Sol. Option (2) 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
. .
CHCl + OH
=~?
~~~
CCl
3
+ H
2
O
?
3
?
CCl ~~~ : CCl + Cl
3 2
Electrophile
62. Carboxylic acids have higher boiling points
than aldehydes, ketones and even alcohols of
comparable molecular mass. It is due to their
(1) Formation of intramolecular H-bonding
(2) More extensive association of carboxylic
acid via van der Waals force of attraction
(3) Formation of carboxylate ion
(4) Formation of intermolecular H-bonding
Answer (4)
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.
63. 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
(1) H
3
C CH
2
? OH and I
2
(2) CH ? CH
3
and I
2

OH
(3) CH
2
? CH
2
? OH and I
2
CH
3
(4) CH
3
OH and I
2

Answer (2)
I
2
64. Which of the following molecules represents
the order of hybridisation sp
2
, sp
2
, sp, sp from
left to right atoms?
(1) HC ~ C ? C ~ CH
(2) CH
2
= CH ? CH = CH
2

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

Answer (3)
Sol.
Number of orbital require in hybridization
= Number of a-bonds around each carbon
atom.
65. Which of the following carbocations is
expected to be most stable?
NO
2

NO
2
(2)
~
(1)
V H
H
V
~
NO
2
~
V H
(3)
~
(4)
H
V
NO
.
Answer (2)
Sol. ?NO
2
group exhibit ?I effect and it decreases
with increase in distance. In option (2)
positive charge present on C-atom at
maximum distance so ?I effect reaching to it
is minimum and stability is maximum.
2
sp sp sp
2
sp
C H 2 = CH?C ~ CH
NEET (UG) - 2018 (Code-EE) CHLAA
?
+ CHI
3

NaOI
CH ? CH
3
C ? CH
3
OH O
Acetophenone
(A)
Spin magnetic moment = 4(4 + 2) = 24 BM
Spin magnetic moment = 2(2 + 2) = 8 BM
Fe
3+
= [Ar] 3d
5
, Unpaired e
?
(n) = 5
Spin magnetic moment = 5(5 + 2) = 35 BM
Spin magnetic moment = 3(3 + 2) = 15 BM
NEET (UG) - 2018 (Code-EE) CHLAA
66. Which of the following is correct with respect
to ? I effect of the substituents? (R = alkyl)
(1) ? NH
2
< ? OR < ? F
(2) ? NH
2
> ? OR > ? F
(3) ? NR
2
< ? OR < ? F
(4) ? NR
2
> ? OR > ? F
Answer (1*)
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 (1),
however option (3) may also be correct answer.
67. The type of isomerism shown by the complex
[CoCl
2
(en)
2
] is
(1) Geometrical isomerism
(2) Ionization isomerism
(3) Coordination isomerism
(4) Linkage isomerism
Answer (1)
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.
68. Which one of the following ions exhibits d-d
transition and paramagnetism as well?
(1) CrO
4
2?
(2) MnO
4
?
2?
(3) Cr
2
O
7
2?
(4) MnO
4

Answer (4)
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
69. Iron carbonyl, Fe(CO)
5
is
(1) Tetranuclear (2) Trinuclear
(3) Mononuclear (4) Dinuclear
Answer (3)
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
Hence, option (3) should be the right
answer.
70. 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
a b c d
(1) iv v ii i
(2) iv i ii iii
(3) i ii iii iv
(4) iii v i ii
Answer (1)
Sol. Co
3+
= [Ar] 3d
6
, Unpaired e
?
(n) = 4
Cr
3+
= [Ar] 3d
3
, Unpaired e
?
(n) = 3
Ni
2+
= [Ar] 3d
8
, Unpaired e
?
(n) = 2
17
NEET (UG) - 2018 (Code-EE) CHLAA
1
,
k[A ]
0
71. The geometry and magnetic behaviour of the
complex [Ni(CO)
4
] are
(1) Square planar geometry and diamagnetic
(2) Square planar geometry and
paramagnetic
(3) Tetrahedral geometry and diamagnetic
(4) Tetrahedral geometry and paramagnetic
Answer (3)
Sol. Ni(28) : [Ar]3d
8
4s
2

? CO is a strong field ligand
Configuration would be :
sp -hybridisation
3
? ? ? ? ? ?
CO CO CO CO
For, four ?CO?-ligands hybridisation would be
sp
3
and thus the complex would be
diamagnetic and of tetrahedral geometry.
CO
Ni
CO
OC
CO
72. The correct difference between first and
second order reactions is that
(1) 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
(2) A first-order reaction can catalyzed; a
second-order reaction cannot be
catalyzed
(3) 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

(4) 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
Answer (3)
Sol. ? For first order reaction,
1 /2
=
0.693
t
k
,
which is independent of initial
concentration of reactant.
? For second order reaction,
1 / 2
= t
which depends on initial concentration of
reactant.
73. Among CaH
2
, BeH
2
, BaH
2
, the order of ionic
character is
(1) BeH
2
< CaH
2
< BaH
2

(2) BeH
2
< BaH
2
< CaH
2

(3) CaH
2
< BeH
2
< BaH
2

(4) BaH
2
< BeH
2
< CaH
2

Answer (1)
Sol. For 2
nd
group hydrides, on moving down the
group metallic character of metals increases
so ionic character of metal hydride increases.
Hence the option (1) should be correct option.
74. In which case is number of molecules of
water maximum?
(1) 18 mL of water
(2) 0.00224 L of water vapours at 1 atm and
273 K
(3) 0.18 g of water
(4) 10
?3
mol of water
Answer (1)
Sol. (1) Mass of water = 18 ? 1 = 18 g
18
N

Molecules of water = mole ? N
A
= A
18
= N
A

0.00224
(2) Moles of water =

22.4
Molecules of water = mole ? N
A
= 10
?4
N
A

0.18
N

(3) Molecules of water = mole ? N
A
=
A
18
= 10
?2
N
A

(4) Molecules of water = mole ? N
A
= 10
?3
N
A
??
= 10
?4

18
? 1.82 V ? 1.5 V
HBrO BrO
4
BrO
3

?
Br
1.0652 V
Br
2

1.595 V
NEET (UG) - 2018 (Code-EE) CHLAA
75. 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)
~
BrO
3

(2) Br
2

(3)
~
BrO
4

(4) HBrO
Answer (4)
+ 1 0
Sol. HBrO Br , E
o
~~~ = 1.595 V
2 HBrO/Br
2
+ 1
+ 5
~ o
HBrO BrO , E
~
~~*
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
Hence, option (4) is correct answer.
76. 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) 200 kJ mol
?1

(2) 800 kJ mol
?1

(3) 100 kJ mol
?1

(4) 400 kJ mol
?1

Answer (2)
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
77. When initial concentration of the reactant is
doubled, the half-life period of a zero order
reaction
(1) Is halved
(2) Is tripled
(3) Is doubled
(4) Remains unchanged
Answer (3)
Sol. Half life of zero order
[A ]
0
2K
t
1/2
will be doubled on doubling the initial
concentration.
78. 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) 16 5 2
(2) 2 16 5
(3) 2 5 16
(4) 5 16 2
Answer (3)
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
t
1/2
~
19
NEET (UG) - 2018 (Code-EE) CHLAA
For FCC lattice : Z = 4, a = 2 2 r
=
~ ~
=
~ ~
~
~
4
4r
4 2
~ ~
~ ~
3
3
2 2 2 r
( ' ( '
3 3
Sol. In real gas equation,
~
~
~
4r
3
79. 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) Low temperature and high pressure
(2) High temperature and high pressure
(3) Low temperature and low pressure
(4) High temperature and low pressure
Answer (1)
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.
80. The correction factor ?a? to the ideal gas
equation corresponds to
(1) Density of the gas molecules
(2) Electric field present between the gas
molecules
(3) Volume of the gas molecules
(4) Forces of attraction between the gas
molecules
Answer (4)
2
'
+ - =
an
P (V nb) nRT
2 ~
V ~
van der Waal?s constant, ?a? signifies
intermolecular forces of attraction.
81. Magnesium reacts with an element (X) to form
an ionic compound. If the ground state
electronic configuration of (X) is 1s
2
2s
2
2p
3
,
the simplest formula for this compound is
(1) Mg
2
X
3
(2) Mg
2
X
(3) MgX
2

(4) Mg
3
X
2
Answer (4)
Sol. Element (X) electronic configuration
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
.
82. 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
3
(1)
3 3
(2)
4 3
(3)
(4)
Answer (2)
Sol. For BCC lattice : Z = 2,
= a
~ =
( ~ ZM
~ 3 ~
d ~ ~
N a
25 C 0
A
BCC
FCC
83. Consider the following species :
CN
+
, CN
?
, NO and CN
Which one of these will have the highest bond
order?
(1) NO (2) CN
+

(3) CN
?
(4) CN
2
4 2
3 2
1
2
d ( " ZM
900 C 0
~ ~
1 3 ~
N a
A
20
1s
2
2s
2
2p
3

OR
NEET (UG) - 2018 (Code-EE) CHLAA
Answer (3)
Sol. NO : ( a1s)
2
, (a* 1s)
2
, ( cy2s)
2
, ( cy* 2s)
2
, ( cy2p
z
)
2
,
(ir2p
x
)
2
= (ir2p
y
)
2
,(1t* 2p
x
)
1
= (~* 2p
y
)
0

BO =
- =
10 5 2.5
2
CN
?
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
, (a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
2

BO = _ ~
10 4
3
2
CN : (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (1r2p
y
)
2
,(a2p
z
)
1

BO =
~ ~ 9 4 2.5
2
CN
+
: (a1s)
2
, (a* 1s)
2
, (a2s)
2
,(a* 2s)
2
, (ir2p
x
)
2

= (ir2p
y
)
2

BO = _ ~
8 4
2
2
Hence, option(3) should be the right answer.
84. Which one is a wrong statement?
(1) Total orbital angular momentum of
electron in 's' orbital is equal to zero
(2) The electronic configuration of N atom is
1
1s
2
2s
2
2p
x
1
2p
y
1
2p
z
(3) An orbital is designated by three quantum
numbers while an electron in an atom is
designated by four quantum numbers
(4) The value of m for d
z
2 is zero
Answer (2)
Sol. According to Hund's Rule of maximum
multiplicity, the correct electronic
configuration of N-atom is
1s
2
2s
2
2p
3

~ Option (2) violates Hund's Rule.
85. Which oxide of nitrogen is not a common
pollutant introduced into the atmosphere both
due to natural and human activity?
(1) N
2
O
5
(2) N
2
O
(3) NO
2

(4) NO
Answer (1)
Sol. Fact
86. 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 ~ CH
(2) CH
3
? CH
3

(3) CH
2
= CH
2

(4) CH
4

Answer (4)
Sol.
CH
4
Br
2
/h v
CH
3
B r
Na/dry ether
Wurtz reaction
CH
3
? CH
3

Hence the correct option is (4)
87. 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
(1) C
2
H
5
OH,
C
2
H
6
,
C
2
H
5
Cl
(2) C
2
H
5
Cl,
C
2
H
6
,
C
2
H
5
OH
(3) C
2
H
5
OH, C
2
H
5
Cl, C
2
H
5
ONa
(4) C
2
H
5
OH, C
2
H
5
ONa, C
2
H
5
Cl
Answer (4)
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)
So the correct option is (4)
(A)
21
CH
3
6?
CCl
3

CCl
3

CH
3
(B)
Zn HCl CH ? CH
3

22
(C
7
H
8
)
(A)
CH
3
? C ? CH
3
+
88. 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) m-bromotoluene
(2) 3-bromo-2,4,6-trichlorotoluene
(3) o-bromotoluene
(4) p-bromotoluene
Answer (1)
Br
(C)
So, the correct option is (1)
89. Identify the major products P, Q and R in the
following sequence of reactions:
CH
3
CH
3
? CH ? CH
3

6+
1, 2?H
Shift
CH
3
CH
2
CH
2
Cl AlCl
3

6+ 6?
(Incipient carbocation)
(R)
(Q)
O
OH
H /H O
+
2
Hydroperoxide
Rearrangement
3
HC?C?O?O?H
Cl
AlCl
3
Sol.
3dl
2 Br
2
A Fe
Now,
Br
CH
3

O
2

(P)
90. Which of the following compounds can form a
zwitterion?
(1) Aniline
(2) Benzoic acid
(3) Acetanilide
(4) Glycine
Answer (4)
Sol.
H
3
N ? CH
2
? COOH
~
(Zwitterion form)
H
2
N ? CH
2
? COO
?
+ CH
3
CH
2
CH
2
Cl
Anhydrous
AlCl
3

NEET (UG) - 2018 (Code-EE) CHLAA
CH
3
? CH ? CH
3

(i) O
2

P Q + R
(ii) H
8
O / A
P Q R
CH
2
CH
2
CH
3
CHO
CH
3
CH
2
? OH
(1)
CH(CH
3
)
2
OH
, CH
3
CH(OH)CH
3 (2)
CH
3
? CO ? CH
3

CH
2
CH
2
CH
3
CHO COOH
(3)
OH
(4)
CH(CH
3
)
2
Answer (4)
Cl
~
H
3
N ? CH
2
? COO
?
pK
,
= 9.60 pK
,
= 2.34
.
CH
3
CH
2
CH
2
? Cl + Al
Cl Cl
Sol
NEET (UG) - 2018 (Code-EE) CHLAA
91. Which of the following statements is correct?
(1) Ovules are not enclosed by ovary wall in
gymnosperms
(2) Horsetails are gymnosperms
(3) Selaginella is heterosporous, while
Salvinia is homosporous
(4) Stems are usually unbranched in both
Cycas and Cedrus
Answer (1)
Sol. ? Gymnosperms have naked ovule.
? Called phanerogams without womb/ovary
92. Pneumatophores occur in
(1) Halophytes
(2) Carnivorous plants
(3) Free-floating hydrophytes
(4) Submerged hydrophytes
Answer (1)
Sol. ? Halophytes like mangrooves have
pneumatophores.
? Apogeotropic (?vely geotropic) roots
having lenticels called pneumathodes to
uptake O
2
.
93. Sweet potato is a modified
(1) Stem
(2) Tap root
(3) Adventitious root
(4) Rhizome
Answer (3)
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
94. Plants having little or no secondary growth
are
(1) Grasses
(2) Conifers
(3) Deciduous angiosperms
(4) Cycads
Answer (1)
Sol. Grasses are monocots and monocots usually
do not have secondary growth.
Palm like monocots have anomalous
secondary growth.
95. Casparian strips occur in
(1) Epidermis
(2) Cortex
(3) Pericycle
(4) Endodermi
Answer (4)
Sol. ? Endodermis have casparian strip on radial
and inner tangential wall.
? It is suberin rich.
96. Secondary xylem and phloem in dicot stem
are produced by
(1) Apical meristems
(2) Phellogen
(3) Vascular cambium
(4) Axillary meristems
Answer (3)
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.
97. Select the wrong statement :
(1) Cell wall is present in members of Fungi
and Plantae
(2) Pseudopodia are locomotory and feeding
structures in Sporozoans
(3) Mushrooms belong to Basidiomycetes
(4) Mitochondria are the powerhouse of the
cell in all kingdoms except Monera
Answer (2)
Sol. Pseudopodia are locomotory structures in
sarcodines (Amoeboid)
98. The experimental proof for semiconservative
replication of DNA was first shown in a
(1) Fungus (2) Plant
(3) Bacterium (4) Virus
Answer (3)
Sol. Semi-conservative DNA replication was first
shown in Bacterium Escherichia coli by
Matthew Meselson and Franklin Stahl.
23
NEET (UG) - 2018 (Code-EE) CHLAAA
99. Select the correct match
(1) Alec Jeffreys

- Streptococcus
pneumoniae
(2) Matthew Meselson - Pisum sativum
and F. Stahl
(3) Alfred Hershey and - TMV
Martha Chase
(4) Francois Jacob and - Lac operon
Jacques Monod
Answer (4)
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
100. Select the correct statement
(1) Franklin Stahl coined the term ?linkage?
(2) Spliceosomes take part in translation
(3) Punnett square was developed by a British
scientist
(4) Transduction was discovered by S. Altman
Answer (3)
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
101. Which of the following pairs is wrongly
matched?
(1) Starch synthesis in pea : Multiple alleles
(2) XO type sex : Grasshopper
determination
(3) ABO blood grouping : Co-dominance
(4) T.H. Morgan : Linkage
Answer (1)
Sol. Starch synthesis in pea is controlled by
pleiotropic gene.
Other options (2, 3 & 4) are correctly
matched.
102. Offsets are produced by
(1) Meiotic divisions
(2) Parthenocarpy
(3) Mitotic divisions
(4) Parthenogenesis
Answer (3)
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)
103. Which of the following flowers only once in its
life-time?
(1) Bamboo species
(2) Mango
(3) Jackfruit
(4) Papaya
Answer (1)
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.
104. Which of the following has proved helpful in
preserving pollen as fossils?
(1) Pollenkitt (2) Oil content
(3) Cellulosic intine (4) Sporopollenin
Answer (4)
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.
24
NEET (UG) - 2018 (Code-EE) CHLAA
105. Which of the following is commonly used as a
vector for introducing a DNA fragment in
human lymphocytes?
(1) Retrovirus
(2) X phage
(3) Ti plasmid
(4) pBR 322
Answer (1)
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.
106. The correct order of steps in Polymerase
Chain Reaction (PCR) is
(1) Extension, Denaturation, Annealing
(2) Denaturation, Extension, Annealing
(3) Annealing, Extension, Denaturation
(4) Denaturation, Annealing, Extension
Answer (4)
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
107. In India, the organisation responsible for
assessing the safety of introducing
genetically modified organisms for public use
is
(1) Indian Council of Medical Research (ICMR)
(2) Research Committee on Genetic
Manipulation (RCGM)
(3) Council for Scientific and Industrial
Research (CSIR)
(4) Genetic Engineering Appraisal Committee
(GEAC)
Answer (4)
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).
108. Use of bioresources by multinational
companies and organisations without
authorisation from the concerned country and
its people is called
(1) Bio-infringement
(2) Biodegradation
(3) Biopiracy
(4) Bioexploitation
Answer (3)
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).
109. 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) Co-667
(2) Lerma Rojo
(3) Sharbati Sonora
(4) Basmati
Answer (4)
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.
110. Select the correct match
(1) Ribozyme - Nucleic acid
(2) T.H. Morgan - Transduction
(3) F
2
? Recessive parent - Dihybrid cross
(4) G. Mendel - Transformation
Answer (1)
Sol. Ribozyme is a catalytic RNA, which is nucleic
acid.
25
NEET (UG) - 2018 (Code-EE) CHLAAA
111. Niche is
(1) all the biological factors in the organism's
environment
(2) the range of temperature that the
organism needs to live
(3) the physical space where an organism
lives
(4) the functional role played by the organism
where it lives
Answer (4)
Sol. Ecological niche was termed by J. Grinnel. It
refers the functional role played by the
organism where it lives.
112. Which of the following is a secondary
pollutant?
(1) CO (2) SO
2

(3) CO
2
(4) O
3

Answer (4)
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
113. World Ozone Day is celebrated on
(1) 5
th
June (2) 16
th
September
(3) 21
st
April (4) 22
nd
April
Answer (2)
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
114. Natality refers to
(1) Death rate
(2) Number of individuals leaving the habitat
(3) Birth rate
(4) Number of individuals entering a habitat
Answer (3)
Sol. Natality refers to birth rate.
? Death rate ? Mortality
? Number of individual
entering a habitat is
? Immigration
? Number of individual
leaving the habital
? Emigration
115. In stratosphere, which of the following
elements acts as a catalyst in degradation of
ozone and release of molecular oxygen?
(1) Carbon (2) Fe
(3) Cl (4) Oxygen
Answer (3)
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
116. 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) Inverted pyramid of biomass
(2) Upright pyramid of numbers
(3) Pyramid of energy
(4) Upright pyramid of biomass
Answer (1)
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.
117. The Golgi complex participates in
(1) Fatty acid breakdown
(2) Respiration in bacteria
(3) Formation of secretory vesicles
(4) Activation of amino acid
Answer (3)
Sol. Golgi complex, after processing releases
secretory vesicles from their trans-face.
118. Which of the following is not a product of light
reaction of photosynthesis?
(1) ATP (2) NADPH
(3) NADH (4) Oxygen
Answer (3)
Sol. ATP, NADPH and oxygen are products of light
reaction, while NADH is a product of
respiration process.
26
NEET (UG) - 2018 (Code-EE) CHLAA
119. Which among the following is not a
prokaryote?
(1) Saccharomyces (2) Nostoc
(3) Mycobacterium (4) Oscillatoria
Answer (1)
Sol. Saccharomyces i.e. yeast is an eukaryote
(unicellular fungi)
Mycobacterium ? a bacterium
Oscillatoria and Nostoc are cyanobacteria.
120. Stomatal movement is not affected by
(1) Temperature (2) O
2
concentration
(3) Light (4) CO
2
concentration
Answer (2)
Sol. Light, temperature and concentration of CO
2

affect opening and closing of stomata while
they are not affected by O
2
concentration.
121. Which of the following is true for nucleolus?
(1) Larger nucleoli are present in dividing
cells
(2) It takes part in spindle formation
(3) It is a membrane-bound structure
(4) It is a site for active ribosomal RNA
synthesis
Answer (4)
Sol. Nucleolus is a non membranous structure
and is a site of r-RNA synthesis.
122. The stage during which separation of the
paired homologous chromosomes begins is
(1) Pachytene (2) Diakinesis
(3) Diplotene (4) Zygotene
Answer (3)
Sol. Synaptonemal complex disintegrates.
Terminalisation begins at diplotene stage i.e.
chiasmata start to shift towards end.
123. The two functional groups characteristic of
sugars are
(1) Hydroxyl and methyl
(2) Carbonyl and phosphate
(3) Carbonyl and methyl
(4) Carbonyl and hydroxyl
Answer (4)
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.
124. Stomata in grass leaf are
(1) Dumb-bell shaped
(2) Rectangular
(3) Kidney shaped
(4) Barrel shaped
Answer (1)
Sol. Grass being a monocot, has Dumb-bell
shaped stomata in their leaves.
125. 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) Hydrilla (2) Banana
(3) Yucca (4) Viola
Answer (3)
Sol. Yucca have an obligate mutualism with a
species of moth i.e. Pronuba.
126. Pollen grains can be stored for several years
in liquid nitrogen having a temperature of
(1) ?120?C (2) ?196?C
(3) ?80?C (4) ?160?C
Answer (2)
Sol. Pollen grains can be stored for several years
in liquid nitrogen at ?196?C (Cryopreservation)
127. Double fertilization is
(1) Fusion of two male gametes of a pollen
tube with two different eggs
(2) Fusion of two male gametes with one egg
(3) Fusion of one male gamete with two polar
nuclei
(4) Syngamy and triple fusion
Answer (4)
Sol. Double fertilization is a unique phenomenon
that occur in angiosperms only.
Syngamy + Triple fusion = Double fertilization
128. Oxygen is not produced during photosynthesis
by
(1) Green sulphur bacteria
(2) Cycas
(3) Nostoc
(4) Chara
Answer (1)
Sol. Green sulphur bacteria do not use H
2
O as
source of proton, therefore they do not evolve
O
2
.
27
NEET (UG) - 2018 (Code-EE) CHLAAA
129. Which of the following elements is responsible
for maintaining turgor in cells?
(1) Magnesium (2) Potassium
(3) Sodium (4) Calcium
Answer (2)
Sol. Potassium helps in maintaining turgidity of
cells.
130. What is the role of NAD
+
in cellular
respiration?
(1) It functions as an enzyme.
(2) It is a nucleotide source for ATP synthesis.
(3) It functions as an electron carrier.
(4) It is the final electron acceptor for
anaerobic respiration.
Answer (3)
Sol. In cellular respiration, NAD
+
act as an electron
carrier.
131. In which of the following forms is iron
absorbed by plants?
(1) Ferric
(2) Free element
(3) Ferrous
(4) Both ferric and ferrous
Answer (1*)
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
++
)
132. Winged pollen grains are present in
(1) Mustard (2) Mango
(3) Cycas (4) Pinus
Answer (4)
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.
133. After karyogamy followed by meiosis, spores
are produced exogenously in
(1) Neurospora
(2) Agaricus
(3) Alternaria
(4) Saccharomyces
Answer (2)
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.
134. Which one is wrongly matched?
(1) Uniflagellate gametes ? Polysiphonia
(2) Gemma cups ? Marchantia
(3) Biflagellate zoospores ? Brown algae
(4) Unicellular organism ? Chlorella
Answer (1)
Sol. ? Polysiphonia is a genus of red algae,
where asexual spores and gametes are
non-motile or non-flagellated.
? Other options (2, 3 & 4) are correctly
matched
135. 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.
28
NEET (UG) - 2018 (Code-EE) CHLAA
a b
(1) (i) (iv)
(2) (ii) (iv)
(3) (iii) (ii)
(4) (iii) (iv)
c d
(iii) (ii)
(iii) (i)
(i) (iv)
(i) (ii)
Answer (4)
Sol. ? Herbarium ? Dried and pressed plant
specimen
? Key

? Identification of various
taxa
? Museum ? Plant and animal
specimen are preserved
? Catalogue ? Alphabetical listing of
species
136. Which of the following is an amino acid
derived hormone?
(1) Epinephrine (2) Estradiol
(3) Ecdysone (4) Estriol
Answer (1)
Sol. Epinephrine is derived from tyrosine amino
acid by the removal of carboxyl group. It is a
catecholamine.
137. Which of the following structures or regions is
incorrectly paired with its functions?
(1) Medulla oblongata : controls respiration
and cardiovascular
reflexes.
(2) Hypothalamus : production of
releasing hormones
and regulation of
temperature,
hunger and thirst.
(3) Limbic system : consists of fibre
tracts that
interconnect
different regions of
brain; controls
movement.
(4) Corpus callosum : band of fibers
connecting left and
right cerebral
hemispheres.
Answer (3)
Sol. Limbic system is emotional brain. It controls
all emotions in our body but not movements.
138. The transparent lens in the human eye is held
in its place by
(1) ligaments attached to the ciliary body
(2) smooth muscles attached to the iris
(3) ligaments attached to the iris
(4) smooth muscles attached to the ciliary
body
Answer (1)
Sol. Lens in the human eye is held in its place by
suspensory ligaments attached to the ciliary
body.
139. Which of the following hormones can play a
significant role in osteoporosis?
(1) Aldosterone and Prolactin
(2) Estrogen and Parathyroid hormone
(3) Progesterone and Aldosterone
(4) Parathyroid hormone and Prolactin
Answer (2)
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.
140. Among the following sets of examples for
divergent evolution, select the incorrect
option :
(1) Forelimbs of man, bat and cheetah
(2) Brain of bat, man and cheetah
(3) Heart of bat, man and cheetah
(4) Eye of octopus, bat and man
Answer (4)
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.
141. Which of the following is not an autoimmune
disease?
(1) Psoriasis
(2) Alzheimer's disease
(3) Rheumatoid arthritis
(4) Vitiligo
Answer (2)
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NEET (UG) - 2018 (Code-EE) CHLAAA
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.
142. 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) b, c and e (2) b, d and e
(3) a, b and c (4) a, c and e
Answer (3)
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)
143. In which disease does mosquito transmitted
pathogen cause chronic inflammation of
lymphatic vessels?
(1) Elephantiasis (2) Ringworm disease
(3) Ascariasis (4) Amoebiasis
Answer (1)
Sol. Elephantiasis is caused by roundworm,
Wuchereria bancrofti and it is transmitted by
Culex mosquito.
144. The similarity of bone structure in the
forelimbs of many vertebrates is an example
of
(1) Homology
(2) Convergent evolution
(3) Analogy
(4) Adaptive radiation
Answer (1)
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.
145. Conversion of milk to curd improves its
nutritional value by increasing the amount of
(1) Vitamin D (2) Vitamin B
12
(3) Vitamin A (4) Vitamin E
Answer (2)
Sol. ? Curd is more nourishing than milk.
? It has enriched presence of vitamins
specially Vit-B
12
.
146. Which one of the following population
interactions is widely used in medical science
for the production of antibiotics?
(1) Commensalism (2) Parasitism
(3) Mutualism (4) Amensalism
Answer (4)
Sol. Amensalism/Antibiosis (0, ?)
? Antibiotics are chemicals secreted by one
microbial group (eg : Penicillium) which
harm other microbes (eg :
Staphylococcus)
? It has no effect on Penicillium or the
organism which produces it.
147. All of the following are included in ? ex-situ
conservation? except
(1) Wildlife safari parks
(2) Botanical gardens
(3) Sacred groves
(4) Seed banks
Answer (3)
Sol. ? Sacred groves ? in-situ conservation.
? Represent pristine forest patch as
protected by Tribal groups.
148. 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) ii i iii iv
(2) iii iv i ii
(3) i iii iv ii
(4) i ii iv iii
Answer (2)
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NEET (UG) - 2018 (Code-EE) CHLAA
Sol. a. Eutrophication iii. Nutrient
enrichment
b. Sanitary landfill iv. Waste disposal
c. Snow blindness i. UV-B radiation
d. Jhum cultivation ii. Deforestation
149. In a growing population of a country,
(1) pre-reproductive individuals are more
than the reproductive individuals.
(2) reproductive and pre-reproductive
individuals are equal in number.
(3) reproductive individuals are less than the
post-reproductive individuals.
(4) pre-reproductive individuals are less than
the reproductive individuals.
Answer (1)
Sol. Whenever the pre-reproductive individuals or
the younger population size is larger than the
reproductive group, the population will be an
increasing population.
150. Which part of poppy plant is used to obtain
the drug ?Smack??
(1) Flowers (2) Roots
(3) Latex (4) Leaves
Answer (3)
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.
151. Hormones secreted by the placenta to
maintain pregnancy are
(1) hCG, hPL, progestogens, prolactin
(2) hCG, hPL, progestogens, estrogens
(3) hCG, hPL, estrogens, relaxin, oxytocin
(4) hCG, progestogens, estrogens,
glucocorticoids
Answer (2)
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.
152. The contraceptive ?SAHELI?
(1) blocks estrogen receptors in the uterus,
preventing eggs from getting implanted.
(2) is an IUD.
(3) increases the concentration of estrogen
and prevents ovulation in females.
(4) is a post-coital contraceptive.
Answer (1)
Sol. Saheli is the first non-steroidal, once a week
pill. It contains centchroman and its
functioning is based upon selective Estrogen
Receptor modulation.
153. The amnion of mammalian embryo is derived
from
(1) ectoderm and mesoderm
(2) mesoderm and trophoblast
(3) endoderm and mesoderm
(4) ectoderm and endoderm
Answer (1)
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.
154. The difference between spermiogenesis and
spermiation is
(1) In spermiogenesis spermatids are formed,
while in spermiation spermatozoa are
formed.
(2) In spermiogenesis spermatozoa from
sertoli cells are released into the cavity of
seminiferous tubules, while in spermiation
spermatozoa are formed.
(3) In spermiogenesis spermatozoa are
formed, while in spermiation spermatids
are formed.
(4) In spermiogenesis spermatozoa are
formed, while in spermiation spermatozoa
are released from sertoli cells into the
cavity of seminiferous tubules.
Answer (4)
31
NEET (UG) - 2018 (Code-EE) CHLAAA
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.
155. Which of the following options correctly
represents the lung conditions in asthma and
emphysema, respectively?
(1) Inflammation of bronchioles; Decreased
respiratory surface
(2) Increased respiratory surface;
Inflammation of bronchioles
(3) Increased number of bronchioles;
Increased respiratory surface
(4) Decreased respiratory surface;
Inflammation of bronchioles
Answer (1)
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.
156. 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
c
ii
iii
ii
iii
Answer (1)
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.
157. 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) iii ii i iv
(2) i iv ii iii
(3) iii i iv ii
(4) iv iii ii i
Answer (3)
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.
158. 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
a b
(1) iii i
(2) i ii
(3) i iii
(4) ii i
32
NEET (UG) - 2018 (Code-EE) CHLAA
c d
iv i
i iv
iii iv
ii iii
a b
(1) iii ii
(2) ii iii
(3) i ii
(4) iv i
Answer (4)
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.
159. 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) iv v ii iii
(2) v iv i ii
(3) iv i ii iii
(4) v iv i iii
Answer (3)
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.
160. Which of the following events does not occur
in rough endoplasmic reticulum?
(1) Protein folding
(2) Cleavage of signal peptide
(3) Protein glycosylation
(4) Phospholipid synthesis
Answer (4)
Sol. Phospholipid synthesis does not take place in
RER. Smooth endoplasmic reticulum are
involved in lipid synthesis.
161. Which of these statements is incorrect?
(1) Enzymes of TCA cycle are present in
mitochondrial matrix
(2) Glycolysis operates as long as it is
supplied with NAD that can pick up
hydrogen atoms
(3) Glycolysis occurs in cytosol
(4) Oxidative phosphorylation takes place in
outer mitochondrial membrane
Answer (4)
Sol. Oxidative phosphorylation takes place in inner
mitochondrial membrane.
162. Nissl bodies are mainly composed of
(1) Proteins and lipids
(2) Nucleic acids and SER
(3) DNA and RNA
(4) Free ribosomes and RER
Answer (4)
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.
163. Which of the following terms describe human
dentition?
(1) Thecodont, Diphyodont, Homodont
(2) Pleurodont, Monophyodont, Homodont
(3) Thecodont, Diphyodont, Heterodont
(4) Pleurodont, Diphyodont, Heterodont
Answer (3)
33
NEET (UG) - 2018 (Code-EE) CHLAAA
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.
164. Select the incorrect match :
(1) Lampbrush ? Diplotene bivalents
chromosomes
(2) Submetacentric ? L-shaped
chromosomes chromosomes
(3) Allosomes ? Sex chromosomes
(4) Polytene ? Oocytes of
chromosomes amphibians
Answer (4)
Sol. Polytene chromosomes are found in salivary
glands of insects of order Diptera.
165. Many ribosomes may associate with a single
mRNA to form multiple copies of a polypeptide
simultaneously. Such strings of ribosomes are
termed as
(1) Polysome
(2) Plastidome
(3) Polyhedral bodies
(4) Nucleosome
Answer (1)
Sol. The phenomenon of association of many
ribosomes with single m-RNA leads to
formation of polyribosomes or polysomes or
ergasomes.
166. According to Hugo de Vries, the mechanism
of evolution is
(1) Multiple step mutations
(2) Phenotypic variations
(3) Saltation
(4) Minor mutations
Answer (3)
Sol. As per mutation theory given by Hugo de
Vries, the evolution is a discontinuous
phenomenon or saltatory phenomenon/
saltation.
167. 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 ii i
(2) ii iii i
(3) i iii ii
(4) iii i ii
Answer (2)
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.
168. All of the following are part of an operon
except
(1) an operator
(2) an enhancer
(3) structural genes
(4) a promoter
Answer (2)
Sol. ? Enhancer sequences are present in
eukaryotes.
? Operon concept is for prokaryotes.
169. AGGTATCGCAT is a sequence from the coding
strand of a gene. What will be the
corresponding sequence of the transcribed
mRNA?
(1) AGGUAUCGCAU
(2) ACCUAUGCGAU
(3) UGGTUTCGCAT
(4) UCCAUAGCGUA
Answer (1)
Sol. Coding strand and mRNA has same nucleotide
sequence except, ?T? ? Thymine is replaced by
?U??Uracil in mRNA.
34
NEET (UG) - 2018 (Code-EE) CHLAA
170. A woman has an X-linked condition on one of
her X chromosomes. This chromosome can be
inherited by
(1) Only daughters
(2) Only grandchildren
(3) Only sons
(4) Both sons and daughters
Answer (4)
Sol. ? Woman is a carrier
? Both son & daughter inherit
X?chromosome
? Although only son be the diseased
171. Which of the following gastric cells indirectly
help in erythropoiesis?
(1) Chief cells (2) Goblet cells
(3) Mucous cells (4) Parietal cells
Answer (4)
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.
172. 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) (iii) (ii) (i)
(2) (i) (iii) (ii)
(3) (i) (ii) (iii)
(4) (ii) (iii) (i)
Answer (4)
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.
173. Which of the following is an occupational
respiratory disorder?
(1) Anthracis (2) Botulism
(3) Silicosis (4) Emphysema
Answer (3)
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.
174. Calcium is important in skeletal muscle
contraction because it
(1) Binds to troponin to remove the masking
of active sites on actin for myosin.
(2) Detaches the myosin head from the actin
filament.
(3) Activates the myosin ATPase by binding to
it.
(4) Prevents the formation of bonds between
the myosin cross bridges and the actin
filament.
Answer (1)
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.
35
NEET (UG) - 2018 (Code-EE) CHLAAA
175. Identify the vertebrate group of animals
characterized by crop and gizzard in its
digestive system
(1) Amphibia (2) Aves
(3) Reptilia (4) Osteichthyes
Answer (2)
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.
176. Ciliates differ from all other protozoans in
(1) using flagella for locomotion
(2) using pseudopodia for capturing prey
(3) having a contractile vacuole for removing
excess water
(4) having two types of nuclei
Answer (4)
Sol. Ciliates differs from other protozoans in
having two types of nuclei.
eg. Paramoecium have two types of nuclei i.e.
macronucleus & micronucleus.
177. Which of the following features is used to
identify a male cockroach from a female
cockroach?
(1) Presence of a boat shaped sternum on the
9
th
abdominal segment
(2) Forewings with darker tegmina
(3) Presence of caudal styles
(4) Presence of anal cerci
Answer (3)
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.
178. Which one of these animals is not a
homeotherm?
(1) Macropus
(2) Camelus
(3) Chelone
(4) Psittacula
Answer (3)
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.
179. Which of the following animals does not
undergo metamorphosis?
(1) Earthworm
(2) Moth
(3) Tunicate
(4) Starfish
Answer (1)
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.
180. Which of the following organisms are known
as chief producers in the oceans?
(1) Dinoflagellates
(2) Cyanobacteria
(3) Diatoms
(4) Euglenoids
Answer (3)
Sol. Diatoms are chief producers of the ocean.
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