Sample paper 1
Question 1
What is the dimensional formula of torque?
A. MLT -2
B. MT -2
D. MLT -1
E. ML3T -2
Correct Answer: C
Explanation:
t = r x F
Where
t is the torque vector;
r is the displacement vector;
Since dimensional formula of force is MLT-2 and that of displacement is L hence, dimensional formula
of torque becomes
MLT-2* L
ML2T-2
Question 2
A car moving at 25 m/s suddenly stops by applying brakes on observing a stoplight. If the time
taken during the process of stopping is 1.5 s, what is the displacement of the car?
A. 17.55 m
C. 16.25 m
D. 15.35 m
E. 17.75 m
Correct Answer: B
This problem can be solved by applying the kinematic equation of motion as shown below
v = u + at ------------ (1)
Where
v is the final velocity of the car;
a is the acceleration the car is undergoing;
t is the time under consideration = 1.5 s;
Since the car finally stops, hence v = 0 m/s. Putting the values in equation 1 we get
0 = 25 + a *1.5
Negative sign is coming because the car finally comes to rest and it is undergoing negative
acceleration. We can neglect the negative sign here. Therefore, equation 2 becomes
a = 25/1.5
Since, we know initial and final velocities and acceleration as well we can use kinematic equation
v2 = u2 - 2as -------------- (3)
to solve for displacement.
Where
that acceleration is negative. Putting the values of v, u, a in equation 3 we get
0 = 252 - 2 ? 16.67 ? S
Or, 252 = 2 ? 16.67 ? S
Or, S = 252/(2*16.67)
Or, S = 18.75m
Hence, B is the correct answer option.
Question 3
Net force acting on a body moving with constant velocity is zero. Which of the following is
A. Body will slow down
B. Body will move faster
C. Body will stop
D. Body will continue moving with same velocity
Correct Answer: D
Explanation:
A body moving with constant velocity or at rest is the condition of static equilibrium. For static
equilibrium, we need balanced force that means all the forces acting on the body balance each other
amongst the first four the options given above. Body cannot accelerate, as acceleration is time rate of
change of velocity. If velocity is remaining constant, the acceleration will become zero. Therefore, E is
an incorrect option.
Question 4
frictionless table as shown below. What would be acceleration of the smaller mass? (Assume
smaller mass is at rest with respect to the larger mass)
A. F/4M
B. F/M
C. F/5M
D. F/3M
E. F/2M
Correct Answer: C
Since the table is frictionless, hence both the bodies will continue moving under the common force F
and will have same acceleration. Since it is brought out in the question that the smaller block is not
moving over the larger block hence, both the blocks will move as a combined body.
The situation above is shown below.
M total = M + 4M
Or, M total = 5M
Common force acting on the bodies is F. Therefore, by Newton's second law of motion we have
F = ma ------------- (1)
F is the force applied;
m is the mass of the body;
a is the acceleration of both the blocks;
Mass of the body here is Mtotal = 5M. Putting the values in equation 1 we get
Or, F/5M = a
Hence, with this acceleration both the smaller and the larger mass will move under the effect of
common force, F.
Therefore, C is the correct answer option.
What would be the moment of inertia of a thin-walled fluorescent lamp having radius of 3 cm
and mass of 50 grams?
A. 4.5 ? 10-5 kgm2
B. 3.5 ? 10-5 kgm2
D. 4.5 ? 10-6 kgm2
E. 4 ? 106 kgm2
Correct Answer: A
Explanation:
I = MR2 ---------------- (1)
Where
I is the moment of inertia;
R is the radius of the lamp = 3 cm;
We have been given mass in grams so, we need to convert it into kg.
1 g = 10-3kg
50g = 50 * 10-3kg
Also, radius is given in cm so we need to convert it into m.
1 cm = 10-2m
3 cm = 3 * 10-2m
3 cm = 0.03 m
I = 0.05 * 0.032
Or, I = 4.5 * 10 -5 kg m2
Therefore, A is the correct answer option.
Consider the figure given below. B is the nodal plane or ground. A body of mass M is kept on a
horizontal plane A at a certain height above the plane B. Plane C is at same height from the
ground (B) as that of plane A. Which of the following statements is correct?
Potential energy of the body is non-zero with respect to the plane A
Potential energy of the body is non-zero with respect to the plane B
A. 1
B. 2
C. 3
E. Both 2 and 3
Correct Answer: C
Explanation:
Potential energy is the energy stored in a body by virtue of its height from the reference plane.
PE = mgh --------------- (1)
Where
m is the mass of the body;
g is the acceleration due to gravity;
The reference plane plays a vital role in defining the potential energy of a body. Most often reference
plane is taken as the ground and hence plane B is the reference plane here. We will discuss potential
energy of the body for all the planes individually.
For plane A, the height of the body is zero as it is kept on the same plane A. Hence, putting the value
PE = 0
Therefore, potential energy of the body is zero with respect to the plane A. Thus, A is an incorrect
option.
Both the planes A and C are at same height above the reference plane. Therefore, potential energy of
Since the plane A is at some height from the reference plane, h will have a non-zero value. Therefore,
potential energy of the body with respect to the ground plane will have some non-zero value. Thus, C
is the correct answer option. In the view of the above, D and E are also incorrect options.
Question 7
x(t) = 23 sin(314t) cm
What is the amplitude and frequency of oscillation of the particle?
A. 2.6 cm, 40 Hz
B. 2.3 cm, 40 Hz
D. 2.6 cm, 50 Hz
E. 3.14 cm, 314 Hz
Correct Answer: C
Explanation:
X(t) = A sin(t) ----------- (1)
Where
X(t) is the displacement of the particle;
A is the amplitude of the oscillation;
Displacement of a particle as given in the question is
X(t) = 23 sin(314t) cm ------------- (2)
Comparing equation 1 with 2 we get
A = 2.3 cm and = 314 rad/s
Where
f is the frequency of the oscillation;
Putting the value of in equation 3 we get
314 = 2f
Or, 314/(2*3.14) = f
( = 3.14)
Or, f = 50 Hz
Hence, C is the correct answer option.
Which of the following do not require any medium to propagate?
A. Sound waves
B. Seismic waves
C. Tides
E. Waves on the surface of water
Correct Answer: D
Explanation:
Sound waves can only propagate in the presence of a medium like solid, liquid or gas. Sound waves
generated due to the fault in tectonic plates of the Earth. Seismic waves originate from a point and
travel through the Earth's layers. Hence, without the presence of Earth's layers seismic waves cannot
propagate. Hence, B is an incorrect option. Tides and waves on the surface of water originate at a
point of disturbance and progress along all directions. Without any medium surrounding the point
tides and surface waves on water is essentially liquid. Therefore, C and E are incorrect options as
well. Light waves do not require any medium to propagate. For example, light waves coming from the
Sun travel most of the distance through the outer space that is devoid of any medium. Therefore, D is
the correct answer option
A cylinder of height 2.5 m is filled completely with water. A hole is made at the bottom of the
cylinder in such a way that water is coming out of it. What is the velocity of water coming out
of the cylinder?
A. 6.4 m/s
C. 9.8 m/s
D. 2.5 m/s
E. 7 m/s
Correct Answer: E
Velocity of water coming out of a cylinder is given as
v2 = 2gh ----------------- (1)
Where
v is the velocity of the water coming out of the cylinder;
g is the acceleration due to gravity = 9.8 m/s2;
Putting the values of h and g in equation 1 we get
v2 = 2 * 9.8 * 2.5
Or, v2 = 49
Hence, E is the correct answer option.
Question 10
An object A of mass 40 kg at 600 C has internal energy of 4000 J. If it is kept in contact with
another object B of mass 40 kg at 650 C having internal energy 4500 J, then which of the
heat transfer to take place)
A. Heat will flow from A to B
B. Heat will flow from B to A
C. No heat will flow between the objects A and B
E. Temperature of B increases
Correct Answer: B
Heat always flows from a hot body to a cooler body. In other words, heat flows from a body at a
higher temperature to a body at a lower temperature. If an object A, at 600 C is kept in contact with
Hence, heat will flow from B to A. Also, as the heat is transferred from B to A, temperature of B will
reduce but temperature of A will increase to reach thermal equilibrium. Thus, B is the correct answer
option and A, C, D and E are incorrect answer options.
Question 11
A. 6 ? 1018
B. 1018
C. 2 ? 1018
D. 2 ? 1019
Correct Answer: A
Explanation:
Charge is given as
Q = ne --------------- (1)
Q is one coulomb of charge;
n is the number of electronic charges;
e is the charge on electron = 1.6 * 10 -19 C;
Putting the values in equation 1 we get
Or, n = 6.25 * 1018
Since n is a number therefore number of electronic charges that make one coulomb of charge is 6 *
1018 . Therefore, A is the correct answer option
Question 12
Three batteries each of 25 V having zero internal resistance are connected in series with a
A. 4 A
B. 0.75 A
C. 6.5 A
D. 0.33 A
Correct Answer: B
Explanation:
Since the batteries are connected in series, equivalent potential of batteries can be found by adding
the individual potential. Let equivalent potential be V eq. Therefore, Veq = 25 + 25 + 25
Or, Veq = 75V
Since internal resistance of the batteries is zero, only resistance to which the equivalent potential is
connected is 100 . Hence, current flowing through the circuit can be found by Ohm's law. Ohm's law
is given as
Where
V is the potential of the cell = 75 V;
I is the current flowing through the circuit;
R is the equivalent resistance of the circuit = 100
75 = I * 100
Or, I = 75/100
Or, I = 0.75 A
Therefore, B is the correct answer option.
Keeping the current same in a circular wire, radius is getting doubled. What will the new value
of magnetic moment be, if the initial magnetic moment is 3 ? 10-5 Am2?
A. 3 ? 10-5 Am2
B. 0.75 ? 10-5 Am2
D. 3.2 ? 10-4 Am2
E. 1.2 ? 10-4 Am2
Correct Answer: E
Explanation:
M = IA ------------ (1)
Where
M is the magnetic moment;
I is the current flowing in the wire;
Since the wire is of circular shape hence, area of the circular loop is given as
A = r2 -------------- (2)
Where
r is the radius of the wire;
A1 and r1 we get
A
2
1 = r1 ----------- (3)
terms of A2and r2 we get
A
2
2 = r2 ------------ (4)
Using the value of r2 in equation 4 we get
A
2
2 = ( 2r1)2 Or, A2 = 4r1 ------------- (5)
given as
A
2
2 = 4r1
Expressing initial and final magnetic moment in equation 1 in terms of initial area and final area we
get
M1 = IA1 ---------------- (7)
And, M2 = IA2 ---------- (8)
M2/ M1 = (IA2)/(IA1) ---------------- (9)
Cancelling out the common term, in 9 we get
M2/ M1 = A2/A1 --------------- (10)
Using equation 6 in 10 we get
Or, M2/ M1 = 4
Therefore, M2 = 4 M1 -------------- (11)
Putting the value of M1 in equation 11 we get M2 = 4 * 3 * 10-5 Or, M2 = 1.2 * 10 -4 Am2
Hence, E is the correct answer option.
Question 14
What are the values of 1 and 2 for the figure if a ray of light is incident at from air to a
medium of refractive index 1.5?
A. sin-1(1/3.7)
C. tan-1(1/3.7)
D. cot-1(1/3)
E. sin-1(1/3)
Correct Answer: E
In the figure given below, the brown arrow represents the incident ray, the green arrow represents the
transmitted or the reflected ray, and the orange arrow represents the refracted ray. OA represents the
normal to the interface between the air and the medium.
Since the incident light is travelling from one medium to another, some part of it will be reflected and
angle of reflection, the green arrow, which is the reflected light will make the same angle with the
normal as made by the incident beam with the normal. Therefore,
1 = 30o
2 can be found by applying Snel 's law. According to Snel 's law we have
?1 sin i = ?2 + sin r ---------------- (1)
Where
?1 is the refractive index of medium 1 (here air = 1);
i is the angle of incidence = 30o ;
?2 is the refractive index of medium 2 = 1.5;
Putting these values in equation 1 we get
Or, 1 * sin 30o = 1.5 * sin r
Or, 1/2 = 1.5 * sin r
Or, 1/(2*1.5) = sin r
Or, r = sin-1 (1/3)
As r is same as 2
. The angle of refraction is sin-1 (1/3).
Hence, E is the correct answer option.
Which of the following gets deflected by electric field?
A. Gamma rays
B. Infrared rays
C. Beta rays
E. X-ray
Correct Answer: C
Explanation:
Electric field can be negative or positive in nature. If the electric field is created by negative charges
field is positive. Depending on the type of the charged particle passing through the region of electric
field, deflection can be either attractive or repulsive. Since gamma rays have no charge they will move
further into the electric field without being deflected. Hence, A is an incorrect option. Infrared rays
consist of uncharged particles and therefore, there would not be any deflection from the normal path.
charge as that on electron. Hence, when beta rays pass through an electric field they will either get
attracted or repelled depending on the nature of the electric field. Thus, C is the correct answer
option. Neutrons are neutral like gamma rays or infrared rays. Hence, they will not show any
deflection while passing through the electric field. Therefore, D is an incorrect option. X-rays consists
deflection. Thus, E is an incorrect option as well.
Question 1
What is the dimensional formula of torque?
A. MLT -2
B. MT -2
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C. ML2T-2D. MLT -1
E. ML3T -2
Correct Answer: C
Explanation:
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Torque is the turning effect of force applied on a body. It is given ast = r x F
Where
t is the torque vector;
r is the displacement vector;
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F is the force vector;Since dimensional formula of force is MLT-2 and that of displacement is L hence, dimensional formula
of torque becomes
MLT-2* L
ML2T-2
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Therefore, C is the correct answer option.Question 2
A car moving at 25 m/s suddenly stops by applying brakes on observing a stoplight. If the time
taken during the process of stopping is 1.5 s, what is the displacement of the car?
A. 17.55 m
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B. 18.75 mC. 16.25 m
D. 15.35 m
E. 17.75 m
Correct Answer: B
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Explanation:This problem can be solved by applying the kinematic equation of motion as shown below
v = u + at ------------ (1)
Where
v is the final velocity of the car;
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u is the initial velocity of the car = 25 m/s;a is the acceleration the car is undergoing;
t is the time under consideration = 1.5 s;
Since the car finally stops, hence v = 0 m/s. Putting the values in equation 1 we get
0 = 25 + a *1.5
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Or, -25 = 1.5a --------------- (2)Negative sign is coming because the car finally comes to rest and it is undergoing negative
acceleration. We can neglect the negative sign here. Therefore, equation 2 becomes
a = 25/1.5
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Or, a = 16.67 m/s2Since, we know initial and final velocities and acceleration as well we can use kinematic equation
v2 = u2 - 2as -------------- (3)
to solve for displacement.
Where
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S is the displacement of the car; We have taken the negative sign for acceleration here due to the factthat acceleration is negative. Putting the values of v, u, a in equation 3 we get
0 = 252 - 2 ? 16.67 ? S
Or, 252 = 2 ? 16.67 ? S
Or, S = 252/(2*16.67)
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Or, S = 625/33.34Or, S = 18.75m
Hence, B is the correct answer option.
Question 3
Net force acting on a body moving with constant velocity is zero. Which of the following is
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incorrect?A. Body will slow down
B. Body will move faster
C. Body will stop
D. Body will continue moving with same velocity
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E. Body will accelerateCorrect Answer: D
Explanation:
A body moving with constant velocity or at rest is the condition of static equilibrium. For static
equilibrium, we need balanced force that means all the forces acting on the body balance each other
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and in this way the net force acting on the body becomes zero. Therefore, only option D is correctamongst the first four the options given above. Body cannot accelerate, as acceleration is time rate of
change of velocity. If velocity is remaining constant, the acceleration will become zero. Therefore, E is
an incorrect option.
Question 4
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Two blocks of masses M and 4 M are moving under the effect of the same force, F on africtionless table as shown below. What would be acceleration of the smaller mass? (Assume
smaller mass is at rest with respect to the larger mass)
A. F/4M
B. F/M
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C. F/5M
D. F/3M
E. F/2M
Correct Answer: C
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Explanation:Since the table is frictionless, hence both the bodies will continue moving under the common force F
and will have same acceleration. Since it is brought out in the question that the smaller block is not
moving over the larger block hence, both the blocks will move as a combined body.
The situation above is shown below.
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As both the bodies are moving together, hence, their combined mass becomesM total = M + 4M
Or, M total = 5M
Common force acting on the bodies is F. Therefore, by Newton's second law of motion we have
F = ma ------------- (1)
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WhereF is the force applied;
m is the mass of the body;
a is the acceleration of both the blocks;
Mass of the body here is Mtotal = 5M. Putting the values in equation 1 we get
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F = 5M * aOr, F/5M = a
Hence, with this acceleration both the smaller and the larger mass will move under the effect of
common force, F.
Therefore, C is the correct answer option.
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Question 5What would be the moment of inertia of a thin-walled fluorescent lamp having radius of 3 cm
and mass of 50 grams?
A. 4.5 ? 10-5 kgm2
B. 3.5 ? 10-5 kgm2
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C. 3.5 ? 10-6 kgm2D. 4.5 ? 10-6 kgm2
E. 4 ? 106 kgm2
Correct Answer: A
Explanation:
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Moment of inertia of thin-walled fluorescent lamp is given asI = MR2 ---------------- (1)
Where
I is the moment of inertia;
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M is the mass of the lamp = 50 g;R is the radius of the lamp = 3 cm;
We have been given mass in grams so, we need to convert it into kg.
1 g = 10-3kg
50g = 50 * 10-3kg
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50g = 0.05 kgAlso, radius is given in cm so we need to convert it into m.
1 cm = 10-2m
3 cm = 3 * 10-2m
3 cm = 0.03 m
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Putting the values of M and R in equation 1 we getI = 0.05 * 0.032
Or, I = 4.5 * 10 -5 kg m2
Therefore, A is the correct answer option.
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Question 6Consider the figure given below. B is the nodal plane or ground. A body of mass M is kept on a
horizontal plane A at a certain height above the plane B. Plane C is at same height from the
ground (B) as that of plane A. Which of the following statements is correct?
Potential energy of the body is non-zero with respect to the plane A
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Potential energy of the body is non-zero with respect to the plane CPotential energy of the body is non-zero with respect to the plane B
A. 1
B. 2
C. 3
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D. Both 1 and 2E. Both 2 and 3
Correct Answer: C
Explanation:
Potential energy is the energy stored in a body by virtue of its height from the reference plane.
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Potential energy is given asPE = mgh --------------- (1)
Where
m is the mass of the body;
g is the acceleration due to gravity;
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h is the height of the body above the reference plane;The reference plane plays a vital role in defining the potential energy of a body. Most often reference
plane is taken as the ground and hence plane B is the reference plane here. We will discuss potential
energy of the body for all the planes individually.
For plane A, the height of the body is zero as it is kept on the same plane A. Hence, putting the value
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of h in equation 1 we getPE = 0
Therefore, potential energy of the body is zero with respect to the plane A. Thus, A is an incorrect
option.
Both the planes A and C are at same height above the reference plane. Therefore, potential energy of
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the body with respect to the plane C is same as that in respect of A. Hence, B is incorrect as well.Since the plane A is at some height from the reference plane, h will have a non-zero value. Therefore,
potential energy of the body with respect to the ground plane will have some non-zero value. Thus, C
is the correct answer option. In the view of the above, D and E are also incorrect options.
Question 7
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Displacement of a particle in SHM is given asx(t) = 23 sin(314t) cm
What is the amplitude and frequency of oscillation of the particle?
A. 2.6 cm, 40 Hz
B. 2.3 cm, 40 Hz
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C. 2.3 cm, 50 HzD. 2.6 cm, 50 Hz
E. 3.14 cm, 314 Hz
Correct Answer: C
Explanation:
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General expression for displacement of a body in SHM is given asX(t) = A sin(t) ----------- (1)
Where
X(t) is the displacement of the particle;
A is the amplitude of the oscillation;
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is the angular frequency;Displacement of a particle as given in the question is
X(t) = 23 sin(314t) cm ------------- (2)
Comparing equation 1 with 2 we get
A = 2.3 cm and = 314 rad/s
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Angular frequency is related to frequency as = 2f ----------- (3)Where
f is the frequency of the oscillation;
Putting the value of in equation 3 we get
314 = 2f
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Or, 314/(2*) = fOr, 314/(2*3.14) = f
( = 3.14)
Or, f = 50 Hz
Hence, C is the correct answer option.
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Question 8Which of the following do not require any medium to propagate?
A. Sound waves
B. Seismic waves
C. Tides
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D. Light wavesE. Waves on the surface of water
Correct Answer: D
Explanation:
Sound waves can only propagate in the presence of a medium like solid, liquid or gas. Sound waves
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cannot propagate in the absence of a medium. Therefore, A is an incorrect option. Seismic waves aregenerated due to the fault in tectonic plates of the Earth. Seismic waves originate from a point and
travel through the Earth's layers. Hence, without the presence of Earth's layers seismic waves cannot
propagate. Hence, B is an incorrect option. Tides and waves on the surface of water originate at a
point of disturbance and progress along all directions. Without any medium surrounding the point
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where the tides or surface waves on water originate, the waves will not propagate. The medium fortides and surface waves on water is essentially liquid. Therefore, C and E are incorrect options as
well. Light waves do not require any medium to propagate. For example, light waves coming from the
Sun travel most of the distance through the outer space that is devoid of any medium. Therefore, D is
the correct answer option
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Question 9A cylinder of height 2.5 m is filled completely with water. A hole is made at the bottom of the
cylinder in such a way that water is coming out of it. What is the velocity of water coming out
of the cylinder?
A. 6.4 m/s
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B. 4 m/sC. 9.8 m/s
D. 2.5 m/s
E. 7 m/s
Correct Answer: E
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Explanation:Velocity of water coming out of a cylinder is given as
v2 = 2gh ----------------- (1)
Where
v is the velocity of the water coming out of the cylinder;
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h is the height to which water is filled= 2.5 m;g is the acceleration due to gravity = 9.8 m/s2;
Putting the values of h and g in equation 1 we get
v2 = 2 * 9.8 * 2.5
Or, v2 = 49
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Or, v = 7 m/sHence, E is the correct answer option.
Question 10
An object A of mass 40 kg at 600 C has internal energy of 4000 J. If it is kept in contact with
another object B of mass 40 kg at 650 C having internal energy 4500 J, then which of the
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following statements is true regarding heat transfer? (Assume boundary of contact can allowheat transfer to take place)
A. Heat will flow from A to B
B. Heat will flow from B to A
C. No heat will flow between the objects A and B
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D. Temperature of A decreasesE. Temperature of B increases
Correct Answer: B
Heat always flows from a hot body to a cooler body. In other words, heat flows from a body at a
higher temperature to a body at a lower temperature. If an object A, at 600 C is kept in contact with
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another object B, at 650 C, then this means that object A is at lower temperature than that of object B.Hence, heat will flow from B to A. Also, as the heat is transferred from B to A, temperature of B will
reduce but temperature of A will increase to reach thermal equilibrium. Thus, B is the correct answer
option and A, C, D and E are incorrect answer options.
Question 11
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How many electronic charges comprise a coulomb of charge?A. 6 ? 1018
B. 1018
C. 2 ? 1018
D. 2 ? 1019
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E. 5 ? 1015Correct Answer: A
Explanation:
Charge is given as
Q = ne --------------- (1)
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WhereQ is one coulomb of charge;
n is the number of electronic charges;
e is the charge on electron = 1.6 * 10 -19 C;
Putting the values in equation 1 we get
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1 = n * 1.6 * 10 -19Or, n = 6.25 * 1018
Since n is a number therefore number of electronic charges that make one coulomb of charge is 6 *
1018 . Therefore, A is the correct answer option
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Question 12
Three batteries each of 25 V having zero internal resistance are connected in series with a
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resistance of 100 . What would be the current in the circuit?A. 4 A
B. 0.75 A
C. 6.5 A
D. 0.33 A
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E. 0.25 ACorrect Answer: B
Explanation:
Since the batteries are connected in series, equivalent potential of batteries can be found by adding
the individual potential. Let equivalent potential be V eq. Therefore, Veq = 25 + 25 + 25
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Or, Veq = 3 * 25 VOr, Veq = 75V
Since internal resistance of the batteries is zero, only resistance to which the equivalent potential is
connected is 100 . Hence, current flowing through the circuit can be found by Ohm's law. Ohm's law
is given as
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V = IR -------------- (1)Where
V is the potential of the cell = 75 V;
I is the current flowing through the circuit;
R is the equivalent resistance of the circuit = 100
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Putting the values of V and R in equation 1 we get75 = I * 100
Or, I = 75/100
Or, I = 0.75 A
Therefore, B is the correct answer option.
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Question 13Keeping the current same in a circular wire, radius is getting doubled. What will the new value
of magnetic moment be, if the initial magnetic moment is 3 ? 10-5 Am2?
A. 3 ? 10-5 Am2
B. 0.75 ? 10-5 Am2
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C. 4 ? 10-5 Am2D. 3.2 ? 10-4 Am2
E. 1.2 ? 10-4 Am2
Correct Answer: E
Explanation:
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Magnetic moment is given asM = IA ------------ (1)
Where
M is the magnetic moment;
I is the current flowing in the wire;
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A is the area of the circular loop;Since the wire is of circular shape hence, area of the circular loop is given as
A = r2 -------------- (2)
Where
r is the radius of the wire;
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Let A1 and r1 be the area and radius of the initial wire. Therefore, expressing equation 2 in terms ofA1 and r1 we get
A
2
1 = r1 ----------- (3)
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Similarly, let A2 and r2 be the area and radius of the final wire. Therefore, expressing equation 2 interms of A2and r2 we get
A
2
2 = r2 ------------ (4)
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Since it is given in the question radius of the final wire is twice that of the initial wire hence, r2 = 2r1Using the value of r2 in equation 4 we get
A
2
2 = ( 2r1)2 Or, A2 = 4r1 ------------- (5)
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Comparing equation 5 with 3 we find that final area can be expressed in terms of initial area and isgiven as
A
2
2 = 4r1
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A2 = 4A1 ------------------ (6)Expressing initial and final magnetic moment in equation 1 in terms of initial area and final area we
get
M1 = IA1 ---------------- (7)
And, M2 = IA2 ---------- (8)
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Both initial and final current in the wire is same. Dividing equation 8 by equation 7 we getM2/ M1 = (IA2)/(IA1) ---------------- (9)
Cancelling out the common term, in 9 we get
M2/ M1 = A2/A1 --------------- (10)
Using equation 6 in 10 we get
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M2/ M1 = 4A2/A1Or, M2/ M1 = 4
Therefore, M2 = 4 M1 -------------- (11)
Putting the value of M1 in equation 11 we get M2 = 4 * 3 * 10-5 Or, M2 = 1.2 * 10 -4 Am2
Hence, E is the correct answer option.
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Question 14
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Consider the figure given below.What are the values of 1 and 2 for the figure if a ray of light is incident at from air to a
medium of refractive index 1.5?
A. sin-1(1/3.7)
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B. cos-1(1/3)C. tan-1(1/3.7)
D. cot-1(1/3)
E. sin-1(1/3)
Correct Answer: E
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Explanation:In the figure given below, the brown arrow represents the incident ray, the green arrow represents the
transmitted or the reflected ray, and the orange arrow represents the refracted ray. OA represents the
normal to the interface between the air and the medium.
Since the incident light is travelling from one medium to another, some part of it will be reflected and
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some will be refracted. As we know from the property of reflection that, angle of incidence is same asangle of reflection, the green arrow, which is the reflected light will make the same angle with the
normal as made by the incident beam with the normal. Therefore,
1 = 30o
2 can be found by applying Snel 's law. According to Snel 's law we have
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?1 sin i = ?2 + sin r ---------------- (1)
Where
?1 is the refractive index of medium 1 (here air = 1);
i is the angle of incidence = 30o ;
?2 is the refractive index of medium 2 = 1.5;
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r is the angle of refraction = 2 ;Putting these values in equation 1 we get
Or, 1 * sin 30o = 1.5 * sin r
Or, 1/2 = 1.5 * sin r
Or, 1/(2*1.5) = sin r
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Or, 1/3 = sin rOr, r = sin-1 (1/3)
As r is same as 2
. The angle of refraction is sin-1 (1/3).
Hence, E is the correct answer option.
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Question 15Which of the following gets deflected by electric field?
A. Gamma rays
B. Infrared rays
C. Beta rays
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D. NeutronsE. X-ray
Correct Answer: C
Explanation:
Electric field can be negative or positive in nature. If the electric field is created by negative charges
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then the electric field is negative and if the electric field is created by positive charges then the electricfield is positive. Depending on the type of the charged particle passing through the region of electric
field, deflection can be either attractive or repulsive. Since gamma rays have no charge they will move
further into the electric field without being deflected. Hence, A is an incorrect option. Infrared rays
consist of uncharged particles and therefore, there would not be any deflection from the normal path.
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Thus, B is an incorrect option. Beta rays are equivalent to electrons i.e. beta rays have the samecharge as that on electron. Hence, when beta rays pass through an electric field they will either get
attracted or repelled depending on the nature of the electric field. Thus, C is the correct answer
option. Neutrons are neutral like gamma rays or infrared rays. Hence, they will not show any
deflection while passing through the electric field. Therefore, D is an incorrect option. X-rays consists
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of neutral particles. Due to this, when they pass through a region of electric field, they undergo nodeflection. Thus, E is an incorrect option as well.