Session: 2023-24 Total Questions: 346

JEE/NEET PHYSICS

5.LAWS OF MOTION

Single Correct Answer Type

1. Two blocks A of 6 kg and B of 4 kg are placed in contact with each other as shown

There is no friction between A and ground and between both the blocks. Coefficient of friction between B
and ground is 0.5. A horizontal force F is applied onA. Find the minimum and maximum values of F, which
can be applied so that both blocks can move combinely without any relative motion between them
a) 10 N, 50 N b) 12 N, 50 N c) 12 N, 75 N d) None of these
-2
2. A man is raising himself and the crate on which stands with an acceleration of 5 ms by a massless rope-
-2
and -pulley arrangement. Mass of the man is 100 kg and that of the crate is 50 kg. If g = 10 ms , then the
tension in the rope is

a) 2250 N b) 1125 N c) 750 N d) 375 N

3. ⃗ ̂ ̂ ⃗ ̂ ̂
A particle has initial velocity v = 3 i + 4 j , and a constant force F = 4 i - 3 j acts on the particle. The
path of the particle is
a) Straight line b) Parabolic c) Circular d) Elliptical
-1
4. Two uniform solid cylinders A and B each of mass 1 kg are connected by a spring of constant 200 Nm at
their axles and are placed on a fixed wedge as shown in the figure. There is no friction between cylinders
and wedge. The angle made by the line AB with the horizontal, in equilibrium, is

a) 0° b) 15° c) 30° d) None of these

-2
5. As shown in the figure, if acceleration of M with respect to ground is 2 ms , then

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 a) Acceleration of m with respect to M is 5 ms-2 b) Acceleration of m with respect to ground is 5 ms-2

c) Acceleration of m with respect M is 2 ms-2 d) Acceleration of m with respect M to ground is 10

-2
ms
6. A block of mass m is placed on another block of massM, which itself is lying on a horizontal surface. The
coefficient of friction between two blocks is μ1 and that between the block of mass M and horizontal
surface isμ2. What maximum horizontal force can be applied to the lower block so that the two blocks
move without separation?

a) (M+m)(μ - μ )g b) (M-m)(μ - μ )g c) (M-m)(μ + μ )g d) (M+m)(μ + μ )g

2 1 2 1 2 1 2 1

7. A 60 kg man stands on a spring scale in a lift. At some instant, he finds that the scale reading has changed
from 60 kg to 50 kg for a while and then comes back to original mark. What should be concluded?
a) The lift was in constant motion upwards b) The lift was constant motion downwards

c) Lift while in motion downwards suddenly stopped d) The lift while in motion upwards suddenly
stopped
8. Two beads A and B move along a semiconductor wire frame as shown in figure. The beads are connected
by an inelastic string which always remains tight. At any instant the speed of A isu, ∠BAC = 45°,
and∠BOC = 75°, where O is the centre of the semicircular arc. The speed of bead B at that instant is

a) 2 u b) u c) u d) 2 u
2 u 3
9. Two trolleys 1 and 2 are moving with accelerations a1 anda2, respectively, in the same direction. A block of
mass m on trolley 1 is in equilibrium from the frame of observer stationary with respect to trolley 2. The
magnitude of friction force on block due to trolley is (assume that no horizontal force other than friction
force is acting on block)

a) m(a - a ) b) ma c) ma d) Data insufficient

1 2 2 1

10. A plumb bob is hung from the ceiling of a train compartment. The train moves on an inclined track of
inclination 30° with horizontal. Acceleration of train up the plane isa = g/2. The angle which the string
supporting the bob makes with normal to the ceiling in equilibrium is
a) 30° b) tan-1 (2/ 3 ) c) tan-1 ( 3/2) d) tan-1 (2)

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11. A coin is placed at the edge of a horizontal disc rotating about a vertical axis through its axis with a
-1
uniform angular speed 2 rad s .The radius of the disc is 50 cm. Find the minimum coefficient of friction
-2
between disc and coin so that the coin does not slip (g = 10 ms )
a) 0.1 b) 0.2 c) 0.3 d) 0.4

12. The system shown in the figure is released from rest. The spring gets elongated

(Neglect friction and masses of pulley, string, and spring)

a) If M > m b) If M > 2m c) If M > m/2 d) For any value of M

13. Find the least horizontal force P to start motion of any part of the system of the three blocks resting upon
one another as shown in the figure. The weights of blocks are A = 300 N, B = 100 N, and C = 200 N.
Between A and B, coefficients of friction is 0.3, between B and C is 0.2 and between C and the ground is 0.2

a) 60 N b) 90 N c) 80 N d) 70 N

14. A man sits on a chair supported by a rope passing over a frictionless fixed pulley. The man who weights
1000 N exerts a force of 450 N on the chair downwards while pulling the rope on the other side. If the
chair weights 250 N, then the acceleration of the chair is
a) 0.45 ms-2 b) Zero c) 2 ms-2 d) 9/25 ms-2

15. A block of metal weighing 2 kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate
-1 -1
of 1 kgs and at a speed of5 ms . The initial acceleration of the block is

a) 5 ms-2 b) 25 ms-2 c) 25 ms-2 d) 5 ms-2

3 4 6 2
16. Block B has mass m and is released from rest when it is on top of wedgeA, which has a mass3m. Determine
the tension in cord CD needed to hold the wedge from moving while B is sliding downA. Neglect friction

a) 2 mgcos θ b) mg cos θ c) mg sin θ d) mgsin 2θ

2 2
17. A small block of mass m rests on a smooth wedge of angle θ. With what horizontal acceleration a should
the wedge be pulled, as shown in the figure, so that the block falls freely?

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 a) gcos θ b) gsin θ c) gcot θ d) gtan θ

18. Two particles A andB, each of massm, are kept stationary by applying a horizontal force F = mg on
particle B as shown in figure. Then

a) 2tan β = tan α b) 2T = 5T c) T 2 = T 5 d) None of these

1 2 1 2

19. A suitcase is gently dropped on a conveyor belt moving at3 ms-1. If the coefficient of friction between the
belt and the suitcase is 0.5, find the displacement of the suitcase relative to conveyor belt before the
-2
slipping between the two is stopped: (g=10 ms )
a) 2.7 m b) 1.8 m c) 0.9 m d) 1.2 m

20. If block B moves towards right with acceleration b, find the net acceleration of block A

a) ̂ ̂ b) ̂ ̂ c) ̂ ̂ d) None of these
b i + 4b j bi +bj b i + 2b j
21. A particle is moving in the x - y plane. At certain instant of time, the components of its velocity and
-1 -1 -2 -2
acceleration are as follows vx = 3 ms , vy = 4 ms , ax = 2 ms and ay = 1 ms . The rate of change of
speed at this moment is
a) 10 ms-1 b) 4 ms-2 c) 5 ms-2 d) 2 ms-2

22. A particle moves in the X-Y plane under the influence of a force such that its linear momentum is
̂ ̂
̇ ̇
p(t) = A[ i cos (kt) - j sin (kt)] where A and k are constant. The angle between the force and the
momentum is
a) 0° b) 30° c) 45° d) 90°

23. The small marble is projected with a velocity of 10 ms-1 in a direction 45° from the horizontal y-direction
on the smooth inclined plane. Calculate the magnitude v of its velocity after 2 s

a) 10 2 ms-1 b) 5 ms-1 c) 10 ms-1 d) 5 2 ms-1

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24. A block of massm, lying on a horizontal plane, is acted upon by a horizontal force P and another forceQ,
inclined at an angle θ to the vertical. The block will remain in equilibrium if the coefficient of friction
between it and the surface is (assume P > Q)

a) (Psin θ - Q)/(mg - cos θ) b) (P - Qsin θ)/(mg + Qcos θ)

c) (Pcos θ + Q)/(mg - Qcos θ) d) (P + Qsin θ)/(mg + Qcos θ)

25. A body of mass m is held at rest at a height h on two smooth wedges of mass M each, which are themselves
at rest on a horizontal frictionless surface. When the mass m is released, it moves down, pushing aside the
wedges. The velocity with which the wedge recede from each other, when m reaches the ground, is

a) 8mgh b) 40mgh×4 c) 32mgh×4 d) None of these

m+2M 5m+6M 32M+9m
26. If block A is moving with an acceleration of 5 ms-2, the acceleration of B w.r.t. ground is

a) 5 ms-2 b) 5 2 ms-2 c) 5 5 ms-2 d) 10 ms-2

27. In the system shown all the surfaces are frictionless while pulley and string are massless. Mass of block A
is 2 m and that of block B is m. Acceleration of block B immediately after system is released from rest is

a) g/2 b) G c) g/3 d) None of these

28. A horizontal force, just sufficient to move a body of mass 4 kg lying on a rough horizontal surface, is
applied on it. The coefficient of static and kinetic friction between the body and the surface are 0.8 and 0.6,
respectively. If the force continuous to act even after the block has started moving, the acceleration of the
-2 -2
block in ms is (g=10 ms )
a) 1/4 b) 1/2 c) 2 d) 4

29. The masses of the blocks A and B are m andM, respectively. Between A and B there is constant frictional
forceF, and B can slide frictionlessly on horizontal surface. A is set in motion with velocity while B is at
rest. What is the distance moved by A relative to B before they move with the same velocity?

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 2 2 2 2
a) mMv0 b) mMv0 c) mMv0 d) mMv0
F(m-M) 2F(m-M) F(m+M) 2F(m+M)
30. A block P of mass m is placed on a horizontal surface. Another block Q of same mass is kept on P and
connected to the wall with the help of a spring of spring constant k as shown in the figure. μs is the
coefficient of friction between P and Q. The blocks move together performing SHM of amplitude A. The
maximum value of the friction force between P and Q is

a) kA b) kA c) Zero d) μ mg
2 s

31. Consider a 14-tyre truck, whose only rear 8 wheels are power driven (means only these 8 wheels can
produce acceleration). These 8 wheels are supporting approximately half of the entire load. If coefficient of
friction between rod and each of the tyres is 0.6, then what could be the maximum attainable acceleration
by this truck?
a) 6 ms-2 b) 24 ms-2 c) 3 ms-2 d) 10 ms-2

32. A block of metal weighing 2 kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate
-1 -1
of 1 kgs and a speed of5 ms . The initial acceleration of the block will be
a) 2.5 ms-2 b) 5 ms-2 c) 10 ms-2 d) 20 ms-2

33. Two blocks m and M tied together with an inextensible string are placed at rest on a rough horizontal
surface with coefficient of frictionμ. The block m is pulled with a variable force F at a varying angle θ with
the horizontal. The value of θ at which the least value of F is required to move the blocks is given by

a) θ = tan-1 μ b) θ > tan-1 μ c) θ < tan-1 μ d) Insufficient data

34. A circular table of radius 0.5 m has smooth diametrical groove. A ball of mass 90 g is placed inside the
2 -1
groove along with a spring of spring constant10 Ncm . One end of the spring is tied to the edge of the
table and the other end to the ball. The ball is at a distance of 0.1 m from the centre when the table is at
2 -1
rest. On rotating the table with a constant angular frequency of 10 rad s , the ball moves away from the
centre by a distance nearly equal to
a) 10-1m b) 10-2m c) 10-3m d) 2×101 m

35. A small block slides without friction down an inclined plane starting from rest. Let sn be the distance
travelled from time t = n - 1 to t = n. Then
sn
is
sn+1
a) 2n-1 b) 2n+1 c) 2n-1 d) 2n
2n 2n-1 2n+1 2n+1
36. A painter of mass M stands on a platform of mass m and pulls himself up by ropes which hang over pulley
as shown. He pulls each rope with force F and moves upward with a uniform accelerationa. Find a,
neglecting the fact that no one could do this for long time

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 a) 4F+(2M+m)g b) 4F+(M+m)g c) 4F-(M+m)g d) 4F-(M+m)g
M+2m M+2m M+m 2M+m
37. ⃗ ̂ ̂
In the figure shown, the acceleration of A is a A = 15 i + 15 j , then the acceleration of B is (A remains in
contact with B)

a) ̂ b) ̂ c) ̂ d) ̂
6i -15 i -10 i -5 i
38. An object moving with a constant acceleration in a non-inertial frame

a) Must have non-zero net force acting on it

b) May have zero net force acting on it

c) May have no force acting on it

d) This situation is practically impossible. (The pseudo force acting on the object has also to be
considered)
39. If the blocks A and B are moving towards each other with acceleration a and b as shown in the figure, find
the net acceleration of block C

a) ̂ ̂ b) ̂ c) ̂ ̂ d) None of these
a i - 2(a + b) j -(a + b) j a i - (a + b) j
40.
A block of mass m is lying on a wedge having inclination angleα = tan
-2
1
5
-1
()
. Wedge is moving with a

constant acceleration a = 2 ms . The minimum value of coefficient of friction μ so that m remains

stationary w.r.t. wedge is

a) 2/9 b) 5/12 c) 1/5 d) 2/5

41. In the following figure, the pulley P1 is fixed and the pulley P2 is movable. If W1 = W2 = 100 N, what is the
angleAP2P1? The pulleys are frictionless and assume equilibrium

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 a) 30° b) 60° c) 90° d) 120°

42. In the system shown in the figure, the friction coefficient between ground and bigger block isμ. There is no
friction between both the blocks. The string connecting both the block is light; all three pulleys are light
and frictionless. Then the minimum limiting value of μ, so that the system remains in equilibrium, is

a) 1 b) 1 c) 2 d) 3
2 3 3 2
43. A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, travelling with a
velocity v m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet
travel independently. The ball hits the ground at a distance of 20 m and the bullet at ball hits the ground at
a distance of 100 m from the foot of the post. The initial velocity v of the bullet is

a) 250 m/s b) 250 2 m/s c) 400 m/s d) 500 m/s

44. For the system shown in the figure, m1 > m2 > m3 > m4. Initially, the system is at rest in equilibrium
condition. If the string joining m4 and ground is cut, then just after the string is cut

Statement I: m1,m23, m3 remain stationary

Statement II: the value of acceleration of all the four blocks can be determined
Statement III: Only m4 remains stationary
Statement IV: Only m4 accelerates
Now, choose the correct options
a) All the statements are correct b) Only I, II and IV are correct

c) Only I and II are correct d) Only II and IV are correct

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45. A block of base 10 cm×10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction
between them is 3. The inclination θ of this inclined plane from the horizontal plane is gradually
increased from 0°. Then,
a) at 0 = 30°, the block will start sliding down the plane

b) The block will remain at the rest on the plane up to certain θ and then it will topple

c) At θ = 60°, the block will start sliding down the plane and continue to do so at higher angles

d) At θ = 60°, the block will start sliding down the plane and on further increasing θ, it will topple at
certain θ
46. Determine the time in which the smaller block reaches other end of bigger block in the figure

a) 4 s b) 8 s c) 2.19 s d) 2.13 s

47. Figure shows the variation of force acting on a body with time. Assuming the body to start from rest, the
variation of its momentum with time is best represented by which plot?

a) b) c) d)

48. A circular road of radius 1000 m has banking angle45°. The maximum safe speed (in ms-1) of a car having a
mass 2000 kg will be, if the coefficient of friction between type and road is 0.5
a) 172 b) 124 c) 99 d) 86

49. System shown in figure is in equilibrium and at rest. The spring and string are massless, now the string is

cut. The acceleration of mass 2 m and m just after the string is cut will be
a) g/2 upward, g downward b) g upward, g/2 downward

c) g upward, 2g downward d) 2g upward, g downward

50. Two blocks of masses M1 and M2 are connected with a string passing over a pulley as shown in figure. The
block M1 lies on a horizontal surface. The coefficient of friction between the block M1 and the horizontal
surface isμ. The system accelerates. What additional mass m should be placed on the block M1 so that the
system does not accelerate?

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 a) M2-M1 b) M2 - M c) M - M1 d) (M - M )μ
μ μ 1 2
μ 2 1

51. A block A of mass 2 kg is placed over another block B of mass 4 kg, which is placed over a smooth
horizontal floor. The coefficient of friction between A and B is 0.4. when a horizontal force of magnitude 10
N is applied on A, the acceleration of blocks A and B are

a) 1 ms-2 and 2 ms-2, respectively b) 5 ms-2 and 2.5 ms-2, respectively

c) Both the blocks will moves together with d) Both the blocks will move together with
-2 -2
acceleration 1/3 ms acceleration, 5/3 ms
52. A house is built on the top of a hill with 45° slope. Due to sliding of material and sand from top to bottom of
hill, the slope angle has been reduced

If coefficient of static friction between sand particles is 0.75, what is the final angle attained by hill?
-1
(tan 0.75 ⋍ 37°)
a) 8° b) 45° c) 37° d) 30°

53. A balloon of mass M is descending at a constant accelerationα. When a mass m is released from the
balloon, it starts rising with the same accelerationα. Assuming that its volume does not change, what is the
value of m?
a) α M b) 2α M c) α+g M d) α+g M
α+g α+g α 2α
54. A trolley A has a simple pendulum suspended from a frame fixed to its desk. A block B is in contact on its
vertical slide. The trolley is on horizontal rails and accelerates towards the right such that the block is just
prevented from falling. The value of coefficient of friction between A and B is 0.5. The inclination of the
pendulum to the vertical is

()
a) tan-1 1
2
b) tan-1 (3) c) tan-1 ( 2 ) d) tan-1 (2)

55. A block of mass M is pulled along a horizontal frictionless surface by a rope of massm. Force P is applied at
one end of rope. The force which the rope exerts on the block is

P a g e | 10
 a) P b) P c) PM d) PM
(M-m) M(m+M) (m+M) (M-m)
56. Three blocks A,B and C of equal mass m are placed one over the other on a frictionless surface (table) as
shown in the figure. Coefficient of friction between any blocks A, B and C isμ. The maximum value of mass
of block D so that the blocks A, B and C move without slipping over each other is

a) 3mμ b) 3m(1-μ) c) 3m(1+μ) d) 3mμ

μ+1 μ μ (1-μ)
57. Two small rings O and O' are put two vertical stationary rods AB andA'B', respectively. One end of an
' '
inextensible thread is tied at pointA . The thread passes through ring O and its other end is tied to ringO.
'
Assuming that ring O moves downwards at a constant at a constant velocity v1, then velocity v2 of the ring
'
O, when ∠AOO = α, is

[ ] [ ] [ ]
2 2 2
a) v 2sin α/2 b) v 2cos α/2 c) v 3cos α/2 d) None of these
1
cos α 1
sin α 1
sin α
58. Statement I A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without
dislodging the dishes from the table.
Statement II For every action there is an equal and opposite reaction.
a) Statement I is true, statement II is true; statement II is a correct explanation for statement I

b) Statement I is true, statement II is true; statement II is not a correct explanation for statement I

c) Statement I is true, statement II is false

d) Statement I is false, statement II is true

59. An ideal liquid of density ρ is pushed with velocity v through the central limb of the tube shown in the
figure. What force does the liquid exert on the tube? The cross sectional areas of the three limbs are equal
to A each. Assume stream-line flow

a) 9 ρ Av2 b) 5 ρ Av2 c) 3 ρ Av2 d) ρ Av2

8 4 2
60. A professor holds an eraser against a vertical chalkboard by pushing horizontally on it. He pushes with a
force that is much greater than it required to hold the eraser. The force friction exerted by the board on the
eraser increases if he
P a g e | 11
 a) Pushes eraser with slightly greater force

b) Pushes eraser with slightly less force

c) Raiser his elbow so that the force he exerts is lightly downward but has same magnitude

d) Lowers his elbow so that the force he exerts is slightly upward but the same magnitude

61. In the following arrangement, the system is initially at rest. The 5 kg block is now released. Assuming the
pulleys and string to be massless and smooth, the acceleration of block C will be

a) Zero b) 2.5 ms-2 c) 10/7 ms-2 d) 5/7 ms-2

62. In the figure shown, all blocks are of equal massm. All surfaces are smooth. The acceleration of the block A
with respect ground is

2
a) 4gsin 2θ b) 4gsin θ
2
c) 4gsin θ d) None of these
2 2
1+3sin θ 1+3sin θ 1+3sin θ
63. A chain of length L is placed on a horizontal surface as shown in the figure. At any instant x is the length of
chain on rough surface and the remaining portion lies on smooth surface. Initiallyx = 0. A horizontal force
P is applied to the chain (as shown in figure). In the duration x chnages from x = 0 tox = L. For chain to
move with constant speed,

a) The magnitude of P should increase with time

b) The magnitude of P should decrease with time

c) The magnitude of P should increase first and then decrease with time

d) The magnitude of P should decrease first and then increase with time

64. Figure shows two blocks, each of massm. The system is released from rest. If acceleration of blocks A and
Bat any instant (not initially) are a1 and a2, respectively, then

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 a) a = a cos θ b) a = a cos θ c) a = a d) None of these
1 2 2 1 1 2

65. In the given diagram, man A is standing on a movable plank while man B is standing on a stationary
platform. Both are pulling the string down such that the plank moves slowly up. As a result of this the
string slips through the hands of the men. Find the ratio of length of the string that slips through the hands
of A and B

a) 3/2 b) 3/4 c) 4/3 d) 2/3

66. A system is pushed by a force Fas shown in figure. All surfaces are smooth except between B andC. Friction
coefficient between B and C isμ. Minimum value of F to prevent block B from down ward slipping is

a) 3
2μ( )
mg ( )
b) 5 mg
2μ ()
c) 5 μmg
2 ()
d) 3 μmg
2
67. A body of mass m is launched up on a rough inclined plane making an angle 45° with horizontal. If the time
of ascent is half of the time of descent, the friction coefficient between plane and body is
a) 2 b) 3 c) 3 d) 4
5 5 4 5
68. In figure, the tension in the rope (rope is light) is

a) (M+m)gsin θ b) (M+m)gsin θ - μ mgcos θ

c) Zero d) (M+m)gcos θ

69. A person is drawing himself up and a trolley on which he stands with some acceleration. Mass of the
person is more than the mass of the trolley. As the person increases his force on the string, the normal
reaction between person and the trolley will

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 a) Increase b) Decrease

c) Remain same d) Cannot be predicted s data is insufficient

70. In the arrangement shown in the figure below at a particular instant the roller is coming down with a
-1 -1
speed of 12 ms and C is moving up with 4 ms . At the same instant, it is also known that w.r.t pulley P,
-1
block A is moving down with speed 3 ms . Determine the motion of block B (velocity) w.r.t. ground

a) 4 ms-1 in downward direction b) 3 ms-1 in upward direction

c) 7 ms-1 in downward direction d) 7 ms-1 in upward direction

71. In the figure shown the velocity of lift is 2 ms-1 while string is winding on the motor shaft with velocity 2
-1 -1
ms and block A is moving downwards with velocity of 2 ms , then find out the velocity of block B

a) 2 ms-1 ↑ b) 2 ms-1 ↑ c) 4 ms-1 ↑ d) None of these

72. A fixed U-shaped smooth wire has a semi-circular bending between A and B as shown in figure. A bead of
mass m moving with uniform speed v through the wire enters the semiconductor bend at A and leaves at
B. The averages force exerted by the bead on the part AB of the wire is

2 2
a) 0 b) 4mv c) 2mv d) None of these
πd πd
73. A block is lying on the horizontal frictionless surface. One end of a uniform rope is fixed to the block which
is pulled in the horizontal direction by applying a force F at the other end. If the mass of the rope is half the
mass of the block, the tension in the middle of the rope will be
a) F b) 2 F/3 c) 3 F/5 d) 5 F/6

74. What is the maximum value of the force F such that the block shown in the arrangement, does not move?

a) 20 N b) 10 N c) 12 N d) 15 N

P a g e | 14
75. In the figure shown, all blocks are of equal massm. All surfaces are smooth, the acceleration of C w.r.t.
ground is

a) 2gsin θcos θ b) gsin θcos θ c) gsin 2θ d) gsin θcos θ +

2 2 2 2
1+3sin θ 1+3sin θ 1+3sin θ 1+3sin θ
76. A box of mass 8 kg is placed on a rough inclined plane of inclinationθ. Its downwards motion can be
prevented by applying an upward pull F and it can be made to slide upwards by applying an upward pull F
and it can be made to slide upwards by applying a force2 F. The coefficient of friction between the box and
the inclined plane is
a) (tan θ)/3 b) 3tan θ c) (tan θ)/2 d) 2tan θ

77. A mass M is suspended by a rope from a rigid support at A as shown in figure. Another rope is tied at the
end B and it is pulled horizontally with a forceF. If the rope AB makes an angle θ with the vertical, then the
tension in the string AB is

a) Fsin θ b) F/sin θ c) Fcos θ d) F/cos θ

78. Two blocks of masses 0.2 kg and 0.5 kg, which are placed 22 m apart on a rough horizontal surface
(μ = 0.5), are acted upon by two forces of magnitude 3 N each as shown in figure at time t = 0. Then, the
time t at which they collide with each other is

a) 1 s b) 2 s c) 2 s d) None

79. A vessel containing water is given a constant acceleration a towards the right, along a straight horizontal
path. Which of the following diagram represents the surface of the liquid

a a a a

(A) (B) (C) (D)

a) A b) B c) C d) D

80. A block is placed on a rough horizontal plane attached with an elastic spring as shown in the figure

Initially spring is unstretched. If the plane is gradually lifted from θ = 0° to θ = 90°, then the graph
showing extension in the spring (x) versus angle (θ) is
P a g e | 15
 a) b) c) d)

81. Blocks A and C stat from rest and move to the right with acceleration aA = 12 t ms-2 andac = 3 ms-2. Here t
is in seconds. The time when block B again comes to rest is

a) 2 s b) 1 s c) 3/2 s d) 1/2 s

82. A passenger is travelling a train moving at40 ms-1. His suitcase is kept on the berth. The driver of train
-1
applies breaks such that the speed of the train decreases at a constant rate to 20 ms in 5 s. What should
be the minimum coefficient of friction between the suitcase and the berth if the suitcase is not to slide
during retardation of the train?
a) 0.3 b) 0.5 c) 0.1 d) 0.2

83. A block of mass m is on an inclined plane of angle θ. The coefficient of friction between the block and the
plane is μ and tan θ>μ. The block is held stationary by applying a force E parallel to the plane. The
direction of force pointing up the plane is taken to be positive. As P is varied from P1 = mg(sin θ-μcos θ)
to
P2 = mg(sin θ+μcos θ), the frictional force f versus P graph will look like

a) b)

c) d)

84. Which of the following is correct, when a person walks on a rough surface

P a g e | 16
 a) The frictional force exerted by the surface keeps him moving

b) The force which the man exerts on the floor keeps him moving

c) The reaction of the force which the man exerts on floor keeps him moving

d) None of the above

85. The system shown in figure is in equilibrium. Masses m1 and m2 are 2 kg and 8 kg, respectively. Spring
-1 -1
constants k1 and k2 are 50 Nm and70 Nm , respectively. If the compression in second spring is 0.5 m.
What is the compression in first spring? (Both springs have natural length initially)

a) 1.3 m b) -0.5 m c) 0.5 m d) 0.9 m

86. Two blocks of masses M1 and M2 are connected with a string which passes over a smooth pulley. The mass
M1 is placed on a rough incline plane as shown in the figure. The coefficient of friction between the block
and the inclined plane isμ. What should be the minimum mass M2 so that the block M1 slides upwards?

a) M = M (sin θ + μcos θ) b) M = M (sin θ + μcos θ)

2 1 2 1

c) M = M1 d) M = M1
2
sin θ+μcos θ 2
sin θ-μcos θ
87. A block of mass m is at rest with respect to a rough incline kept in elevator moving up with accelerationa.
Which of following statements is correct?

a) The contact force between block and incline is parallel to the incline

b) The contact force between block and incline is of the magnitude m(g + a)

c) The contact force between block and incline is perpendicular to the incline

d) The contact force is of magnitude mgcos θ

88. A block of mass 15 kg is resting on a rough inclined plane as shown in figure. The block is tied by a
horizontal string which has a tension of 50 N. The coefficient of friction between the surfaces of contact is

P a g e | 17
 a) 1/2 b) 2/3 c) 3/4 d) 1/4

89. The figure represents a light inextensible string ABCDE in which AB = BC = CD = DE and to which are
attached masses M, m and M at the points B, C andD, respectively. The system hangs freely in equilibrium
with ends A and E of the string fixed in the same horizontal line. It is given that tan α = 3/4
andtan β = 12/5. Then the tension in the string BC is

a) 2 mg b) (13/10)mg c) (3/10)mg d) (20/11)mg

90. A system is shown in the figure. A man standing on the block is pulling the rope. String slips through the
-1
hands of man with velocity 2 ms w.r.t. man. The velocity of the block will be (assume that the block does
not rotate)

a) 3 ms-1 b) 2 ms-1 c) 1/2 ms-1 d) 1 ms-1

91. A block of mass 0.1 kg is held against a wall by applying a horizontal force of 5 N on the block. If the
coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on
the block is
a) 2.5 N b) 0.98 N c) 4.9 N d) 0.49 N

92. Two blocks M and m are arranged as shown in the diagram. The coefficient of friction between the blocks
1
is μ1 = 0.25 and between the ground and M is μ2 = . If M = 8 kg, then find the value of m so that the
3
system will remain at rest

a) 4/3 kg b) 8/9 kg c) 1 kg d) 8/5 kg

93. A block of mass 4 kg is pressed against the wall by a force of 80 N as shown in the figure. Determine the
value of friction force and block’s acceleration (take μs = 0.2, μk = 0.15)

P a g e | 18
 a) 8 N, 0 ms-2 b) 32 N, 6 ms-2 c) 8 N, 6 ms-2 d) 32 N, 2 ms-2

94. A rope of length 4 m having mass 1.5 kg per metre lying on a horizontal friction less surface is pulled at
one end by a force of 12 N. What is the tension in the rope at a point 1.6 m from the other end?
a) 5 N b) 4.8 N c) 7.2 N d) 6 N

95. ⃗ ̂ ̂ -1
A particle of mass 2 kg moves with an initial velocity of v = 4 i + 4 j ms . A constant force of
⃗ ̂
F = -20 j N is applied on the particle. Initially, the particle was at (0, 0). The x-coordinate of the particle
when its y-coordinates again becomes zero is given by
a) 1.2 m b) 4.8 m c) 6.0 m d) 3.2 m

96. A body of mass M is resting on a rough horizontal plane surface, the coefficient of friction being equal toμ.
Att = 0, a horizontal force F = F0t starts acting on it, where F0 is a constant. Find the time T at which the
motion starts?
a) μMg/F b) Mg/μF c) μF /Mg d) None of these
0 0 0

97. Two blocks of masses 3 kg and 2 kg are placed side by side on an incline as shown in the figure. A force
F = 20 N is acting on 2 kg block along the incline. The coefficient of friction between the block and the
incline is same and equal to 0.1. find the normal contact force exerted by 2 kg block on 3 kg block

a) 18 N b) 30 N c) 12 N d) 27.6 N

98. A triangular prism of mass M with a block of mass m placed on it is released from rest on a smooth
inclined plane of inclination θ. The block does not slip on the prism. Then

a) The acceleration of the prism is gcos θ

b) The acceleration of the prism is gtan θ

c) The minimum coefficient of friction between the block and the prism is μ = cot θ
min

d) The minimum coefficient of friction between the block and the prism is μ = tan θ
min

99. If block A is moving horizontally with velocity vA, then find the velocity of block B at the instant as shown
in the figure

P a g e | 19
 hvA xvA xvA hvA
a) b) c) d)
2 2 2 2 2 2 2 2
2 x +h x +h 2 x +h x +h
100. Three forces are acting on a particle of mass m initially in equilibrium. If the first two forces (R1 and R2)
are perpendicular to each other and suddenly the third force (R3) is removed, then the acceleration of the
particle is
a) R3 b) R1+R2 c) R1-R2 d) R1
m m m m
101. A block of base 10 cm×10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction
between them is 3. The inclination θ of this inclined plane from the horizontal plane is gradually
increased from 0°. Then
a) At θ = 30°, the block will start sliding down the plane

b) The block will remain at rest on the plane up to certain θ and then it will topple

c) At θ = 60°, the block will start sliding down the plane and continue to do so at higher angles

d) At θ = 60°, the block will start sliding down the plane and on further increasing θ, it will topple at
certain θ
102. A bead of mass m is attached to one end of a spring of natural length R and spring
( 3+1)mg
constantK = . The other end of the spring is fixed at a point A on a smooth vertical ring of
R
radius R as shown in the figure. The normal reaction at B just after it is released to move is

a) mg/2 b) 3 mg c) 3 3 mg d) 3 3 mg
2
103. ⃗
In the shown arrangement below, if acceleration of B is a , then find the acceleration of A

a) asin α b) acot θ c) atan θ d) (sin αcot θ + cos α)

104. In the figure, the block of mass M is at rest on the floor. The acceleration with which a boy of mass m
should climb along the rope of negligible mass so as to lift the block from the floor is

P a g e | 20
 ( )
a) M -1 g
m ( )
b) M -1 g
m
c) M g
m
d) > M g
m
105. In the figure shown, blocks A and B move with velocities v1 and v2 along horizontal direction. Find the ratio
of v1/v2

a) sin θ1 b) sin θ2 c) cos θ2 d) cos θ1

sin θ2 sin θ1 cos θ1 cos θ2
106. An object moving with constant velocity in a non-inertial; frame of reference

a) Must have non-zero net force acting on it

b) May have zero net force acting on it

c) Must have zero net force acting on it

d) May have non-zero net force acting on it (Consider only the real forces)

107. A heavy uniform chain lies on a horizontal table top. If the coefficient of friction between the chain and the
table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of
the table is
a) 20 % b) 25 % c) 35 % d) 15 %

108. Two objects A andB, each of massm, are connected by a light inextensible string. They are restricted to
move on a frictionless ring of radius R in a vertical plane (as shown in the figure). The objects are released
from rest at the position shown. Then the tension in the cord just after release is

a) Zero b) mg c) 2 mg d) mg/ 2

109. Three equal weights A, B, C of mass 2 kg each are hanging on a string passing over a fixed frictionless
pulley as shown in the figure. The tension in the string connecting weights B and C is

P a g e | 21
 a) Zero b) 13 N c) 3.3 N d) 19.6 N

110. Three blocks A, B and C are suspended as shown in the figure. Mass of each of blocks A and B ism. If system
is in equilibrium, and mass of C is M, then

a) M > 2 m b) M = 2 m c) M < 2 m d) None of these

111. A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hands of
a person standing in the car. The car is moving with constant acceleration a directed horizontally as shown
in the figure. Other end of the string is pulled with constant acceleration a vertically. The tension in the
string is equal to

a) 2 2 b) 2 2 c) 2 2 d) m(g + a)
m g +a m g +a - ma m g +a + ma
112. In the arrangement shown in the figure, the block of mass m = 2kg lies on the wedge of mass M = 8 kg.
the initial acceleration of the wedge, if the surfaces are smooth, is

a) 3g ms-2 b) 3 3g ms-2 c) 3g ms-2 d) g ms-2

23 23 23 23
113. A given object takes n times more time to slide down 45° rough inclined plane as it takes to slide down a
perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline is
a) 1 b) 1- 1 c) 1 - 1 d) 1 2
2 2 2
1-n n n 2-n
114. In the situation shown in the figure, all the strings are light and inextensible and pullies are light. There is
no friction at any surface and all blocks are of cuboidal shape. A horizontal force of magnitude F is applied
to right most free end of string in both cases shown in the figure. At the instant shown, the tension in all
strings are non-zero. Let the magnitude of acceleration of large blocks (of massM) in figure (a) and figure
(b) be a1 and a2, respectively. Then,

P a g e | 22
 a) a = a ≠ 0 b) a = a = 0 c) a > a d) a < a
1 2 1 2 1 2 1 2

115. In the figure, a block of weight 60 N is placed on a rough surface. The coefficient of friction between the
block and the surface is 0.5. What minimum can be the weight W such that the block does not slip on the
surface?

a) 60 N b) 60 N c) 30 N d) 30 N
2 2
116. The two particles of mass m each are tied at the ends of a light string of length 2a. The whole system is
kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance ‘a’ from
the center P (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a
small but constant force F. As a result, the particles move towards each other on the surface. The
magnitude of acceleration, when the separation between them becomes 2x, is

a) F a b) F x 2 2
c) F x d) F a -x
2m a2-x2 2m a2-x2 2m a 2m x
117. n Balls each of mass m impinge elastically each second on a surface with velocityu. The average force
experienced by the surface will be
a) mnu b) 2 mnu c) 4 mnu d) mnu/2

118. A man pulls himself up the 30° incline by the method shown. If the combined mass of the man and cart is
100 kg, determine the acceleration of the cart if the man exerts a pull of 250 N on the rope. Neglect all
friction and the mass of the rope, pulleys, and wheels

a) 4.5 ms-2 b) 2.5 ms-2 c) 3.5 ms-2 d) 1.5 ms-2

119. A block compartment containing gas is moving with some acceleration in horizontal direction. Neglect
effect of gravity. Then the pressure in the compartment is
a) Same everywhere b) Lower in front side c) Lower in rear side d) Lower in upper side

120. ( ̂ ̂
)-1
A particle of mass 2 kg moves with an initial velocity of 4 i +2 j ms on the x - y plane. A force

P a g e | 23
 ⃗ ̂ ̂
F = (2 i - 8 j )N acts on the particle. The initial position of the particle is (2 m, 3 m). Then for y = 3 m,
a) Possible value of x is only x = 2 m b) Possible value of x is not only x = 2 m, but there
exists some other value of x also
c) Time taken is 2 s d) All of the above

121. A monkey of mass 40 kg climbs on a massless rope of breaking strength 600 N. The rope will break if the
monkey
a) Climbs up with a uniform speed of 5 ms-1 b) Climbs up with an acceleration of 6 ms-2

c) Climbs down with an acceleration of 4 ms-2 d) Climbs down with a uniform speed of 5 ms-2

122. Two blocks A and B of masses 2m and m, respectively, are connected by a massless and inextensible string.
The whole system is suspended by a massless spring as shown in the figure. The magnitudes of
acceleration of A and B , immediately after the spring is cut, are respectively

a) g, g/2 b) g/2, g c) g, g d) g/2, g/2

123. A rope is stretched between two boats at rest. A sailor in the first boat pulls the rope with a constant force
of 100 N. first boat with the sailor has a mass of 250 kg whereas the mass of second boat is double of this
mass. If the initial distance between the boats was 100 m, the time taken for two boats to meet each other
is (neglect water resistance between boats and water)

a) 13.8 s b) 18.3 s c) 3.18 s d) 31.8 s

124. A system of two blocks, a light string, and a light and frictionless pulley is arranged as shown in the figure.
The coefficient of friction between fixed incline and 10 kg block is given by μs = 0.27 and μk = 0.20. If the
system is released from rest, then find the acceleration of 10 kg block

a) Zero b) 0.114 ms-2 c) 0.228 ms-2 d) 2.97 ms-2

125. A uniform chain is placed at rest on a rough surface of base length l and height h on an irregular as shown.
Then, the minimum coefficient of friction between the chain and the surface must be equal to

P a g e | 24
 a) μ = h b) μ = h c) μ = 3h d) μ = 2h
2l l 2l 3l
126. A particle of small mass m is joined to a very heavy body by a light string passing over a light pulley. Both
bodies are free to move. The total downward force on the pulley is
a) >> mg b) 4 mg c) 2 mg d) mg

127. A wooden block of mass M resting on a rough horizontal floor is pulled with a force F at an angle ϕ with
the horizontal. If μ is the coefficient of kinetic friction between the block and the surface, then the
acceleration of the block is
a) F (cos ϕ-μsin ϕ) - μg b) μF cos ϕ
M M
c) F (cos ϕ+μsin ϕ) - μg d) F sin ϕ
M M
128. A plank is held at an angle α to the horizontal on two fixed supports A andB. The plank can slide against the
supports (without friction) because of the weightMgsin α. Acceleration and direction in which a man of
mass m should move so that the plank does not move are

( )
a) gsin α 1+ m down the incline
M ( )
b) gsin α 1+ M down the incline
m

( )
c) gsin α 1+ m up the incline
M ( )
d) gsin α 1+ M up the incline
m
129. A block of mass m1 lies on the top of fixed wedge as shown in figure (a) and another block of mass m2 lies
on top of wedge which is free to move as shown in figure (b). At timet = 0, both the blocks are released
from rest from a vertical height h above the respective horizontal surface on which the wedge is placed as
shown. There is no friction between block and wedge in both the figures. Let T1 and T2 be the time taken by
the blocks, respectively, to just reach the horizontal surface

a) T > T b) T < T c) T = T d) Data insufficient

1 2 1 2 1 2

130. Two blocks A and B of masses 6 kg and 3 kg rest on a smooth horizontal surface as shown in the figure. If
coefficient of friction between A and B is 0.4, the maximum horizontal force which can make them move
without separation is

P a g e | 25
 a) 72 N b) 40 N c) 36 N d) 20 N

131. Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A 15 kg
weight is attached to the rope at the midpoint which now no longer remains horizontal. The minimum
tension required to completely straighten the rope is
a) 15 kg b) 12/2 kg c) 5 kg d) Infinitely large

132. In an arrangement shown below, the acceleration of block A an B are

a) g/3, g/6 b) g/6, g/3 c) g/2, g/2 d) 0, 0

133. For the situation shown in figure, the block is stationary w.r.t. incline fixed in an elevator. The elevator is
having an acceleration of 5a0 whose components are shown in the figure. The surface is rough and
coefficient of static friction between the incline and block isμs. Determine the magnitude of force exerted
g
by incline on the block. (take a0 = and θ = 37°, μs = 0.2)
2

a) mg b) 9mg c) 3mg × 41 d) 13 mg
10 25 25 2
134. A system is shown in the figure. Assume that cylinder remains in contact with the two wedge, then the
velocity of cylinder is

a) -1 b) 13u ms-1 c) 3 u ms-1 d) 7 ms-1

19-4 3 ms
2
135. If the resultant of all the external forces acting on a system of particles is zero, then form an inertial frame,
one can surely say that
a) Linear momentum of the system does not change in time

b) Kinetic energy of the system does not change in time

c) Angular momentum of the system does not change in time

d) Potential energy of the system does not change in time

136. The upper half of an inclined plane with inclination ϕ is perfectly smooth while the lower half is rough. A

P a g e | 26
 body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the
lower half is given by
a) 2tan ϕ b) tan ϕ c) 2sin ϕ d) 2cos ϕ

137. A horizontal force of 25 N is necessary to just hold a block stationary against a wall. The coefficient of
friction between the block and the wall is 0.4. The weight of the block is

a) 2.5 N b) 20 N c) 10 N d) 5 N

138. Inside a horizontally moving box, an experimenter finds that when an object is placed on a smooth
-2
horizontal table and is released, it moves with an acceleration of10 ms . In this box if 1 kg body is
suspended with a light string the tension in the string in equilibrium position. (w.r.t experimenter) will be
-2
(Take g = 10 ms )
a) 10 ms-2 b) 10 2 ms-2 c) 20 ms-2 d) Zero

139. In the figure shown, all blocks are of equal massm. All surfaces are smooth, the acceleration of B w.r.t.
ground is

a) 2gsin 2θ b) 4gsin 2θ c) 2gsin θ d) 4gsin θ

2 2
1+3sin θ 1+3sin θ 1+3sin θ 1+3sin θ
140. A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The
bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at
rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the
new equilibrium position of the bead, where the bead can stays at rest with respect to the wire, from the y-
axis is
a) a b) a c) 2a d) a
gk 2gk gk 4gk
141. In the given figure, the mass m2 starts with velocity v0 and moves with constant velocity on the surface.
During motion the normal reaction between the horizontal surface and fixed triangle block m1 isN. Then
during motion

a) N = (m + m )g b) N = m g c) N < (m +m )g d) N > (m + m )g
1 2 1 1 2 1 2

142. Two wooden blocks are moving on a smooth horizontal surface such that the mass m remains stationary
with respect to block of mass M as shown in the figure. The magnitude of force P is

P a g e | 27
 a) (M+m)gtan β b) gtan β c) mgcos β d) (M+m)g cosec β

143. In the figure shown, a person wants to rise a block lying on the ground to a height h. In both the cases, if
time required is same then in which case he has to exert more force. Assume pulleys and strings light

a) (i) b) (ii)

c) Same in both d) Cannot be determined

144. A block A has a velocity of 0.6 ms-1 to the right, determine the velocity of cylinder B

a) 1.2 ms-1 b) 2.4 ms-1 c) 1.8 ms-1 d) 3.6 ms-1

145. In the figure, the string does not slip on pulleyP, but pulley P is free to rotate about its own axis. Block A is
displaced towards left, then pulley P

a) Rotates clockwise and translates b) Rotates anticlockwise and translates

c) Only translates d) Only rotates (clockwise or anticlockwise)

146. A trolley T of mass 5 kg on a horizontal smooth surface is pulled by a load of 2 kg through a uniform rope
-2
ABC of length 2 m and mass 1 kg. as the load falls from BC = 0 to BC = 2 m, its acceleration (in ms )
changes from

a) 20 to 30 b) 20 to 30 c) 20 to 30 d) None of these
6 6 8 8 5 6
147. In the above problem, contact force between man and the crate is

a) 2250 N b) 1125 N c) 750 N d) 375 N

148. If the coefficient of friction between all surfaces (for the shown diagram) is 0.4, then find the minimum
force F to have equilibrium of the system

P a g e | 28
 a) 62.5 N b) 150 N c) 31.25 N d) 50 N

149. In the arrangement shown, by what acceleration the boy must go up so that 100 kg block remains
-2
stationary on the wedge? The wedge is fixed and friction is absent everywhere (take g = 10 ms )

a) 2 ms-1 b) 4 ms-2 c) 6 ms-2 d) 8 ms-2

150. A pendulum of mass m hangs from a support fixed to a trolley. The direction of the string when the trolley
rolls up a plane of inclination α with acceleration a0 is

a) θ = tan-1 α
()
b) θ = tan-1 a0
g

()
c) θ = tan-1 g
a0 (
d) θ = tan-1 a0+gsin α
gcos α )
151. Blocks A and B in the figure are connected by a bar of negligible weight. Mass of each block is 17 kg and
μA = 0.2 and μB = 0.4, where μA and μB are the coefficients of limiting friction between blocks and plane,
-2
calculate the force developed in the bar (g=10 ms )

a) 150 N b) 75 N c) 200 N d) 250 N

152. A block of mass m is attached with a massless spring of force constantk. The block is placed over a rough
inclined surface for which the coefficient of friction is 0.5 M is released from rest when the spring was
unstretched. The minimum value of M required to move the block m up the plane is (neglect mass of string
and pulley and friction in pulley)

a) m/2 b) m/3 c) m/4 d) None of these

P a g e | 29
153. Two bodies of masses 4 kg and 6 kg are attached to the ends of a string passing over a pulley. The 4 kg
mass is attached to the table top by another string. The tension in this string T1 is equal to (take g = 10
-2
ms )

a) 20 N b) 25 N c) 10.6 N d) 10 N

154. In two pulley-particle systems (i) and (ii), the acceleration and force imparted by the string on the pulley
and tension in the strings are (a1,a2), (N1,N2) and (T1,T2), respectively. Ignoring friction in all contacting
surfaces
Study the following statements :

a1 T N a
I. = 1 (ii) 1 < 1 (iii) 1 > 1 (iv) 1 < 1
a2 T2 N2 a2
Now mark correct answer:
a) Relations (ii) and (iii) always follow b) Relations (ii) and (iv) always follow

c) Only relation (i) always follows d) Only relation (iv) always follows

155. Two blocks A and B of masses m and2 m, respectively, are held at rest such that the spring is in natural
length. Find out the accelerations of both the blocks just after release

a) g↓, g↓ b) g ↓, g ↑ c) 0, 0 d) g↓, c
3 3
156. A circular disc with a groove along its diameter is placed horizontally. A block of mass 1kg is placed as
shown. The co-efficient of friction between the block and all surfaces of groove in contact is μ = 2/5. The
2
disc has an acceleration of 25 m/s . Find the acceleration of the block with respect to disc

a) 10 m/s2 b) 5 m/s2 c) 20 m/s2 d) 1 m/s2

P a g e | 30
157. An inclined plane makes an angle 30° with the horizontal. A groove (OA) of length = 5m cut in the plane
makes an angle 30° withOX. A short smooth cylinder is free to slide down under the influence of gravity.
-2
The time taken by the cylinder to reach from A to O is (g = 10 ms )

a) 4 s b) 2 s c) 3 s d) 1 s

158. Three arrangement of a light spring balance are shown in the following figure. The reading of the spring
scales in three arrangements are, respectively,

a) 20 g, 20 g, 10 g b) 20 g, 20 g, 40 g c) Zero, 20 g, 10 g d) Zero, 20 g, 40 g
3 3
159. In the given figure, the blocks are at rest and a force 10 N act on the block of 4 kg mass. The coefficient of
static friction and the coefficient of kinetic friction are μs = 0.2 and μk = 0.15 for both the surface in
contact. The magnitude of friction force acting between the surface of contact between the 2 kg and 4 kg
block in this situation is

a) 3 N b) 4 N c) 3.33 N d) Zero

160. A flat plate moves normally with a speed v1 towards a horizontal jet of water of uniform area of cross-
section. The jet discharges water at the rate of volume V per second at a speed of v2. The density of water is
ρ. Assume that water splashes along the surface of the plate at right angles to the original motion. The
magnitude of the force acting on the plate due to the jet of water is
a) ρVv
1
b) ρV(v + v )
1 2
c) ρV v2
v1+v2 1 []
d) ρ V (v +v )2
v2 1 2
161. An intersteller spacecraft far away from the influence of any star or planet is moving at high speed under

P a g e | 31
 the influence of fusion rockets (due to thrust exerted by fusion rockets, the spacecraft is acceleration).
Suddenly the engine malfunctions and stops. The spacecraft will
a) Immediately stop, throwing all of the occupants to the front

b) Begin slowing down and eventually come to rest

c) Keep moving at constant speed for a while, and then begin to slow down

d) Keep moving forever with constant speed

162. Velocity of point A on the rod 2 ms-1 (leftwards) at the instant shown in the figure. The velocity of the point
B on the rod at this instant is

a) 2 ms-1 b) 1 ms-1 c) 1 ms-1 d) 3 ms-1

3 2 3 2
163. Two skaters have weight in the ratio 4:5 and are 9 m apart, on a smooth frictionless surface. They pull on a
rope stretched between them. The ratio of the distance covered by them when they meet each other will
be
a) 5:4 b) 4:5 c) 25:16 d) 16:25

164. If the acceleration of wedge in the shown arrangement a ms-2 towards left, then at this instant acceleration
of the block (magnitude only) would be

a) 4 a ms-2 b) a 17-8cos α ms-2 c) ( 17 )a ms

-2 d) 17cos α ×a ms-2
2
165. Two bodies A and B each of mass m are placed on a smooth horizontal surface. Two horizontal force F and
2 F are applied on the blocks A and B, respectively, as shown in the figure. The block A does not slide on
block B. Then the normal reaction acting between the two blocks is (assume no friction between the
blocks)

a) F b) F/2 c) F d) 3F
3
166. A vehicle is moving with a velocity v on a curved road of width b and radius of curvature R. For
counteracting the centripetal force on the vehicle, the difference in elevation required in between the
outer and inner edges of the rod is

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 a) v2b/Rg b) vb/Rg c) vb2/Rg d) vb/R2g

167. A body of mass 2 kg has an initial velocity of 3 ms-1 along OE and it is subjected to a force of 4 N in a
direction perpendicular toOE. The distance of body from O after 4 s will be

a) 12 m b) 20 m c) 8 m d) 48 m

168. The acceleration of the block B in the following figure, assuming then surfaces and the pulleys P1 and P2 are
all smooth is

a) F b) 3F c) F d) 3F
4m 13m 2m 17m
169. A lift is moving down with an acceleration a. A man in the lift drops a ball inside the lift. The acceleration of
the ball as observed by the man standing stationary on the ground are, respectively,
a) a,g b) (g-a);g c) a,a d) g,g

170. A wooden box is placed on a table. The normal force on the box from the table isN1. Now another identical
box is kept on first box and the normal force on lower block due to upper block is N2 and normal force on
lower block by the table isN3. For this situation mark out the correct statement (s)
a) N = N = N b) N < N = N c) N = N < N d) N = N > N
1 2 3 1 2 3 1 2 3 1 2 3

171. Two identical particles A andB, each of mass m, are interconnected by a spring of stiffness k. If the particles
B experience a force F and the elongation of the spring is x, the acceleration of particles B relative to
particle A is equal to

a) F b) F-kx c) F-2kx d) kx
2m m m m
172. A solid block of mass 2 kg is resting inside a cube as shown in the figure. The cube is moving with a
⃗ ̂ ̂ -1
velocity v = 5 i + 2 j ms . If the coefficient of friction between the surface of cube and block is 0.2, then
the force of friction between the block and cube is

a) 10 N b) 4 N c) 14 N d) Zero

173. A unidirectional force F varying with time t as shown in the figure acts on a body initially at rest for a short
duration2 T. Then the velocity acquired by the body is
P a g e | 33
 a) πF0T b) πF0T c) F0T d) Zero
4m 2m 4m
174. An object is suspended from a spring balance in a lift. The reading is 240 N when the lift is at rest. If the
spring balance reading now changes to 220 N, then the lift is moving
a) Downward with constant speed b) Downward with decreasing speed

c) Downward with increasing speed d) Upward with increasing speed

175. A ball of mass m moving with a velocity u rebounds from a wall. The collision is assumed to be elastic and
the force of interaction between the ball and wall varies as shown in the figure. Then the value of F0 is

a) mu/T b) 2 mu/T c) 4 mu/T d) mu/2 T

176. In the figure shown, the block of mass m is at rest relative to the wedge of mass Mand the wedge is at rest
with respect to ground. This implies that

a) Net force applied by m on M is mg b) Normal force applied by m on M is mg

c) Force of friction applied by m on M is mg d) None of the above

177. For the pulley system shown, each of the cables at A and B is given a velocity of 2 ms-1 in the direction of
the arrow. Determine the upward velocity v of the load m

a) 1.5 ms-1 b) 3 ms-1 c) 6 ms-1 d) 4.5 ms-1

178. The maximum value of mass of block C so that neither A nor B moves is (Given that mass of A is 100 kg and
that of B is 140 kg. Pulleys are smooth and friction coefficient between A and b and between B and
-2
horizontal surface isμ = 0.3). take g = 10 ms

P a g e | 34
 a) 210 kg b) 190 kg c) 185 kg d) 162 kg

179. Three light strings are connected at the pointP. A weight W is suspended from one of the string. End A of
string AP and end B of string PB are fixed as shown. In equilibrium, PB is horizontal and PA makes an angle
of 60° with the horizontal. If the tension in PB is 30 N, then the tension in PA and weight W are,
respectively, given by

a) 60 N; 30 N b) 60/ 3;30 3 N c) 60 N;30 3 N d) 60 3;30 3 N

180. If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame,
one can surely say that
a) Liner momentum of the system does not change in time

b) Kinetic energy of the system does not change in time

c) Angular momentum of the system does not change in time

d) Potential energy of the system does not change in time

181. In order to raise a mass of 100 kg, a man of mass 60 kg fastens a rope to its and passes the rope over a
smooth pulley. He climbs the rope with acceleration 5 g/4 relative to the rope. The tension in the rope is
-2
(take g = 10 ms )

a) 4875 N b) 4875 N c) 4875 N d) 4875 N

8 2 4 6

Multiple Correct Answers Type

182. A 20 kg black is placed on top of 50 kg block as shown. An horizontal force F acting on A cause an
-2 -2
acceleration of 3 ms to A and 2 ms to B as shown. For this situation, mark out the correct statement (s)

P a g e | 35
 a) The friction force between A and B is 40 N b) The net force acting on A is 150 N

c) The value of F is 190 N d) The value of F is 150 N

183. A block is pressed against a vertical wall as shown in the figure

a) It is most easier to slide the block along 4

b) It is most difficult to slide the block along 1

c) It is equally easier or difficult to slide the block in any direction

d) It is most difficult to slide the block along 3

184. A 10 kg block is placed on top of 40 kg block as shown. A horizontal force F acting on B causes an
-2
acceleration of 2 ms toB. For this situations mark out the correct statement(s)

a) The acceleration of A may be 2 ms-2 or less than 2 ms-2

b) The acceleration of A must also be 2 ms-2

c) The coefficient of friction between the blocks may be 0.2

d) The coefficient of friction between the blocks must be 0.2 only

185. Two blocks of masses m1 and m2 are connected through a massless inextensible string. Block of mass m1 is
placed at the fixed rigid inclined surface while the block of mass m2 hanging at the other end of the string,
which is passing through a fixed massless frictionless pulley shown in the figure. The coefficient of static
friction between the block and the inclined plane is 0.8. the system of masses m1 and m2 is released from
rest

a) The tension in the string is 20 N after releasing the system

b) The contact force by the inclined surface on the block is along normal to the inclined surface

P a g e | 36
 c) The magnitude of contact force by the inclined surface on the block m is 20 3 N
1

d) None of these

186. Coefficient of friction between the two blocks is 0.3. Whereas the surface AB is smooth

a) Acceleration of the system of masses is 88/15 ms-2

b) Net force acting on 3 kg mass is greater than that on 2 kg mass

c) Tension T > T
2 1

d) Since 10 kg mass is acceleration downwards, so net force acting on it should be greater than any of the
two blocks shown in the figure
187. A body of mass 5 kg is suspended by the strings making angles 60° and 30° with the horizontal as shown in
-2
the figure(g = 10 ms ). Then

a) T = 25 N b) T =25 N c) T = 25 3 N d) T = 25 3 N
1 2 1 2

188. The spring balance A reads 2 kg with a block m suspended from it. A balance B reads 5 kg when a beaker
filled with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging
mass is inside the liquid as shown in figure. In this situation

a) The balance A will read more than 2 kg

b) The balance B will read more than 5 kg

c) The balance A will read less than 2 kg and B will read more than 5 kg

d) The balances A and B will read 2 kg and 5 kg respectively

189. In the figure, if F = 4 N, m = 2 kg, M = 4 kg, then

P a g e | 37
 a) The acceleration of m w.r.t. ground is 2 ms-2 b) The acceleration of m w.r.t. ground is 1.2 ms-2
3
c) Acceleration of M is 0.4 ms-2 d) Acceleration of M w.r.t. ground is 2 ms-2
3
190. Two blocks A and B of masses 5 kg and 2 kg, respectively, connected by spring of force constant= 100
-1
Nm are placed on an inclined plane of inclination 30° as shown in figure. If the system is released from
rest

a) There will be no compression or elongation in the spring if all surfaces are smooth

b) There will be elongation in the spring if A is rough and B is smooth

c) Maximum elongation in the spring 35 cm if all surfaces are smooth

d) There will be elongation in the spring if A is smooth and B is rough

191. ⃗ ⃗ ⃗ ⃗
Two rough blocks A and B, A placed over B, move with acceleration a A and a B,, velocities v A and v B by
⃗ ⃗
the action of horizontal forces F A and F B, respectively. When no friction exists between the blocks A and
B,

a) v = v b) a = v c) Both a. and b. d) FA = FB
A B A B
mA mB
192. In the figure, a man of true mass M is standing on a weighing machine placed in a cabin. The cabin is joined
by a string with a body of massm. Assuming no friction, and negligible mass of cabin and weighing
machine, the measured mass of man is (normal force between the man and the machine is proportional to
the mass)

a) Measured mass of man is Mm b) Acceleration of man is mg

(M+m) (M+m)

P a g e | 38
 c) Acceleration of man is Mg d) Measured mass of man is M
(M+m)
193. A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at t = 0. At this instant
of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at
t = 0 along the horizontal string AB, with the speed v. Friction between the bead and the string may be
neglected. Let tp and tQ be the respective times taken by P and Q to reach the point B. Then

a) t < t b) t = t
P Q P Q

c) t > t d) tP = Length of arc ACB

P Q
tQ Length of AB
194. A ship of mass 3×107 kg, initially at rest, is pulled by a force of 5×104 N through a distance of 3 m.
Assuming that the resistance due to water is negligible, the speed of the ship is
a) 1.5 ms-1 b) 60 ms-1 c) 0.1 ms-1 d) 5 ms-1

195. A reference frame attached to the Earth

a) In an inertial frame by definition

b) Cannot be an inertial frame because the Earth in revolving round the Sun

c) Is an inertial frame because Newton’s laws are applicable in thus frame

d) Cannot be an inertial frame because the Earth is rotating about its own axis

196. Which of the following statement (s) can be explained by Newton’s second law of motion?

a) To stop of heavy body (say truck), much greater force is needed than to stop a light body (say
motorcycle), in the same time, if they are moving with the same speed
b) For a given body, the greater the speed, the greater the opposing force needed to stop the body in a
certain time
c) To change the momentum (given), the force required is independent of time

d) The same forces acting on two different bodies for same time cause the same change in momentum in
the bodies
197. A gardner waters the plants by a pipe of diameter 1 mm. The water comes out at the rate or 10 cm3s-1. The
reactionary force exerted on the hand of the gardner is
a) Zero b) 1.27 ×10-2 N c) 1.27 ×10-4 N d) 0.127 N

198. A block of mass 2 kg rests on a rough inclined plane making an angle of 30° with horizontal. The coefficient
of static friction between the block and the plane is 0.7. The frictional force on the block is
a) 9.8 N b) 0.7×9.8× 3 N c) 9.8× 3 N d) 0.7×9.8 N

199. A block is resting over a rough horizontal floor. Att = 0, a time-varying force starts acting on it, the force is
described by equationF = kt, where k is constnst and t is in second. Mark the correct statement (s) for this
situation

P a g e | 39
 a) Curve 1 shows acceleration-time graph b) Curve 2 shows acceleration-time graph

c) Curve 3 shows velocity-time graph d) Curve 4 shows displacement-time graph

200. A golf ball of mass 0.05 kg placed on a tee, is struck by a golf club. The speed of the gold ball as it leaves the
-1
tee is 100 ms , the time of contact between them is 0.02 s. If the force decreases to zero linearly with time,
then the force at the beginning of the contact is
a) 5000 N b) 250 N c) 200 N d) 100 N

201. The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in
equilibrium, the angle θ shopuld be

a) 0° b) 30° c) 45° d) 60°

202. An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the insect and
the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle α
with the vertical, the maximum possible value of α is given by

a) cot α = 3 b) tan α = 3 c) sec α = 3 d) cosec α = 3

203. Which of the following are correct?

a) A parachutist of weight W strikes the ground with his legs and comes to rest with an upward
acceleration of magnitude 3 g. force exerted on him by ground during landing is 4 W
b) Two massless spring balances are hung vertically in series from a fixed point and a mass M kg is
attached to the lower end of the lower spring balance. Each spring balance reads M kgf
A rough vertical broad has an acceleration a along the horizontal direction so that a block of mass m
c) pressing against its vertical side does not fall. The coefficient of friction between the block and the
broad is greater than g/a
d) A man is standing at a spring platform. If man jumps away from the platform the reading of the spring
balance first increases and then decreases to zero
204. A block of mass m is placed in contact with one end of a smooth tube of massM. A horizontal force F acts in
the tube in each case (i) and (ii). Then,

P a g e | 40
 a) a = 0 and a = F in (i) b) a = a = F in (i)
m m
M
M M
M+m
c) a = a = F d) Force on m is mF in (ii)
m
in (ii)
M
M+m M+m
205. Two blocks A and B of masses mA and mB have velocity v and d2v, respectively, at a given instant. 000000A
horizontal force F acts on the blockA. There is no friction between ground and block B and coefficient of
friction between A and B isμ. The friction

a) On A supports its motion b) On B opposes its motion relative to A

c) On B opposes its motion d) Opposes the motion of both

206. Mark the correct statement (s) regarding friction

a) Friction force can be zero, even through the contact surface is rough

b) Even though there is no relative motion between surfaces, frictional force may exist between them

c) The expression f = μ N or f = μ N are approximate expression

L s k k

d) The expression f = μ N tells that the directions of f and N are the same
L s L

207. Figure shows the displacement of particle going along the X-axis as a function of time. The force acting on
the particle is zero in the region

a) AB b) BC c) CD d) DE

208. A string of negligible mass going over a clamped pulley of mass m supports a block of M as shown in figure.
The force on the pulley by the clamp is given by

a) 2 Mg b) 2 Mg c) ( (M+m)2+m)g d) ( (M+m)2+M2 )g
209. A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 ms-1. A plumb bob

P a g e | 41
 is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is
a) Zero b) 30° c) 45° d) 60°

210. A block of mass m is placed on a wedge. The wedge can be accelerated in four manners marked as (1), (2),
(3) and (4) as shown. If the normal reactions in situations (1), (2), (3) and (4) are N1,N2,N3 and N4,
respectively, and acceleration with which the block slides on the wedge in the situations are b1,b2,b3 and b4,
respectively, then

a) N > N > N > N b) N > N > N > N c) b > b > b > b d) b > b > b > b
3 1 2 4 4 3 1 2 2 3 4 1 2 3 1 4

211. A 3 kg block of wood is on a level surface where μs = 0.25 and μk = 0.2. A force of 7 N is being applied
horizontally to the block. Mark the correct statement (s) regarding this situation
a) If the block is initially at rest, it will remain at rest and friction force will be about 7 N

b) If the block is initially moving, then it will continue its motion forever if force applied is in the direction
of motion of the block
c) If the block is initially moving and the direction of applied force is same as that of motion of block, then
-2
block moves with an acceleration of 1/3 ms along its initial direction of motion
d) If the block is initially moving and direction of applied force is opposite to that of initial motion of block,
then block decelerates, comes to a stop, and starts moving in the opposite direction
212. 80 railway wagons all of same mass 5×103 kg are pulled by an engine with a force of 4×105N. The tension
in the coupling between 30th and 31st wagon from the engine is
a) 25×104 N b) 40 ×104 N c) 20 ×104 N d) 32 ×104 N

213. The figure shows a block of mass m placed on a smooth wedge of massM. Calculate the minimum value of
'
M and tension in the string, so that the block of mass m will move vertically downward with acceleration
-2
10 ms

a) The value of M' is Mcot θ

1-cot θ
b) The value of M' is Mtan θ
1-tan θ
c) The value of tension in the string is mg
tan θ
d) The value of tension is , Mg
cot θ
P a g e | 42
214. A block of mass 0.1 kg is held against a wall applying a horizontal force of 5 N on the block. If the
coefficient of friction between the block and the wall is 0.5, the magnitude of the friction force acting on
the block is
a) 2.5 N b) 0.98 N c) 4.9 N d) 0.49 N

215. The string shown in the figure is passing over small smooth pulley rigidly attached to trolleyA. If the speed
of trolley is constant and equal to vA towards right, speed and magnitude of acceleration of block B at the
instant shown in figure are

2
a) v = v , a = 0 b) a = 0 c) v = 3 v d) a = 16vA
B A B B B
5 A B
125
216. In the arrangement shown in the figure, the ends P and Q of an unstretchable string move downwards with
uniform speedU. Pulleys A and B are fixed

Mass M moves upwards with speed

a) 2 Ucos θ b) U/cos θ c) 2 U/cos θ d) Ucos θ

217. Suppose a body, which is acted on by exactly two forces, is accelerated. For this situation, mark the
incorrect statement (s)
a) The body can’t move with constant speed b) The velocity can never be zero

c) The vector sum of two forces can’t be zero d) The two forces must act in the same line

218. Seven pulleys are connected with the help of three light strings as shown in the figure below. Consider
P3,P4,P5 as light pulleys and pulleys P6 and P7 have masses m each. For this arrangement, mark the correct
statement (s)

a) Tension in the string connecting P ,P , and P is zero

1 3 4

b) Tension in the string connecting P ,P and P is mg/3

1 3 4

c) Tensions in all the three strings are same and equal to zero

P a g e | 43
 d) Acceleration of P is g downwards and that of P is g upwards
6 7

219. If a dipole is situated in a non uniform field,

⃗ ⃗
a)
∑ F =0,∑⃗ τ =0 b)
∑ F ≠0,but∑⃗ τ =0
⃗ ⃗
c)
∑ F =0,but∑⃗ τ ≠0 d)
∑ F ≠0,∑⃗ τ ≠0
220. A man of mass M is standing on a board of mass m. The friction coefficient between the board and the floor
is μ, figure. The maximum force that the man can exert on the rope so that the board does not move is

a) μ(m + M)g b) μ(m+M)g c) μ(m+M)g d) None of these

μ+1 μ-1
221. The ring shown in the figure is given a constant horizontal acceleration(a0 = g/ 3). Maximum deflection
of the string from the vertical is θ0, then

a) θ = 30° b) θ = 60°
0 0

At maximum deflection, tension in string is equal

c) At maximum deflection, tension in string is equal d)
2mg
to mg to
3
222. The accelerations of a particle as observed from two different frames S1 and S2 have equal magnitudes of 2
-2
ms
a) The relative acceleration of the frame may either be zero or 4 ms-2

b) Their relative acceleration may have any value between 0 and 4 ms-2

c) Both the frames may be stationary with respect to earth

d) The frames may be moving with same acceleration in same direction

223. In the figure, the blocksA, B, and C of mass m each have acceleration a1,a2, and a3 , respectively. F1 and F2
are external forces of magnitude 2 mg and mg, respectively, then

P a g e | 44
 a) a ≠ a ≠ a b) a = a ≠ a c) a > a > a d) a ≠ a = a
1 2 3 1 2 3 1 2 3 1 2 3

224. During paddling of a bicycle, the force of friction exerted by the ground on the two wheels is such that it
acts
a) In the backward direction on the front wheel and in the forward direction on the rear wheel

b) In the forward direction on the front wheel and in the backward direction on the rear wheel

c) In the backward direction on both the front and the rear wheels

d) In the forward direction on both the front and the rear wheels

225. A man tires to remain in equilibrium by pushing with his hands and feet against two parallel walls. For
equilibrium,

a) The forces of friction at the two walls must be equal

b) Friction must be present on both walls

c) The coefficient of friction must be the same between both walls and the man

d) None of the above

226. In the figure, a small block is kept on m, then

a) The acceleration of m w.r.t. ground is F b) The acceleration of m w.r.t. ground is zero

m
c) The time taken by m to separate from M is 2lM d) The time taken by m to separate from M is 2lM
F F
227. A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle θ with the
horizontal. A horizontal force of 1 N acts on the block through its center of mass as shown in the figure.
2
The block remains stationary if (take g = 10 m/s )

P a g e | 45
 a) θ = 45°

b) θ > 45° and a frictional force acts on the block towards P

c) θ > 45° and a frictional force acts on the block towards Q

d) θ < 45° and a frictional force acts on the block towards Q

Assertion - Reasoning Type

This section contain(s) 0 questions numbered 228 to 227. Each question contains STATEMENT 1(Assertion)
and STATEMENT 2(Reason). Each question has the 4 choices (a), (b), (c) and (d) out of which ONLY ONE is
correct.

a) Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1

b) Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1

c) Statement 1 is True, Statement 2 is False

d) Statement 1 is False, Statement 2 is True

228

Statement 1: In the figure shown below, ground is smooth and masses of both the blocks are different.
Net force acting on each of the block is not same
Statement 2: Acceleration of both will be different

229

Statement 1: The work done in bringing a body down from the top to the base along a frictionless
incline plane is the same as the work done in the bringing it down the vertical side
Statement 2: The gravitational force on the body along the inclined plane is the same as that along the
vertical side
230

Statement 1: Force is required to move a body uniformly along a circle

Statement 2: When the motion is uniform, acceleration is zero

231

Statement 1: If the net external force on the body is zero, then its acceleration is zero

Statement 2: Acceleration does not depend on force

P a g e | 46
232

Statement 1: Frictional forces are conservative forces.

Statement 2: Potential energy can be associated with frictional forces.

233

Statement 1: Angle of repose is equal to angle of limiting friction

Statement 2: When the body is just at the point of motion, the force of friction in this stage is called as
limiting friction
234

Statement 1: It is easier to pull a heavy object than to push it on a level ground

Statement 2: The magnitude of frictional force depends on the nature of the two surfaces in contact

235

Statement 1: A body in equilibrium has to be at rest only

Statement 2: A body in equilibrium may be moving with a constant speed along a straight line path

236

Statement 1: A block of mass m is placed on a smooth inclined plane of inclination θ with the
horizontal. The force exerted by the plane on the block has a magnitude mgcos θ
Statement 2: Normal reaction always cats perpendicular to the contact surface

237
-2.
Statement 1: The acceleration of body sliding down a smooth plane of inclination 30 °is 5ms

Statement 2: Acceleration is given by a =μg sin θ.

238

Statement 1: Use of ball bearings between two moving parts of machine is a common practice

Statement 2: Ball bearings reduce vibrations and provide good stability

239

Statement 1: Inertia is the property by virtue of which the body is unable to change by itself the state
of rest only
Statement 2: The bodies do not change their state unless acted upon by an unbalanced external force

240

Statement 1: Coefficient of friction can be greater than unity

Statement 2: Force of friction is dependent on normal reaction and ratio of force of friction and normal
reaction cannot exceed unity
241

Statement 1: On a rainy day, it is difficult to drive a car or bus at high speed

P a g e | 47
 Statement 2: The value of coefficient of friction is lowered due to wetting of the surface

242

Statement 1: Block A is moving on horizontal surface towards right under action of force F. All surfaces
are smooth. At the instant shown, the force exerted by block A on block B is equal to net
force on block B

Statement 2: From Newton’s third law, the force exerted by block A on B is equal in magnitude to force
exerted by block B on A
243

Statement 1: An electric fan continues to rotate for some time after the current is switched off

Statement 2: It is because of inertia of rest

244

Statement 1: A block is lying stationary as on inclined plane and coefficient of friction isμ. Friction on
block is μ mgcos θ
Statement 2: Contact force on block is mg

245

Statement 1: Linear momentum of a body changes even when it is moving uniformly in a circle

Statement 2: In uniform circular motion velocity remain constant

246

Statement 1: A concept of pseudo forces is valid both for inertial as well as non-inertial frame of
reference
Statement 2: A frame accelerated with respect to an inertial frame is a non-inertial frame

247

Statement 1: A frame of reference which is moving with uniform velocity is non inertial frame of
reference.
Statement 2: A reference frame in which Newton’s laws of motion are applicable is non – inertial.

248

Statement 1: The apparent weight of a body in an elevator moving with some downward acceleration
is less than the actual weight of body
Statement 2: The part of the weight is spent in producing downward acceleration, when body is in
elevator
249

Statement 1: When a bicycle is in motion, the force of friction exerted by the ground on the two wheels
is always in forward direction

P a g e | 48
 Statement 2: The frictional force acts only when the bodies are in contact

250

Statement 1: A table cloth can be pulled from a table without dislodging the dishes

Statement 2: To every action there is an equal and opposite reaction

251

Statement 1: A rocket in flight is a system of varying mass.

Statement 2: The rocket fuel is being consumed continuously.

252

Statement 1: Aeroplanes always fly at low altitudes

Statement 2: According to Newton’s third law of motion, for every action there is an equal and
opposite reaction
253

Statement 1: A body subjected to three concurrent forces cannot be in equilibrium

Statement 2: If large number of concurrent forces acting on the same point, then the point will be in
equilibrium, if sum of all the forces is equal to zero
254

Statement 1: Use of ball bearings between two moving parts of machine is a common practice.

Statement 2: Ball bearings reduce vibrations and provide good stability.

255

Statement 1: A cyclist always bends inward while negotiating a curve

Statement 2: By bending, cyclist lowers his centre of gravity

256

Statement 1: The acceleration of a body down a rough inclined plane is greater than the acceleration
due to gravity
Statement 2: The body is able to slide on a inclined plane only when its acceleration is greater than
acceleration due to gravity
257

Statement 1: A reference frame attached to the earth is an inertial frame of reference

Statement 2: Newton’s laws can be applied in this frame of reference

258

Statement 1: When ball of a mass m hits normally a wall with a velocity v and rebounds with same
Velocity v, impulse imparted to the ball is 2mv.
Statement 2: Impulse= change in linear momentum.

259

P a g e | 49
 Statement 1: Linear momentum of a body changes even when it is moving uniformly in a circle

Statement 2: Force required to move a body uniformly along a straight line is zero

260

Statement 1: A player lowers his hands while catching a cricket ball and suffers less reaction force

Statement 2: The time of catch increases when cricketer lowers its hand while catching a ball

261

Statement 1: The maximum speed with which a vehicle can go round a level curve of diameter 20 m
-1
without skidding is 10 ms , given μ = 0.1
Statement 2: It follows from v ≤ μrg

262

Statement 1: A string can never remain horizontal, when loaded at the middle, however large the
tension may be
Statement 2: W W
For horizontal spring, angle with vertical, θ = 90°⇒T = = =∞
2cos θ 2cos 90°
263

Statement 1: The greater the rate of the change in the momentum vector, the greater the force applied

Statement 2: ⃗ ⃗
dp
Newton’s second law is F =
dt
264

Statement 1: The driver of a moving car sees a wall in front of him. To avoid collision, he should apply
brakes rather than taking a turn way from the wall
Statement 2: Friction force is needed to stop the car or taking a turn on a horizontal road

265

Statement 1: In high jump, it hurts less when an athlete lands on a heap of sand

Statement 2: Because of greater distance and hence greater time over which the motion of an athlete is
stopped, the athlete experience less force when lands on heap of sand
266

Statement 1: Friction is a self adjusting force

Statement 2: Friction does not depend upon mass of the body

267

Statement 1: Pulling (figure a) is easier than pushing (figure b) on a rough surface

Statement 2: Normal reaction is less in pulling than in pushing

268

P a g e | 50
 Statement 1: A particle is found to be at rest when seen from a frame S1 and moving with a constant
velocity when seen from another frameS2. We can say both the frames are inertial
Statement 2: All frames moving uniformly with respect to an internal frame are themselves internal

269

Statement 1: Newton’s third law applies is applicable only when bodies are in motion

Statement 2: Newton’s third law applies to all types of forces, e.g. gravitational, electric or magnetic
forces etc.
270

Statement 1: When the lift moves with uniform velocity the man in the lift will feel weightlessness

Statement 2: In downward accelerated motion of lift, apparent weight of a body decreases

271

Statement 1: A bullet is fired from a rifle. If the rifle recoils freely, the kinetic energy of the rifle is more
than that of the bullet
Statement 2: In the case of rifle bullet system the law of conservation of momentum violates

272

Statement 1: Mass is a measure of inertia of the body in linear motion

Statement 2: Greater the mass, greater is the force required to change its state of rest or of uniform
motion
273

Statement 1: Two bodies of masses M and m(M > m) are allowed to fall from the same height if the air
resistance for each be the same then both the bodies will reach the earth simultaneously
Statement 2: For same air resistance, acceleration of both the bodies will be same

274

Statement 1: Force is required to move a body uniformly along a circle

Statement 2: When the motion is uniform, acceleration is zero

275

Statement 1: It is not possible to drive a car on a slippery road

Statement 2: Friction always opposes motion

276

Statement 1: The velocity of a body at the bottom of an inclined plane of given height is more when it
slides down the plane, compared to, when it rolling down the same plane
Statement 2: In rolling down a body acquires both, kinetic energy of translation and rotation

277

Statement 1: The slope of momentum versus time curve give us the acceleration

Statement 2: Acceleration is given by the rate of change of momentum

P a g e | 51
278

Statement 1: The value of dynamic friction is less than the limiting friction

Statement 2: Once the motion has started, the inertia of rest has been overcome

279

Statement 1: Frictional heat generated by the moving ski is the chief factor which promotes sliding in
skiing while waxing the ski makes skiing more easy
Statement 2: Due to friction energy dissipates in the form of heat as a result it melts the snow below it.
Wax is water repellent
280

Statement 1: Two particles are moving towards each other due to mutual gravitational attraction. The
momentum of each particle will increase
Statement 2: Rate of change of momentum depends upon Fext

281

Statement 1: A man in a closed cabin falling freely does not experience gravity

Statement 2: Inertial and gravitational mass have equivalence

282

Statement 1: A reference frame attached to earth is an inertial frame of reference

Statement 2: The reference frame which has zero acceleration is called a non inertial frame of
reference
283

Statement 1: Friction is a self adjusting force

Statement 2: The magnitude of static friction is equal to the applied force and its direction is opposite
to that of the applied force
284

Statement 1: Moment of inertia is same as inertia

Statement 2: Moment of inertia of a body represents rotational inertia of the body

Matrix-Match Type

This section contain(s) 0 question(s). Each question contains Statements given in 2 columns which have to be
matched. Statements (A, B, C, D) in columns I have to be matched with Statements (p, q, r, s) in columns II.

285. The system shown below is initially in equilibrium. Masses of the blocks A, B, C, D, and E are, respectively,
3 kg, 3 kg, 2 kg, 2 kg and 2 kg, Match the conditions in Column I with the effect in Column II

P a g e | 52
 Column-I Column- II

(A) After spring 2 is cut, tension is string AB (p) Increases

(B) After spring 2 is cut, tension in string CD (q) Decreases

(C) After string between C and pulley is cut, (r) Remain constant
tension in string AB
(D) After string between C and pulley is cut, (s) Zero
tension in string CD
CODES :

A B C D

a) c b b,d b

b) b b,d b c

c) b,d b c b

d) b c b b,d

286. Column I gives four different situations involving two blocks of mass m1 and m_2 placed in different ways
on smooth horizontal surface as shown. In each of the situations, horizontal forces F1 and F2 are applied or
blocks of mass m1 andm2, respectively and also m2F1 < m1F2. Match the statements in Column I with
corresponding results in Column II
Column-I Column- II

(A)

Both the block are connected by the massless

(p)
( )
m1m2 F1 F2
-
m1+m2 m1 m2

inelastic string. The magnitude of tension in

the string is
(B)

Both the blocks are connected by the massless

(q)
(
m1m2 F1 F2
+
m1+m2 m1 m2 )
inelastic string. The magnitude of tension in
the string is
(C) (r)
( )
m1m2 F2 F1
-
m1+m2 m2 m1

The magnitude of normal reaction between

the blocks is

P a g e | 53
 (D) (s)
m1m2 ( F1+F2
m1+m2 )
The magnitude of normal reaction between
the blocks is
CODES :

A B C D

a) c b c b

b) b c a c

c) c b a b

d) b c b c

287. Coefficient of friction between the block and the surface in each of the given figures is 0.4. Match Column I
with that of column II

Column-I Column- II

(A) Force of friction is zero in (p) Fig i.

(B) Force of friction is 2.5 N in (q) Fig ii.

(C) Acceleration of the block is zero in (r) Fig iii.

(D) Normal force is not equal to 2g in (s) Fig iv.

CODES :

A B C D

a) A,b,c,d c,d a,c b,d

b) c,d a,c b,d a,b,c,d

c) a,c b,d a,b,c,d c,d

d) b,d a,b,c,d c,d a,c

288. There is no friction anywhere in the system shown in figure. The pulley is light. The wedge is free to move
on a frictionless surface. A horizontal force F is applied on the system in such a way that m does not slide
on M or both move together with some common acceleration. Given M > 2 m

P a g e | 54
 Match the entries of Column I with that of Column II
Column-I Column- II

(A) Pseudo force acting on m as seen from the (p) mF

Equal to
frame of M is m+M
(B) Pseudo force acting on M as seen from the (q) mF
Greater than
frame of m is m+M
(C) Normal force (for θ = 45°) between m and M (r) Less than mgsin θ
is
(D) Normal force between ground and M is (s) Greater than mgsin θ

CODES :

A B C D

a) A,c b,c b,c b,d

b) b,c b,c b,d a,c

c) b,c b,d a,c b,c

d) b,d a,c b,c b,c

289. In the figure shown, a block of mass m is released from rest when spring was in its natural length. The
pulley also has mass m but it is friction less. Suppose the value of m is such that finally it is just able to lift
the block M up after releasing it

Column-I Column- II

(A) Weight of ‘m’ required to just lift ‘M’ (p) M

3 g
2
(B) Tension in the rod, when ‘m’ is in equilibrium (q) Mg

(C) Normal force acting on M when m is in (r) M

g
equilibrium 2
(D) Tension in the string when displacement of m (s) 2 mg
is maximum possible
CODES :

A B C D

a) a b c d

b) b,a c d a

c) c c c b,d

d) a,c b c,d a,b

P a g e | 55
290. For the situation shown in figure in Column I, the statements regarding friction forces are mentioned,
while in Column II some information related to friction forces are given
Match the entries of Column I with the entries of Column II

Column-I Column- II

(A) Total friction force on 3 kg block is (p) Towards right

(B) Total friction force on 5 kg block is (q) Towards left

(C) Friction force on 2 kg block due to 3 kg block (r) Zero

is
(D) Friction force on 3 kg block due to 5 kg block (s) Non-zero
is
CODES :

A B C D

a) c a,d b,d b,d

b) a,d b,d b,d c

c) b,d b,d c a,d

d) b,d c a,d b,d

291. Column I describes the motion of the object and one or more of the entries of Column II may be the cause
of motions described in Column I. Match the entries of Column I with the entries of column II
Column-I Column- II

(A) An object is moving towards east (p) Net force acting on the object must be towards
east
(B) An object is moving towards east with (q) At least one force must act towards east
constant acceleration
(C) An object is moving towards east with varying (r) No forces may act towards east
acceleration
(D) An object is moving towards east with (s) No force may act on the object
constant velocity
CODES :

A B C D

a) A,c a,c c,d c,d

b) c,d a,c a,c c,d

c) a,c c,d c,d a,c

d) c,d c,d a,c a,c

292. Coefficient of friction between the masses 2 m and m is 0.5. All other surface are frictionless and pulleys

P a g e | 56
 are massless. Column I gives the different values of m1 and Column II gives the possible acceleration of 2 m
andm. Match the columns

Column-I Column- II

(A) m1 = 2m (p) Accelerations of 2 m and m are same

(B) m1 = 3 m (q) Accelerations of 2 m and m are different

(C) m1 = 4 m (r) Acceleration of 2 m is greater than m

(D) m1 = 6 m (s) Acceleration of m is less than 0.6 g

CODES :

A B C D

a) A,d a,d b,c,d b,c,d

b) a,d b,c,d b,c,d a,d

c) b,c,d b,c,d a,d a,d

d) b,c,d a,d a,d b,c,d

293. A block is attached to an unstretched vertical spring and released from rest. As a result of this block comes
down due to its weight, stops momentarily, and then bounces back. Finally the block starts oscillating up
and down

During oscillations, match Column I with Column II:

Column-I Column- II

(A) When the block is at its maximum downward (p) Acceleration is in upward direction
displacement position (may be known as
extreme position)
(B) When the block is at its equilibrium position (q) Acceleration is in downward is in downward
direction
(C) When the block is somewhere between (r) Acceleration is zero
equilibrium position and downward extreme
position
(D) When the block is above equilibrium position (s) Velocity may be in upward or in downward
but below the initial unstretched position direction
CODES :

A B C D
P a g e | 57
 a) C,d a,d b,d a

b) a c,d a,d b,d

c) a,d b,d a c,d

d) b,d a c,d a,d

294. A horizontal force F pulls a ring of mass m1 such that θ remains constant with time. The ring is constrained
to move along a smooth rigid horizontal wire. A bob of mass m2 hangs from m1 by an inextensible light
string. Then match the entries of Column I with that of Column II

Column-I Column- II

(A) F (p) (m1 + m2)g

(B) Force acting on m2 is (q) m2gsec θ

(C) Tension in the string is (r) F

m2
m1+m2
(D) Force acting on m1 by the wire is (s) (m1+m2)gtan θ
CODES :

A B C D

a) d c b a

b) c b a d

c) b a d c

d) a d c b

295. For the situation shown in the figure below, match the entries of Column I with the entries of Column II

Column-I Column- II

(A) If F = 12 N, then (p) There is relative motion between A and B

(B) If F = 15 N, then (q) There is relative motion between B and C

(C) If F = 25 N, then (r) There is relative motion between C and the

ground
(D) If F = 40 N, then (s) Relative motion is not there at any of the
surface
CODES :
P a g e | 58
 A B C D

a) B,c a,b,c c c

b) a,b,c c c b,c

c) c c b,c a,b,c

d) c b,c a,b,c c

296. In the system shown in figure, masses of the blocks are such that when the system is released, acceleration
of pulley P1 is a upwards and acceleration of block 1 is a1 upwards. It is found that acceleration of block 3 is
same as that of 1 both in magnitude and direction

a1
Given thata1 > a > . Match the following
2
Column-I Column- II

(A) Acceleration of 2 (p) 2a + a1

(B) Acceleration of 4 (q) 2a - a1

(C) Acceleration of 2 w.r.t. 3 (r) Upwards

(D) Acceleration of 2 w.r.t. 4 (s) Downwards

CODES :

A B C D

a) A,d d c b,c

b) b,c a,d d c

c) d c b,c a,d

d) c b,c a,d d

297. For the situation shown in figure, in Column I, the statements regarding friction forces are mentioned,
while in Column II some information related to friction forces are given

Match the entries of Column I with the entries of Column II

Column-I Column- II

P a g e | 59
 (A) Total friction force on 4 kg block is (p) Towards right

(B) Total friction force on 2 kg block is (q) Towards left

(C) Friction force on 6 kg block due to 2 kg block (r) Zero

is
(D) Total Friction force on 6 kg block is (s) Non-zero

CODES :

A B C D

a) A,c a,c a d

b) c,d a b,c a,d

c) b,d c a,d c

d) a,c a c d

298. For the figure shown, both the pulley are massless and frictionless. A force F (of any possible magnitude) is
applied in horizontal direction. There is no friction between M and ground m1 and m2 are the coefficients
of friction as shown between the blocks. Column I gives the different relations between m1 andm2, and
Column II is regarding the motion ofM. Match the columns:

Column-I Column- II

(A) If μ1 = μ2 = 0 (p) May accelerate towards right

(B) If μ1 = μ2 ≠ 0 (q) May accelerate towards right

(C) If μ1 > μ2 (r) Does not accelerate

(D) If μ1 < μ2 (s) May or may not accelerate

CODES :

A B C D

a) A,b c,d c,d d

b) d,c a,c c a

c) a,d b,d a, b

d) c c b,d a,d

299. When the system shown in figure is released, A accelerates downwards

P a g e | 60
 Column-I Column- II

(A) Acceleration of B (p) Towards left

(B) Acceleration of C w.r.t B (q) Towards right

(C) Acceleration of A w.r.t. C (r) At some angle θ with horizontal

(0 < θ < 90°)
(D) Acceleration of B w.r.t. A (s) At some angle θ with vertical (0 < θ < 90°)

CODES :

A B C D

a) A,b c d a

b) b a c,d c,d

c) a,b c a,d a

d) c,d a c b

Linked Comprehension Type

This section contain(s) 52 paragraph(s) and based upon each paragraph, multiple choice questions have to be
answered. Each question has atleast 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

Paragraph for Question Nos. 300 to -300

A body of mass 10 kg is lying on a rough horizontal surface. The coefficient of friction between the body and
horizontal surface is 0.577. When the horizontal surface is inclined gradually, the body just begins to slide at a
certain angle α. This is called angle of repose. When angle of inclination is increased further, the body slides
down with some acceleration

300. The minimum force required just to slide the block on the horizontal surface is

a) 57.7 N b) 100 N c) 100 kg d) 57.7 kg

Paragraph for Question Nos. 301 to - 301

A force that acts on a body for a very short time is called impulsive force. Impulse measures the effect of the
force. It is the product of force and time for which the force acts. Impulse is measured by the change in
momentum of the body. For a given change in momentum, Favg×t = constant. By increasing the tie (t) of impact,
we can reduce the average force Fav
-2
Read the above passage carefully and answer the following question (g = 10 ms )

301. If the impact lasts for 0.1s, force exerted by the impinging ball on the ground is

a) 45.2 N b) 45.2 kg-wt c) 42.5 N d) 42.5 kg-wt

P a g e | 61
Paragraph for Question Nos. 302 to - 302

In the system shown in the figure, m1 > m2. System is held at rest by threadBC. Now thread BC is burnt. Answer
the following:

302. Before burning the thread, what are the tension in spring and threadBC, respectively?

a) m g, m g b) m g, m g - m g c) m g, m g d) m g, m g + m g
1 2 1 1 2 2 1 1 1 2

Paragraph for Question Nos. 303 to - 303

For problems-4-6
Three blocksA, B, and C having masses 1 kg, 2 kg, and 3 kg, respectively, are arranged as shown in figure. The
pulleys P and Q are light and frictionless. All the blocks are resting on a horizontal floor and the pulleys are held
such that strings remain just taut. At momentt = 0, a force F = 40t N starts acting on pulley P along vertically
-2
upward direction as shown in the figure. Take g = 10 ms

303. Regarding the times when blocks lose contact with ground, which is correct?

a) A loses contact at t = 2 s b) C loses contact at t = 1.5 s

c) A and B lose contact at the same time d) All three blocks lose contact at the same time

Paragraph for Question Nos. 304 to - 304

-1
A ball of mass 200 gm is thrown with a speed 20 ms . The ball strikes a bat and rebounds along the same line at
-1
a speed 40 ms . Variation of the interaction force, as long s the ball remains in contact with the bat, is as shown

P a g e | 62
in the figure

304. Maximum force F0 exerted by the hat on the ball is

a) 4000 N b) 5000 N c) 3000 N d) 2500 N

Paragraph for Question Nos. 305 to - 305

In the system shown in the figure, mA = 4 m, mB = 3 m, and mc = 8 m. friction is absent everywhere. String is
inextensible and light. If the system is released from rest, then

305. The tension in the string is

a) 1.5 mg b) 5.8 mg c) 4.7 mg d) 3.2 mg

Paragraph for Question Nos. 306 to - 306

Block B rests on a smooth surface. The coefficient of static friction between A and B isμ = 0.4. When F = 30 N,
then

306. Acceleration of upper block is

a) 3/2 ms-2 b) 6/7 ms-2 c) 4/3 ms-2 d) 3/7 ms-2

Paragraph for Question Nos. 307 to - 307

Study the following diagram and answer the following questions accordingly. Neglect all friction and the masses
of the pulleys

P a g e | 63
307. What is the tension in the string?

a) 700 N b) 450 N c) 500 N d) 900 N

11 11 11 11

Paragraph for Question Nos. 308 to - 308

A monkey of mass m clings to a rope slung over a fixed pulley. The opposite end of the rope is tried to a weight
of mass M lying on a horizontal table. The coefficient of friction between the weight and the table isμ. Find the
acceleration of weight and the tension of the rope for two cases

The monkey moves downwards with respect to the rope with an acceleration b

308. The acceleration of weight is

a) 2m(g+b)-μMg n b) m(g+b)-μMg c) m(g+b)-3μMg d) m(g-b)-μMg

M+2m 2(M+m) M+3m M+m

Paragraph for Question Nos. 309 to - 309

Block A weights 4 N and block B weights 8 N. The coefficient of kinetic friction is 0.25 for all surfaces. Find the
force F to slide B at constant speed when

309. A rests on B and moves with it (figure a)

a) 2 N b) 3 N c) 4 N d) 5 N

Paragraph for Question Nos. 310 to - 310

Block A of mass m and block B of mass 2m are placed on a fixed triangular wedge by means of a massless
inextensible string and a frictionless pulley as shown in the figure. The wedge in inclined at 45° to the
horizontal on both sides. The coefficient of friction between block A and the wedge is 2/3 and that between
block B and the wedge is 1/3. If the system of A and B is released from rest, find the following

P a g e | 64
310. Acceleration of A is

a) g b) Zero c) g d) g
3 2 7 2 3

Paragraph for Question Nos. 311 to - 311

A block of mass 10 kg is kept on a rough floor. Coefficient of friction between floor and block are μs = 0.4
andμk = 0.3. forces F1 = 5 N and F2 = 4 N are applied on the block as shown in the figure

311. Determine the magnitude to friction force

a) 31 N b) 26 N c) 41 N d) 36 N

Paragraph for Question Nos. 312 to - 312

Block A has mass 40 kg and B has mass 15 kg and F is 500 N parallel to smooth inclined plane. The system is
moving together

312. The acceleration of the system is

a) 45 ms-2 b) 23 ms-2 c) 13 ms-2 d) 8 ms-2

11 11 7 3

Paragraph for Question Nos. 313 to - 313

A 10 kg block rests on a 5 kg bracket as shown in the figure. The 5 kg bracket rests on a horizontal frictionless
surface. The coefficients of friction between the 10 kg block and the bracket on which in rests are μs = 0.40 and
μs = 0.30

313. The maximum force F that can be applied if the 10 kg block is not to slide on the bracket is

a) 32 N b) 24 N c) 18 N d) 48 N

P a g e | 65
Paragraph for Question Nos. 314 to - 314

A sufficiently long plank of mass 4 kg is placed on a smooth horizontal surface. A small block of mass 2 kg is
placed over the plank and is being acted upon by a time-varying horizontal forceF = (0.5 t), where F is in
newton and t is in seconds as shown in the figure. The coefficient of friction between the plank and the block is
given asμa = μk = μ. At ti9me t = 12 s, the relative slipping between the plank and the block is just likely to
occur

314. The coefficient of friction μ is equal to

a) 0.10 b) 0.15 c) 0.20 d) 0.30

Paragraph for Question Nos. 315 to - 315

Three blocksA, B, and C of masses3 M, 2M, and M are suspended vertically with the help of springsPQ, andTU,
and a string RSas shown. If acceleration of blocks A, B, and C are a1,a2, and a3 , respectively, then

315. The value of acceleration a3 at the moment spring PQ is cut is

a) g, downward b) g, upwards

c) More than g, downwards d) Zero

Paragraph for Question Nos. 316 to - 316

In the figure, all the pulleys and strings are massless and all the surfaces are frictionless. A small block of mass
-2
m is placed on fixed wedge (take g = 10 ms )

P a g e | 66
316. The tension in the string attached to m is

a) 40 N b) 10 N c) 20 N d) 5 N

Paragraph for Question Nos. 317 to - 317

In the shown arrangement, both pulleys and the string are massless and all the surfaces are frictionless

Given m1 = 1 kg, m2 = 2 kg, m3 = 3 kg

317. Find the tension in the string

a) 120 N b) 240 N c) 130 N d) None of these

7 7 7

Paragraph for Question Nos. 318 to - 318

A plank A of mass M rests on a smooth horizontal surface over which it can move without friction. A cube B of
mass m lies on the plank at one edge. The coefficient of friction between the plank and the cube isμ. The size of
cube is very small in comparison to the plank

318. At what force F applied to the plank in the horizontal direction will the cube begin to slide towards the
other end of the plank?
a) F > μ(m + M)g b) F > 0.5μ(m + M)g c) F = 0.5μ(m+M)g d) F = μ(m + M)g

Paragraph for Question Nos. 319 to - 319

In the arrangement shown in the figure, all pulleys are smooth and massless. When the system is released from
-2 -2
the rest, acceleration of blocks 2 and 3 relative to 1 are 1 ms downwards and 5 ms downwards, respectively.
Acceleration of block 3 relative to 4 is zero
P a g e | 67
319. Find the absolute acceleration of block 1

a) 2 ms-2 upwards b) 1 ms-2 downwards c) 3 ms-2 upwards d) 1.5 ms-2 downwards

Paragraph for Question Nos. 320 to - 320

A sphere of mass 500 g is attached to a string of length 2 m, whose other end is fixed to a ceiling. The sphere is
made to describe a circle of radius 1 m in a horizontal plane

320. Find the period of revolution for the sphere

a) π 10 s b) π 5 s c) 2π 10 s d) π 5 s

Paragraph for Question Nos. 321 to - 321

A ball of mass m is suspended from a rope of lengthL. It describes a horizontal circle of radius r with speedv.
The rope makes and angle θ with the vertical

321. Find the tension in the rope

( ) ( ) ( ) ( )
2 2 2 2 2 2 2 2
a) (mg)2+ mv b) (mg)2- mv c) (mg)2- mv d) (mg)2+ mv
2r r 2r r

Paragraph for Question Nos. 322 to - 322

A small block of mass m is placed over a long plank of massM. Coefficient of friction between them isμ. Ground
is smooth. At t = 0, m is given a velocity v1 and M a velocity v2( > v1) as shown. After this M is maintained at
constant acceleration a( < μg)

P a g e | 68
Initially there will be some relative motion between the block and the plank, but after some time relative
motion will cease and velocities of both will become same

322. Find the time t0 when velocities of both block and plank becomes same

a) v2-v1 b) v2+v1 c) v2-v1 d) v2+v1

μg+a μg-a μg-a μg+a

Paragraph for Question Nos. 323 to - 323

Two blocks of masses m1 and m2 are connected with a light spring of force constant k and the whole system is
kept on a frictionless horizontal surface. The masses are applied forces F1 and F2 as shown. At any time, the
blocks have same acceleration a0 but in opposite directions. Now answer the following

323. The value of a0 is

a) F1-F2 b) F1-F2 c) F1+F2 d) F1+F2

m1+m2 m1-m2 m1-m2 m1+m2

Paragraph for Question Nos. 324 to - 324

A block of mass 4 kg is pressed against a rough wall by two perpendicular horizontal forces F1 and F2 as shown
in the figure. Coefficient of static friction between the block and wall is 0.6 and that of kinetic is 0.5

324. For F1 = 300 N and F2 = 100 N, find the direction and magnitude of friction force acting on the block

a) 180 N, vertically upwards

b) 40 N vertically upwards

()
c) 107.7 N making an angle of tan-1 2 with the horizontal in the upward direction
5

()
d) 91.6 N, making an angle of tan-1 2 with the horizontal in the upward direction
5

Paragraph for Question Nos. 325 to - 325

A system of two blocks and a light string are kept on two inclined faces (rough) as shown in the figure below.
All the required data are mentioned in the diagram. Pulley is light and frictionless (Take g = 10

P a g e | 69
 -2 3
ms ,sin 37° = )
5

325. If the system is released from rest, then the acceleration of the system is

a) 7 ms-2 b) Zero c) 47 ms-2 d) 2.25 ms-2

15 15 15

Paragraph for Question Nos. 326 to - 326

A system of two blocks is placed on a rough horizontal surface as shown in the figure below. The coefficient of
static and kinetic friction at two surfaces are shown. A force F is horizontally applied on the upper block as
shown
Let f1,f2 represent the frictional forces between upper and lower surfaces of contact, respectively, and a1,a2
represent the acceleration of 3 kg an d2 kg block, respectively

326. If F is gradually increasing force then which of the following statement (s) would be true?

a) For a particular value of F( < F ) there is no motion at any of the contact surface
0

b) The value of F is 10 N
0

c) As F increase beyond F ,f increases and continues to increase until it acquires its limiting value
0 1

d) All of the above

Paragraph for Question Nos. 327 to - 327

Two smooth blocks are placed at a smooth corner as shown. Both the blocks are having massm. We apply a
force F on the small blockm. Block A prsses block B in the normal direction, due to which passing force on
vertical wall will increase, and pressing force on the horizontal wall decreases, as we increases F(θ = 37° with
horizontal)
As soon as the pressing force on the horizontal wall by block B becomes zero, it will lose contact with ground. If
the value4 of F further increase, block B will accelerate in the upward direction and simultaneously block A will
move towards right

P a g e | 70
327. What is minimum value of F to lift block B from ground?

a) 25 mg b) 5 mg c) 3 mg d) 4 mg
12 3 4 3

Paragraph for Question Nos. 328 to - 328

Two containers of sand are arranged like the block as shown. The containers alone have negligible mass; the
sand in them has a total mass Mtot; the sand in the hanging container H has mass m

To measure the magnitude a of the acceleration of the system, a larger number of experiments carried out
where m varies from experiment to experiment but Mtot does not; that is sand is shifted between the containers
before each trial

328. Which of the curves in graph correctly gives the acceleration magnitude as a function of the ratio m/Mtot
(vertical axis is for acceleration)
a) 1 b) 2 c) 3 d) 4

Paragraph for Question Nos. 329 to - 329

Two bodies A and B of masses 10 kg and 5 kg are placed very slightly separated as shown in the figure. The
coefficient of friction between the floor and the blocks are asμa = 0.4. Block A is pushed by an external forceF.
The value of F can be changed. When the welding between block A and ground breaks, block A will start
pressing block B and when welding of B also breaks, block B will start pressing the vertical wall

329. If F = 20 N, with how much force does block A press blockB?

a) 10 N b) 20 N c) 30 N d) Zero

P a g e | 71
Paragraph for Question Nos. 330 to - 330

A string of length l is fixed at one end and carries a mass m at the other end. The string makes 2/π rps around a
vertical axis throughout the fixed end so that the mass moves in horizontal circle

330. What is the tension in string?

a) m l b) 16m l c) 4m l d) 2m l

Paragraph for Question Nos. 331 to - 331

2
A time-varying force F = 6t - 2t N, at t = 0 stats acting on a body of mass 2 kg initially at rest, where t is in
second. The force is withdrawn just at the instant when the body comes to rest again. We can see that at t = 0,
the force F = 0. Now answer the following:

331. Find the duration for which the force acts on the body

a) 2 s b) 3 s c) 3.5 s d) 4.5 s

Paragraph for Question Nos. 332 to - 332

For the system shown in the figure, there is no friction anywhere. Masses m1 and m2 can move up or down in
the slots cut in massM. Two non-zero horizontal forces F1 and F2 are applied as shown. The pulleys are massless
and frictionless. Given m1 ≠ m2

332. According to the above passage, which is correct?

a) It is not possible for the entire system to be in equilibrium

b) For some values of F and F , it is possible that entire system is in equilibrium

1 2

c) It is possible that F1 and F2 are applied in such a way that m1 and m2 remain in equilibrium but M does
not
d) None of the above
P a g e | 72
Paragraph for Question Nos. 333 to - 333

A mass M is suspended as shown in the figure. The system

is in equilibrium. Assume pulleys to be massless. K is the force constant of the spring

333. The extension produced in the spring is given by

a) 4 Mg/K b) Mg/K c) 2 Mg/K d) 3 Mg/K

Paragraph for Question Nos. 334 to - 334

On a stationary block of mass 2 kg, a horizontal, a horizontal force f starts acting at t = 0 whose variation with
time is shown in the adjoining diagram. Coefficient of friction between the block and ground is 0.5. Now answer
the following questions:

334. Find the time when acceleration of the block is zero

a) At 5 s only b) At 10 s only

c) Both at 5 s and 10 s d) At a time after t = 10 s only

Paragraph for Question Nos. 335 to - 335

-1
A long conveyer belt moves with a constant velocity of 8 ms . Two blocks A and B each of mass 2 kg are placed
gently on the belt with B on A. Initial velocity of both blocks is zero. Coefficient of friction between A and belt is
0.1. There is no friction between A andB. Length of A is 4 m

335. Find the time when B falls offA. Initially B is on right end ofA. Ignore the dimensions of B

P a g e | 73
 a) 1 s b) 3 s c) 2 s d) 4 s

Integer Answer Type

336. A block is moving on an inclined plane making an angle 45° with the horizontal and the coefficient of
friction is μ. The force required to just push it up the inclined plane is 3 times the force required to just
prevent it from sliding down. If we define N = 10μ, then N is

337. You are designing an elevator for a hospital. The force exerted on a passenger by the floor of the elevator is
not to exceed 1.60 times the passenger’s weight. The elevator accelerates upward with constant
-1
acceleration for a distance of 3.0 m and then starts to slow down. What is the maximum speed (inms ) of
the elevator?
338. A block A of mass m is palced over a plank B of mass 2m. Plank B is placed over a smooth horizontal
surface. The coefficient of friction between A and B is 0.4. Block A is given a velocity v0 towards right. Find
-2
acceleration (in ms ) of B relative to A

339. Block A is given an acceleration 12 ms-2 towards left as shown in figures. Assuming the block B always
-2
remains horizontal, find the acceleration (in ms ) of B

340. A block is placed on an inclined plane moving towards right horizontally with an accelerationa0 = g. The
length of the plane AC = 1 m. Friction is absent everywhere. Find the time taken (in second) by the block
to reach from C to A

341. A block of mass m = 2 kg is resting on a rough inclined plane of inclination 30° as shown in figure. The
coefficient of friction between the block and the plane isμ = 0.5. What minimum force F (in newton)
should be applied perpendicular to the plane, so that block does not slip on the plane?

P a g e | 74
342. A rod AB of length 2 m is hinged at point A and its other end B is attached to a platform on which a block of
mass m is kept. Rod rotates about point A maintaining angle θ = 30° with the vertical in such a way that
platform remains horizontal and revolves on the horizontal circular path. If the coefficient of static friction
-1
between the block and platform is μ = 0.1, then fin d the maximum angular velocity in rad s of rod so
-2
that block does not slip on the platform (g = 10 ms )

343. BlockB, of mass mB = 0.5 kg, rests on blockA, with mass mA = 1.5 kg, which in turn is on a horizontal
tabletop (as shown in figure). The coefficient of kinetic friction between block A an the tabletop is μk = 0.4
and the coefficient of static friction between block A and block B isμs = 0.6. A light string attached to block
C is suspended from the other end of the string. What is the largest mass mC (in kg) that block C can have
so0 that blocks A and B still slide together when the system is released from rest?

344. Figure represents a painter in a crate which hangs alongside a building. When the painter of mass 100 kg
pulls the rope, the force exerted by him on the floor of the crate is 450 N, If the crate weights 25 kg, find
-2
the acceleration (in ms ) of the painter

345. The elevator shown in figure is descending with an acceleration of2 ms-2. The mass of the block A = 0.5 kg.
Find the force (in Newton) exerted by the block A on the block B

346. A blockA, of weightW, slides down an inclined plane S of slope 37° at a constant velocity, while the plankB,
also of weightw, rests on top of A. The plank B is attached by a cord to the top of the plane. The coefficient
of kinetic friction μ is the same between the surfaces A and B and between S and A. Determine the value of
P a g e | 75
1/μ

P a g e | 76
 5.LAWS OF MOTION

: ANSWER KEY :

1) c 2) b 3) b 4) c 101) b 102) d 103) d 104) b

5) c 6) d 7) d 8) a 105) c 106) a 107) a 108) d

9) b 10) b 11) b 12) d 109) b 110) c 111) c 112) b

13) a 14) c 15) d 16) c 113) c 114) b 115) c 116) b

17) c 18) c 19) c 20) a 117) b 118) b 119) b 120) b

21) d 22) d 23) c 24) b 121) b 122) b 123) b 124) a

25) c 26) c 27) c 28) c 125) b 126) b 127) c 128) b

29) d 30) b 31) c 32) a 129) a 130) c 131) d 132) d

33) a 34) b 35) c 36) c 133) d 134) d 135) a 136) a

37) d 38) a 39) a 40) b 137) c 138) b 139) c 140) d

41) d 42) c 43) d 44) b 141) a 142) a 143) a 144) c

45) d 46) c 47) c 48) a 145) c 146) a 147) d 148) c

49) a 50) b 51) d 52) c 149) c 150) d 151) a 152) a

53) b 54) d 55) c 56) d 153) a 154) d 155) a 156) a

57) a 58) b 59) b 60) c 157) b 158) a 159) d 160) d

61) b 62) b 63) a 64) d 161) d 162) b 163) a 164) b

65) c 66) b 67) b 68) c 165) d 166) a 167) b 168) b

69) a 70) d 71) d 72) b 169) b 170) c 171) c 172) d

73) d 74) a 75) a 76) a 173) d 174) c 175) c 176) a

77) b 78) c 79) c 80) a 177) a 178) d 179) c 180) a

81) d 82) b 83) a 84) c 181) c 1) a,b,c 2) a,b 3)

b,c 4) a,b,c
85) b 86) a 87) b 88) a
5) a,b,c 6) a,d 7) b,c 8)
89) b 90) d 91) b 92) c b,c

93) a 94) b 95) d 96) a 9) a,d 10) a,b,c,d 11) a,c 12) a
97) c 98) d 99) c 100) a 13) c 14) b,d 15) a,b,d 16) d

P a g e | 77
17) a 18) b,c,d 19) b 20) c 45) a 46) d 47) b 48) b

21) a 22) a,b,c,d 23) a,c,d 24) 49) a 50) d 51) a 52) a
a,b,c
53) a 54) a 55) d 56) d
25) a,b,c 26) a,c 27) d 28) c
57) a 1) a 2) d 3) c
29) a,c 30) a,b,c 31) a 32) 4) a
a,c
5) c 6) d 7) b 8) a
33) b 34) c,d 35) b 36)
a,b,d 9) b 10) a 11) c 12) b

37) a,c 38) b,d 39) b 40) 13) c 14) d 15) b 1) d

a,d 2) c 3) b 4) b

41) b,c,d 42) a,c 43) a 44) d 5) a 6) a 7) b 8) d

45) b,d 46) a,c 1) c 2) c 9) d 10) b 11) b 12) c

3) c 4) c 13) a 14) b 15) c 16) d
5) d 6) b 7) b 8) a 17) d 18) a 19) d 20) a
9) a 10) a 11) c 12) e 21) c 22) d 23) c 24) b
13) c 14) a 15) d 16) c 25) c 26) b 27) d 28) c
17) a 18) c 19) d 20) d 29) a 30) d 31) b 32) d
21) c 22) e 23) b 24) a 33) a 34) a 35) c 36) c
25) a 26) e 27) c 28) c 1) 5 2) 6 3) 6 4) 2
29) d 30) d 31) a 32) b 5) 1 6) 8 7) 1 8) 5
33) a 34) a 35) a 36) a 9) 2 10) 4 11) 4
37) b 38) a 39) d 40) d

41) b 42) e 43) e 44) d

5.LAWS OF MOTION

: HINTS AND SOLUTIONS :

1 (c) 4N
N2 = 4g - Ncos 37° = 40 - (ii)
5
3N Nsin 37° - f = 4a (iii)
F - Nsin 37° = 6a ⇒ F - = 6a (i)
5 From (i) and (iii): F - f = 10a
⇒F - mN2 = 10a (iv)

[
⇒ F - m 40-
4N
5 ]
= 10a (v)

Put the value of N from (i) in (v) and also put the

P a g e | 78
 5F-60 F = (M+m)g(μ1 + μ2)
value of m to get a =
42 7 (d)
Now to start motion: a > 0 ⇒F > 12 N
So the minimum force F to just start the motion is The reading of the spring scale is the normal
12 N reaction between man and spring scale figure
Now maximum F will be when N2 just becomes
zero
Then from (ii): N = 50 N
From (i) and (ii), we get F = 75 N (by putting
N2 = 0
If we apply F > 75 N, then B will start sliding up
on A, but we do not want this As the reading is decreasing, it means normal
3 (b) reaction is decreasing. Firstly, the lift must be
moving upwards with constant velocity and
For constant acceleration if initial velocity makes decelerated to rest
an angle with acceleration, then path will be 8 (a)
parabolic
4 (c) As in figure ucos 45° = v

In equilibrium, if θ is the required angle, Cylinder

A:mgsin 60° = kxcos (60°-θ)

cos 60°
Cylinder
or v = 2u
B:mgsin 30° = kxcos (30°+θ)=kxsin (60°-θ) on
9 (b)
solving θ = 30°
5 (c) Since m is in equilibrium w.r.t. observer,
-2 acceleration of m should also be a2. So net friction
If acceleration of M is 2 ms , then acceleration of
-2 force (as there is no other horizontal force on m)
m e.r.t. M will be 2 ms
acting on m should be mass ×acceleration = ma2
10 (b)

Tsin θ - mgsin 30° = ma

⇒ Tsin θ = mgsin 30° + mg/2 (i)
Acceleration of m w.r.t. ground Tcos θ = mgcos 30° (ii)
2 2 -2 Dividing (i) by (ii), we get
am = 2 +2 +2×2×2cos (180-37°) = 8/5 ms
6 (d)

Here the force applied should be such that friction

force acting on the upper block of m should not be
more than the limiting friction( = μ1mg). Let the
system moves with accelerationa. Then for whole
system, 2
tan θ =
F - μ2(M+m)g = (M + m)a (i) 3
For block of mass m, 11 (b)
f1 = ma or μ1mg = ma or a = μ1g (ii)
f1 = μmg, friction will provide the necessary
Form Eq. (i) and (ii), we get
P a g e | 79
 2
centripetal force f = mω r

2
f ≤ fl ⇒ mω r ≤ μmg
2r 2
ω 2 ×50/100 17 (c)
⇒μ ≥ = ⇒ μ ≥ 0.2
g 10
12 (d) g
tan θ = ⇒a = gcot θ
a
Let spring does not get elongated, then net pulling
force on the system is Mg + mg - mg or simplyMg.
Total mass being pulled isM + 2m. Hence,
acceleration of the system is
Mg
a=
M+2m 18 (c)
Now sincea < g, there should be an upward force
on M so that its acceleration becomes less than g. From figure, T2cos θ = mg, T2sin θ = mg
It means there is some tension developed in the
string. Hence, for any value of M spring will be
elongated
13 (a)

fl1 = μ1mAg = 0.3×300 = 90 N T1sin α = mg 2sin 45°

fl2 = μ2(mA+mB)g = 0.2(300+100) = 80 N T1sin a = mg ⇒
cos α
=
2mg
sin α mg
1 2
tan α = ⇒cos α =
2 5
2
T1 = 2 mg ⇒T1 = mg 5
5
lf3 = μ3(mA + mB + mC)g T1
= 5 ⇒ 2T1 = 5 T2
= 0.1(300+100+200) = 60 N T2 2
14 (c) 19 (c)

Let the tension in the rope beT. Let the Acceleration of suitcase till the slipping
acceleration of the man and chair to a upwards continuous is
1000 f
For man: T + 450 - 1000 = a a = max
10 m
or T = 550 + 100a (1) μmg -2
a= = μg = 0.05×10 = 5 ms
For chair: T - 450 - 250 = ( )
250
10
a
m
Slipping will continue till its velocity also becomes
-1
or T = 700 + 25a (ii) 3 ms
-2
From (i) and (ii), we get a = 2 ms ∴ v = u + at
15 (d) or 3 = 0 + 5t or t = 0.6 s
in this time, displacement of suitcase
dm
Force = V 1 2 1
dt s1 = at = ×5×(0.6)2 = 0.9 m
2 2
16 (c)
and displacement of belt, s2 = vt = 3×0.6 = 1.8
T = Nsin θ and N = mgcos θ m
mg Displacement of suitcase with respect to belt
T = mgcos θsin θ = sin 2θ
2 s1 - s2 = 0.9 m. this displacement will be opposite
P a g e | 80
 to the direction of motion of belt P-Qsin θ
μ(mg+Qcos θ) = P - Qsin θ ⇒ μ =
20 (a) mg+Qcos θ
25 (c)

At any instant, velocity of two wedges would be of

same magnitude but it opposite directions. This
can be concluded from conservation of
momentum or by symmetry
4
From constraint theory, vM = vm
3
Acceleration of A in horizontal direction = the From energy conservation,
acceleration of B = b rightwards MvM
2
mvm
2
32 mgh
Acceleration of A in vertical direction =the ×2 + - 0 = mgh ⇒vM =
2 2 32M+9m
acceleration of A with respect to b in upwards So the velocity with which wedges recede away
direction = a = 4b form each other is
̂ ̂
Hence, net acceleration of A = b i + 4b j 32mgh×4
2vM =
21 (d) 32M+9m
26 (c)
2 2
Speed v = vx +vy
By virtual work method:
Rate of change of speed -2
Acceleration of B w.r.t. A will be 10 ms
dv dv
2vx x +2vy y downward. Apart from this, B also has an
dv dt dt
= -1
acceleration 5 ms in horizontal direction along
dt 2
2 v +vy
2
x
with A, so net acceleration of B is
vxax+vyay 2 2 -2
= 10 +5 = 100+25 = 125 = 5 5 ms
2 2
v +v
x y 27 (c)
22 (d)
T = 2ma
̂ ̂
dp ̇ ̇
Force F = = -k Asin (kt) i - k Acos (kt) j
dt

̂ ̂
̇ ̇
p = Acos (kt) i - Asin (kt) j mg - T = ma
a = g/3
Since, F ∙ p = 0
28 (c)
∴ Angle between F and p should be 90°
The minimum force required to just move a body
23 (c) will bef1 = μsmg. After the motion is started, the
friction will becomes kinetic. So the force which is
-1
Given (V = 10 ms ) responsible for the increase in velocity of the
V g 10 10 block is
After 2s: Vx = - ×2 ⇒Vx = - ×2
2 2 2 2 F = (μs-μk)mg = (0.8-0.6)×4×10 = 8 N
10 -1 10 -1
Vx = - ms and Vy = - ms F 8 -2
2 2 So, a = = = 2 ms
m 4
100 100 1 1 -1
V= + = 10 + = 10 ms 29 (d)
2 2 2 2
24 (b) Free-body diagram (figure)

Frictional force = μR = μ(mg+Qcos θ) and

horizontal push= P - Qsin θ
For equilibrium, we have
P a g e | 81
 Equations of motion: F 5 -2
a= = = 2.5 ms
F m 2
aB = (in +x direction)
M 33 (a)
F
aA = (in –x direction) Minimum effort is required by pulling a block at
m
Relative acceleration of A w.r.t. B: the angle of friction
⃗ ⃗ ⃗ F F
a A.B = a A - a B = - - = -F
m M ( )m+M
mM
34 (b)
2 -1 4 -1
K = 10 N cm = 10 Nm . Let the ball move
(along –x direction)
distance x away from the centre as shown in
Initial relative velocity of A w.r.t. B
figure
uA.B = v0
Final relative velocity of A w.r.t. B = 0
2 2
Using v = u + 2as
2
2 F(m+M) Mmv0
0 = v0 - 2 S ⇒ S=
mM 2F(m+M)
30 (b)
2
kx = mw (0.1+x)
Angular frequency of the system, 4 90 ( 2)2
⇒10 x = × 10 ×(0.1 + x)
1000
k k -2
ω= = Solve to get x ≈ 10 m
m+m 2m
35 (c)
2
Maximum acceleration of the system will be, ω A th
Distance travelled in t second is,
kA
or . This acceleration to the lower block is
2m 1
st = u + at - a
provided by friction. 2

Hence, fmax = mamax Given, u = 0

2
= mω A = m ( )
kA
2m
=
kA
2
∴
sn
=
1
an- a
2
=
2n-1
sn+1 1 2n+1x
31 (c) a(n+1)- a
2
Here friction force would be responsible to cause 36 (c)
the acceleration of truck., here maximum friction
Mg
force can be f = μ×< where M→ mass of
2
entire truck
μMg
This is the net force acting on tyre, so Ma =
2
0.6×10 -2
⇒a = = 3 ms
2 From figure,
32 (a) 2F + N - Mg = Ma
The water jet striking the block at the rate of 1 kg 2F - mg - N = ma
-1 -1 4 F - (M+m)g = (M+m)a
s at a speed of 5 ms will exert a force on the
4F-(M+m)g
block a=
M+m
dm
F=v = 5×1 = 5 N 37 (d)
dt
Under the action of this force of 5 N, the block of From constraint, the acceleration of both block
mass 2 kg will move with an acceleration given by and wedge should be same in a direction

P a g e | 82
 perpendicular to the inclined plane as shown in N = mgcos α - masin α
figure Now f = μN = macos α + mgsin α
acos α+gsin α
μ=
gcos α-asin α
a+gtan α 5
μ= =
g-atan α 12
41 (d)

Let the tension in the strings AP2 and P2P1 beT.

Considering the force on pulley P1, we get
(aA)⊥ = (aB)⊥, aAX = 15, aAY = 15 T = W1. Further, let ∠AP1P1 = 2θ
aAXcos 53° - aAYcos 37° = aBcos 53° Resolving tensions in horizontal and vertical
-1 ⃗ ̂ directions and considering the forces on pulley P2,
or aB = -5 ms or a B = -5 i
38 (a) we get 2Tcos θ = W2
or 2 W1cos θ = W2 or cos θ = 1/2
⃗ or θ = 60° So ∠AP2P1 = 2θ = 120°
Let a 0 be the acceleration of choosen non-inertial
frame of reference w.r.t some inertial frame of 42 (c)
⃗
reference and a 1 be the acceleration of the object In equilibrium (figure)
in non-inertial frame

T = mg, N = 3 mg and f = 2T = 2 mg
⃗ In limiting case f < fmax
For a 1 to be non-zero, the net force acting on
object (including pseudo force) must be non-zero 2
2mg < μN ⇒2mg ≤ 3μ mg ⇒ μ ≥
39 (a) 3
43 (d)

Time taken by the bullet and ball to strike the

ground is

2h 2×5
t= = = 1s
g 10
From length constraint l1 + l2 + l3 + l4 = C
'' '' '' ''
l1 + l2 + l3 + l4 = 0 Let v1 and v2 are the velocities of ball and bullet
(-a-b) + 0 + (-a-b) + c = 0 after collision. Then applying
c = 2a + 2b
From wedge constraints, acceleration of C is right x = vt
side isa. Acceleration of C w.r.t. ground
We have, 20 = v1×1
̂ ̂
= a i - 2(a + b) j
40 (b) or v1 = 20 m/s

FBD of m in frame of wedge 100 = v2×1 or v2 = 100 m/s

Now, from conservation of linear momentum

before and after collision we have,

0.01v = (0.2×20) + (0.01×100)

On solving, we get
P a g e | 83
 v = 500 m/s For 2 kg block, 10 - 6 = 2a1
For 8 kg block, 6 = 8a2
44 (b)
-2 3 -2
⇒ a1 = 2 ms ,a2 = ms
Before cutting the string, the tension in string 4
joining m4 and the ground is Acceleration of 2 kg block relative to 8 kg block is
T = (m1+m2-m3-m4)g and the spring force in the 5 -2
aref = a1 - a2 = ms
4
spring joining m3 and m4 is T + m4g. As the string
1 5 2
is cut, the spring forces do not change instantly, so Using the equation of motion, 3 = × r
2 4
just after cutting the string the equilibrium of t = 2.19 s
m1,m2 and m3 would be maintained but m4 47 (c)
accelerates in upward direction with acceleration
given by From 0 to 2 s: at any time t, F = 10 t
T+m4g-m4g ⇒a = F/m = 10t/m
a= v t 2
10t 5t
45 (d)
m4 ⇒
∫ 0
dv =
∫ 0
m
dt ⇒v =
m
2
Momentum: P = mv = 5t
Condition of sliding is -1
At t = 2 s, P = 5(2)2 = 20 kg ms , v = 20/m
From 2 to 4 s; F = 40 - 10 t
mgsin θ>μ mgcos θ v t
40-10t 1
∫ ∫
2
dv = dt ⇒ v = [40t - 40 - 5t ]
or tan θ>μ 20/m 2
m m
2
P = mv = 40t - 40 - 5t
ortan θ> 3 …(i) 48 (a)
condition of toppling is
Rg (tan θ-μ)
Vmax =
1-μtan θ
49 (a)

Initially under equilibrium of mass m

T = mg
Torque of mgsin θ about O > torque of mgcos θ
Now, the string is cut. Therefore, T = mg force is
about.
decreased on mass m upward and downwards on

∴ (mgsin θ) ( )
15
2
>(mgcos θ) ( )
10
2
mass 2m.

mg
∴ am = = g (downwards)
2 m
or tan θ> ….(ii)
3
mg g
and a2m = = (upwards)
With increase in value of θ, condition of sliding is 2m 2
satisfied first.
50 (b)
46 (c)
For the equilibrium of block of mass M1:
Friction between 2 kg and 8 kg blocks is kinetic in Frictional force, f =tension in the string, T
nature, so Where T = f = μ(m+M1)g (i)
F = m×2g = 0.3×2×10 = 6N For the equilibrium of block of mass M2:
T = M2g (ii)
Form (i) and (ii), we get μ(m+M1)g = M2g
M
m = 2 - M1
μ
P a g e | 84
51 (d) Let a be the common acceleration of the system
Here T = Ma (for block)
If the blocks move together, P - T = Ma (for rope)
F 10 5 -1
a= = = ms
mA+mB 6 3
mBF 20 P - Ma = ma
fB (frictional force on B)= mBa = = N
mA+mB 3 or P = (m + M)a or a = P/(m + M)
fBmax = μmAg = 0.4×2×10 = 8 N MP
now T = Ma =
As fBmax > fB, the blocks will not be separated and M+m
move together with common acceleration 5/3 56 (d)
-2
ms
Maximum acceleration of B or C can be mg so that
52 (c)
they do not slip with each other or on A
As sand particles are sliding down, the slope of For the system of (A+B+C)
the hill gets reduced. The sand particle stops T = 3 ma = 3 μ mg
coming down when component of gravity force For D:
alone hill is balanced by limiting friction force Mg - T = Ma
mgsin θ = μsmgcos θ 3μm
⇒Mg - 3μ mg = Mμg ⇒M =
-1
⇒ θ = tan (μs) ≅ 37° where θ is the new slope 1-μ
57 (a)
angle of hill
53 (b)
[ ]
2
2sin (α/2)
v2cos α + v1cos α = v1 ⇒ v2 = v1
cos α
Suppose F = upthrust due to buoyancy
58 (b)
Then while descending, we find
Mg - F = Ma (i) The cloth can be pulled out without dislodging the
When ascending, we have dishes from the table due to law of inertia, which
F - (M-m)g = (M - m)a (ii) is Newton’s first law. While, the statement II is
Solving Eqs. (i) and (ii), we get m =
2α
α+g[ ]
M true, but it is Newton’s third law.

54 (d) 59 (b)

In the free-body diagram of B (figure (a)) v

Velocity of liquid through inclined limbs =
2
Rate of change of momentum of the liquid is

[ () ]
2
2 v 5 2
ρAv + 2 ρA cos 60° = ρAv
2 4
60 (c)

As the eraser is at rest w.r.t. board, friction

N = mBa (i) between two is static in nature
f = mBg For figure (a) and (b), the friction force is same as
that of gravity force as shown in figure
μN = mBg (ii)
g -2
Form (i) and (ii), a = = 20 ms
μ
FBD of bob:
Tsin θ = ma (iii)
Tcos θ = mg (iv)
From (iii) and (iv) For (c), f = F2 + Mg > Mg
a -1
For (d) Mg - F2 < Mg as angle by which arm is
tan θ = ⇒ θ = tan (2)
g tilted is very small, so F2 would be small
55 (c) 61 (b)
P a g e | 85
 For A:
5g - T = 5(2C)
8 5 -2
For C:2T - 8g = 8C ⇒ C = = ms
14 7
62 (b)

1. Acceleration of block A downwards w.r.t.

ground

Let the initial length of string passing over pulley

2. Acceleration of block B w.r.t. inclined 2 l1 + l2 (i)
plane After displacement x and y mentioned above, the
3. Acceleration of block C w.r.t. ground right lengths becomes
⃗ ⃗ (l1-2x) + (l2 + y - x) (ii)
side. b + c acceleration of B w.r.t.
ground Equating (i) and (ii), we get y = 3x
Applying Newton’s law on system along Length of string that slips through A is
horizontal direction, we have y + x = 4x
mc + m(c-bcos θ) = 0 (i) Length of string that slip[s through A is
Applying Newton’s law on (A + B) along the y + x = 4x and the through B is y = 3x
inclined plane, 4x 4
Required ratio =
2mgsin θ = m(b-ccos θ) + masin θ 3x 3
2gsin θ = b - ccos θ + asin θ (ii) 66 (b)
From wedge constraint between A and B,
Horizontal acceleration of the system is
a = bsin θ (iii)
F F
4gsin θ a= =
From Eq. (i), (ii) and (iii), b = 2
2m+m+2m 2m
1+3sin θ
63 (a)

For chain to move with constant speed, P needs to Let N be the normal reaction between B and C.
be equal to frictional force on the chain. As the Free-body diagram of C gives
length of chain on the rough surface increases. 2
N = 2ma = F
Hence, the friction force fk = πkN increases 5
64 (d) Now B will nor slide downwards if μN ≥ mBg

Let at any time, their velocities are v1 and v2,

2
5 ( )
Or μ F ≥ mg or F ≥
5
2μ
mg

respectively, then v1 = v2cos θ 5

Or Fmin = mg
dθ 2μ
Differentiating: a1 = a2cos θ - v2sin θ
dt 67 (b)
Hence, none of them is correct
[Note: Option (a) is correct initially, because 2s 2s 1
tA = ⇒ tD = , t = tD
initially v2 = 0] aA aD A 2
65 (c) 2s 1 2s
=
gsin θ+μgcos θ 2 gsin θ-μgcos θ
Let the plank move up by x, then pulley 2 will 4gsin θ - 4μgcos θ = gsin θ + μgcos θ
move down by x. Let the end of string C moves 3 3
3gsin θ = 5μgcos θ or μ = tan θ =
down by a distance y 5 5
( ∵ θ = 45°)

P a g e | 86
68 (c) ⇒ v1 + 2v2 + v3 = 0
Take downward as positive and upward as –ve
N1 = mgcos θ and f1 = μ mgcos θ
So +12 + 2(-4) + v3 = 0
-1
v3 =velocity of pulley P = -4 ms
-1
= 4ms In upward direction
⃗ ⃗
v AP = - v BP ⇒ VAP = -(vB-vP)
-1
vB = vP - vAP = -4 - (3) = -7 ms
-1
i.e., block B is moving up with speed 7 ms
71 (d)
N2 = Mgcos θ and f2 = μ Mgcos θ
⃗ -1 ⃗ ⃗ ⃗
v B, l = 4 ms ↑, V B, l = V B, g - V l, g
-1 ⃗ -1 ⃗ -1
⇒4 ms = V Bg, - 2 ms ; V Bg,. = 6 ms ↑
72 (b)

Choosing the positive x - y axis as shown in the

figure, the momentum of the bead at A
⃗ ⃗
Equation of motion are is p i = +m v . The momentum of the bead at B is
T - f1 = mgsin θ = ma (i) ⃗ ⃗
p f = -m v
Mgsin θ - T - f2 = Ma (ii)
Solving Eqs. (i) and (ii), we get T = 0
69 (a)

2T - (M+m)g = (M+m)a (i)

T - mg + N = ma (ii)

Therefore, the magnitude of the change in

momentum
⃗ ⃗ ⃗ ⃗
Between A and B is ∆ p = p f - p i = -2m v
i.e., ∆p = 2 mv along the positive x-axis
the time taken by the bead to reach from A to B is
From (i) and (ii), we get πd/2 πd
∆t = =

( )
v 2v
m-M
N= (g+a) > 0 Therefore, the average force exerted by the bead
2
As m > M, if T increase, a increase and if a on the wire is

( )
2
increase N increases ∆p πd 4mv
Fav = = 2mv/ =
70 (d) ∆p 2v πd
73 (d)
Using constraint theory
The acceleration of block-rope system is
F
a=
(M+m)
Where M is the mass of block and m is the mass of
rope
So the tension in the middle of the rope will be
M+(m/2)F
T = {M + (m/2)}a =
M+m
Given that m = M/2

l1 + 2l2 + l3 = constant
P a g e | 87
 ∴T = [ M+(M/4)
M+(M/2) ]
F=
5F
6
74 (a)

Free body diagram (FBD) of the block (shown by

a dot) is shown in figure.

Here Tsin θ - F or T = F/sin θ

78 (c)

F-f F-μm1g -2
a1 = = = 10 ms
m1 m1
F-μm2g -2
For vertical equilibrium of the block, a2 = = 1 ms
m2
F 1 2 1 2
N = mg + Fsin 60°= 3g + 3 ….(i) ∴ s = arealt = [10+1]t ⇒ t = 2s
2 2 2
79 (c)
For no motion, force of friction
Due to acceleration in forward direction, vessel is
f ≥ Fcos 60° in an accelerated frame therefore a Pseudo force
will be exerted in backward direction. Therefore
or °μ N ≥ Fcos 60°
water will be displaced in backward direction

or
1
2 3 ( 3 g+
3F
2 )≥
F
2
80 (a)

x = 0, till mgsin θ < μmgcos θ and gradually x

F -1
or g ≥ or F ≤ 2 g or 20 N will increase. At angle θ > tan (μ)
2
kx + μmgcos θ = mgsin θ
Therefore, maximum value of F is 20 N. mgsin θ-μ mgcos θ
or x =
k
Here k is the force constant of spring
81 (d)
75 (a)
From constraint relations we can see that the
Applying Newton’s law on system along
acceleration of block B in upward direction is

( )
horizontal direction, we have
a +a
mc + m(c-bcos θ) = 0 (i) aB = C A with proper sings
2

( )
bcos θ
c= 3-12t
2 So aB = = 1.5 - 6t
2
76 (a)
dv
or B = 1.5 - 6t or ∫0 dvB = ∫10(1.5-6t) dt
vB

During downward motion: dt

2
F = mgsin θ - mgcos θ or vB = 1.5t - 3t or vB = 0 at t = 1/2 s
During upward motion: 82 (b)
2F = mgsin θ + μmgcos θ -2
Solving above two equations, we get Retardation of train = 20/4 = 5 ms
m = (tan θ)/3 It acts in the backward direction. Fictious force on
77 (b) suitcase= 5m Newton, wherer m is the mass of
suitcase. In act6s in the forward direction. Due to
In figure, the point B is in equilibrium under the this force, the suitcase has a tendency to slide
action of T, F and Mg forward. If suitcase is not to slide, then 5m =
Newton, where m is the mass of suitcase. In acts
P a g e | 88
 in the forward direction. Due to this force, the
suitcase has a tendency to slide forward. If
suitcase is not to slide, then 5m = force f of
friction
5
or 5m = m mg or m = = 0.5
10
83 (a) f = m(g+a)sin θ, N = m (g+a)cos θ
2 2
Net contact force: FC = f +N = m(g + a)
When 88 (a)
P = mg (sin θ-μcos θ) The string is under tension, Hence there is
f = μmgcos θ (upwards) limiting friction between the block and the plane
figure
when P = mgsin θ

f=0

and when P = mg(sin θ+μcos θ)

f = μ mgcos θ (downwards)

Hence, friction is first positive, then zero and then

negative.
∑F x
=0
⇒ μN + 50cos 45° = 150sin 45° (i)
85 (b)
∑F y =0
As the springs have natural length initially, if one ⇒ N = 50sin 45° + 150cos 45° (ii)
spring is compressed, the other must be Solving (i) and (ii), we get m = 1/2
expanded. Hence, the compression will be 89 (b)
negative
The free-body diagram of m2 [figure]
T + F2 = 80 N and F2 = 70×0.5 = 35 N
∴ T = 80 - 35 = 45 N

12 5
tan β = ∴ cos β =
5 13
T1cos β + T2cos β = mg (i)
T1sin β = T2sin β (ii)
FBD of m1 (figure)
∴ T1 = T2 = T
T + F1 = m1g or F1 = -25 N
mg 13
-25 -25 ∴ 2Tcos β = mg ⇒ T = ⇒T = mg
∴ X1 = = = -0.5 m 2cos β 10
k1 50
90 (d)
Therefore, compression in first spring is -0.5 m
(negative sing indicates that it is extension) Let v1 be the velocity of block and v2 be the
86 (a) velocity of end A of the string, w.r.t. man

For upward acceleration of M1:

M2g ≥ M1gsin θ + μM1gcos θ
⇒ (M2) = M1(sin θ + μcos θ)
min
87 (b)
2 2
Figure, Fc = f +N = m(g + a)
P a g e | 89
 dl5 -1 dl dl So, fL = 0.2×64 = 32 N
= v2 = 2 ms (given) , 1 = 2 = -v1
dt dt dt As 4g < 80sin 37°, friction force will act
Now l1 + l2 + l3 + l4 + l5 =constant downwards. Net applied force in upward
dl1 dl dls direction (excluding friction force) is
⇒ + 2 +0+0+ =0
dt dt dt 80sin 37° - 40 = 48 - 40 = 8 N
v 2 -1
As Fapplied in vertical direction is less than fL, block
⇒ - v1 - v1 + v2 = 0 ⇒ v1 = 2 = = 1 ms
2 2 won’t move in vertical direction and value of
91 (b) static friction force is f = 8 N
94 (b)
Limiting friction F1 = μsR = 0.5×(5) = 2.5 N
As in the figure, mass of the rope: m = 4×1.5 = 6
R kg
-2
Acceleration: a = 12/6 = 2 ms
R 5N

mg

Mass of part 1 as in the figure 7.567

Since downward force is less than limiting friction m1 = 1.6×1.5 = 2.4 kg
therefore block is at rest so the static force of
friction will work on it
Fs = downward force = Weight
= 0.1×9.8 = 0.98 N
T = m1a = 2.4×2 = 4.87 N
92 (c)
95 (d)
In the free-body diagram of m [figure (a)],
⃗ ̂
T = mg (i) ⃗
a = F = -10 j (ms )
-1 2
[No friction will act between M and m m
Displacement in y-direction
1 2 1 2
y = ut + at ⇒0 = 4×t×- ×10×t
2 2
4 4
t = s ⇒ x = 4t = 4× = 3.2 m
5 5
96 (a)
In the free-body diagram of M (figure (b))
f = μ2N = 3T (ii) f l = mMg. If motion does not start, then
and T + Mg = N (iii) f = F = F0t
From Eq. (iii), N = (m+M)g Motion will start when f = f1
From Eq. (ii), μ2(m+M)g = 3 mg
μM 1/3×8
m= 2 = = 1 kg
(3-μ2) (3-1/3) μMg
⇒F0T = μMg ⇒T =
93 (a) F0
97 (c)
The FBD of the block is as shown in the figure
The free-body diagrams of two blocks are shown
in figure. Under the assumption that blocks are
moving together,
F + 2gsin 37° + 3 gsin 37° - f1 - f2 = 5a
Where f1 = μ×3gcos 37°
And f2 = μ×2gcos 37°
N = 80cos 37° = 64 N

P a g e | 90
 46 -2
⇒a = ms a 10 2
5 ⇒tan θ = = =
b 15 3
For 3 kg block, N + 3gsin 37° - f1 = 3a ⇒N = 12
⇒ θ = 33.69°
N
i.e., toppling starts at θ = 33.69°
98 (d)
and angle of repose
-1 -1
As the block does not slip on prism, the combined = tan (μ) = tan ( 3) = 60°
acceleration of the prism is a = gsin θ It mean the block will remain at rest on the plane
up to certain angle θ and then it will topple
102 (d)

Extension in the string is

x = AB - R = 2Rcos 30° - R = ( 3-1)R
Spring force:
F = kx =
( 3+1)mg ×( 3-1)R = 2mg
R
mgsin θ is the pseudo force on m
2
N + mgsin θ + sin θ = mg or N = mgcos θ
And for no slipping, mgsin θcos θ ≥ μN
2
mgsin θcos θ ≤ μmgcos θ or μ ≥ tan θ
99 (c)
2 2
From figure L = 2h - 2y + x +h
From the figure, we have
Differentiating the equation, we get
3 3mg
dy x d xVA N = (F+mg)cos 30° =
= ⇒VB = 2
dy 2 h +x dt
2 2 2
2 h +x
2
103 (d)

Direction of acceleration of B is along the fixed

incline, and the of A is alonmg horizontal towards
left

100 (a)
2
R1+R2
2 From diagram, acceleration of B is represented by
R
a=
m
= 3
m
[ 2
∴ R3= R1+R2
2
] AB while its horizontal and vertical components
are shown by AO and OB, respectively.
101 (b)
Acceleration of A is represented by AC
For rotational equilibrium about point “P”, ⃗
O C = a(sin αcot θ + cos α)
mgsin θ ()
b
2
= mgcos θ ()
a
2
104 (b)

Equation of motion for M:

P a g e | 91
 Since M is stationary
T - Mg = 0 ⇒T = mg From FBD of B:
Since the boy moves up with an acceleration a Mg - Tcos 45° = ma (2)
T - mg = ma ⇒T = m(g + a) From (i) and(ii), we get T = mg/ 2
Equating Eqs. (i) and (ii), we obtain 109 (b)
Mg = m(g+a)
⇒ a= ( )
M
m
-1 g, the block M can be lifted
For A:T - 2g = 2a (i)
For B:T1 + 2g - T = 2a (ii)
105 (c) For C:2g - T1 = 2a (iii)
Adding (i) and (ii), we get
dl1 dl
From figure l1 + l2 = C or + 2 =0 T1 = 4a (iv)
dt dt
From Eq. (iii) and (iv), we get
2g - 4a = 2a
or a = g/3

v1 cos θ2
-v1cos θ1 + v2cos θ2 = 0 or =
v2 cos θ1
106 (a)

Velocity of object w.r.t. non-inertial frame is g

constant and hence w.r.t. some inertial frame of From Eqs. (iv) and (v), T1 = 4× = 13 N
3
reference it is changing, hence it is acceleration. 110 (c)
So net force acting on the object must be non-zero
107 (a) At equation 2Tcos θ = Mg, T = mg

Let the total mass of the chain be M and mass of

the hanging part be m1. Then the mass of the part
placed on table will be m2 = M - m1
Here weight of the hanging part will be balanced
by the friction force acting on the upper part, i.e.
m1g = μm2g solve to get (m1/M)×100 = 20% Mg M
cos θ = = ⇒cos θ < 1
2mg 2m
108 (d)
M
< 1 ⇒ M < 2m
As shown in figure (a) and (b) from FBD of A 2m
Tcos 45° = ma (i) 111 (c)

(Check figure in the frame of the car)

P a g e | 92
 ' 2s 2s
∴ t = nt ⇒ =n
g(sin θ-μcos θ) gsin θ
∴ n g (sin θ-μcos θ) = gsin θ
2

When θ = 45°,sin θ = cos θ = 1/ 2

( )
Solving, we get μ = 1- 2
1
n
114 (b)
Applying Newton’s law perpendicular to string
From FBD it is obvious that net force on each
a
mgsin θ = macos θ ⇒ tan θ = block is zero in horizontal direction. So
g
a1 = a2 = 0
Applying Newton’s law along string
T - mgcos θ - masin θ = ma
2 2
⇒T = m g +a + ma
112 (b)

If initial acceleration of M towards right is A, 115 (c)

thewn we can show that acceleration of m w.r.t. M
Frictional force:
down the incline is
F = mR = 0.5×mg = 0.5×60 = 30 N
3A
a = A(1+cos θ) = ( ∵ θ = 60°) Now F = T1 = T2cos 45° or 30 = T2cos 45°
2
and W = T2sin 45°
FBD of block m (w.r.t. M) is shown below:
Solving them, we get W = 30 N
116 (b)

F = 2Tcos θ

F
⇒T =
2cos θ

FBD of M (figure)

Equation of motion:
1 3
For m:mg 3 + mA× - T = m A (i) Magnitude of acceleration of the particle
2 2 2
1 Tsin θ
N + mA 3 = mg (ii) =
2 2 m
For M:T + N 3 = MA (iii) Ftan θ F x
2 = =
3 3g -1
2m 2m a -x2
2

From Eq. (i), (ii) and (iii) A = ms

23
113 (c) 117 (b)

Change in momentum of one ball = 2 mu, time

1 2 1 2 2s
From s = ut + at = 0 + at ,t = taken = 1 s
2 2 a
From smooth plane a = gsin θ
'
For rough plane, a = g(sin θ - μcos θ)

P a g e | 93
 -2 -2
Total chnage in momentum or a1 = 0.4 ms and a2 = 0.2 ms
Fav =
Time taken the relative acceleration: a2 = 0.2 ms
2

n(2 mu) 1 2
= = 2 mnu Using S = ut + at we get
1 2
118 (b) 2
100 = (1/2)×0.6×t or t = 18.3 s
3T-mgsin θ 124 (a)
a=
m
Let us first assume that the 4 kg block is moving
3×250-100×10×sin 30° -2
= = 2.5 ms down, then different forces acting on two blocks
100
would be like as shown in the figure. (Normal to
119 (b)
inclines forces are not shown in figure)
The pressure on the rear side would be more due
to fictitious force (acting in the opposite direction
of acceleration) on the rear face. Consequently the
pressure in the front side would be lowered

120 (b)
To have the motion, he friction force f should be
̂ ̂ ⃗ ⃗ ̂ ̂ equal to limiting value
⃗
u = 4 i + 2 j , a = F = i -4 j 4
m i.e., fL = μsmgcos 37 = 0.27×10g× = 2.16 g
Let at any time, the coordinate be (x, y) 5
1 2 here, the 4 kg block is not able to pull the 10 kg
x - 2 = uxt + at block up the incline as 4g < 10gsin 37 + fL , so
2
1 2 1 2 system won’t move in the direction that we
⇒x - 2 = 4t + t and y - 3 = 2t - 4t
2 2 assumed. So if there is a chance of motion of
2
⇒y - 3 = 2t - 2t system, it can only move down the incline and
When y = 3m, t = 0, 1 s; when t = 0, x = 2 m system will move only if the net pulling force
When t = 1 s, x = 6.5 m down the incline is greater than zero. For down
121 (b) the incline motion, the FBD is as shown in the
figure
To move up with acceleration a, themonkey will,
push the rope downwards with a force of
Tmin = mg + 40 amax;600 = 400 + 40 amax
200 -2
a¬max = 5 ms
40
So rope will break if the monkey climbs up with
-2
acceleration 6 ms
For a to be non-zero, i.e., positive
122 (b)
10gsin 37 > fL + 4g
Which is not, so the system is moving neither
down the incline, nor up the incline and so the
system remains at rest
125 (b)
aA = g/2
For equilibrium
aB = g
∫μdmgcos θ ≥ ∫dmgsin θ
123 (b) or ∫μλ dlgcos θ ≥ ∫λ dlgsin θ

The force of 100 N acts on both the boats

250 a1 = 100 and 500a2 = 100

P a g e | 94
 In first case, acceleration of m1 will be a1 = gsin θ
μ
∫dlcos θ ≥ ∫dlsin θ down the inclined plane. In second case,
(∴sin θ=
dy
dl
,cos θ=
dx
dl )
or μ∫dx ≥ ∫dy or μl ≥ h
126 (b)

2m Mg 2mg
T= = ≅ 2 mg
m+M m Acceleration of m2 w.r.t. incline is
1+
M m gsin θ+m2acos θ
Hence, total downward force is 2T = 4 mg a2 = 2 ⇒ a2
m2
= gsin θ + acos θ
Since a2 > a1, so T2 < T1
130 (c)

Maximum frictional force on block B is

μmBg = 0.4×3×10 = 12 N
12 -1
Hence, maximum acceleration = = 4 ms
3
Hence, maximum force
127 (c) F = (mA+mB)a = (6+3)×4 = 36 N
Aliter: We can also apply the formula discussed in
N = Mg - Fsin ϕ previous problem by putting μ2 = 0 and μ1 = 0.4
From figure 131 (d)

When a string is fixed horizontally (by champing

its free ends) and loaded at the middle, then for
the equilibrium of point P

Fcos ϕ-μ(Mg-Fsin ϕ)
a=
M
128 (b)

Free-body diagram of man and plank is given 2Tsin θ = W

below figure W
i.e., T =
2sin θ
Tension in the string will be maximum when sin θ
is minimum, i.e., θ = 0° or sin θ = 0 and
thenT = ∞. However, as every string can bear a
maximum finite tension (lesser than breaking
strength), this situation cannot be realized
For plank to be at rest, applying Newton’s second practically. We conclude that a string can never
law to plank along the incline remain horizontal when loaded at the middle
Mgsin α = f (i) howsoever great may be the tension applied
And applying Newton’s second law to man along 132 (d)
the incline
mgsin α + f = ma (ii) For b (figure)

( )M mg - 2T = ma
⇒a = gsin α 1+ down the incline
m
129 (a)

P a g e | 95
 As it also remain in contact with wedge B
For A, usin 30° = V,cos 30° - V,sin 30°
mg sin 30° usin 30°
T- = m(2a) Vy = Vx +
2 cos 30° cos 30°
Solving a = 0 Vy = 3utan 30° = 3 u
133 (d) 2
⇒V = Vx +Vy = 7u
2

The FBD of block from lift frame is as shown in Method-2

figure. From given data, as m(g+a0)sin θ > 2 In the frame of A
ma0cos θ

3usin 30° = Vycos 30°

Vy = 3utan 30° = 3 u and Vx = 2u
2 2
⇒ V = Vx +Vy = 7 u
135 (a)

So friction force acts upwards Since there in no resultant external force, linear
f = m(g+a0)sin θ - 2ma0cos θ momentum of the system remains constant
9g 4mg mg
= - = 136 (a)
10 5 10
9mg
N = m(g+a0)cos θ + 2ma0sin θ = For first half acceleration = gsin ϕ
5 Therefore, velocity after travelling half distance
18mg 9mg 2
v = 2(gsin ϕ)l (i)
As fL = μsN = = > f so static
50 25
friction
Reaction force,
2 2 mg 1 2 mg 13
R = f +N = +9 =
5 4 2
Alternative solution:
⃗ ̂ ̂ ̂ For second half, acceleration = g(sin ϕ - μkcos ϕ)
Net force F - mg i = m(a0 j - 2a0 i ) 2 2
So 0 = v + 2g(sin ϕ - μkcos ϕ)l (ii)
⃗
( ̂
⇒ F = m -2a0 i +(a0+g) j ⇒ F
̂
) ⃗
Solving (i) and (ii), we get μk = 2tan ϕ
̂
(
= m -g i +
3g ̂
2
j) 137 (c)

Given horizontal force F = 25 N and coefficient of

( )
2
2 3g 13mg friction between block and wall (μ) = 0.4
⇒F = m g + =
2 2
We know that at equilibrium horizontal force
134 (d) provides the normal reaction to the block against
Method-1:Velocity components perpendicular to the wall. Therefore, normal reaction to the block
the comtact surface remain same (R) = F = 25 N
As cylinder will remain in contact with wedge A, We also know that weight of the block (W) =
(figure) Frictional force= μR = 0.4×25 = 10 N
Vx = 2u 138 (b)
-2
Acceleration of box = 10 ms
Inside the box forces acting on bob (see figure)
P a g e | 96
 T= (mg2)+(ma)2 = 10 2 N Rsin β = ma (ii)
139 (c) Rcos β = mg (iii)
From Eqs. (ii) and (iii), we get
2 2
aB, g = b +c +2bccos (180-θ) a = gtan β

( )
2 Putting the value of a in (i), we get
bcos θ
+bbcos θ(-cos θ)
2
= b+ P = (M+m)gtan β
2
2 143 (a)
cos θ 2 b 2
= b 1+ -cos θ = 1+3sin θ
4 2 1 2
Since, h = at , a should be same in both cases,
2gsin θ 2
=
1+3sin θ because h and t are same in both cases as given
140 (d) In figure (i), F1 - mg = ma ⇒F1 = mg + ma
mg+ma
Nsin θ=mg In figure (ii), 2F2 - mg = ma F2 =
2
∴ F1 > F2
Ncos θ=ma

g
tan θ =
a

a dy
cos θ = = tan (90°-θ)- =2k x
8 dx

144 (c)
'
Tx(Hanged part) = 2Tx (Sliding part)
' -1
∴ x = 3x ⇒ x = 3×0.6 = 1.8 ms
145 (c)

If we take two points 1 and 2 on the string near

the pulley P as shown, then velocities of both
points 1 and 2 will be same. Hence, P does not
rotate but only translate

a
∴x=
2k g
146 (a)
141 (a)
Initial force = 2g = 20 N
As m2 moves with constant velocity, there is no
Force 20 20 -2
acceleration in the centre of mass. Net force Initial acceleration= = = ms
Mass 5+1 6
should be zero. For this N = m1g + m2g Final force=(load+mass of thread)×g
142 (a) = (2+1)×10 = 30 N
30 -2
Acceleration of the system ∴ final acceleration= ms
6
P
a= (i) 147 (d)
M+m
The FBD of mass m is shown in the figure Same solution for both
P a g e | 97
 -2
m1 = 100 kg, m2 = 50 kg, a = 5 ms If the plane makes an angle θ with horizontal,
T + N - m1g = m1a, T - N - m2g = m2a thentan θ = 8/15. If R is the normal reaction
Solving these : T = -1125 N and N = 375 N R = 170 gcos θ = 170×10× ( ) 15
17
= 1500 N

Force of friction on A = 1500×0.2 = 300 N

Force of friction ob B = 1500×0.4 = 600 N
Considering the two blocks as a system, the net
force parallel to the plane is
= 2×170gsin θ - 300 - 600 = 1600 - 900 = 700 N
700 35 -2
∴ Acceleration = = ms
148 (c) 340 17
Consider the motion of A alone
In the absence of friction, we can find that 15 kg 35
170 gsin θ - 300 - P = P 170×
will accelerate downwards and 25 kg upwards. So 17
various forces acting on these will be as shown (where P is pull on the bar)
N2 = N1 = F, fl = μN1 = 0.4F, fl = μN2 = 0.4 F P = 500 - 350 = 150 N
1 2

152 (a)

Let m start moving down and extension produced

in spring be x at any time. Value of x required to
move the block m is

For 15 kg: T + fl + 15 g ⇒T + 0.4 F = 15 g (i)

1

For 25 kg: 2T = fl + fl + 25g

1 2

⇒ 2T = 0.8 F + 25g (ii)

Solve (i) and (ii) to get F = 31.25 N
149 (c)

For block to be stationary, T = 800 N kx = μ mgcos θ + mgsin θ

4 3
⇒kx = 0.5 mg + mg = mg
5 5
Fore minimum M, it will stop after producing
extension in the spring x
1 2 1
Mgx = kx ⇒Mg = kx
2 2
If man moves up by acceleration ‘a
1 m
T - mg = ma ⇒Mg = mg ⇒M =
-2
2 2
800 - 500 = 50α a = 6ms 153 (a)
150 (d)
T = 60 N, T = T1 + 40
As in figure
Tsin θ = ma0 + mgsin α

⇒60 = T1 + 40⇒T1 = 20 N
154 (d)
Tcos θ = mgcos α
(m2-m1)g (m2-m1)g
a +gsin α a1 = and a2 =
tan θ = 0 (m1+m2) m1
gcos α
151 (a)
P a g e | 98
 a1 m1 1 a Spring balance reads the tension in the string

( )
Hence = = ⇒ 1 <1
a2 (m1+m2) m
1+ 2
a2 connected to its hook side. As the spring balance
m1 is light, the tension in the string on its either side
2m1m2g is same. Now the only thing that remains to be
As T1 = and T2 = m2g
(m1+m2) found is the tension in the string which could be
T 2m1 found easily by using Newton’s second law
2

( )
Hence 1 = = 159 (d)
T2 (m1+m2) m
1+ 2
m1 The minimum value of F required to be applied on
T1 the blocks to move is 0.2×(2+4)×10 = 12 N.
will depend upon the values of m1 and m2
T2 since the applied force is less than the minimum
N = 2T1, N2 = 2T2 value of force required to move the blocks
T together, the blocks will remain stationary
So the relation of N1/N2 will, be same as 1 160 (d)
T2

( )
155 (a)
dp dm
Force acting on plate, F = =v
In this case, spring force is zero initially. FBD of A dt dt
and B are shown below dm
Mass of water reaching the plate per sec =
dt

V
= Avρ = A(v1+v2)ρ = (v + v2)ρ
v2 1

( v = v1 + v2 = velocity of water coming out of

jet w.r.t. plate)
Tension in the string and spring will be zero just
after release
156 (a) [A=Area of cross section of jet=
V
v2 ]
Making FBD of block with respect to disc
dm V
Let A be the acceleration of block with respect to ∴F = v = (v1+v2)ρ×(v1+v2)
dt v2

[]
disc
V
=ρ (v +v )2
v2 1 2

161 (d)

Due to malfunctioning of engine, the process of

rocket fusion stops and hence net force
experienced by spacecraft becomes zero.
N1 = mg Afterwards the spacecraft continues to move with
N2 = m asin θ constant speed
m acos θ-μN2-μN1 2
162 (b)
A= = 10m/s
m
dl
157 (b) Here y is constant = vB
dt
Acceleration of cylinder down the plane is
a = (gsin 30°)(sin 30°) = 10 ( )( )
1 1
2 2
= 2.5 ms
-1

2s 2×5
Time taken t = = = 2s
a 2.5
158 (a)

P a g e | 99
 2 2 2 dl dx F 4 -2
l = y + x ⇒ 2l = 2x a= = = 2 ms
dt dt m 2
x -1 Displacement along OE, s1 = vt = 3×4 = 12 m
⇒ vB = vA = vAcos 60° = 1 ms
l Displacement perpendicular to OE
163 (a) 1 2 1
s2 = at = ×2×(4)2 = 16m
2 2
Acceleration of the skaters will be in the ratio
The resultant displacement
F F
: or 5:4 2 2
s = s1+s2 = 144+256 = 400 = 20 m
4 5
1 2 168 (b)
Now according to the problem, s = 0 + at
2
As in figure
s a 5
We get 1 = 1 =
s2 a2 4
164 (b)

If the wedge moves leftwards by x, then the block

moves down the wedge by 4x, i.e. w.r.t. wedge the
block comes sown by 4x l1 + l2 + l3 = C
' ' '
l1 + l2 + l3 = 0
-VB + VA - VB + VA - VB = 0
3VB = 2VA ⇒ 3aB = 2aA
Applying Newton’s law on A and B
So, acceleration of block w.r.t. wedge= 4a along F - 2T = 2 maA, 3T = 2maB
the incline plane of wedge 3F
Solve to get aB +
Acceleration of wedge with respect to ground= a, 13m
alonm gleft. So acceleration of block w.r..t ground 169 (b)
is the vector sum of two vectors shown in the
⃗ ⃗ ⃗ ⃗
figure. That is a b,l = a b - a l = (g-a)↓ ⇒ a b = g↓
| |
⃗ 2
a BG = a +(4a)2+2×a×4a×cos (π-α)
170 (c)
-2
= ( 17-8cos α)a ms Let the weight of each block be W (figure)
165 (d)

F
Acceleration of two mass system is a =
2m
leftwards. FBD of block A is shown in below

mF N3
Ncos 60° - F = ma = So, N1 = N2 =
2m 2
Solving, we get N = 3 F 171 (c)
166 (a)
Equation of motion for A (figure)
h vb
2 kx
2
tan θ = v /Rg ⇒
2
= v /Rg ⇒h = kx = ma ⇒ a =
b Rg m
'
167 (b) For B:F - T = ma
' F-kx
⇒ a =
The acceleration of the body perpendicular to OE m
is
P a g e | 100
 F-2kx Now 2 VP = VB + VP
⇒ The relative acceleration = ar = |a -a| =
'
2 1

m VB+VP 2+1 -1
∴ VP = 1
= = 1.5 ms
2
2 2
172 (d) 178 (d)

As there is no tendency of relative slipping Maximum friction that can be obtained between A
between the block and cube, the friction force is and B is f1 = μmAg = (0.3)(100)(10) = 300 N
zero and maximum
173 (d) Friction between B and ground is
f2 = μ(mA+mB)g = (0.3)(100+140)(10) = 720
From 0 toT, area is positive and from T to 2 T, N
area is negative. Net area is zero. Hence no change Drawing free-body diagrams of A, B and C in
in momentum occurs limiting case
174 (c)

Acceleration is to be downward which is possible

in option (c)
175 (c)
Equilibrium of A gives
Area under the force- time graph is impulse, and T1 = f1 = 300 N (1)
impulse is change in momentum Equilibrium of B gives
Area of graph=change in momentum 2T1 + f1 + f2 = T2
1 4mu or T2 = 2(300) + 300 + 720 = 1620 N (2)
⇒ TF0 = 2 mu ⇒F0 =
2 T and equilibrium of C gives mCg = T2
176 (a) or 10 mC = 1620 or mC = 162 kg
179 (c)

N = mgcos θ, f = mgsin θ
Net force applied by M on m (or m on M):
F = N +f
2 2
Tcos 60° = 30 N ⇒T = 60 N
= (mgcos θ) +(mgsin θ) = mg
2 2
Tsin 60° = T2 = W ⇒W = 60 3 = 30 3 N
2
177 (a)
180 (a)
-1
VA = 2 ms (towards right) On the system of particle if,
V -1
∴ VP = A = 1 ms (upwards)
1
2
-1
VA = 2 ms (towards left)
∑F ext
=0

then Psystem = constant

No other conclusions can be drawn.

181 (c)

Let T be the tension in the rope and a the

acceleration of rope. The absolute acceleration of
man is, therefore,
5g
4 ( )
-a . Equations of motion for

P a g e | 101
 mass and man gives: 10g - 2f = 15a ⇒10×10 - 2×6 = 15a
-2
T - 100g = 100a (i) ⇒a = 88/15ms
T - 60g = 60 ( )
5g
4
-a (ii) T2 = 10g - 10a = 10×10 - 10×
88
15
= 41.3 N

4875 88
Solving (i) and (ii), we get T = N T1 = f + 2a = 6 + 2× = 17.7 N
4 15
182 (a,b,c) Clearly T2 > T1
2. This is correct because of greater mass of
As the acceleration of A and B are different, it
3 kg since acceleration is same for both
means there is relative motion between A and B.
3. This is incorrect, because net force acting
The free-body diagram of A and B can be drawn as
on 10 kg mass is greater due to its larger
below
mass, not due to its acceleration
downward
187 (a,d)
For A, F - f = MaA = 50×3
For B, f = maB = 20×2 ⇒f = 40 N, F = 190 N
183 (a,b)

Because mg acts downwards which makes sliding

along 4 to be easiest and along 4 to be difficult
most
184 (b,c) T2cos 60° = T1cos 30° (i)
And T2sin 60° + T1sin 30° = 5g (ii)
From (i) and (ii)
T1 = 25 N and T2 = 25 3 N
188 (b,c)
F = 10a1 + 40a2
-2
⇒100 = 10a1 + 40×2 ⇒ a12 ms Force of upthrust will be there on mass m shown
in figure, so A weighs less than 2 kg. Balance will
show sum of load of beaker and reaction of
upthrust so it is reads nore than 5 kg

-2 189 (b,c)
So acceleration of A must be 2 ms for given
conditions to be satisfied
f ≤ fl ⇒ 20 ≤ μmAg ( )
Here F > μ, mg 1+
m
M
, so slipping will occur

⇒ 20 ≤ μ×10g ⇒ μ ≥ 0.2 between the blocks

-2
Hence, m can be greater or equal to 0.2 Fro mF - μkmg = m.a ⇒ a = 1.2 ms
185 (a,b,c) For Mμkmg = MA ⇒ A = 0.4 ms
-2

Here m1gsin 30° = m2g = 20 N, so there is no 190 (a,d)

tendency of motion in any direction If initially acceleration of A is greater than that of
Hence there is no friction on m1. Contact force will B, then there will be extension and if that of B is
be only normal force greater than A, then there will be compression in
186 (a,b,c) the spring. Otherwise length of spring will remain
same
1. Let acceleration of each block be a
191 (a,b,c,d)
10g - T2 = 10a, T2 - T1 - f = 3a
T1 - f = 2a, where f = 0.3×2g = 6 N When friction between the blocks becomes zero,
From above equations the relative sliding between the block will be
stopped hence; vA = vB and aA = aB. Also when

P a g e | 102
 the friction becomes zero, only forces to move the limiting case)
blocks are FA and FB Option (d) is correct
F F 196 (a,b,d)
aA = aB ⇒ A = B
mA mB
⃗ ⃗
192 (a,c) dp
Newton’s second law is F = , which itself
dt
Mg - T = Ma (i) explains the validity of the given statements
T = ma (ii) 197 (d)

v 3 -1
Rate of flow water = 10 cm s
t
-6 3 -1
= 10×10 m s
3
10 kg
Density of water ρ = 3
m
Mg
Solving (i) and (ii), a =
M+m Cross-sectional area of pipe A = π(0.5×10 )
-3 2

FBD of man
Mmg dv mv Vρv ρv v
Mg - N = Ma ⇒N = Force = m = = = ×
(M+m) dt t t t At
193 (a)
() ( )
2
v l v
= ∴v=
At A the horizontal speeds of both the masses is t A At
the same. The velocity of Q remains the same in
horizontal as no force is acting on the horizontal F=
(10×10 ) ×10
-6 2 3

π×(0.5×10 )
-3 2
direction. But in case of P as shown at any
intermediate position, the horizontal velocity first = 0.127 N
increases (due toNsin θ), reaches a maximum
value at O and then decreases. Thus it always 198 (a)
remains greater than v, Therfore, tP < tQ
Since, μ mgcos θ > mgsin θ
Force of friction is f = mgsin θ
199 (b,c,d)

For some time, the block won’t move due to

friction force. When F > fL, the motion of block
starts
194 (c)
4
F 5×10 5 -3 -2
a= = 7 = ×10 ms
m 3×10 3
F-fk kt-fk dv kt-fk
5 -3 -1 a= = ⇒ =a=
v = 2as = 2× ×10 ×3 = 0.1 ms m m dt m
3 2
kt fkt
2 2 3
195 (b,d) kt kt
-fkt -fkt -
2 ds 2 6 2
v= ⇒ = ⇒ s=
Option (a) is wrong since Earth is an accelerated m dt m m
frame and hence cannot be an inertial frame 200 (b)
Option (b) is correct
dv 100
Option (c) is incorrect; strictly speaking as Earth F=m = 0.05× = 250 N
dt 0.02
is accelerated reference frame (earth is treated as
a reference frame for practical examples and 201 (c)
Newton’s law’s are applicable to it only as a
P a g e | 103
 For equilibrium in vertical direction for body B we g
or g < μsa or μsa > g or μs >
have a
d The jumping away of the man involved upward
acceleration. It means an upward force acts on
man during jumping. Then from third law, a
downward force acts on platform due to which
reading first increases
204 (a,c,d)

2 mg = 2Tcos θ = 2(mg)cos θ In first case, m will remain at rest.aM = F/M

T = mg (at equilibrium) In second case, both will accelerate
1 am = aM = F/(M + m)
∴ cos θ = ⇒ θ = 45°
2 In second case, force on
202 (a) m = mam = mF/(M + m)
The two forces acting on the insect are mg and N. 205 (a,b,c)
Let us resolve mg into two components:
Friction on A and B acts as shown
mgcos α balances N
mgsin α is balanced by the frictional force

From the figure it is clear that friction on A

supports it motion and on B opposes its motion.
And friction always opposes relative motion
206 (a,b,c)
N = mgcos α For (i); Consider a block at rest on a rough surface
f = mgsin α and no force (horizontal) is acting on it. Now
But f = μN = μ mgcos α friction force on it would be zero. For (ii):
1 Consider a heavy block, under the application of
⇒ cot α = ⇒ cot α = 3
μ small force F which is not sufficient to cause its
203 (a,b,c,d) motion, so friction force is static in nature and
a block doesn’t move

For (iii): Refer to concepts and formulae

For (iv): Friction force and normal force always
F - mg = ma
act perpendicular to each other
F = m(g+a) = 4mg = 4W
207 (a,c)
b Think of Newton’s third law of motion
In region AB and CD, slope of the graph is
constant i.e. velocity is constant. It means no force
acting on the particle in this region

208 (d)

c mg < fmax Force on the pulley are

or mg < μsR or mg < μsma

P a g e | 104
 F = F1+F2
2 2 3 4
And b2 = g+ a
5 5
= ( (m+M)2+M2 )g

209 (c)
4 4
Similarly for this case get N3 = mg + ma
5 5
3 3
b3 = g+ a
5 5

2
mv
FBD of bob is Tsin θ =
R
and Tcos θ = mg
2
v (10)2
tan θ = =
Rg (10)(10)
tan θ = 1
4 4
or θ = 45° Similarly for this case, get N4 = mg - ma
5 5
210 (a,c)
3 3
And b4 = g- a
5 5
211 (a,b,c)

If the block is at rest, then force applied has to be

greater than limiting friction force for its motion
to begin
fL = μsmg = 0.25×3 g = 7.5 N < Fapplied
Balancing forces perpendicular to incline So friction is static in nature and its value would
N1 = mgcos 37° + masin 37° be equal to applied force, i.e., 7 N. if the body is
initially moving, then kinetic friction is present
4 3
N1 = mg + ma (fk = μk mg = 6 N), acting opposite to direction
5 5
And along incline, mgsin 37° - macos 37° = mb1 of motion
3 4 As F > fk the block is accelerated with an

[ ]
b1 = g - a
5 5 F-fk 1 -2
acceleration of a = = ms and hence its
m 3
speed is continuously increasing
If applied force is opposite to direction of motion,
then block is under deceleration of

a =- [ ]
F-fk
m
13 -2
= - ms and hence after some time
3
block stops and kinetic friction vanishes but
applied force continuous to act
4 3 But as F < fL, the block remains at rest and
Similarly for this case get n2 = MG - MA
5 5 friction force acquires the value equal to applied
P a g e | 105
 force, i.e. friction is static in nature
212 (a)

Total mass of 80 wagons

3 5
= 80×5×10 = 4×10 kg
5
F 4×10 -2
Acceleration, a = = 5 = 1 ms
M 4×10
Therefore, the friction force is 0.98 N. Hence,
Tension in the coupling between 30th and 31st
option (b) is the correct option
wagon will be due to mass of remaining 50
215 (c,d)
wagons. Now, mass of remaining 50 wagons
3 4
m = 50×5×10 kg = 25×10 kg
4
∴ Required tension, T = mg = 125×10 ×1
4
= 25×10 N

213 (a,c)
' '
M g - T = M a (i)
T = Ma (ii) (y-h) + x2+h2 = l or dy + x dx
=0
dt 2
x +h
2 dt

dy x dx dy 3
=- 2 2 ⇒ = - ( - vA)
dt x +h dt dt 5
3
VB = vA (i)
5
2y 2 2
d vAh 2 16
2 = ⇒aB = vA 3
'
Mg dt (x +h )
2 2 3/2 (5)
M g = a(M+M ) ⇒ a =
' '
(M+M') 16 2
aB = v (ii)
125 A
216 (b)

Let the velocity of the block M be v upwards, then

vcos θ = U ⇒ v = U/cos θ
217 (a,b,d)

masin θ = mgcos θ→ so that normal force is zero 1. Since the body is accelerated, it can’t have
a = gcot θ constant velocity, but it can have constant
Mg
'
' ' speed
gcot θ = ⇒ cot θM + cot θM = M
(M+M') 2. If acceleration of the body is opposite to
cot θM + cot θM = M
' ' the velocity, then at some instant its
' Mcot θ velocity will become zero
M = , T = Ma = Mgcot θ = Mg/tan θ
(1-cot θ) 3. As the body is accelerated, net force on it
214 (b) can’t be zero
4. Forces may act at some angle also
The magnitude of the frictional force f has to 218 (a,c)
balance the weight 0.98 N acting downwards
fl = 0.5×5 = 2.5 N First of all draw FBD ofP3. Let tensions, in three
f < Fl strings be T1, T2 and T3, respectively
2T1 - T1 = 0×a ⇒ T1 = 0

P a g e | 106
 Tcos θ0 = mg …(i)
Tsin θ0 = ma0 …(ii)
Dividing Eq. (ii) by Eq. (i), we get
a
tan θ0 = ⇒ θ0 = 30°
g
mg 2mg
T= =
Now draw FBD of P4 and P5 cos 30° 3
2T1 - T2 = 0 ⇒ T2 = 0 222 (b,c,d)
2T2 - T3 = 0 ⇒ T2 = T3 = 0 Acceleration of particle w.r.t. frame S1:
⃗ ⃗ ̂
a p- a s = 2 n
1

Acceleration of particle w.r.t. frame S2:

⃗ ⃗ ̂
a p - a s = 2m
2

̂ ̂
Where m and n are unit vectors in any
So forces acting on P6 and P7 will be that of gravity directions. Now relative acceleration of frames:
and they will be in free fall. Hence, acceleration of ⃗ (⃗
a s = a s = 2 n -m ) ̂ ̂
each of them will be g downwards 2 1

Its magnitude can have Any value between 0 to 4

219 (b,d)
-2 ̂ ̂
ms depending upon the direction of m and n
⃗ 223 (a,c)
In a non uniform field, ∑ F ≠ 0. When dipole is
⃗ 2mg-mg
aligned with field ∑ τ = 0 and when dipole is not a1 = =g
⃗ m
aligned, ∑ τ ≠ 0
mg+mg-mg
a2 = = g/2
220 (b) 2m
2mg-mg
a3 = = g/3
As in clear from figure 3m
Clearly a1 > a2 > a3
R + T = (m + M)g
224 (a)
R = (m+M)g - T
We give power to real wheel, so fr5iction on rear
The system will not move till wheel acts in forward direction. Front wheel is a
free wheel on which friction acts in backward
T ≤ F or T ≤ μR direction. Net friction is in forward direction, due
to which cycle accelerates
T ≤ μ[(m+M)g- T]
225 (d)
μ(m+M)g
T≤ The horizontal forces on the man must balance,
μ+1
i.e., the forces exerted by the two walls on him
μ(m+M)g must be equal
∴ Fmax = The vertical forces can balance even if the forces
μ+1
of friction on the two walls are unequal. The
221 (a,d) torques due to the forces of friction about his
centre of mass must balance. This requires
friction on both walls
226 (b,d)

Acceleration of M, a = ()
F
M

P a g e | 107
 1F 2 2Ml frictional forces.
l= t ⇒t =
2M F
234 (b)
227 (a,c)
Statement 1 is practical experience based; so it is
f = 0, ifsin θ = cos θ⇒θ = 45°
true. Statement 2 is also true but is not the correct
f towards Q,sin θ > cos θ
explanation of Statement 1. Correct explanation is
⇒θ > 45°
“ there is increase in normal reaction when the
f towards P,sin θ < cos θ
object is pushed and there is decreases in normal
⇒θ < 45°
reaction when object is pulled”
235 (a)

In equilibrium, net force on body is zero,

therefore, its acceleration a is zero. If the body is
at rest, it will remain at rest. If the body is moving
with a constant speed along a straight line path, it
will continue to do so
228 (c) 236 (a)

Here acceleration of both will be same, but their In the direction of normal reaction, net
masses are different. Hence, net force acting on acceleration zero. Hence, forces in this direction
each of them will not be same will be balanced.
229 (c) Hence N = mgcos θ
237 (a)
Work done in moving an object against
gravitational force (conservative force) depends Acceleration of body sliding down a smooth plane
only on the initial and final position of the object, inclination θ is given by
not upon the path taken. But gravitational force
on the body along the inclined plane is not same 1 -2
α = g sinθ = gsin 30˚=10× =5ms
as that along the vertical and it varies with the 2
angle of inclination 238 (c)
230 (c) Bearings are used to reduce friction
This is because the direction of motion is 239 (e)
changing continuously. Hence the velocity is Inertia is the property by virtue of which the body
changing and acceleration is being produced. is unable to change by itself not only the state of
Assertion is true but reason is false rest, but also the state of motion
231 (c) 240 (c)
According to Newton’s second law Coefficient of frictionμ = tan θ. The value of tan θ
Force may exceed unity
Acceleration = i.e. If net external force on 241 (a)
Mass
the body is zero then acceleration will be zero
On a rainy day, the roads are wet. Wetting of
232 (d) roads lowers the coefficient of friction between
the tyres and the road. Therefore, grip of car on
If a body is moved in a closed path the net work the road reduces and thus chances of skidding
done is zero. Gravity and an electrostatic field in increases
vacuum are conservative. But any form of friction 242 (d)
prevents the field from being conservative. Also
potential energy cannot be associated with The FBD of block A is as follows:

P a g e | 108
 direction, due to inertia. Hence force of friction on
both the wheels acts in backward direction
250 (b)

The force exerted by B on A is N (normal According to law of inertia (Newton’s first law),
reaction). The force acting on A are N (horizontal) when cloth is pulled from a table, the cloth come
and mg (weight downwards) in state of motion but dishes remains stationary
Hence Statement I is false due to inertia. Therefore when we pull the cloth
243 (c) from table the dishes remains stationary

Assertion is true, but the reason is false. The fan 251 (a)
continue to rotate due to inertia of motion
The fuel is consumed continuously when the
244 (a) rocket if flying. Hence, the rocket in a flight is a
system of varying mass.
Contact force is the sum of friction and normal
reaction 252 (a)
245 (c)
The wings of the aeroplane pushes the external
In uniform circular motion, the direction of air backward and the aeroplane move forward by
motion changes, therefore velocity changes reaction of pushed air. At low altitudes density of
air is high and so the aeroplane gets sufficient
As P = mv therefore momentum of a body also force to move forward
changes in uniform circular motion
253 (e)
246 (d)
A body subjected to three concurrent forces is
Pseudo force is applied only for non-inertial frame found to in equilibrium if sum of these forces is
247 (d) equal to zero

A frame of reference which is at rest or which is ⃗ ⃗ ⃗

i.e. F 1 + F 2 + F 3…… = 0
moving with a uniform velocity along a straight
line is called intertial frame of reference. But the 254 (c)
frame is which Newton’s laws of motion are
applicable is an intertial frame Ball bearings, also known as anti-friction bearings
are small metallic or ceramic spheres used to
248 (c) reduce friction between shafts and axles in a
number of applications.
The apparent weight of a body in an elevator
moving with downward acceleration a is given by 255 (c)
W = m(g - a)
The purpose of bending is to acquire centripetal
249 (e) force for circular motion. By doing so component
of normal reaction will counter balance the
When a bicycle is in motion, two cases may arise :
centrifugal force
(i) When the bicycle is being pedalled. In this case,
the applied force has been communicated to rear 256 (d)
wheel. Due to which the rear wheel pushes the
earth backwards. Now the force of friction acts in Acceleration down a rough inclined plane
the forward direction on the rear wheel but front a = g(sin θ - μcos θ) and this is less than g
wheel moves forward due to inertia, so force of 257 (d)
friction works on it in backward direction
(ii) When the bicycle is not being pedalled : Reference frame attached to Earth is not an
In this case both the wheels move in forward inertial frame of reference because Earth is

P a g e | 109
 revolving about the Sun, as well as it is rotating θ = 90°,cos θ=cos 90°=0
about its own axis
258 (a) W
∴T = =∞
2cos 90°
We know that
Both the assertion and reason are true and latter
Inpulse =change in linear momentum is correct explanation of the former

=final momentum-initial momentum 263 (a)

= mv-m(-v) =2mv Statement II is correct, as it represents Newton’s

⃗ ⃗
259 (b) dp
second law as F = , from this only we can
dt
In uniform circular motion of a body the speed ⃗
dp
remains constant but velocity changes as say for greater value of , force applied has to
dt
direction of motion changes be more
264 (b)
As linear momentum = mass × velocity, therefore
linear momentum of a body changes in a circle Force needed when breaks are applied
2
On the other hand, if the body is moving mv
f1 = ma =
d
uniformly along a straight line then its velocity
(v: initial speed, d: distance from wall)
remains constant and hence acceleration is equal
When turn is taken
to zero. So force is equal to zero 2
mv
f2 = ma =
260 (a) d
Hence, breaks must be applied
By lowering his hand player increases the time of 265 (a)
catch, by doing so he experience less force on his
hand because F ∝ 1/dt ∆P
F= . If ∆t is more, then F will be less
∆t
261 (a) 266 (d)
-1
v = μrg = 0.1×10×10 = 10 ms Only static friction is a self adjusting force. This is
because force of static friction is equal and
Both the assertion and reason are true, and opposite to applied force (so long as actual
reason is correct explanation of assertion motion does not start). Frictional force = μmg i.e.
262 (a) friction depends on mass
267 (d)
As is clear from figure
Due to change in normal reaction, pulling is easier
268 (b)

By the definition of inertial and nom-inertial

frame
269 (e)

2Tcos θ = W According to third law of motion it is impossible

to have a single force out of mutual interaction
W
T= between two bodies, whether they are moving or
2cos θ
at rest. While, Newton’s third law is applicable for
For the string to become horizontal, all types of forces

270 (e)

P a g e | 110
 For uniform motion apparent weight = Actual 276 (a)
weight for downward accelerated motion
In sliding down, the entire potential energy of
271 (d) body is converted only into translational energy.
While in rolling motion, some part of potential
Law of conservation of linear momentum is energy is converted into kinetic energy of rotation
correct when no external force acts. When bullet and rest into kinetic energy of translation.
is fired from a rifle then both should possess equal Therefore, in sliding motion, the velocity acquired
2
P by the body is more
momentum but different kinetic energy, E =
2m 277 (d)
∴ Kinetic energy of the rifle is less than that of
bullet because E ∝ 1/m dp
F= = Slope of momentum-time graph
dt
272 (a)
i.e. Rate of change of momentum = Slope of
According to Newton’s second law of motion momentum – time graph = force
F
a= i.e. magnitude of the acceleration
m 279 (a)
produced by a given force is inversely
proportional to the mass of the body. Higher is the Once the ski is in motion, it melts the snow below
mass of the body, lesser will be the acceleration it and hence skiing can be performed. To make
produced i.e. mass of the body is a measure of the skiing easier, wax has been put on bottom surface
opposition offered by the body to change a state, to ski as wax is water repellent and hence reduces
when the force is applied i.e. mass of a body is the the friction between the ski and film of water
measure of its inertia 280 (a)

273 (d) Due to attraction force, their velocities increase;

hence, momentum also increases. For individual
The force acting on the body of mass M are its particle, gravitational attractive force will be
weight Mg acting vertically downwards and air external force
resistance F acting vertically upward 281 (a)
Mg-F F
∴ Acceleration of the body, a = = g- mgrav. g - N = minertial. a For freely falling a = g.
M M
Now, M > m, therefore, the body with larger Since mgrav = minert⇒N = 0
mass will have greater acceleration and it will
reach the ground first 282 (d)
274 (b) An inertial frame of reference is one which has
When a body is moving in a circle, its speed zero acceleration and in which law of inertia hold
remains same but velocity changes due to change good i.e. Newton’s law of motion are applicable
in the direction of motion of body. According to equally. Since earth is revolving around the sun
first law of motion, force is required to change the and earth is rotating about its own axis also, the
state of a body. As in circular motion the direction forces are acting on the earth and hence there will
of velocity of body is changing so the acceleration be acceleration of earth due to these factors. That
cannot be zero. But for a uniform motion is why earth cannot be taken a inertial frame of
acceleration is zero (for rectilinear motion) reference

275 (b) 283 (d)

Both the statements are true but reason is not a Static friction alone is a self adjusting force and
correct explanation of assertion. Here, friction not all types of friction. Assertion is false, reason
causes motion is true

284 (a)
P a g e | 111
 Assertion is false, but reason is true. Moment of
inertia is not inertia, but rotational inertia
N= ( )
m1m2 F2 F1
-
m1+m2 m2 m1
287 (c)
285 (a)
1. Force of friction is zero in (a) and (c)
(i),(ii) After spring 2 is cut, tension in string AB
because block has no tendency to move
will not change
2. Force of friction is 2.5 N in (b) and (d)
(TCD) = 4 mg
i because applied force in horizontal
m +m -m -m
(TCD)f = mDg + mD. m +m
A B C D
+m +m
.g direction in both is 2.5 N
A B C D 3. Acceleration is zero in all cases
= 2mg 1+ ( ) 1
5
= 2.4 mg (iv) Normal force is not equal of 2g in (c)
and (d) because some extra vertical
Hence TCD decreases force is also acting
(iii), (iv) After string between C and pulley is cut, 288 (a)
tension in string AB will become zero
(TCD) = (mD+mE)g = 4 mg Acceleration of the whole system towards right:
i
F
Acceleration of C and D blocks is a=
M+m
(mC+mD)g + mEg = (mC+mD).a
6mg 3
a= = g, (TCD) + mCg = mCa
4mg 2 f

(TCD)f = 2m 32 g - 2mg = mg
The tension decrease F - mgsin θ = macos θ
286 (d) F
⇒ F - mgsin θ = m cos θ
M+m
1. Let a be acceleration of two block system (M+m)mgsin θ
towards right, then ⇒F =
M+m-mcos θ
F2-F1 Pseudo force on m as seen from the frame of M:
a=
m1+m2 mF
Fs =ma =
1
m+F
= mgsin θ ( m
M+m(1-cos θ) )
< mgsin θ
F2 - T = m2.a
Pseudo force ion M as seen from the frame of m:
Solving T = (
m1m2 F2 F1
m1+m2 m2 m1
+ ) Fs2 = Ma =
MF
>
m+F m+M
mF
( )
2. Replace F1 by –F1 in result of (i),
= mgsin θ ( M
)< mgsin θ
T=
m1m2 F2 F1
m1+m2 m2 m1 ( )
-
M+m(1-cos θ)
Now mgcos θ - N = masin θ ⇒
3. Let a be acceleration of two block system N = mgcos θ - masin θ
towards left, then Hence N is less than mgcos θ. Hence, it will also be
F -F less than mgsin θ, because θ = 45°
a = 2 1 , F2 - N = m2a Applying equation on ‘m’ in horizontal direction:
m1+m2
Fcos θ - Nsin θ = ma
F
⇒ Fcos θ - Nsin θ = m
M+m

(
mF (M+m)cos θ-m
)
( )
⇒N =
m1m2 F1 F2 M+m msin θ
Solving, N = +
m1+m2 m1 m2 Put θ = 45°
4. Replacing F1 by –F1 in result of (iii)
⇒ N=
mF M+m- 2m
M+m (
m )
>
mF
m+M
P a g e | 112
 Normal force between ground and M will Hence, (ii)→(c)
be(M + m)g. It is greater than mgsin θ. It is also 291 (b)
mF mF
greater than because is less than For (i), it is not mentioned whether the object is
M+m M+m
mgsin θ accelerated or moving with constant velocity. So
289 (c) nothing can be predicted with surety
If no net force is acting along east, then also it can
Let the maximum downward displacement of m is move with constant velocity, and if no force is
x0., then acting at all, then also it can move with constant
1 2 velocity
kx = mgx0 ⇒ x0 = 2 mg/k
2 0 For (ii) and (iii): As the object is accelerated
To lift the block (M):kx0 = Mg ⇒ 2 mg = Mg (weather uniform or non-uniform) a force must
⇒ mg = Mg/2 act on the object in such a manner that a
Hence (i)-(c) component or whole of the force would be along
(ii) When m is in equilibrium east, and also the net force must be towards east
For (iv): It is moving with constant velocity, so net
force must be zero that implies no force may act
on the object
292 (a)

Maximum possible acceleration of

m:a0 = μg = 0.5 g
3M
kx = mg, T = 2kx + mg = 3mg = g So (d) matches with all (i), (ii), (iii) and (iv)
2
Let us assume that m and 2m move together with
acceleration a:
m1g
a=
3m+m1
m1g
If a = a0 ⇒ = 0.5g ⇒m1 = 3m
3m+m1
Hence (ii)-(a) So 3m is the maximum value of m1 such that both
(iii) N + kx = Mg move together
M M 1. m1 = 2m < 3m, hence (i)→(a, d)
N = Mg - kx = Mg - g= g
2 2
2. m1 = 3m hence (ii)→(a, d)
Hence (iii)-(c)
3. m1 = 4m > 3m, hence (iii)→(b, c, d)
1. Tension= kx0 = Mg = 2 mg
4. m1 = 6m > 3m, hence (iv)→(b, c, d)
Hence (iv)-(b, d)
290 (d) 293 (b)

In figure, 3 is the equilibrium position where

velocity is maximum and acceleration is zero. 1
and 2 are the extreme positions where velocity is
zero and acceleration is maximum. 1 is the
fl = 0.2×2 g = 4 N unstretched position
1

fl = 0.1×5 g = 4 N When the block is at position 3, then mg = kx. So

2

fl = 0.1×10 g = 4 N net force is zero, hence acceleration is zero. But

velocity may be either in upward or downward
32

Friction on 3 kg block is towards left and non-

direction
zero. Hence
Hence (ii)-(c, d)
(i)→b, d
fl < fl Hence 5 kg block will not move. So net
2 3

friction on 5 kg will be zero

P a g e | 113
 12-10 1
F - f3 = (2+3+5)a ⇒ a =
-2
= ms
When the block is between position 3 and 2, then 10 10
kx > mg. So net force is in upward direction, 2
f1 = 2a = N ⇒ f2 = 12 - f1 - 3a = 11 N
hence acceleration is in upward direction. But 5
velocity may be either in upward or downward f3 = 10 N
direction For F = 15 N, the situation is similar
Hence (iii)-(a, d) For relative motion to start between B and C,
But if the block is at position 2, then velocity is f2 ≥ fL
2

zero and acceleration is in upward direction F - f3 = 10a and F - f2 = 5a

Hence (i)-(a)
When the block is between position 3 and 1, f2 = F - 5a = F - 5 [ ]
F-f3
10
=
F+f3
2
mg > kx. So net force is in downward direction, F+10
> 15
hence acceleration is in downward direction. But 2
velocity may be either in upwards or downward ⇒F > 20 N (Condition for relative motion to start
direction between B and C)
Hence (iv)-(b, d) For relative motion to start between A and B
294 (a) 3
f1fL = 8 N
1

F = (m1+m2)a (i) F - f1 - f2 = 3a and f1 = 2a

Tsin θ = m2a (ii) f1 = 2 [ ] F-15
5
>8

f > 35 N (condition for relative motion between

A and B)
296 (b)

Let the accelerations of various blocks are as

Tcos θ = m2g ⇒T = m2gsec θ (iii) shown. Pulley P2 will have downward acceleration
From (ii) and (iii), a = gtan θ a
Put in (i), F = (m1+m2)gtan θ
m2F
Net force acting on m2 = m2a =
m1+m2
Force acting on m1 by wire:
m1g + Tcos θ = m1g + m2g
295 (c)

Let f1,f2,f3 represent the friction forces between

three contact surfaces A - B, B - C and C- ground,
respectively. Limiting val.ues of friction forces at a1+a2
Now a = ⇒a2 = 2a - a1 > 0
three surfaces are 8 N, 15 N, and 10 N respectively 2
For relative motion between C and Ground, the So acceleration of 2 is upwards
minimum force needed is F = 10 N Hence, (i)→(b, c)
For F = 12 N, all the three blocks move together -a +a
And a = 1 4 ⇒a4 = 2a + a1 >0
with same acceleration i.e., a1 = a2 = a3 = a 2
So acceleration of 4 is downwards
Hence (ii)→ (a, d)

P a g e | 114
 Acceleration of 2 w.r.t. 3:
a2/3 = a2 - a3 = a2 - a1 = 2(a-a2) < 0
This is downwards, hence (iii)→(d)
Acceleration of 2 w.r.t. 4: Acceleration of B is towards right, hence (i)→b.
a 2 = a2—a4 = 4a > 0 Acceleration of C w.r.t. B is towards left
4 Hence (ii)→a
This is upwards. Hence, (iv)→(c) Acceleration of A w.r.t.
297 (c) ⃗ ⃗ ⃗ ̂ ( ) ̂ ̂ ̂
C: a A/C = a A - a C = - a j - -a i = a i - a j as
fl = 0.2×4 g = 8 N shown below
1

fl = 0.4×6 g = 24 N
2

fl = 0.5×12g = 60N
3

Hence (iii)→(c, d), Similarly (iv)→(c, d)

300 (d)

Minimum force required just to slide the block =

force of static friction

f = μR = μ mg = 0.577×10×10 = 57.7 N

301 (c)

As impulse = F×t

4.25 = F×0.1
Here only 4 kg will accelerate, 2 kg and 6 kg will
remain at rest 4.25
F= = 42.5 N (downwards)
298 (d) 0.1

1. If m1 = m2 = 0, then there is no force on 302 (b)

M in horizontal direction. So M does not Before burning BC, the Free-Body Diagrams are
accelerate shown in the figure
Hence (i)-(c)
2. If m1 = m2 ≠ 0

f1 < μ1 mg, f2 < μ2mg

f1 And f2 will be of same magnitude at any time T2 = T1 + m2g (i)
whether the blocks slip on larger block or not, so kx = T2 = m1g (ii)
net force on M is zero. Hence, M does not Where x is extension in the spring. Just after
accelerate. So (ii)-(c) burning, T1 will become zero, but T2 will remain
3. m1 > m2, here f1 > f2, hence (iii)-(b, d) same
4. m1 < m2, here f1 < f2, hence (iv)-(a, d) T2 = m2g = m2a
299 (b) (m1-m2)g
⇒ a=
m2
The direction of acceleration of various blocks are
As T2 remain same, acceleration of block A will
as shown
still remain zero
P a g e | 115
303 (b) Using ∑Fx = max, we get
T - N = 8 ma1 (iv)
T3 = 20 t, T1 = T2 = 10 t
We have four unknowns T, N, a1 and a2. Solving
For A to lose contact: 10t = 1 g ⇒t = 1 s
these four equations, we get
For B to lose contact: 10t = 2g ⇒t = 2 s
g g 5
For C to lose contact :20t = 3g ⇒t = 1.5 s a1 = and a2 = ,a1 + a2 = g
8 2 8
T -1g
aA = 1 Velocity of A when B loses contact g
1 Thus, acceleration of A is in horizontal direction
2 2
8

∫ ∫ (10t-g) dt = 5 ms 5g
-1
V1 = aAdt = and in vertical direction
1 1 8
10×2-10 -2 g
At t = 2s, aB = 0, aA = = 10 ms Acceleration of B is in horizontal direction
1 2
-2
aA/B = aA - aB = 10 - 0 = 10 ms g
(leftwards) and acceleration of C is in
304 (a) 8
horizontal direction (rightwards)
Area under F - t graph = change in momentum 306 (b)
1 200
⇒ F0(6×10 ) =
-3
[40+20] ⇒ F0
2 1000 For upper block
-2
= 4000 N amax = mg = 4 ms and fmax = 40 N
305 (a) 1. When F = 30 N, as F < fmax
So both blocks will move together
Let acceleration of block C be a1 (rightwards) and
F 30 6 -2
acceleration of block B be a2 (leftwards) ∴ a= = = ms
M+m 35 7
Then, acceleration of A will be (a1 + a2) 2. When F = 250 N
downwards and a1 rightwards For upper block: 250 - 40 = 10 a1
Free-body diagram of A is shown ion the figure 210 = 10a1 ⇒ a1 = 21 ms
-2

For lower block :

40 8 -2
a2 = = ms
25 5
307 (d)

From constraint relations, we can see that

3TXB = 2TXA
Using ∑Fx = max and ∑Fy = may, we get 3 3
XA = XB ⇒ aA = aB
N = 4m (a1) (i) 2 2
And 4mg - T = 4m(a1 + a2) (ii) So let aB = a, then aA = 1.5 a
Free-body diagram of B (showing horizontal Writing equation of motion:
forces only) is shown in the figure From block A,
2T = 70aA = 105a = 3×35a
2
35a = T (i)
3
Using ∑Fx = max, we get From block B,
T = 3ma2 (iii) 300 - 3T = 35aB = 35a (ii)
Free body diagrams of C (showing horizontal Solving Eqs. (i) and (ii), we get
forces only) is shown in figure 2T
300 - 3T =
3
⇒ 900 - 9T = 2T ⇒ 900 = 11 T

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 900 45 -2
T= N 500 - 55 gsin 30° = 55 a ⇒ a = ms
11 11
180 -2 120 -2
aA = ms and aB = ms
77 77
308 (d)

As the monkey moves downwards with respect to N - 15g = 15 asin 30°

[ ]
rope with an accelerationb, its absolutre 45 1 265×15
acceleration isa + b, where a is the acceleration N = 15 10+ × =
11 2 22
of rope. Therefore, equations of motion are 45 3
mg - T = m(a + b) (i) f = 15acos 30° = 15×
11 2
T - μMg = Ma (ii) For A not to slide on B:f ≤ fl

( )
Putting the value of T from Eq. (ii) into Eq. (i), we 15×45 3 265×15
get ⇒ × ≤μ
11 2 22
(m-μM)g = (M+m)a + mb 9 3
m(g-b)-μMg ⇒ m≥ = 0.294
⇒ 53
(M+m) = a
2 313 (b)
Mmg-Mmb-μM g
T = μMg +
M+m
2 2
μM g+μMmg+Mmg-Mmb-μM g
=
M+m
Mm(μg+g-b)
=
M+m
309 (b)
Assuming systems move together there is no
fl = 0.25×4 = 1 N, fl = 0.25×(4+8) = 3 N sliding, acceleration of the system
1 2

F = fl = 3 N for constant velocity F F

a=
2
(5+10) 15 (i)
=
310 (b)
FBD of m
T - mgsin 45° - fl = ma, 2 mg sin 45° - T - fl
1

= 2 ma
2 FBD of M:f - F = ma = 10 ( )
F
15
⇒ f = F 1+ ( )
10
15
5
= F (ii)
3
If there is no sliding, F ≤ μSN

F
5
3[] ≤ 0.4×10×10 ⇒F ≤ 24 N

F 24 -2
From (i) a = = = 1.6 ms
15 15
Where fl = μ22mgcos 45° = 2mgcos 45°/3
1 314 (c)
And fl = μ22mgcos 45° = 2mgcos 45°/3

( )
2
m
g 0.5t = μ mg 1+ , t = 12 s
Now we get a = - M
9 2
⇒ μ = 0.2
This is negative, which is not possible. Hence
315 (d)
a=0
311 (c) Force in spring can’t change abruptly whereas
2 2 tension in string can change
Net force F = F1+F2 = 41 N
When PQ is cut, no effect on the forces acting on C,
fl = 0.4×10 g = 40 N, fk = 0.3×10 g = 30 N hence its acceleration remains zero
Net force is less than fl, hence 316 (d)
Required friction force=applied force= 41 N
312 (a)
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 mgsin 30 - T = ma
50 - 5 = 10 a
⃗ ⃗ ⃗
a m, M = a m - a M = μg - ( F-μmg
M )
a = 4.5 ms
-2
μ(m+M)g-F
=
0+4.5 -1 M
aP = = 2.25 ms
4
2 If the cube falls the plank, it will cover a distance l
1 2 2l 2lM
-l = a t ⇒t = =
2 mM aM,m F-μ(m+M)g
319 (a)

Acceleration of different objects is shown in

317 (a) figure. All the acceleration are w.r.t. ground

Free-body diagrams

Constraint relations:
x1 = x2 + x3
a1 = a2 + a3 Given :a3 = a4 (i)
Equation of motion: a2 + a1 = 1 (ii)
T - N = m1a1 (i) a3 + a1 = 5 (iii)
N = m2a1 (ii) From figure, we can write
m2g - T = m2a2 (iii) a +a
-a5 = 2 3 (iv)
m3g - T = m3a3 (iv) 2
Using above equation, we can calculate the values a -a
a1 = 4 5 (v)
318 (d) 2
Solving the above equations, we get
Free-body diagrams: -2 -2
a1 = 2 ms , a2 = -1 ms , a3 = a4 = 3 ms
-2

320 (c)
2 2
h = l -r = 1 m
Let both blocks move together h 1 2π
Time period:T0 = 2π = 2π = s
F g 10 10
Acceleration of blocks, a =
(m+M) 321 (d)
f=m ( ) F
m+M Tcos θ = mg, Tsin θ = mv /r
2

If both the blocks moves together, f ≤ μmg Squaring and adding both, we get the answer
mF 322 (c)
≤ μ mg
(m+M)
f
F ≤ μ(m+M)g For t < t0:f = μ mg a1 = = μg
m
If the begins to slide then, F = μ(m+M)g
f
am = = μg (towards+x direction)
m
F-μmg
aM = (towards +x direction)
M

Velocity of block at any time:

P a g e | 118
 vb = v1 + a1t possible direction of motion, it means net applied
⇒vb = v1 + μ gt force is not enough to cause the motion of system
Velocity of plank at any time:vp = v2 + at or to overcome the limiting friction force
At t = t0, both velocities are same 326 (d)
v2-v1 From the data given we can find the limiting
v1 + μ gt0 = v2 + at0 ⇒ t0 =
μg-a friction force for the two surfaces
323 (b)

Let x be the compression in spring at any time

fL = 0.5×3×10 = 15 N
1
F1 - kx = m1a0,F2 - kx = m2a0
fL = 0.2×5×10 = 10 N
2
F -F
Solve to get b:a0 = 1 2 For F < fL
m1-m2 2
Both the blocks remain at rest and f1 = F, f2 = f1
m F -F m
and kx = 1 2 1 2 anda1 = a2 = 0. For F > fL and F is less than a
m1-m2 2

Just after m2 is removed certain value say F1, the motion starts at lower
kx F -m a F surface but both the blocks continue to move with
a2 = = 2 2 0 = 2 - a0 same acceleration. The friction on lower surface
m2 m2 m2
becomes kinetic in nature
324 (c)
F-fk -2
Here, a = a1 = a2 = 2
ms
The force acting on the block areF1,F2, mg, normal 5
contact force and friction force. Here frictional F - f1 = 3 a and f1 - fk = 2a
force won’t act along vertical direction as the 2F+3fk
component of resultant force along the surface All these equations give f = 2
for relative
1
5
acting on body is not along vertical direction and motion to start between two bodies, f1 ≥ fL .
1
direction of the friction force is either opposite to
F ≥ 30 N. So minimum value of F to cause relative
the motion of block (direction of acceleration of
motion between blocks is 30 N. For f = 12 N,
resultant force along the surface if it is not
f1 = 7.8 N
moving)
327 (c)
N1 = 300 N
So, fL = mN1 = 0.6×300 = 180 N For equilibrium of block A
Resultant of 4 g and F2 is 107.7 N making an angle

of tan ()
-1 2

5
with the horizontal. As force applied

F2 = 100 N is less than fL, the block doesn’t move

and friction is static in nature

F = Nsin θ
N = F/ sin θ
To lift block B from ground
F
Ncos θ ≥ mg ⇒ cos θ ≥ mg
sin θ
f = 107.7 N making an angle of tan
-1 2

5 ()
with the
F ≥ mgtan θ = mg ()
3
4
horizontal in upward direction
3
325 (b) So, Fmin = mg
4
As acceleration is coming negative for both the 328 (a)

P a g e | 119
 a= ( )
m
Mtot
g
T1 = T + Mg = 2 Mg

Kx = 2 T1 or x =
2T1
=
4Mg
K K
334 (c)

fl = fk = 0.5×2 g = 10 N
Initially F = 20 N > fl, so the block will start
accelerating immediately
329 (d)

If F = 20 N, 10 kg block will not move and it will

not press 5 kg block So N = 0

At any time t:F = 20 - 2t

20-2t-fk 10-2t
Acceleration :a = = (i)
m m
10-2t
For a = 0, = 0 ⇒t = 5 s (ii)
2
dv 10-2t
330 (b) From (i) = = 5-t
dt 2
v 1 2

ω=
2
2π = 4 rad s
-1
∫ dv =
∫ (5-1) dt ⇒ v = 5t - t
π 0 0
2
2
T = mω l = m(4)2l = 16ml Let us see when the velocity becomes zero, For
331 (d) this:
2
t
5t - = 0 ⇒ t = 10 s
Initially velocity:u = 0 2
t

∫ ∫ (6t-2t )dt We see that at t = 10 s, alsoF = 0. So the block

2
I= Fdt ⇒ m(v-u) =
0 has no tendency to move. Hence, acceleration is
2 2 3 zero at this time. Now the block will not move
⇒ 2v = 3t - t
3 from t = 10 s to 15 s, because for this magnitude
Put v = 0, we gte t = 4.5 s of F < 10 N. so block will remain at rest from
332 (a) t = 10 s to 15 s or acceleration is zero from 10 to
15 s
Let m1 and m2 do not accelerate up or down, then
335 (c)
F1 = m1g, F2 = m2g. But m1 ≠ m2 , so F1 ≠ F2
Hence net horizontal force on M isF1 - F2. So M Since there4 is no friction between A and B, B will
cannot be in equilibrium. If M acceelrates remain at rest
horizontally, then m1 and m2 also accelerate
horizontally
333 (a) fk = u(2+2)g = 0.1(2+2)g = 4 N
fk 4 -1
T = Mg Acceleration of A:a = = = 23 ms
m 2
1 2
For B to fall off A:S = ut + at
2
1 2
⇒4 = 0×t + 2t ⇒ t = 2s
2
336 (5)

P a g e | 120
 mg μmg
F1 = +
2 2
mg μmg
F2 = -
2 2

' ' ' '

l3 + l4 + l5 + l6 = C ⇒ l3 + l4 + l5 + l6 = 0
-1
aP - aB - aB - aB = 0 ⇒ aP = 3aB ⇒ aB = 2 ms
340 (1)
F1 = 3F2
1 Drawing free-body diagram of block with respect
1 + μ = 3 - 3μ⇒4μ = 2⇒μ =
2 to plane. Acceleration of the block up the plane is
N = 10μ⇒N = 5
337 (6)

Consider the forces on the person:

a=
mgcos 37°-mgsin 37°
m
4 3
= g - = 2 ms
5 5
-2
( )
1 2
Applying, s = at we get
2
2s 2×1
∑F y = may ⇒ t=
a
=
2
= 1 sec

n - mg = ma 341 (8)
-2
n = 1.6 mg so a = 0.60 g = 6 ms
2 2 Since mgsin 37° > μ mgcos 37°, the block has a
v = u + 2as
2 2 tendency to slip downwards
⇒ v = 0 + 2×6×3
-1 Let F be the minimum force applied on it, so that it
v = 6 ms
does not slip. Then,
338 (6)
N = F + mgcos 37°
Force of friction between the two will be ∴ mgsin 37° = μN = μ(F + mgcos 37°)
maximum i.e., mgsin 37°
or F = - mgcos 37°
μmg μ

()
μmg. Retardation of A is aA = = μg (2)(10)(3/5) 4
m = - (2)(10) = 8N
μmg μg 0.5 5
And acceleration of B is aB = = 342 (1)
2m 2
Acceleration of B relative to A is 2
N = mg, μN = mrω
3μg
aBA = aA + aB =
2
1 3g
Substituting, μ = ; aBA =
2 4
339 (2)
' '
l1 + l2 = C ⇒ l1 + l2 = 0 2 μg 0.1×10
μmg = m 2sin θω ⇒ ω = =
⇒ - ap + (12-ap) = 0 sin θ 2×sin 30°
-2
⇒ ap = 6 ms
-1
= 1 rad s
343 (5)

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 Denote the common magnitude of the maximum 2N-(M-m)g
a=
acceleration asa. For block A to remain at rest M-m
with respect to block A to remain at rest with 2×450-(100-25)×10 900-750 -2
= = = 2 ms
respect to block B, a ≤ μsg. to be largest. The 100-25 75
tension in the cord is then 345 (4)
T = (mA + mB)a + μkg(mA+mB)
Let A apply a force R on B
= (mA+mB)(a + μkg)
This tension is related to the mass mC (largest) by
T = mC(g-a).Solving for mC yields

mC =
(mA+mB)(μs+μk) =
(1.5+0.5)(0.6+0.4)
Then B also applies an opposite force R on A as
1-μs 1-0.6
shown in figure
= 5 kg
For A:
344 (2)
mg - R = ma
Let M = mass of painter =10 kg ⇒R = m(g-a) = 0.5[10-2] = 4 N
m = mass of crate=25 kg 346 (4)
Let FA be the action force exerted by painter on
Since A tends to slip down, frictional forces act on
crate, reaction force exerted by crate on man it from both sides up the plane
N = FA = 450 N '
Let N be the reaction of the plank on A and N be
The free-body diagram of painter is shown in the mutual normal action-reaction between A and
figure (b) B
From the free-body diagram of A
' '
N + mgcos α = N and mgsin α = μ(N + N )
From the free-body diagram of B

Therefore, equation of motion of painter is

N + T - Mg = Ma (i)
The equation of motion of whole system is ''
N6 = mgcos α
2T - (M+m)g = (M + m)a (ii) '
mgsin α + μN = T
Multiplying (i) by 2, we get
∴ 2 mgcos α = N
2N + 2T - 2Mg = 2Ma (iii)
and mgsin α = μ(2 mgcos α + mgcos α)
Subtracting (ii) from (iii), we get
1 1 3
2N - 2Mg + (M+m)g = (2M-M-m)a or μ = tan α = × = 0.25 or 1/m=4
3 3 4
or 2N - (M-m)g = (M - m)a

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