JB AND KARP VIDYA SANKUL

Subject : Physics Paper Set : 1

Standard : 11 H Date : 25-11-2024
Total Mark : 200 Time : 0H:0M

(Assume uniform angular acceleration)

.............. Physics - Section A (MCQ) ..............
(A) 10 (B) 20
(1) A block of mass 5 kg is on a rough horizontal surface and is (C) 30 (D) 40
at rest. Now a force of 24 N is imparted to it with negligible (11) A rocket with a lift-off mass 2 × 104 kg is blasted upwards
impulse. If the coefficient of kinetic friction is 0.4 and with an intial acceleration of 5 m s−2 . The initial thrust of
g = 9.8 m/s2 , then the acceleration of the block is ........ m/s2 the blast is (Take g = 10 m s−2 )
(A) 0.26 (B) 0.39 (A) 2 × 105 N (B) 3 × 105 N
(C) 0.69 (D) 0.88 (C) 4 × 105 N (D) 5 × 105 N
(2) Three forces F⃗1 = (2î + 4ĵ) N ; F⃗2 = (2ĵ − k̂) N and (12) A body of mass m = 3.513 kg is moving along the x - axis
F⃗3 = (k̂ − 4î − 2ĵ) N are applied on an object of mass 1kg with a speed of 5.00 ms−1 . The magnitude of its momentum
at rest at origin. The position of the object at t = 2s will be is recorded as (Consider significant figures)
(A) (−2 m, −6 m) (B) (−4 m, 8 m) (A) 17.57 (B) 17.6

(C) (3 m, 6 m) (D) (2 m, −3 m) (C) 17.565 (D) 17.56

(13) Two blocks of masses 5 kg and 2 kg are connected by a
(3) The dimensions of K in the equation W = 1
Kx2 is
2 spring of negilible mass and placed on a frictionless
(A) M 1 L0 T −2 (B) M 0 L1 T −1
horizontal surface. An impulse gives a velocity of 7 m/s to
(C) M 1 L1 T −2 (D) M 1 L0 T −1 the heavier block in the direction of the lighter block. The
velocity of the centre of mass is ....... m/s
(4) A particle of mass m is tied to string of length l and whirled
in vertical circle. The difference of tension and kinetic (A) 30 (B) 20
energy at highest and lowest position of circular path will be (C) 10 (D) 5
(A) 2 mg, 5 mgl (B) 5 mg, 3 mgl (14) A body of mass m is moving in a circle of radius r with a
constant speed v. The force on the body is mvr and is
2
(C) 6 mg, 2 mgl (D) 3 mg, 5 mgl
directed towards the centre. What is the work done by this
(5) Vector product of two vectors 2î + ĵ and î + 2ĵ is force in moving the body over half the circumference of the
(A) k̂ + ĵ (B) î + ĵ circle
(A) mv (B) Zero
2

(C) 3k̂ (D) 2î πr 2

(6) The displacement of a particle starting from rest (at t = 0) is (C) mv 2

(D) mv
2
πr
r2 2

given by s = 6t2 − t3 . The time in seconds at which the  

particle will attain zero velocity again, is ..... sec (15) A particle moves from the point 2.0î + 4.0ĵ m, at t = 0
 
(A) 2 (B) 4 with an initial velocity 5.0î + 4.0ĵ ms−1 . It is acted upon
(C) 6 (D) 8 by
 a constant  force which produces a constant acceleration
4.0î + 4.0ĵ ms−2 . What is the distance of the particle
(7) 0.4î + 0.8ĵ + ck̂ represents a unit vector when c is
√ from the origin at time 2 s
(A) −0.2 (B) 0.2 √
√ (A) 15 m (B) 20 2 m
(C) 0.8 (D) 0 √
(C) 5 m (D) 10 2 m
(8) A hollow sphere of mass 1 kg and radius 10 cm is free to
(16) A circular plate of uniform thickness has diameter 56 cm. A
rotate about its diameter. If a force of 30 N is applied
circular part of diameter 42 cm is removed from one edge.
tangentially to it, its angular acceleration is (in rad/s2 )
What is the position of the centre of mass of the remaining
(A) 5000 (B) 450 part ........ cm.
(C) 50 (D) 5 (A) 3 (B) 6
(9) An elevator weighing 6000 kg is pulled upward by a cable (C) 9 (D) 12
with an acceleration of 5 ms−2 . Taking g to be 10 ms−2 , (17) What is the unit vector perpendicular to the following
then the tension in the cable is ........... N vectors 2î + 2ĵ − k̂ and 6î − 3ĵ + 2k̂
(A) 6000 (B) 9000
(A) î+10√
ĵ−18k̂
(B) î−10√
ĵ+18k̂

(C) 60000 (D) 90000

5 17 5 17

(10) When a ceiling fan is switched on, it makes 10 rotations in (C) î−10√
ĵ−18k̂
5 17
(D) î+10√
ĵ+18k̂
5 17
the first (18) A man is pulling on a rope attached to a block on a smooth
3 sec . How many rotations will it make in the next 3 sec horizontal table. The tension in the rope will be the same at

1
 all points arrangement
√ willbe located at  
√
(A) If and only if the rope is not accelerated (A) 2 a, 0.95a
3
(B) 0.95a, 3
4 a

(B) If and only if the rope is massless

(C) 3a a
4 , 2 (D) a2 , 3a
4
(C) If either the rope is not accelerated or is massless (27) Five masses are placed in a plane as shown in figure. The
(D) Always coordinates of the centre of mass are nearest to.

(19) An engine pump is used to pump a liquid of density ρ

continuously through a pipe of cross-sectional area A. If the
speed of flow of the liquid in the pipe is v, then the rate at
which kinetic energy is being imparted to the liquid is
(A) 12 Aρv 3 (B) 12 Aρv 2
(C) 1
2 Aρv (D) Aρv
(20) A space craft of mass M is moving with velocity V and
suddenly explodes into two pieces. A part of it of mass m
becomes at rest, then the velocity of other part will be
(A) MM−m
V
(B) MM+m
V

(C) mV
M −m (D) (M +m)V
m
(A) 1.2, 1.4 (B) 1.3, 1.1

(21) The vector that must be added to the vector î − 3ĵ + 2k̂ and (C) 1.1, 1.3 (D) 1.0, 1.0
3î + 6ĵ − 7k̂ so that the resultant vector is a unit vector (28) Dimensions of µ0 ε0 , where symbols have their usual
1
along the y−axis is meaning, are
(A) 4î + 2ĵ + 5k̂ (B) −4î − 2ĵ + 5k̂ (A) [LT −1 ] (B) [L−1 T ]
(C) 3î + 4ĵ + 5k̂ (D) Null vector (C) [L−2 T 2 ] (D) [L2 T −2 ]
(22) A 20 kg block is initially at rest on a rough horizontal surface. (29) A block rests on a rough inclined plane making an angle of
A horizontal force of 75 N is required to set the block in 30o with the horizontal. The coefficient of static friction
motion. After it is in motion, a horizontal force of 60 N is between the block and the plane is 0.8. If the frictional force
required to keep the block moving with constant speed. The on the block is 10 N , the mass of the block (in kg) is (take
coefficient of static friction is g = 10 m/s2 )
(A) 0.38 (B) 0.44
(A) 2 (B) 4
(C) 0.52 (D) 0.6
(C) 1.6 (D) 2.5
(23) Due to a force of (6î + 2ĵ)N the displacement of a body is (30) Which of the following statements is true
(3î − ĵ)m, then the work done is.....J
(A) In elastic collisions, the momentum is conserved but not
(A) 16 (B) 12 in inelastic collisions
(C) 8 (D) 0 (B) Both kinetic energy and momentum are conserved in
(24) A child with mass m is standing at the edge of a disc with elastic as well as inelastic collisions
moment of inertia I, radius R, and initial angular velocity ω . (C) Total kinetic energy is not conserved but momentum is
See figure given below. The child jumps off the edge of the conserved in inelastic collisions
disc with tangential velocity v with respect to the ground.
The new angular velocity of the disc is (D) Total kinetic energy is conserved in elastic collisions but
momentum is not conserved in elastic collisions
(31) The position x of a particle with respect to time t along
x−axis is given by x = 9t2 − t3 where x is in metres and t in
seconds. What will be the position of this particle when it
achieves maximum speed along the x direction ?..........m
q q (A) 54 (B) 81
(A) Iω 2 −mv2
(B) (I+mR2 )ω 2 −mv2
I I (C) 24 (D) 32
(32) Which of the following must be known in order to
2
(C) Iω−mvR
I (D) (I+mR )ω−mvR
I
determine the power output of an automobile?
(25) Two bodies moving towards each other collide and move
away in opposite directions. There is some rise in (A) Final velocity and height
temperature of bodies because a part of the kinetic energy (B) Mass and amount of work performed
is converted into
(A) Heat energy (B) Electrical energy (C) Force exerted and distance of motion

(C) Nuclear energy (D) Mechanical energy (D) Work performed and elapsed time of work
(26) Four particle of masses m, 2m, 3m and 4m are arranged at (33) A block of mass 10 kg accelerates uniformly from rest to a
the corners of a parallelogram with each side equal to a and speed of 2 m/s in 20 sec. The average power developed in
one of the angle between two adjacent sides is 60o . The time interval of 0 to 20 sec is .............. W
parallelogram lies in the x − y plane with mass m at the (A) 10 (B) 1
origin and 4m on the x− axis. The centre of mass of the (C) 20 (D) 2

2
 (34) The moment of inertia of a regular circular disc of mass (43) A block of mass 10 kg is kept on a rough inclined plane as
0.4 kg and radius 100 cm about an axis perpendicular to the shown in the figure. A force of 3 N is applied on the block.
plane of the disc and passing through its centre is ...... kg m2 The coefficient of static friction between the plane and the
(A) 0.2 (B) 0.02 block is 0.6. What should be the minimum value of force P ,
such that the block does not move downward? (take
(C) 0.002 (D) 2 g = 10 ms−2 ) ........ N
(35) A wagon weighing 1000 kg is moving with a velocity
50 km/h on smooth horizontal rails. A mass of 250 kg is
dropped into it. The velocity with which it moves now is .........
km/hour
(A) 2.5 (B) 20
(C) 40 (D) 50

.............. Physics - Section B (MCQ) ..............

(A) 32 (B) 18
(36) A wheel of mass 10 kg has a moment of inertia of
(C) 23 (D) 25
160 kg − m2 about its own axis, the radius of gyration will
be ....... m (44) A wheel is rotating about an axis through its centre at
(A) 10 (B) 8 720 rpm. It is acted on by a constant torque opposing its
motion for 8 seconds to bring it to rest finally. The value of
(C) 6 (D) 4 torque (in N − m) is (Given I = 24π kg − m )
2

(37) An object of mass 10 kg is projected from ground with (A) 48 (B) 72

speed 40 m/s at an angle 60◦ with horizontal. The rate of
change of momentum of object one second after projection (C) 96 (D) 120
in SI unit is ........ [Take g = 9.8 m/s2 ] (45) When a particle moves in a uniform circular motion. It has
(A) 73 (B) 98 (A) Radial velocity and radial acceleration
(C) 176 (D) 140 (B) Tangential velocity and radial acceleration
(38) The angular velocity of a particle rotating in a circular orbit (C) Tangential velocity and tangential acceleration
100 times per minute is
(D) Radial velocity and tangential acceleration
(A) 1.66 rad/s (B) 10.47 rad/s
(46) A particle is simultaneously acted by two forces equal to
(C) 10.47 deg/s (D) 60 deg/s 4 N and 3 N . The net force on the particle is
(39) A body of mass 1.0 kg is falling with an acceleration of (A) 7 N (B) 5 N
10 m/sec2 . Its apparent weight will be ......... kg wt
(g = 10 m/sec2 ) (C) 1 N (D) Between 1 N and 7 N
(A) 1.0 (B) 2.0 (47) For the given figure, if block remains in equilibrium position
then find frictional force between block and wall ........ N
(C) 0.5 (D) 0
(40) A flywheel has moment of inertia 4 kg − m2 and has kinetic
energy of 200 J. Calculate the number of revolutions it
makes before coming to rest if a constant opposing couple
of 5 N − m is applied to the flywheel .......... rev
(A) 12.8 (B) 24
(A) 100 (B) 50
(C) 6.4 (D) 16
(C) 200 (D) None
(41) The initial velocity of a body moving along a straight line is
7 m/s. It has a uniform acceleration of 4 m/s2 . The distance (48) A solid sphere of radius R has moment of inertia I about its
covered by the body in the 5th second of its motion is..........m geometrical axis. It is melted into a disc of radius r and
(A) 25 (B) 35 thickness t. If it’s moment of inertia about the tangential axis
(which is perpendicular to plane of the disc), is also equal to
(C) 50 (D) 85 I, then the value of r is equal to
(42) A block of mass 8 kg is at rest on a rough inclined plane as
shown in figure. The magnitude of net force exerted by the
surface on the block will be ......... N

(A) √2
15
R (B) √2 R
5
√
(C) √3
15
R (D) √3 R
15
(A) 0 (B) 48 (49) A particle of mass m travelling along x− axis with speed v0
(C) 64 (D) 80 shoots out 1/3rd of its mass with a speed 2v0 along y− axis.

3
 The velocity
 ofremaining piece is  
(A) v0 3
2 î − ĵ (B) v0
2 3î − ĵ
   
(C) v0 1
2 î − 3ĵ (D) v0
2 î + 3ĵ
(50) A horizontal bridge is built across a river. A student standing
on the bridge throws a small ball vertically upwards with a
velocity 4 ms−1 . The ball strikes the water surface after 4 s.
The height of
bridge above water surface is ...... m ( Take
g = 10 ms−2
(A) 68 (B) 56
(C) 60 (D) 64

4
 JB AND KARP VIDYA SANKUL
Subject : Physics Paper Set : 1
H Date : 25-11-2024
Standard : 11
Total Mark : 200 (Answer Key) Time : 0H:0M

Physics - Section A (MCQ)

1-D 2-B 3-A 4-C 5-C 6-B 7-B 8-B 9-D 10 - C

11 - B 12 - B 13 - D 14 - B 15 - B 16 - C 17 - C 18 - C 19 - A 20 - A
21 - B 22 - A 23 - A 24 - D 25 - A 26 - B 27 - C 28 - D 29 - A 30 - C
31 - A 32 - D 33 - B 34 - A 35 - C

Physics - Section B (MCQ)

36 - D 37 - B 38 - B 39 - D 40 - C 41 - A 42 - D 43 - A 44 - B 45 - B
46 - D 47 - B 48 - A 49 - A 50 - D

5
 JB AND KARP VIDYA SANKUL
Subject : Physics Paper Set : 1
H Date : 25-11-2024
Standard : 11
Total Mark : 200 (Solutions) Time : 0H:0M

(5) Vector product of two vectors 2î + ĵ and î + 2ĵ is

.............. Physics - Section A (MCQ) ..............
(A) k̂ + ĵ (B) î + ĵ
(1) A block of mass 5 kg is on a rough horizontal surface and is (C) 3k̂ (D) 2î
at rest. Now a force of 24 N is imparted to it with negligible
impulse. If the coefficient of kinetic friction is 0.4 and Solution:(Correct Answer:C)
g = 9.8 m/s2 , then the acceleration of the block is ........ m/s2 (2î × ĵ) × (î + 2ĵ) = 4k̂ − k̂ = 3k̂
(A) 0.26 (B) 0.39
(6) The displacement of a particle starting from rest (at t = 0) is
(C) 0.69 (D) 0.88 given by s = 6t2 − t3 . The time in seconds at which the
particle will attain zero velocity again, is ..... sec
Solution:(Correct Answer:D)
(A) 2 (B) 4
(d) Net force = Applied force -Friction force
ma = 24 − µ mg (C) 6 (D) 8
= 24 − 0.4 × 5 × 9.8 = 24 − 19.6
⇒ a = 4.4 2 Solution:(Correct Answer:B)
5 = 0.88 m/s
(b) v = ds
dt = 12t − 3t
2

(2) Three forces F⃗1 = (2î + 4ĵ) N ; F⃗2 = (2ĵ − k̂) N and Velocity is zero for t = 0 and t = 4 sec
F⃗3 = (k̂ − 4î − 2ĵ) N are applied on an object of mass 1kg
at rest at origin. The position of the object at t = 2s will be (7) 0.4î + 0.8ĵ + ck̂ represents a unit vector when c is
√
(A) (−2 m, −6 m) (B) (−4 m, 8 m) (A) −0.2 (B) 0.2
√
(C) (3 m, 6 m) (D) (2 m, −3 m) (C) 0.8 (D) 0

Solution:(Correct Answer:B)
Solution:(Correct Answer:B) q
(b) (b) (0.4) + (0.8) + c2 = 1
2 2
⃗ ⃗ ⃗ √
⃗a = F1 +F12 +F3 = −2î + 4ĵ == > 0.16 + 0.64 + c2 = 1== > c = 0.2
1 2
S = 2 at
= 12 (−2î + 4ĵ)(2)2 (8) A hollow sphere of mass 1 kg and radius 10 cm is free to
= −4î + 8ĵ rotate about its diameter. If a force of 30 N is applied
tangentially to it, its angular acceleration is (in rad/s2 )
(3) The dimensions of K in the equation W = 1
2 Kx2 is (A) 5000 (B) 450
(A) M 1 L0 T −2 (B) M 0 L1 T −1
(C) 50 (D) 5
(C) M 1 L1 T −2 (D) M 1 L0 T −1
Solution:(Correct Answer:B)
Solution:(Correct Answer:A) (b)
h i Use τ =
Iα
(a) W = 12 kx2 ⇒ [k] = [W ]
= M L2 T −2
= [M T −2 ] 10


10 2
[x2 ] L2 30 100 = 23 (1) 100 α
α = 450 rad/s2
(4) A particle of mass m is tied to string of length l and whirled
in vertical circle. The difference of tension and kinetic (9) An elevator weighing 6000 kg is pulled upward by a cable
energy at highest and lowest position of circular path will be with an acceleration of 5 ms−2 . Taking g to be 10 ms−2 ,
(A) 2 mg, 5 mgl (B) 5 mg, 3 mgl then the tension in the cable is ........... N
(A) 6000 (B) 9000
(C) 6 mg, 2 mgl (D) 3 mg, 5 mgl
(C) 60000 (D) 90000
Solution:(Correct Answer:C)
√
TL = mg + m( 5g1)
2
Solution:(Correct Answer:D)
ℓ
TL = 6mg (d)T = m (g + a) = 6000 (10 + 5) = 90000 N
√
TH + mg = m( ℓg1)
2

TH = mg − mg = 0 (10) When a ceiling fan is switched on, it makes 10 rotations in

TL − TH = √6mg the first
KL = 12 m( √5g1)2 3 sec . How many rotations will it make in the next 3 sec
KH = 12 m( g1)2 (Assume uniform angular acceleration)
KL − KH ⇒ m2 (5g1 − g1) (A) 10 (B) 20
⇒ 2mg1 (C) 30 (D) 40

6
 Solution:(Correct Answer:C) (b)Work done by centripetal force is always zero, because
In first three seconds, angle rotated θ = 2π × 10rad force and instantaneous displacement are always
Using, θ = ω0 t + 12 αt2 perpendicular.
→→
−
∴ 2π × 10 = 0 + 21 α × 32 = 92 α . . . (i) W = F .− s = F s cos θ = F s cos(90◦ ) = 0
For the rotation of fan in next three second, the total time  
of revolutions (15) A particle moves from the point 2.0î + 4.0ĵ m, at t = 0
=3 + 3 = 6s  
Let total number of revolutions = N with an initial velocity 5.0î + 4.0ĵ ms−1 . It is acted upon
Then angle of revolutions, θ′ = 2πN rad by a constant
  force which produces a constant acceleration
∴ 2πN = 0 + 21 α × 62 = 18α ...(ii) 4.0î + 4.0ĵ ms−2 . What is the distance of the particle
Dividing (ii) by (i), we get from the origin at time 2 s
N = 40 √
(A) 15 m (B) 20 2 m
No. of revolutions in last three seconds √
= 40 − 10 = 30 revolutions (C) 5 m (D) 10 2 m
Solution:(Correct Answer:B)
(11) A rocket with a lift-off mass 2 × 104 kg is blasted upwards → 
−   
with an intial acceleration of 5 m s−2 . The initial thrust of S = 5î + 4ĵ 2 + 21 4ĵ + 4ĵ 4
the blast is (Take g = 10 m s−2 ) = 10î + 8ĵ + 8î + 8ĵ
(A) 2 × 105 N (B) 3 × 105 N −
→
r2 − −
→
r1 = 18î + 16ĵ
→
−
r2 = 20î +
(C) 4 × 105 N (D) 5 × 105 N √20ĵ
|−
→
r | = 20 2
f
Solution:(Correct Answer:B)
(16) A circular plate of uniform thickness has diameter 56 cm. A
Here, m = 2 × 104 kg
circular part of diameter 42 cm is removed from one edge.
Initial acceleration , a = 5ms−2 What is the position of the centre of mass of the remaining
Initial thrust = upthrust required to impart part ........ cm.
acceleration a + upthrust required to overcome (A) 3 (B) 6
gravitational pull
(C) 9 (D) 12
∴ F = m(a + g)


= 2 × 104 kg (5 + 10)ms−2 = 3 × 105 N Solution:(Correct Answer:C)
Let mass of plate per unit area = m

2
(12) A body of mass m = 3.513 kg is moving along the x - axis = M1 = π 56 2
m = π(28) m
2

with a speed of 5.00 ms−1 . The magnitude of its momentum = M1 = π 42

2 2
m = π(21) m
is recorded as (Consider significant figures) 2
Mass of remaining Part = M − M1
(A) 17.57 (B) 17.6 C.M of whole disc = O i.e at origin
(C) 17.565 (D) 17.56 C.M of removed Plate = r1 = 28 − 21 = 7cm
C.M of remaining Portion = r2
Solution:(Correct Answer:B) hence
p = mv = 3.513 × 5.00 = 17.565 kgms−1 M × O = M1 r1 + (M − M1 ) r2
since result should have only 3 significant digits M1 r1 h= − (m − im1 ) r2
p = 17.6 kgms−1 (M1 r1 )
r2 = −(m−m
h 1)
i
π(21)2 m
= −π(28)2 −π(21)2 m × 7
(13) Two blocks of masses 5 kg and 2 kg are connected by a h i
spring of negilible mass and placed on a frictionless (21)2
= (−343) ×7
horizontal surface. An impulse gives a velocity of 7 m/s to r2 = −9cm
the heavier block in the direction of the lighter block. The
velocity of the centre of mass is ....... m/s (17) What is the unit vector perpendicular to the following
(A) 30 (B) 20 vectors 2î + 2ĵ − k̂ and 6î − 3ĵ + 2k̂
(C) 10 (D) 5 (A) î+10√
ĵ−18k̂
5 17
(B) î−10√
ĵ+18k̂
5 17

Solution:(Correct Answer:D) (C) î−10√

ĵ−18k̂
5 17
(D) î+10√
ĵ+18k̂
5 17
(d) Solution:(Correct Answer:C)
vcm = m1mv11 +m 2 v2
+m2
5(7)+2(0)
(c) A
⃗ = 2î + 2ĵ − k̂ and B

⃗ = 6î − 3ĵ + 2k̂
  
= 7 = 5 m/s C = A × B = 2î + 2ĵ − k̂ × 6î − 3ĵ + 2k̂
⃗ ⃗ ⃗

(14) A body of mass m is moving in a circle of radius r with a î ĵ k̂

constant speed v. The force on the body is mv 2 2 −1 = î − 10ĵ − 18k̂
r and is
2
=
directed towards the centre. What is the work done by this 6 −3 2
force in moving the body over half the circumference of the Unit vector perpendicular to both A
⃗ and B
⃗
circle î−10 ĵ−18
= √12 +102 +182 k̂

(A) mv (B) Zero

2

πr 2 î−10√
ĵ−18k̂
= 5 17
(C) mv 2
(D) πr 2
(18) A man is pulling on a rope attached to a block on a smooth
r2 mv 2

Solution:(Correct Answer:B) horizontal table. The tension in the rope will be the same at

7
 all points Solution:(Correct Answer:A)
→→
−
(A) If and only if the rope is not accelerated (a)Work done = F .− s
= (6î + 2ĵ) · (3î − ĵ) = 6 × 3 − 2 × 1 = 18 − 2 = 16 J
(B) If and only if the rope is massless
(C) If either the rope is not accelerated or is massless (24) A child with mass m is standing at the edge of a disc with
(D) Always moment of inertia I, radius R, and initial angular velocity ω .
See figure given below. The child jumps off the edge of the
Solution:(Correct Answer:C) disc with tangential velocity v with respect to the ground.
The tension in the rope will be same at all points if it is The new angular velocity of the disc is
massless or if the string has mass, the tension will be the
same in the string when it is not accelerating.

(19) An engine pump is used to pump a liquid of density ρ

continuously through a pipe of cross-sectional area A. If the
speed of flow of the liquid in the pipe is v, then the rate at q q
which kinetic energy is being imparted to the liquid is (A) Iω 2 −mv2
(B) (I+mR2 )ω 2 −mv2
I I
(A) 12 Aρv 3 (B) 12 Aρv 2
(I+mR2 )ω−mvR
(C) 1
(D) Aρv (C) Iω−mvR
(D)
2 Aρv
I I

Solution:(Correct Answer:A) Solution:(Correct Answer:D)

(a)Energy supplied to liquid per second by the pump = From conservation

of linear momentum
2 t = 2
t
I + mR2 ω = mvR + Iω ′
2
1 mv 2 1 V ρv
 
= 12 A × tl × ρ × v 2 tl = v ω′ =
(I+mR2 )ω−mvR
= 12 A × v × ρ × v 2 = 12 Aρv 3
I

(20) A space craft of mass M is moving with velocity V and (25) Two bodies moving towards each other collide and move
suddenly explodes into two pieces. A part of it of mass m away in opposite directions. There is some rise in
becomes at rest, then the velocity of other part will be temperature of bodies because a part of the kinetic energy
is converted into
(A) MM−m
V
(B) MM+m
V
(A) Heat energy (B) Electrical energy
(C) mV
M −m (D) (M +m)V
m (C) Nuclear energy (D) Mechanical energy
Solution:(Correct Answer:A)
Solution:(Correct Answer:A)
(a)After explosion m mass comes at rest and let Rest
(M − m) mass moves with velocity v. (a) There is rise in temperature so, mechanical energy is
By the law of conservation of momentum M V = (M − m)v converted into heat energy
⇒ v = (MM−m)
V

(26) Four particle of masses m, 2m, 3m and 4m are arranged at

(21) The vector that must be added to the vector î − 3ĵ + 2k̂ and the corners of a parallelogram with each side equal to a and
3î + 6ĵ − 7k̂ so that the resultant vector is a unit vector one of the angle between two adjacent sides is 60o . The
along the y−axis is parallelogram lies in the x − y plane with mass m at the
origin and 4m on the x− axis. The centre of mass of the
(A) 4î + 2ĵ + 5k̂ (B) −4î − 2ĵ + 5k̂
arrangement
 willbe located at  
√ √
(C) 3î + 4ĵ + 5k̂ (D) Null vector (A) 3
(B) 3
2 a, 0.95a 0.95a, 4 a
Solution:(Correct Answer:B)


(C) 3a
4 , 2
a
(D) a2 , 3a
(b) Unit vector along y axis = ĵ so the required vector 4

= ĵ − [(î − 3ĵ + 2k̂) + (3î + 6ĵ − 7k̂)] Solution:(Correct Answer:B)

= − 4î − 2ĵ + 5k̂
Therefore the coordinates of centre of mass are.
xαn = m1 x1m+m 2 x2 +m3 x3 +m4 x4
=
(22) A 20 kg block is initially at rest on a rough horizontal surface. 1 +m2 +m3 +m4
M (o)+2M ( 2 )+3M ( 3a
a
2 )+aM (a)
A horizontal force of 75 N is required to set the block in M +2M +3M +4M
motion. After it is in motion, a horizontal force of 60 N is Ma
+ 9M A
2 + 2
8M a
⇒ xcm = 2 10M = 9M a
required to keep the block moving with constant speed. The 10M
⇒ xcm = 10 9a
coefficient of static friction is
ycn = m1 y1m+m 2 y2 +m3 y3 +m4 y4
=
(A) 0.38 (B) 0.44 ( √ 1 +m
) 2 +m +m)
(3√ 4
3a 3a
M (0)+2M +3M +4M (0)
(C) 0.52 (D) 0.6
2 2
M +2M +3M +4M
√
5 3M a √
5 3M a
Solution:(Correct Answer:A) ycm = 10M = 20M 2
√
(a)Coefficient of friction µs = Fl
= 75
= 75 ⇒ ycm = 43a
20×9.8 = 0.38
Therefore, the coordinates of the centre of mass are.
R mg
√ 
(23) Due to a force of (6î + 2ĵ)N the displacement of a body is (xcm · ycm ) = 10 , 4
9a 3a

(3î − ĵ)m, then the work done is.....J

(A) 16 (B) 12 (27) Five masses are placed in a plane as shown in figure. The
(C) 8 (D) 0 coordinates of the centre of mass are nearest to.

8
 collision. The kinetic energy is not conserved in inelastic
collision but the total energy is conserved in all type of
collisions.

(31) The position x of a particle with respect to time t along

x−axis is given by x = 9t2 − t3 where x is in metres and t in
seconds. What will be the position of this particle when it
achieves maximum speed along the x direction ?..........m
(A) 54 (B) 81
(C) 24 (D) 32

Solution:(Correct Answer:A)
(A) 1.2, 1.4 (B) 1.3, 1.1 Give : x = 9t2 − t3 ... (i)
dt = dt (9t − t ) = 18t − 3t .
Speed v = dx d 2 3 2
(C) 1.1, 1.3 (D) 1.0, 1.0
F or maximum speed, dt = 0 ⇒ 18 − 6t = 0
dv

Solution:(Correct Answer:C) ∴ t = 3s.

(c) ∴ xmax = 81m − 27m = 54 m. F rom x = 9t2 − t3
xcm = M1 x1 +M2 x2 +M
∑3 x3 +M4 x4 +Ms x5
m
(32) Which of the following must be known in order to
= 1×6+2×2+3×(0)+4×2+5×1
1+2+3+4+5 determine the power output of an automobile?
= 4+8+5 = 17
15
= 1.133
15
(A) Final velocity and height
Ycm = M1 y1 +M2 y2 +M∑5 y3 +M4 y4 +M5 y5
m (B) Mass and amount of work performed
= 1×0+2×0+3×2+4×2×5×1
1+2+3+4+5 (C) Force exerted and distance of motion
= 6+8+5 = 19
15 = 1.26
(D) Work performed and elapsed time of work
15
So, (xcm , ycm ) = (1.1, 1.3).

(28) Dimensions of µ0 ε0 ,
1
where symbols have their usual Solution:(Correct Answer:D)
meaning, are (d)
(A) [LT −1 ] (B) [L−1 T ] Power is defined as the rate of doing work. For the
automobile, the power output is the amount of work done
(C) [L−2 T 2 ] (D) [L2 T −2 ]
(overcoming friction) divided by the length of time in which
Solution:(Correct Answer:D) the work was done.
(d) C = √µ10 ε0 ⇒ µ01ε0 = c2 = [L2 T −2 ]
(33) A block of mass 10 kg accelerates uniformly from rest to a
(29) A block rests on a rough inclined plane making an angle of speed of 2 m/s in 20 sec. The average power developed in
30o with the horizontal. The coefficient of static friction time interval of 0 to 20 sec is .............. W
between the block and the plane is 0.8. If the frictional force (A) 10 (B) 1
on the block is 10 N , the mass of the block (in kg) is (take
(C) 20 (D) 2
g = 10 m/s2 )
(A) 2 (B) 4 Solution:(Correct Answer:B)
(C) 1.6 (D) 2.5 Average power
Net work done
Pav = Total time taken
Solution:(Correct Answer:A) Net work done = change in kinetic
(a) Angle of repose α = tan−1 (µ) = tan−1 (0.8) = 38.6◦ energy = Final energy − initial energy
Angle of inclined plane is given θ = 30◦ . = 12 × 10 × 22 = 20J . . . (1)
It means block is at rest therefore, Average power = 20 20 = 1 watt
Static friction = component of weight in downward
direction = mg sin θ = 10 N (34) The moment of inertia of a regular circular disc of mass
∴ m = 9×sin 10
30◦ = 2 kg 0.4 kg and radius 100 cm about an axis perpendicular to the
plane of the disc and passing through its centre is ...... kg m2
(30) Which of the following statements is true
(A) 0.2 (B) 0.02
(A) In elastic collisions, the momentum is conserved but not
in inelastic collisions (C) 0.002 (D) 2

(B) Both kinetic energy and momentum are conserved in Solution:(Correct Answer:A)
elastic as well as inelastic collisions Moment of inertia of a circular disc
(C) Total kinetic energy is not conserved but momentum is = 12 M R2 = 12 × 0.4 × 1 × 1 = 0.2 kg − m2
conserved in inelastic collisions
(35) A wagon weighing 1000 kg is moving with a velocity
(D) Total kinetic energy is conserved in elastic collisions but 50 km/h on smooth horizontal rails. A mass of 250 kg is
momentum is not conserved in elastic collisions dropped into it. The velocity with which it moves now is .........
Solution:(Correct Answer:C) km/hour
The law of conservation of momentum is true in all type of (A) 2.5 (B) 20
collisions, but kinetic energy is conserved only in elastic (C) 40 (D) 50

9
 Solution:(Correct Answer:C) 5 = 4α
(c) According to principle of conservation of linear α = 5/4rad/sec2
momentum wt = 0
1000 × 50 = 1250 × v⇒ v = 40 km/hr wt2 = wi2 + 2αθ

0 = 100 + 2 −5 4 ×θ
θ = 40 radians
.............. Physics - Section B (MCQ) ..............
So number of revolution = θ
2π
40
= 2π = 6.4
(36) A wheel of mass 10 kg has a moment of inertia of 6.4 revolution
160 kg − m2 about its own axis, the radius of gyration will
be ....... m
(41) The initial velocity of a body moving along a straight line is
(A) 10 (B) 8
7 m/s. It has a uniform acceleration of 4 m/s2 . The distance
(C) 6 (D) 4 covered by the body in the 5th second of its motion is..........m
(A) 25 (B) 35
Solution:(Correct Answer:D)
mass = 10kg (C) 50 (D) 85
M l = 160kg.m2 about axis
radius of gyration =? Solution:(Correct Answer:A)
I = MK2 (a) Sn = u + a2 [2n − 1]
K is radius of gyration S5th = 7 + 42 [2 × 5 − 1] = 7 + 18 = 25 m.
K2 = [(160)/(10)] = 16
K = 4m (42) A block of mass 8 kg is at rest on a rough inclined plane as
shown in figure. The magnitude of net force exerted by the
(37) An object of mass 10 kg is projected from ground with surface on the block will be ......... N
speed 40 m/s at an angle 60◦ with horizontal. The rate of
change of momentum of object one second after projection
in SI unit is ........ [Take g = 9.8 m/s2 ]
(A) 73 (B) 98
(C) 176 (D) 140
Solution:(Correct Answer:B)
(b) Force = ∆p
∆t , force remains constant = mg
⇒ 10 × 9.8 ⇒ 98 N (A) 0 (B) 48
At t = 1, particle is at its maximum height. (C) 64 (D) 80

(38) The angular velocity of a particle rotating in a circular orbit Solution:(Correct Answer:D)
100 times per minute is Net force applied by block on the inclined plane is equal to
(A) 1.66 rad/s (B) 10.47 rad/s the weight of the body.
(C) 10.47 deg/s (D) 60 deg/s N =p mg cos θ and fs = mg sin θ
F = f2s + N2 = mg
Solution:(Correct Answer:B)
(b)ω = 2πn = 2π×100
60 = 10.47 rad/s (43) A block of mass 10 kg is kept on a rough inclined plane as
shown in the figure. A force of 3 N is applied on the block.
(39) A body of mass 1.0 kg is falling with an acceleration of The coefficient of static friction between the plane and the
10 m/sec2 . Its apparent weight will be ......... kg wt block is 0.6. What should be the minimum value of force P ,
(g = 10 m/sec2 ) such that the block does not move downward? (take
(A) 1.0 (B) 2.0 g = 10 ms−2 ) ........ N
(C) 0.5 (D) 0
Solution:(Correct Answer:D)
(d) R = m (g − a) = m (10 − 10) = zero

(40) A flywheel has moment of inertia 4 kg − m2 and has kinetic

energy of 200 J. Calculate the number of revolutions it
makes before coming to rest if a constant opposing couple
of 5 N − m is applied to the flywheel .......... rev
(A) 12.8 (B) 24 (A) 32 (B) 18
(C) 6.4 (D) 16 (C) 23 (D) 25

Solution:(Correct Answer:C) Solution:(Correct Answer:A)

kE = 200 F or equilibrium of the block net f orce
1 2
2 Iwi = 200 should be zero. Hence we can write.
wi = 200×2 = 400 mg sin θ + 3 = p + f riction
2
I 4 = 100
wi = 10rad/sec mg sin θ + 3 = p + µmg cos θ.
ζ = Tα Af ter solving, we get, p = 32N.

10
(44) A wheel is rotating about an axis through its centre at
720 rpm. It is acted on by a constant torque opposing its
motion for 8 seconds to bring it to rest finally. The value of
torque (in N − m) is (Given I = 24π kg − m )
2

(A) 48 (B) 72
(C) 96 (D) 120

Solution:(Correct Answer:B) (A) √2 R (B) √2 R

15 5
60 rev. per sec
ω1 = 720 √
= 2π × 12 rad /sec (C) √3
15
R (D) √3 R
15
ω2 = 0
t = 8sec Solution:(Correct Answer:A)
5 MR = 2 Mr + Mr
2 2 1 2 2
α = retardation = 24π8 = 3πrad/sec
2

Torque τ = Iα = π × 3π = 72.N − m
24 or 52 MR2 = 32 Mr2
∴ r = √215 R

(45) When a particle moves in a uniform circular motion. It has

(49) A particle of mass m travelling along x− axis with speed v0
(A) Radial velocity and radial acceleration shoots out 1/3rd of its mass with a speed 2v0 along y− axis.
(B) Tangential velocity and radial acceleration The velocity
 ofremaining piece is  
(A) v0 3
2 î − ĵ (B) v0
3î − ĵ
(C) Tangential velocity and tangential acceleration 2
   
(D) Radial velocity and tangential acceleration (C) v0 1
2 î − 3ĵ (D) v0
2 î + 3ĵ

Solution:(Correct Answer:A)
Solution:(Correct Answer:B)
mu⃗1 = m1 ⃗v1 + m
 2⃗v2
An object moving in a circle is accelerating. Accelerating m
objects are objects which are changing their velocity - mv0 î = 3 2v0 ĵ + 2m 3 ~
v2
either the speed (i.e., magnitude of the velocity vector) or v⃗2 = 32 v0 î − v0 ĵ
the direction. An object undergoing uniform circular motion
is moving with a constant speed. (50) A horizontal bridge is built across a river. A student standing
on the bridge throws a small ball vertically upwards with a
velocity 4 ms−1 . The ball strikes the water surface after 4 s.
(46) A particle is simultaneously acted by two forces equal to
The height of
bridge above water surface is ...... m ( Take
4 N and 3 N . The net force on the particle is
g = 10 ms−2
(A) 7 N (B) 5 N
(A) 68 (B) 56
(C) 1 N (D) Between 1 N and 7 N
(C) 60 (D) 64
Solution:(Correct Answer:D) Solution:(Correct Answer:D)
(d) If two vectors Aand
⃗ ⃗ are given then the resultant
B S = ut + 12 at2
Rmax = A + B = 7N and Rmin = 4 − 3 = 1N −H = 4 × 4 − 12 × 10 × 42
i.e. net force on the particle is between 1 N and 7 N. −H = 16 − 80
−H = −64
(47) For the given figure, if block remains in equilibrium position H = 64 m
then find frictional force between block and wall ........ N

(A) 100 (B) 50

(C) 200 (D) None

Solution:(Correct Answer:B)
F = N = 1000N fL = 1000 × 0.1 = 100N
Freq = 50N(mg) ∵ freq < fL
∴ freq = 50N

(48) A solid sphere of radius R has moment of inertia I about its

geometrical axis. It is melted into a disc of radius r and
thickness t. If it’s moment of inertia about the tangential axis
(which is perpendicular to plane of the disc), is also equal to
I, then the value of r is equal to

11