1

Yakeen NEET 2.0 (2026)

Physics by Saleem Sir
KPP 25 Laws of Motion

Ist, IInd and IIIrd LAWS OF MOTION: 6. A particle moves in 𝑥 − 𝑦 plane under the
1. An object with mass 500 g moves along x-axis with influence of a force 𝐹⃗ such that its linear
speed v = 4 x m/s. The force acting on the object momentum is 𝑝⃗(𝑡) = 𝑖̂cos(𝑘𝑡) − 𝑗̂sin(𝑘𝑡). If k is
is: [April 7, 2025 (II)] constant, the angle between 𝐹⃗ and 𝑝⃗ will be:
(1) 8 N (2) 5 N [April 5, 2024 (II)]
(3) 6 N (4) 4 N (1) /2
(2) /6
2. A body of mass 2 kg moving with velocity of (3) /4
vin = 3iˆ + 4 ˆj ms–1 enters into a constant force field (4) /3
of 6N directed along positive z-axis. If the body
remains in the field for a period of 5/3 seconds, then 7. A wooden block, initially at rest on the ground, is
velocity of the body when it emerges from force pushed by a force which increases linearly with
field is. [April 8, 2025 (II)] time t. Which of the following curve best describes
(1) 4iˆ + 3 ˆj + 5kˆ acceleration of the block with time:
(2) 3iˆ + 4 ˆj + 5kˆ [April 4, 2024 (I)]

(3) 3iˆ + 4 ˆj − 5kˆ

(4) 3iˆ + 4 ˆj + 5kˆ (1) (2)

3. A balloon and its content having mass M is moving

up with an acceleration 'a'. The mass that must be
released from the content so that the balloon starts
moving up with an acceleration '3a' will be (Take
'g' as acceleration due to gravity) (3) (4)
[Jan. 28, 2025 (II)]
3Ma 3Ma
(1) (2)
2a − g 2a + g
2Ma 2Ma 8. A cricket player catches a ball of mass 120 g
(3) (4)
3a + g 3a − g moving with 25 m/s speed. If the catching process
is completed in 0.1 s then the magnitude of force
4. A player caught a cricket ball of mass 150 g moving exerted by the ball on the hand of player will be
at a speed of 20 m/s. If the catching process is (in SI unit): [Feb 1, 2024 (II)]
completed in 0.1 s, the magnitude of force exerted (1) 30 (2) 24
by the ball on the hand of the player is: (3) 12 (4) 25
[April 8, 2024 (I)]
(1) 150 N (2) 3 N 9. A spherical body of mass 100 g is dropped from a
(3) 30 N (4) 300 N height of 10 m from the ground. After hitting the
ground, the body rebounds to a height of 5 m. The
5. A wooden block of mass 5 kg rests on a soft
impulse of force imparted by the ground to the body
horizontal floor. When an iron cylinder of mass
is given by: (given, g = 9.8 m/s²)
25 kg is placed on the top of the block, the floor
[Jan 30, 2024 (I)]
yields and the block and the cylinder together go
(1) 4.32 kg ms–1
down with an acceleration of 0.1 ms–2. The action
(2) 43.2 kg ms–1
force of the system on the floor is equal to:
[April 5, 2024 (I)] (3) 23.9 kg ms–1
(1) 297 N (2) 294 N (4) 2.39 kg ms–1
(3) 291 N (4) 196 N
 2

10. A body of mass 1000 kg is moving horizontally 16. Figure (a), (b), (c) and (d) show variation of force
with a velocity 6 m/s. If 200 kg extra mass is added, with time. [Feb. 1, 2023 (II)]
the final velocity (in m/s) is:
[Jan 27, 2024 (I)]
(1) 6 (2) 2
(3) 3 (4) 5 (A)

11. A bullet 10 g leaves the barrel of gun with a

velocity of 600 m/s. If the barrel of gun is 50 cm
long and mass of gun is 3 kg, then value of impulse
supplied to the gun will be:
[April 13, 2023 (I)]
(1) 12 Ns
(2) 6 Ns (B)
(3) 36 Ns
(4) 3 Ns

12. Three forces F1 = 10 N, F2 = 8 N, F3 = 6 N are acting

on a particle of mass 5 kg. The forces F2 and F3 are
applied perpendicularly so that particle remains at
rest. If the force F1 is removed, then the (C)
acceleration of the particle is:
[April 12, 2023 (I)]
(1) 2 ms–2
(2) 0.5 ms–2
(3) 4.8 ms–2
(4) 7 ms–2

13. An average force of 125 N is applied on a machine (D)

gun firing bullets each of mass 10 g at the speed of
250 m/s to keep it in position. The number of
bullets fired per second by the machine gun is:
[April 11, 2023 (I)] (1) Fig. (A) (2) Fig. (D)
(1) 5 (2) 50 (3) Fig. (C) (4) Fig. (B)
(3) 100 (4) 25
17. The figure represents the momentum time (p – t)
14. A body of mass 500 g moves along x-axis such that curve for a particle moving along an axis under the
it's velocity varies with displacement x according to influence of the force. Identify the regions on the
graph where the magnitude of the force is
the relation v = 10 x ms−1 the force acting on the
maximum and minimum respectively?
body is: [April 11, 2023 (II)] If (t3 – t2) < t1. [Jan 30, 2023 (I)]
(1) 166 N
(2) 25 N
(3) 125 N
(4) 5 N

15. At any instant the velocity of a particle of mass

500 g is (2tiˆ + 3t 2 ˆj)ms−1 . If the force acting on the
particle at t = 1 s is (iˆ + xjˆ)N . Then the value of x
(1) c and a
will be: [April 8, 2023 (I)] (2) b and c
(1) 3 (2) 4 (3) c and b
(3) 6 (4) 2 (4) a and b
 3

18. A machine gun of mass 10 kg fires 20 g bullets at 23. A block of metal weighing 2 kg is resting on a
the rate of 180 bullets per minute with a speed of frictionless plane (as shown in figure). It is struck
100 m s–1 each. The recoil velocity of the gun is: by a jet releasing water at a rate of 1 kgs–1 and at a
[Jan 30, 2023 (II)] speed of 10 ms–1. Then, the initial acceleration of
(1) 0.02 m/s the block, in ms–2, will be:
(2) 2.5 m/s [Jan 29, 2023 (I)]
(3) 1.5 m/s
(4) 0.6 m/s

19. Force acts for 20 s on a body of mass 20 kg, starting

from rest, after which the force ceases and then (1) 3 (2) 6
body describes 50 m in the next 10 s. The value of (3) 5 (4) 4
force will be: [Jan 29, 2023 (II)]
(1) 40 N (2) 5 N 24. A man of 60 kg is running on the road and suddenly
(3) 20 N (4) 10 N jumps into a stationary trolly car of mass 120 kg.
Then the trolly car starts moving with velocity
2 m s–1. The velocity of the running man was
20. In two different experiments, an object of mass 5
_______ ms–1. When he jumps into the car.
kg moving with a speed of 25–1 hits two different [June 28, 2022 (I)]
walls and comes to rest within (i) 3 second, (ii) 5
second, respectively. 25. A block of mass 2 kg moving on a horizontal
Choose the correct option out of the following: surface with speed of 4 ms–1 enters a rough surface
[Jan 28, 2022 (I)] ranging from x = 0.5 m to x = 1.5 m. The retarding
(1) Impulse and average force acting on the object force in this range of rough surface is related to
will be same for both the cases. distance by F = –kx where k = 12 Nm–1. The speed
(2) Impulse will be same for both the cases but the of the block as it just crosses the rough surface will
average force will be different. be: [June 28, 2022 (II)]
–1
(3) Average force will be same for both the cases (1) Zero (2) 1.5 ms
but the impulse will be different. (3) 2.0 ms–1 (4) 2.5 ms–1
(4) Average force and impulse will be different
for both the cases. 26. A batsman hits back a ball of mass 0.4 kg straight
in the direction of the bowler without changing its
21. A balloon has mass of 10 g in air. The air escapes initial speed of 15 ms–1. The impulse imparted to
from the balloon at a uniform rate with velocity the ball is _______.
[June 26, 2022 (II)]
4.5 cm/s. If balloon shrinks in 5 s completely. Then,
the average force acting on that balloon will be
(in dyne). [July 28, 2022 (I)] 27. (
A force on an object of mass 100 g is 10iˆ + 5 ˆj N.)
(1) 3 (2) 9 the position of that object at t = 2s is ( aiˆ + bjˆ ) m
(3) 12 (4) 18
after starting from rest. The value of a/b will
22. A ball of mass 0.15 kg hits the wall with its initial be______. [June 26, 2022 (I)]

speed of 12 ms–1 and bounces back without

28. An object of mass 5 kg is thrown vertically upwards
changing its initial speed. If the force applied by the
from the ground. The air resistance produces a
wall on the ball during the con-tact is 100 N.
constant retarding force of 10 N throughout the
Calculate the time duration of the contact of ball
motion. The ratio of time of ascent to the time of
with the wall. [July 26, 2022 (II)]
descent will be equal to: [Use g = 10 ms–2]
(1) 0.018 s [June 24, 2022 (II)]
(2) 0.036 s (1) 1 : 1 (2) 2: 3
(3) 0.009 s
(3) 3: 2 (4) 2 : 3
(4) 0.072 s
 4

29. The initial mass of a rocket is 1000 kg. Calculate at 34. A spaceship in space sweeps stationary
what rate the fuel should be burnt so that the rocket interplanetary dust. As a result, its mass increases
is given an acceleration of 20 ms–2. The gases come dM(t )
out at a relative speed of 500 ms–1 with respect to at a rate = bv 2 (t ) where v(t) is its
dt
the rocket : [Use g = 10 m/s2] instantaneous velocity. The instantaneous
[Aug. 26, 2021 (I)] acceleration of the satellite is:
(1) 6.0 × 102 kg s–1 [Sep 05, 2020 (II)]
(2) 500 kg s–1 bv 3
(3) 10 kg s–1 (1) −bv3 (t ) (2) −
(4) 60 kg s–1 M(t )
2bv3 bv3
(3) − (4) −
30. A particle of mass M originally at rest is subjected M(t ) 2M(t )
to a force whose direction is constant but
magnitude varies with time according to the 35. A small ball of mass m is thrown upward with
relation velocity u from the ground. The ball experiences a
  t − T 2  resistive force mkv2 where v is its speed. The
F = F0 1 −    maximum height attained by the ball is:
  T   [Sep 04, 2020 (II)]
Where F0 and T are constants. The force acts only 1 ku 2
1  ku 2 
(1) tan −1 (2) ln 1 + 
k  2 g 
for the time interval 2T. The velocity v of the
2k g
particle after time 2T is:
[Aug. 27, 2021 (II)] 1 −1 ku 2 1  ku 2 
(1) 2 F0 T/M (2) F0 T/2M (3) tan (4) ln 1 + 
k 2g 2k  g 
(3) 4F0 T/3M (4) F0 T/3M

31. A force F = (40iˆ + 10 ˆj)N acts on a body of mass 36. A ball is thrown upward with an initial velocity v0
from the surface of the earth. The motion of the ball
5 kg. If the body starts from rest, its position vector
is affected by a drag force equal to mv2 (where m
r at time = 10 s, will be: is mass of the ball, v is its instantaneous velocity
[July 25, 2021 (II)]
and  is a constant). Time taken by the ball to rise
(1) (100iˆ + 400 ˆj)m to its zenith is: [10 April 2019 I]
(2) (100iˆ + 100 ˆj)m 1   
(1) tan −1  v0 
(3) (400iˆ + 100 ˆj)m g  g 
(4) (400iˆ + 400 ˆj)m 1   
(2) sin −1  v0 
g  g 
32. A boy pushes a box of mass 2 kg with a force
( )
F = 20iˆ + 10 ˆj N on a frictionless surface. If the (3)
1 
ln 1 +
 
v0 
g  g 
box was initially at rest, then ________m is
displacement along the x-axis after 10 s. 1  2 
[Feb. 26, 2021 (I)] (4) tan −1  v0 
2g  g 
33. A particle moving in the xy plane experiences a
velocity dependent force F = k (vyiˆ + vx ˆj ) ,where 37. A ball is thrown vertically up (taken as + z-axis)
from the ground The correct momentum-height
vx and vy are the x and y components of its velocity (p-h) diagram is:
v . If a is the acceleration of the particle, then [9 April 2019 I]
which of the following statements is true for the
particle?
[Sep 06, 2020 (II)] (1) (2)
(1) quantity v  a is constant in time
(2) F arises due to a magnetic field
(3) kinetic energy of particle is constant in time
(3) (4)
(4) quantity 𝑣⃗ . 𝑎⃗ is constant in time
 5

38. A particle of mass m is moving in a straight line 42. This question has Statement 1 and Statement 2. Of
with momentum p. Starting at time t = 0, a force the four choices given after the Statements, choose
F = kt acts in the same direction on the moving the one that best describes the two Statements.
particle during time interval T so that its Statement 1: If you push on a cart being pulled by
momentum changes from p to 3p. Here k is a a horse so that it does not move, the cart pushes you
constant. The value of T is: back with an equal and opposite force.
[11 Jan. 2019 II] Statement 2: The cart does not move because the
k p force described in statement 1 cancel each other.
(1) 2 (2) 2 [Online May 26, 2012]
p k
(1) Statement 1 is true, Statement 2 is true,
2k 2p Statement 2 is the correct explanation of
(3) (4)
p k Statement 1.
(2) Statement 1 is false, Statement 2 is true.
39. A particle of mass m is acted upon by a force F (3) Statement 1 is true, Statement 2 is false.
(4) Statement 1 is true, Statement 2 is true,
R
given by the empirical law F = 2 v(t ) . If this law Statement 2 is not the correct explanation of
t Statement 1.
is to be tested experimentally by observing the
motion starting from rest, the best way is to plot: 43. Two fixed frictionless inclined planes making an
[Online April 10, 2016]
angle 30° and 60° with the horizontal are shown in
(1) log v(t) against 1/t the figure. Two blocks A and B are placed on the
(2) v(t) against t2 two planes. What is the relative vertical
(3) log v(t) against 1/t2 acceleration of A with respect to B? [2010]
(4) log v(t) against t

40. A body of mass 5 kg is under the action of a

constant force F = Fxiˆ + Fy ˆj . Its velocity at t = 0 s
is v = (6iˆ − 2 ˆj) m/s, and at t = 10 s its velocity is
v = +6 ˆj m/s. The force F is: (1) 4.9 ms–2 in horizontal direction
[Online April 11, 2014] (2) 9.8 ms–2 in vertical direction
(3) Zero
 3 4 
(1) (−3 ˆj + 4 ˆj)N (2)  − iˆ + ˆj  N (4) 4.9 ms–2 in vertical direction
 5 5 
3 4  44. A ball of mass 0.2 kg is thrown vertically upwards
(3) (3iˆ − 4 ˆj)N (4)  iˆ − ˆj  N
5 5  by applying a force by hand. If the hand moves
0.2 m while applying the force and the ball goes
41. A particle of mass m is at rest at the origin at time upto 2 m height further, find the magnitude of the
t = 0. It is subjected to a force F(t) = F0e–bt in the x force. (Consider g = 10 m/s2).
[2006]
direction. Its speed v(t) is depicted by which of the (1) 4 N (2) 16 N
following curves? (3) 20 N (4) 22 N
[2012]

45. A player caught a cricket ball of mass 150 g moving

(1) at a rate of 20 m/s. If the catching process is
completed in 0.1s, the force of the blow exerted by
the ball on the hand of the player is equal to
[2006]
(2) (1) 150 N (2) 3 N
(3) 30 N (4) 300 N

46. A particle of mass 0.3 kg subject to a force F = –kx

(3) with k = 15 N/m. What will be its initial
acceleration if it is released from a point 20 cm
away from the origin? [2005]
(4) (1) 15 m/s2 (2) 3 m/s2
(3) 10 m/s2 (4) 5 m/s2
 6

47. A block is kept on a frictionless inclined surface Motion of Connected Bodies, Pulley and Equilibrium
with angle of inclination ‘’. The incline is given of Forces
an acceleration ‘a’ to keep the block stationary. 51. A body of mass m is suspended by two strings
Then a is equal to [2005] making angles 1 and 2 with the horizontal ceiling
with tensions T1 and T2 simultaneously. T1 and T2
are related by T1 = 3T2 , the angles 1 and 2 are
[April 4, 2025 (I)]
3mg
(1) 1 = 30° 2 = 60° with T2 =
4
(1) g cosec  mg
(2) g / tan  (2) 1 = 60° 2 = 30° with T2 =
2
(3) g / tan  3mg
(4) g (3) 1 = 45° 2 = 45° with T2 =
4
4mg
48. A rocket with a lift-off mass 3.5 × 104 kg is blasted (4) 1 = 30° 2 = 60° with T2 =
upwards with an initial acceleration of 10 m/s2. 5
Then the initial thrust of the blast is
[2003] 52. A body of mass 1 kg is suspended with the help of
(1) 5
3.5 × 10 N two strings making angles as shown in figure.
(2) 7.0 × 105 N Magnitudes of tensions T1 and T2, respectively, are
(3) 14.0 × 105 N (in N): [April 2, 2025 (II)]
(4) 1.75 × 105 N

49. Three forces start acting simultaneously on a

particle moving with velocity, v . These forces are
represented in magnitude and direction by the three
sides of a triangle ABC. The particle will now
move with velocity [2003] (1) 5,5 3
(2) 5 3,5
(3) 5 3,5 3
(4) 5, 5

53. A massless spring gets elongated by amount x1

(1) less than v under a tension of 5N. Its elongation is x2 under the
(2) greater than v tension of 7N. For the elongation of (5x1 - 2x2), the
(3) | v | in the direction of the largest for BC tension in the spring will be
(4) v , remaining unchanged [Jan. 23, 2025 (II)]
(1) 15 N (2) 20 N
50. A solid sphere, a hollow sphere and a ring are (3) 11 N (4) 39 N
released from top of an inclined plane (frictionless)
54. A light unstretchable string passing over a smooth
so that they slide down the plane. Then maximum
light pulley connects two blocks of masses m1 and
acceleration down the plane is for (no rolling)
[2002]
m2. If the acceleration of the system is g/8, then the
(1) solid sphere ratio of the masses m2/m1 is:
[April 9, 2024 (I)]
(2) hollow sphere
(1) 9:7
(3) ring
(2) 4:3
(4) all same (3) 5:3
(4) 8:1
 7

55. A 1 kg mass is suspended from the ceiling by a rope 58. All surfaces shown in figure are assumed to be
of length 4m. A horizontal force 'F' is applied at the frictionless and the pulleys and the string are light.
mid point of the rope so that the rope makes an The acceleration of the block of mass 2 kg is:
angle of 45° with respect to the vertical axis as [Jan 30, 2024 (I)]
shown in figure. The magnitude of F is:
[April 9, 2024 (II)]

(1) g (2) g/3

(3) g/2 (4) g/4

59. Three blocks A, B and C are pulled on a horizontal

smooth surface by a force of 80 N as shown in
10 figure
(1) N (2) 1 N
2
1
(3) N (4) 10 N The tensions T1 and T2 in the string are respectively
10  2 [Jan 30, 2024 (II)]
(1) 40 N, 64 N (2) 60 N, 80 N
56. A light string passing over a smooth light pulley (3) 88 N, 96 N (4) 80 N, 100 N
connects two blocks of masses m1 and m2
(where m2 > m1). If the acceleration of the system 60. As shown in figure, a 70 kg garden roller is pushed
is g / 2 , then the ratio of the masses m1/m2 is: with a force of 𝐹⃗ = 200 N at an angle of 30° with
[April 6, 2024 (I)] horizontal. The normal reaction on the roller is
2 −1 1+ 5 _______. (Given g = 10 ms–2).
(1) (2) [Jan 31, 2024 (I)]
2 +1 5 −1
1+ 5 3 +1
(3) (4)
2 −1 2 −1

57. Three blocks M1, M2, M3 having masses 4 kg, 6

kg and 10 kg respectively are hanging from a
smooth pully using rope 1, 2 and 3 as shown in (1) 800 2 N (2) 600 N
figure. The tension in the rope 1, T1 when they are (3) 800 N (4) 200 3 N
moving upward with acceleration of 2 ms–2
is_________ N (if g = 10 m/s2). 61. A block of √3 kg is attached to a string whose other
[April 5, 2024 (I)]
end is attached to the wall. An unknown force F is
applied so that the string makes an angle of 30°
with the wall. The tension T is:
(Given g = 10 ms–2)
[Jan 30, 2023 (II)]

3
(1) 20 N (2) 25 N
(3) 10 N (4) 15 N
 8

62. Given below are two statements: 66. A uniform metal chain of mass m and length 'L'
Statement-I: An elevator can go up or down with passes over a massless and frictionless pulley. It is
uniform speed when its weight is balanced with the released from rest with a part of its length 'l' is
tension of its cable. hanging on one side and rest of its length 'L – l' is
Statement-II: Force exerted by the floor of an hanging on the other side of the pulley. At a certain
elevator on the foot of a person standing on it is
point of time, when l = L/x, the acceleration of the
more than his/her weight when the elevator goes
down with increasing speed. chain is g/2. The value of x is _______.
[July 28, 2022 (II)]
In the light of the above statements, choose the
correct answer from the options given below:
[Jan 24, 2023 (I)]
(1) Both statement I and statement II are false
(2) Statement I is true but Statement II is false
(3) Both Statement I and Statement II are true
(4) Statement I is false but Statement II is true

63. As per given figure, a weightless pulley P is

attached on a double inclined frictionless surface.
The tension in the string (massless) will be (1) 6 (2) 2
(if g = 10 m/s2). [Jan 24, 2023 (I)] (3) 1.5 (4) 4

67. A monkey of mass 50 kg climbs on a rope which

can withstand the tension (T) of 350N. If monkey
initially climbs down with an acceleration of 4 m/s2
(1) (4 )
3 +1 N and then climbs up with an acceleration of 5m/s2.
Choose the correct option (g = 10m/s2)
(2) 4( 3 + 1) N [July 26, 2022 (I)]

4( 3 − 1) N
(1) T = 700N while climbing upward
(3)
(2) T = 350N while going downward
(4) (4 3 − 1) N (3)
(4)
Rope will break while climbing upward
Rope will break while going downward
64. Two bodies of masses m1 = 5 kg and m2 = 3 kg are
68. Three masses M = 100 kg, m1 = 10 kg and m2 = 20
connected by a light string going over a smooth
light pulley on a smooth inclined plane as shown in kg are arranged in a system as shown in figure. All
the figure. The system is at rest. The force exerted the surface are frictionless and strings are
by the inclined plane of the body of mass m1 will inextensible and weightless. The pulleys are also
be: [Take g = 10 ms–2] weightless and frictionless. A force F is applied on
[July 29, 2022 (II)] the system so that the mass m2 moves upward with
an acceleration of 2 ms–2. The value of F is:
(Take g = 10 ms–2).
[July 26, 2022 (I)]

(1) 30 N (2) 40 N
(3) 50 N (4) 60 N

65. A block 'A' takes 2s to slide down a frictionless

incline of 30° and length 'l', kept inside a lift going
up with uniform velocity 'v'. If the incline is (1) 3360 N
changed to 45°, the time taken by the block, to slide (2) 3380 N
down the incline, will be approximately: (3) 3120 N
[July 27, 2022 (II)]
(1) 2.66 s (2) 0.83 s (4) 3240 N
(3) 1.68 s (4) 0.70 s
 9

69. Two masses M1 and M2 are tied together at the two 73. A mass of 10 kg is suspended vertically by a rope
ends of a light inextensible string that passes over a of length 5 m from the roof. A force of 30 N is
frictionless pulley. When the mass M2 is twice that applied at the middle point of rope in horizontal
of M1. The acceleration of the system is a1. When direction. The angle made by upper half of the rope
the mass M2 is thrice that of M1. The acceleration with vertical is θ = tan–1(x × 10–1). The value of x is
of the system is a2. The ratio a1/a2 will be: _______. (Given g = 10 m/s2).
[July 26, 2022 (II)] [June 27, 2022 (II)]

74. A person is standing in an elevator. In which

situation, he experiences weight loss?
[June 26, 2022 (I)]
(1) When the elevator moves upward with
constant acceleration
(2) When the elevator moves downward with
constant acceleration
(1) 1/3 (2) 2/3 (3) When the elevator moves upward with
(3) 3/2 (4) 1/2 uniform velocity
(4) When the elevator moves downward with
70. Four forces are acting at a point P in equilibrium as uniform velocity
shown in figure. The ratio of force F1 to F2 is 1 : x
where x =________. 75. In the arrangement shown in figure a1, a2, a3 and a4
[July 25, 2022 (I)]
are the acceleration of masses m1, m2, m3 and m4
respectively. Which of the following relation is true
for this arrangement?
[June 26, 2022 (II)]

71. For a free body diagram shown in the figure, the

four forces are applied in the ‘x’ and ‘y’ directions.
What additional force must be applied and at what
angle with positive x-axis so that the net
acceleration of body is zero?
[July 25, 2022 (II)]

(1) 4a1 + 2a2 + a3 + a4 = 0

(2) a1 + 4a2 + 3a3 + a4 = 0
(3) a1 + 4a2 + 3a3 + 2a4 = 0
(4) 2a1 + 2a2 + 3a3 + a4 = 0

76. A system of 10 balls each of mass 2 kg are

(1) 2 N, 45°
connected via massless and un-stretchable string.
(2) 2 N, 135° The system is allowed to slip over the edge of a
2 smooth table as shown in figure. Tension on the
(3) N, 30° string between the 7th and 8th ball is _______ N
3
(4) 2 N, 45° when 6th ball just leaves the table.
[June 26, 2022 (II)]

72. A block of mass M placed inside a box descends

vertically with acceleration ‘a’. The block exerts a
force equal to one-fourth of its weight on the floor
of the box. The value of ‘a’ will be _______.
[June 29, 2022 (II)]
(1) g/4 (2) g/2
(3) 3g/4 (4) g
 10

77. A block of mass 200 g is kept stationary on a 81. A steel block of 10 kg rests on a horizontal floor as
smooth inclined plane by applying a minimum shown. When three iron cylinders are placed on it
horizontal force F = x N as shown in figure. The as shown, the block and cylinders go down with an
value of x = _______. acceleration 0.2 m/s2. The normal reaction R' by the
[June 25, 2022 (II)] floor if mass of the iron cylinders are equal and of
20 kg each, is _______N.
[Take g = 10 m/s2] [July 20, 2021 (I)]

78. A block of mass m slides on the wooden wedge,

which in turn slides backward on the horizontal (1) 716 (2) 686
surface. The acceleration of the block with respect (3) 714 (4) 684
to the wedge is: Given m = 8 kg, M = 16 kg. Assume
all the surfaces shown in the figure to be 82. Consider a frame that is made up of two thin
frictionless. [Sep. 1, 2021 (II)] massless rods AB and AC as shown in the figure. A
vertical force 𝑃⃗⃗ of magnitude 100 N is applied at
point A of the frame. Suppose the force is 𝑃⃗⃗
resolved parallel to the arms AB and AC of the
frame. The magnitude of the resolved component
along the arm AC is xN. The value of x, to the
4 6 nearest integer, is ________.
(1) g (2) g
3 5 [Given: sin(35°) = 0.573, cos(35°) = 0.819,
3 2 sin(110°) = 0.939, cos(110°) = –0.342]
(3) g (4) g
5 3 [March 16, 2021 (I)]

79. A car is moving on a plane inclined at 30º to the

horizontal with an acceleration of 10 ms–2 parallel
to the plane upward. A bob is suspended by a string
from the roof of the car. The angle in degrees which
the string makes with the vertical is______.
(Take g = 10 ms–2)
[Aug. 31, 2021 (I)]
83. A person standing on a spring balance inside a
80. The boxes of masses 2 kg and 8 kg are connected stationary lift measures 60 kg. The weight of that
by a massless string passing over smooth pulleys. person if the lift descends with uniform downward
Calculate the time taken by box of mass 8 kg to acceleration of 1.8 m/s2 will be _______N.
strike the ground starting from rest. [g =10 m/s2] [Feb. 26, 2021 (I)]
(use g = 10 m/s2) [Aug. 27, 2021 (II)]
84. A mass of 10 kg is suspended by a rope of length
4 m, from the ceiling. A force F is applied
horizontally at the mid-point of the rope such that
the top half of the rope makes an angle of 45° with
the vertical. Then F equals: (Take g = 10 ms–2 and
the rope to be massless)
[9 Jan. 2019 II, 7 Jan. 2020 (II)]
(1) 100 N
(2) 90 N
(1) 0.34 s (2) 0.2 s (3) 70 N
(3) 0.25 s (4) 0.4 s (4) 75 N
 11

85. Two blocks of mass M1 = 20 kg and M2 = 12 kg are 89. A block of mass M is pulled along a horizontal
connected by a metal rod of mass 8 kg. The system frictionless surface by a rope of mass m. If a force
is pulled vertically up by applying a force of 480 N P is applied at the free end of the rope, the force
as shown. The tension at the mid-point of the rod exerted by the rope on the block is:
is: [April 22, 2013] [2003]
Pm
(1)
M+m
Pm
(2)
M−m
(3) P
PM
(4)
(1) 144 N (2) 96 N M+m
(3) 240 N (4) 192 N
90. A light spring balance hangs from the hook of the
86. A block of mass m is connected to another block of other light spring balance and a block of mass M kg
mass M by a spring (massless) of spring constant k. hangs from the former one. Then the true statement
The blocks are kept on a smooth horizontal plane. about the scale reading is [2003]
Initially the blocks. are at rest and the spring is (1) both the scales read M kg each
unstretched. Then a constant force F starts acting (2) the scale of the lower one reads M kg and of
on the block of mass M to pull it. Find the force on the upper one zero
the block of mass m. [2007] (3) the reading of the two scales can be anything
MF mF but the sum of the reading will be Mkg
(1) (2)
(m + M) M (4) both the scales read M/2 kg each
(M + m)F mF
(3) (4) 91. A lift is moving down with acceleration a. A man
m (m + M) in the lift drops a ball inside the lift. The
acceleration of the ball as observed by the man in
87. Two masses m1 = 5 kg and m2 = 4.8 kg tied to a the lift and a man standing stationary on the ground
string are hanging over a light frictionless pulley. are respectively [2002]
What is the acceleration of the masses when left (1) g, g (2) g – a, g – a
free to move ? (g = 9.8 m/s2) [2004]
(3) g – a, g (4) a, g

92. When forces F1, F2, F3 are acting on a particle of

mass m such that F2 and F3 are mutually
perpendicular, then the 2 particle remains
stationary. If the force F1 is now removed then the
acceleration of the particle is [2002]
(1) F1/m
(1) 5 m/s2 (2) 9.8 m/s2 (2) F2F3/mF1
(3) 0.2 m/s2 (4) 4.8 m/s2 (3) (F2–F3)/m
(4) F2/m
88. A spring balance is attached to the ceiling of a lift.
A man hangs his bag on the spring and the spring 93. Three identical blocks of masses m = 2kg are drawn
reads 49 N, when the lift is stationary. If the lift by a force F = 10 2 N with an acceleration of
moves downward with an acceleration of 5 m/s2, 0.6 ms–2 on a frictionless surface, then what is the
the reading of the spring balance will be tension (in N) in the string between the blocks B
[2003]
and C? [2002]
(1) 24 N
(2) 74 N
(3) 15 N
(4) 49 N (1) 9.2 (2) 3.4
(3) 4 (4) 9.8
 12

94. One end of a massless rope, which passes over a 99. A 2 kg brick begins to slide over a surface which is
massless and frictionless pulley P is tied to a hook inclined at an angle of 45° with respect to
C while the other end is free. Maximum tension that horizontal axis. The co-efficient of static friction
the rope can bear is 360 N. With what value of between their surfaces is:
maximum safe acceleration (in ms–2) can a man of [April 4, 2024 (II)]
60 kg climb on the rope? [2002] (1) 1 (2) 1 / 3
(3) 0.5 (4) 1.7

100. Consider a block and trolley system as shown in

figure. If the coefficient of kinetic friction between
the trolley and the surface is 0.04, the acceleration
(1) 16 (2) 6 of the system in ms–2 is:
(3) 4 (4) 8 (Consider that the string is massless and
unstretchable and the pulley is also massless and
FRICTION frictionless): [Feb. 1, 2024 (I)]
95. A cubic block of mass m is sliding down on an
inclined plane at 60° with an acceleration g/2, of,
the value of coefficient of kinetic friction is
[April 7, 2025 (I)]
(1) 3 −1 (2) 3/2
3
(3) 2 /3 (4) 1 −
2 (1) 3 (2) 4
(3) 2 (4) 1.2
96. A block of mass 25 kg is pulled along a horizontal
surface by a force at an angle 45° with the 101. A coin is placed on a disc. The coefficient of
horizontal. The friction coefficient between the friction between the coin and the disc is . If the
block and the surface is 0.25. The displacement of distance of the coin from the center of the disc is r,
5 m of the block is: the maximum angular velocity which can be given
[April 4, 2025 (II)]
to the disc, so that the coin does not slip away, is:
(1) 970 J (2) 735 J [Jan. 31, 2024 (I)]
(3) 245 J (4) 490 J
 g
(1) (2)
97. A given object takes n times the time to slide down rg r
45° rough inclined plane as it takes the time to slide g r
down an identical perfectly smooth 45° inclined (3) (4)
r g
plane. The coefficient of kinetic friction between
the object and the surface of inclined plane is:
[April 8, 2024 (II)] 102. In the given arrangement of a doubly inclined plane
1 two blocks of masses M and m are placed. The
(1) 1 − (2) 1 – n2 blocks are connected by a light string passing over
n2
an ideal pulley as shown. The coefficient of friction
1
(3) 1− (4) 1 − n2 between the surface of the plane and the blocks is
n2 0.25. The value of m, for which M = 10 kg will
move down with an acceleration of 2 m/s2, is:
98. A heavy box of mass 50 kg is moving on a (take g = 10 m/s2 and tan37° = 3/4)
horizontal surface. If co-efficient of kinetic friction [Jan. 31, 2024 (I)]
between the box and horizontal surface is 0.3 then
force of kinetic friction is:
[April 5, 2024 (II)]
(1) 14.7 N
(2) 147 N
(3) 1.47 N (1) 6.5 kg (2) 4.5 kg
(4) 1470 N (3) 2.25 kg (4) 9 kg
 13

103. A block of mass 5 kg is placed on a rough inclined 107. A bullet of mass 0.1 kg moving horizontally with
surface as shown in the figure. If F 1 is the force speed 400 ms–1 hits a wooden block of mass 3.9 kg
required to just move the block up the inclined kept on a horizontal rough surface. The bullet gets
embedded into the block and moves 20 m before
plane and F 2 is the force required to just prevent
coming to rest. The coefficient of friction between
the block from sliding down, then the value of
the block and the surface is ________.
F1 − F 2 is: [Use g = 10 m/s2]. (Given g = 10 ms2)
[Jan. 31, 2024 (II)] [April 8, 2023 (II)]
(1) 0.50 (2) 0.90
(3) 0.65 (4) 0.25

108. A block of mass 5 kg is placed at rest on a table of

rough surface. Now, if a force of 30 N is applied
in the direction parallel to surface of the table, the
(1) 25 3 N (2) 5 3 N block slides through a distance of 50 m in an
interval of time 10 s. Coefficient of kinetic friction
(3)
5 3
N (4) 10 N is: (given, g = 10 ms–2)
2 [Feb. 1, 2023 (I)]
(1) 0.50 (2) 0.60
104. A block of mass m is placed on a surface having (3) 0.75 (4) 0.25
vertical cross section given by y = x2/4. If
coefficient of friction is 0.5, the maximum height 109. As shown in the figure a block of mass 10 kg lying
above the ground at which block can be placed on a horizontal surface is pulled by a force F acting
without slipping is: at an angle 30°, with horizontal. For s = 0.25, the
[Jan. 30, 2024 (II)] block will just start to move for the value of F:
(1) 1/4 m (2) 1/2 m [Given g = 10 ms–2] [Feb. 1, 2023 (I)]
(3) 1/6 m (4) 1/3 m

105. Given below are two statements:

Statement (I): The limiting force of static friction
depends on the area of contact and independent of
materials. (1) 20 N (2) 33.3 N
Statement (II): The limiting force of kinetic (3) 25.2 N (4) 35.7 N
friction is independent of the area of contact and
depends on materials. 110. A body of mass 10 kg is moving with an initial
In the light of the above statements, choose the speed of 20 m/s. The body stops after
most appropriate answer from the options given 5 s due to friction between body and the floor. The
below: value of the coefficient of friction is:
[Jan. 27, 2024 (II)] (Take acceleration due to gravity g = 10 ms–2)
(1) Statement I is correct but Statement II is [Jan.29, 2023 (I)]
incorrect (1) 0.2
(b) Statement I is incorrect but Statement II is (2) 0.3
correct (3) 0.5
(c) Both Statement I and Statement II are (4) 0.4
incorrect
(d) Both Statement I and Statement II are correct 111. A block of mass m slides down the plane inclined
at angle 30° with an acceleration g/4. The value of
106. A coin placed on a rotating table just slips when it coefficient of kinetic friction will be:
is placed at a distance of 1 cm from the centre. If [Jan. 29, 2023 (I)]
the angular velocity of the table is halved, it will 2 3 +1 1
just slip when placed at a distance of _______ from (1) (2)
2 2 3
the centre: [April 11, 2023 (I)]
(1) 2 cm (2) 1 cm 3 2 3 −1
(3) (4)
(3) 8 cm (4) 4 cm 2 2
 14

112. The time taken by an object to slide down 45°

rough inclined plane is n times as it takes to slide
down a perfectly smooth 45° incline plane. The
coefficient of kinetic friction between the object
and the incline plane is _______.
[Jan. 29, 2023 (II)]
1 1
(1) (2) 1− 2 (1) 10 N (2) 20 N
1 − n2 n
(3) 25 N (4) 45 N
1 1
(3) 1 + 2 (4) 1 − 2
n n 117. A block of mass 40 kg slides over a surface, when
a mass of 4 kg is suspended through an inextensible
113. Consider a block kept on an inclined plane
massless string passing over frictionless pulley as
(inclined at 45°) as shown in the figure. If the force
required to just push it up the incline is 2 times the shown below. The coefficient of kinetic friction
force required to just prevent it from sliding down, between the surface and block is 0.02. The
the coefficient of friction between the block and acceleration of block is _______.
inclined plane (μ) is equal to: (Given g = 10 ms–2)
[Jan. 25, 2023 (II)] [June 29, 2022 (II)]

(1) 1 ms–2 (2) 1/5 ms–2

(3) 4/5 ms–2 (4) 8/11 ms–2
(1) 0.33 (2) 0.60
(3) 0.25 (4) 0.50 118. A hanging mass M is connected to a four times
bigger mass by using a string-pulley arrangement.
114. A bag is gently dropped on a conveyor belt moving
at a speed of 2 m/s. The coefficient of friction as shown in the figure. The bigger mass is placed
between the conveyor belt and bag is 0.4 Initially, on a horizontal ice-slab and being pulled by 2Mg
the bag slips on the belt before is stops due to force. In this situation tension in the string is x/5 mg
friction. The distance travelled by the bag on the for x = ________. Neglect mass of the string and
belt during slipping motion is: [Take g = 10 m/s–2] friction of the block (bigger mass) with ice slab.
[July 27, 2022 (I)] (Given g = acceleration due to gravity)
(1) 2 m (2) 0.5 m [June 28, 2022 (I)]
(3) 3.2 m (4) 0.8 m

115. A block of mass M slides down on a rough inclined

plane with constant velocity. The angle made by the
incline plane with horizontal is θ. The magnitude of
the contact force will be: [July 27, 2022 (II)] 119. A system of two blocks of masses m = 2 kg and
(1) Mg M = 8 kg is placed on a smooth table as shown in
(2) Mg cos  figure. The coefficient of static friction between
(3) Mg sinθ + Mg cosθ two blocks is 0.5. The maximum horizontal force F
(4) Mg sinθ 1 + μ that can be applied to the block of mass M so that
the blocks move together will be:
[June 27, 2022 (I)]
116. A 2 kg block is pushed against a vertical wall by
applying a horizontal force of 50 N. The coefficient
of static friction between the block and the wall is
0.5. A force F is also applied on the block vertically
upward (as shown in figure). The maximum value
of F applied, so that the block does not move (1) 9.8 N (2) 39.2 N
upward, will be: [June 30, 2022 (I)] (3) 49 N (4) 78.4 N
 15

120. A disc with a flat small bottom beaker placed on it 125. A boy of mass 4 kg is standing on a piece of wood
at a distance R from its centre is revolving about an having mass 5kg. If the coefficient of friction
axis passing through the centre and perpendicular between the wood and the floor is 0.5, the
maximum force that the boy can exert on the rope
to its plane with an angular velocity ω. The
so that the piece of wood does not move from its
coefficient of static of the disc is μ. The beaker will place is ________N.
revolve with disc if: (Round off to the Nearest Integer)
[June 25, 2022 (II)] [Take g = 10 ms–2]
g g [March 17, 2021 (II)]
(1) R  2 (2) R  2
2 
g g
(3) R  2 (4) R  2
2 

121. A block of mass 10 kg starts sliding on a surface

with an initial velocity of 9.8 ms–1. The coefficient
of friction between the surface and block is 0.5. The
distance covered by the block before coming to rest 126. A body of mass 1 kg rests on a horizontal floor with
in [use g = 9.8 ms–2] which it has a coefficient of static friction
1
. It is
[June 24, 2022 (I)] 3
(1) 4.9 m (2) 9.8 m desired to make the body move by applying the
(3) 12.5 m (4) 19.6 m minimum possible force F N. The value of F will
be.
122. When a body slides down from rest along a smooth (Round off to the Nearest Integer)
inclined plane making an angle of 30° with the [Take g = 10 ms–2].
[March 17, 2021 (II)]
horizontal, it takes time T. When the same body
slides down from the rest along a rough inclined
plane making the same angle and through the same 127. As shown in the figure, a block of mass √3 kg is
distance, it takes time αT, where α is a constant kept on a horizontal rough surface of coefficient of
greater than 1. The coefficient of friction between 1  α2 − 1 
friction   . The critical force to be
1   2 −1  x  α2 
the body and the rough plane is  
x  2  applied on the vertical surface as shown at an angle
where x =________ 60° with horizontal such that it does not move, will
[Sep. 1, 2021 (II)] be 3x. The value of x will be:
3 1
123. The coefficient of static friction between two [g = 10 m/s2; sin60 = ; cos60 = ]
2 2
blocks is 0.5 and the table is smooth. The maximum [Feb. 26, 2021 (I)]
horizontal force that can be applied to move the
blocks together is ______ N. (Take: g = 10 ms–2) =
1
[Aug. 26, 2021 (II)] 3 3 m = 3kg

128. An inclined plane is bent in such a way that the

x2
vertical cross-section is given by y = where y is
124. A body of mass 'm' is launched up on a rough 4
inclined plane making an angle of 30°with the in vertical and x in horizontal direction. If the upper
horizontal. The coefficient of friction between the surface of this curved plane is rough with
x coefficient of friction μ = 0.5, the maximum height
body and plane is if the time of ascent is half in cm at which a stationary block will not slip
5 downward is ________cm.
of the time of descent. The value of x is ______ [Feb. 24, 2021 (I), 2003 S]
[July 20, 2021 (II)]
 16

129. The coefficient of static friction between a wooden 134. A block kept on a rough inclined plane, as shown
block of mass 0.5 kg and a vertical rough wall is in the figure, remains at rest upto a maximum force
0.2. The magnitude of horizontal force that should 2 N down the inclined plane. The maximum
be applied on the block to keep it adhere to the wall external force up the inclined plane that does not
will be __________ N. [g = 10 ms–2] move the block is 10 N. The coefficient of static
[Feb. 24, 2021 (I)] friction between the block and the plane is:
[Take g = 10 m/s2]
130. An insect is at the bottom of a hemispherical ditch [12 Jan. 2019 II]
of radius 1 m. It crawls up the ditch but starts
slipping after it is at height h from the bottom. If
the coefficient of friction between the ground and
the insect is 0.75, then h is (g = 10 ms–2)
[Sep. 06, 2020 (I)]
(1) 0.20 m (2) 0.45 m (1) 3/2 (2) 3/4
(3) 0.60 m (4) 0.80 m (3) 1/2 (4) 2/3

131. A block starts moving up an inclined plane of 135. Two masses m1 = 5 kg and m2 = 10 kg connected
inclination 30° with an initial velocity of 𝑣0. It by an inextensible string over a frictionless pulley,
comes back to its initial position with velocity 𝑣0/2. are moving as shown in the figure. The coefficient
The value of the coefficient of kinetic friction of friction of horizontal surface is 0.15. The
between the block and the inclined plane is close to minimum weight m that should be put on top of m2
1/1000. The nearest integer to I is ________. to stop the motion is: [2018]
[Sep. 03, 2020 (II)]

132. A block of mass 5 kg is (i) pushed in case (A) and

(ii) pulled in case (B), by a force F = 20 N, making
an angle of 30° with the horizontal, as shown in the
figures. The coefficient of friction between the
block and floor is µ = 0.2 . The difference between
the accelerations of the block, in case (B) and case (1) 18.3 kg (2) 23.3 kg
(A) will be: (g = 10 ms–2) (3) 43.3 kg (4) 10.3 kg
[12 April 2019 II]

136. A body of mass 2 kg slides down with an

acceleration of 3 m/s2 on a rough inclined plane
having a slope of 30°. The external force required
to take the same body up the plane with the same
(1) 0.4 ms–2 (2) 3.2 ms–2 acceleration will be: (g = 10 m/s2)
[Online April 15, 2018]
(3) 0.8 ms–2 (4) 0 ms–2
(1) 4 N (2) 14 N
(3) 6 N (4) 20 N
133. Two blocks A and B masses mA = 1 kg and
mB = 3 kg are kept on the table as shown in figure. 137. A rocket is fired vertically from the earth with an
The coefficient of friction between A and B is 0.2 acceleration of 2g, where g is the gravitational
and between B and the surface of the table is also acceleration. On an inclined plane inside the rocket,
0.2. The maximum force F that can be applied on B making an angle 0 with the horizontal, a point
horizontally, so that the block A does not slide over object of mass m is kept. The minimum coefficient
the block B is: [Take g = 10 m/s2] of friction µmin between the mass and the inclined
[10 April 2019 II] surface such that the mass does not move is:
[Online April 9, 2016]
(1) tan2
(2) tan
(3) 3 tan
(1) 8 N (2) 16 N (4) 2 tan
(3) 40 N (4) 12 N
 17

138. Given in the figure are two blocks A and B of

weight 20 N and 100 N, respectively. These are
being pressed against a wall by a force F as shown.
If the coefficient of friction between the blocks is (1)
0.1 and between block B and the wall is 0.15, the
frictional force applied by the wall on block B is:
[2015]

(2)

(1) 120 N (2) 150 N (3)

(3) 100 N (4) 80 N

139. A small ball of mass m starts at a point A with speed

v0 and moves along a frictionless track AB as
shown. The track BC has coefficient of friction µ.
The ball comes to stop at C after travelling a (4)
distance L which is:
[Online April 11, 2014]

142. A body starts from rest on a long inclined plane of

slope 45°. The coefficient of friction between the
body and the plane varies as µ = 0.3 x, where x is
distance travelled down the plane. The body will
have maximum speed (for g = 10 m/s2) when x =
[Online April 22, 2013]
(1) 9.8 m (2) 27 m
2h v02 h v02 (3) 12 m (4) 3.33 m
(1) + (2) +
 2g  2g
143. The minimum force required to start pushing a
h v02 h v2
(3) + (4) + 0 body up rough (frictional coefficient µ) inclined
2 g 2 2g plane is F1 while the minimum force needed to
prevent it from sliding down is F2. If the inclined
140. A block A of mass 4 kg is placed on another block plane makes an angle  from the horizontal such
B of mass 5 kg, and the block B rests on a smooth that tan  = 2µ then the ratio of F1/F2 is
horizontal table. If the minimum force that can be [2011]
applied on A so that both the blocks move together (1) 1 (2) 2
is 12 N, the maximum force that can be applied to (3) 3 (4) 4
B for the blocks to move together will be:
[Online April 9, 2014] 144. The upper half of an inclined plane with inclination
(1) 30 N (2) 25 N  is perfectly smooth while the lower half is rough.
(3) 27 N (4) 48 N A body starting from rest at the top will again come
to rest at the bottom if the coefficient of friction for
141. A block is placed on a rough horizontal plane. A the lower half is given by [2005]
time dependent horizontal force F = kt acts on the (1) 2 cos  (2) 2 sin 
block, where k is a positive constant. The (3) tan  (4) 2 tan 
acceleration - time graph of the block is:
[Online April 25, 2013]
 18

145. Consider a car moving on a straight road with a 150. A car of 800 kg is taking turn on a banked road of
speed of 100 m/s. The distance at which car can be radius 300 m and angle of banking 30°. If
stopped is [µk = 0.5] [2005] coefficient of static friction is 0.2 then the
(1) 1000 m (2) 800 m maximum speed with which car can negotiate the
(3) 400 m (4) 100 m turn safely: (g = 10 m/s2, √3 = 1.73)
[April 6, 2024 (II)]
146. A block rests on a rough inclined plane making an (1) 70.4 m/s
angle of 30° with the horizontal. The coefficient of (2) 51.4 m/s
static friction between the block and the plane is (3) 264 m/s
0.8. If the frictional force on the block is 10 N, the (4) 102.8 m/s
mass of the block (in kg) is
(take g = 10 m/s2) [2004] 151. A man carrying a monkey on his shoulder does
(1) 1.6 (2) 4.0 cycling smoothly on a circular track of radius 9m
(3) 2.0 (4) 2.5 and completes 120 revolutions in 3 minutes. The
magnitude of centripetal acceleration of monkey is
147. A marble block of mass 2 kg lying on ice when (in m/s2): [April 5, 2024 (II)]
given a velocity of 6 m/s is stopped by friction in
(1) zero
10 s. Then the coefficient of friction is
[2003] (2) 16 2 ms–2
(1) 0.02 (2) 0.03 (3) 4 2 ms–2
(3) 0.04 (4) 0.06 (4) 57600 2 ms–2

CIRCULAR MOTION, BANKING OF ROAD 152. A ball of mass 0.5 kg is attached to a string of
148. A car of mass 'm‘ moves on a banked road having length 50 cm. The ball is rotated on a horizontal
radius 'r' and banking angle θ. To avoid slipping circular path about its vertical axis. The maximum
from banked road, the maximum permissible speed tension that the string can bear is 400 N. The
of the car is v0. The coefficient of friction μ between maximum possible value of angular velocity of the
the wheels of the car and the banked road is ball in rad/s is: [Feb. 1, 2024 (I)]
[Jan. 24, 2025 (I)] (1) 1600 (2) 40
v02+ rg tan  (3) 1000 (4) 20
(1)  =
rg − v02 tan 
153. If the radius of curvature of the path of two particles
v02 + rg tan 
(2)  = of same mass are in the ratio 3 : 4, then in order to
rg + v02 tan  have constant centripetal force, their velocities will
v02 − rg tan  be in the ratio of: [Jan. 29, 2024 (I)]
(3)  = (1) 1: 3 (2) 3 :1
rg + v02 tan 
(3) 3:2 (4) 2 : 3
v2 − rg tan 
(4)  = 0
rg − v02 tan  154. A train is moving with a speed of 12 m/s on rails
which are 1.5 m apart. To negotiate a curve radius
149. A circular table is rotating with an angular velocity 400 m, the height by which the outer rail should be
of  rad/s about its axis (see figure). There is a raised with respect to the inner rail is
smooth groove along a radial direction on the table. (Given, g = 10 m/s2):
A steel ball is gently placed at a distance of 1 m on [Jan. 27, 2024 (I)]
the groove. All the surface are smooth. If the radius (1) 6.0 cm (2) 5.4 cm
of the table is 3 m, the radial velocity of the ball (3) 4.8 cm (4) 4.2 cm
w.r.t. the table at the time ball leaves the table is
x 2 m/s, where the value of x is _______ 155. A stone of mass 900 g is tied to a string and moved
[April 8, 2024 (II)] in a vertical circle of radius 1 m making 10 rpm.
The tension in the string, when the stone is at the
lowest point is:
(if 2 = 9.8 and g = 9.8 m/s2)
[Jan. 29, 2024 (II)]
(1) 17.8 N (2) 8.82 N
(3) 97 N (4) 9.8 N
 19

156. A vehicle of mass 200 kg is moving along a

levelled curved road of radius 70 m with angular
velocity of 0.2 rad/s. The centripetal force acting on (1) (2)
the vehicle is: [April 13, 2023 (I)]
(1) 560 N (2) 2800 N
(3) 14 N (4) 2240 N

157. A child of mass 5 kg is going round a merry-go- (3) (4)

round that makes 1 rotation in 3.14 s. The radius of
the merry-go-round is 2 m. The centrifugal force on
the child will be [April 6, 2023 (II)]
(1) 80 N (2) 50 N 162. A stone tide to a spring of length L is whirled in a
vertical circle with the other end of the spring at the
(3) 100 N (4) 40 N
centre. At a certain instant of time, the stone is at
its lowest position and has a speed u. The
158. A car is moving on a horizontal curved road with
magnitude of change in its velocity, as it reaches a
radius 50 m. The approximate maximum speed of
position where the string is horizontal, is
car will be, if friction between tyres and road is
√𝑥 (𝑢2 − 𝑔𝐿). The value of x is: [June 27, 2022 (II)]
0.34. [take g = 10 ms–2]
[Jan. 29, 2023 (I)] (1) 3 (2) 2
(1) 3.4 ms –1
(2) 22.4 ms –1 (3) 1 (4) 5
(3) 13 ms–1 (4) 17 ms–1
163. A ball is released from rest from point P of a
smooth semi-spherical vessel as shown in figure.
159. A car is moving on a circular path of radius 600 m
The ratio of the centripetal force and normal
such that the magnitudes of the tangential
reaction on the ball at point Q is A while angular
acceleration and centripetal acceleration are equal.
position of point Q is α with respect to point P.
The time taken by the car to complete first quarter Which of the following graphs represent the correct
of revolution, if it is moving with an initial speed relation between A and α when ball goes from Q to
of 54 km/hr is t(1 – e–/2)s. The value of t is _____. R? [June 26, 2022 (I)]
[Jan. 29, 2023 (II)]

160. A car is moving with a constant speed of 20 m/s in

a circular horizontal track of radius 40 m. A bob is
suspended from the roof of the car by a massless
string. The angle made by the string with the (1) (2)
vertical will be: (Take g = 10 m/s2)
[Jan. 25, 2023 (I)]
(1) /6 (2) /2
(3) /4 (4) /3
(3) (4)
161. A smooth circular groove has a smooth vertical
wall as shown in figure. A block of mass m moves
against the wall with a speed v. Which of the
following curve represents the correct relation 164. A curved in a level road has a radius 75 m. The
between the normal reaction on the block by the maximum speed of a car turning this curved road
wall (N) and speed of the block (v)? can be 30 m/s without skidding. If radius of curved
[July 29, 2022 (I)] road is changed to 48 m and the coefficient of
friction between the tyres and the road remains
same, then maximum allowed speed would be
_____m/s. [June 25, 2022 (II)]
 20

165. A boy ties a stone of mass 100 g to the end of a 2 (1) Statement I is correct and statement II is
m long string and whirls it around in a horizontal incorrect
plane. The string can withstand the maximum (2) Statement I is incorrect and statement II is
tension of 80 N. If the maximum speed with which correct
𝐾
the stone can revolve is rev./min. The value of K (3) Both statement I and statement II are true
π
is (Assume the string is massless and unstretchable) (4) Both statement I and statement II are false
[June 24, 2022 (I)]
(1) 400 (2) 300 170. A small bob tied at one end of a thin string of length
(3) 600 (4) 800
1 m is describing a vertical circle so that the
166. A stone of mass m, tied to a string is being whirled maximum and minimum tension in the string are in
in a vertical circle with a uniform speed. The the ratio 5 : 1. The velocity of the bob at the highest
tension in the string is position is _________ m/s. (Take g = 10 m/s2)
[June 24, 2022 (II)] [Feb. 25, 2021 (I)]
(1) the same throughout the motion.
(2) minimum at the highest position of the circular 171. A disc rotates about its axis of symmetry in a
path. hoizontal plane at a steady rate of 3.5 revolutions
(3) minimum at the lowest position of the circular per second. A coin placed at a distance of 1.25 cm
path. from the axis of rotation remains at rest on the disc.
(4) minimum when the rope is in the horizontal
The coefficient of friction between the coin and the
position.
disc is (g = 10 m/s2)
[Online April 15, 2018]
167. A particle of mass m is suspended from a ceiling
through a string of length L. The particle moves in (1) 0.5 (2) 0.7
L (3) 0.3 (4) 0.6
a horizontal circle of radius r such that r = . The
√2
speed of particle will be: 172. A conical pendulum of length 1 m makes an angle
[Aug. 26, 2021 (II)]
 = 45° w.r.t. Z-axis and moves in a circle in the
(1) rg (2) 2rg
XY plane. The radius of the circle is 0.4 m and its
(3) 2 rg (4) rg / 2 centre is vertically be-low O. The speed of the
pendulum, in its circular path, will be:
168. A block of 200 g mass moves with a uniform speed (Take g = 10 ms-²)
in a horizontal circular groove, with vertical side [Online April 9, 2017]
walls of radius 20 cm. If the block takes 40 s to
complete one round, the normal force by the side
walls of the groove is
[March 16, 2021 (I)]
(1) 9.859 × 10–2 N
(2) 9.859 × 10–4 N
(3) 6.28 × 10–3 N
(4) 0.0314 N

169. Statement I: A cyclist is moving on an unbanked (1) 0.4 m/s

road with a speed of 7 kmh–1 and takes a sharp (2) 4 m/s
circular turn along a path of radius of 2m without (3) 0.2 m/s
reducing the speed. The static friction coefficient (4) 2 m/s
is 0.2. The cyclist will not slip and pass the curve.
(g = 9.8 m/s²) 173. A particle is released on a vertical smooth
Statement II: If the road is banked at an angle of semicircular track from point X so that OX makes
45°, cyclist can cross the curve of 2m radius with
the speed of 18.5 kmh–1 without slipping. angle  from the vertical (see figure). The normal
In the light of the above statements, choose the reaction of the track on the particle vanishes at
correct answer from the options given below. point Y where OY makes angle  with the
[March 16, 2021 (II); 2002 (S)] horizontal. Then: [Online April 19, 2014]
 21

176. For a particle in uniform circular motion, the

acceleration a at a point P(R,) on the circle of
radius R is (Here  is measured from the x-axis)
[June 25, 2022 (II)]
2 2
v v
(1) − cos iˆ + sin  ˆj
R R
2
v v2
(2) − sin iˆ + cos ˆj
R R
1
(2) sin  = cos 
2
(1) sin  = cos  v v2
2 (3) − cos iˆ − sin ˆj
R R
2 3
(3) sin  = cos  (4) sin  = cos  2
v ˆ v ˆ 2
3 4 (4) i+ j
R R
174. A body of mass 'm' is tied to one end of a spring and
177. An annular ring with inner and outer radii R1 and
whirled round in a horizontal plane with a constant
R2 is rolling without slipping with a uniform
angular velocity. The elongation in the spring is
angular speed. The ratio of the forces experienced
1 cm. If the angular velocity is doubled, the
by the two particles situated on the inner and outer
elongation in the spring is 5 cm. The original length
parts of the ring, F1/F2 is [2005]
of the spring is: 2
[Online April 23, 2013] (1) (R1/R2) (2) R2/R1
(1) 15 cm (2) 12 cm (3) R1/R2 (4) 1
(3) 16 cm (4) 10 cm
178. Which of the following statements is FALSE for a
175. A point P moves in counter-clockwise direction on particle moving in a circle with a constant angular
a circular path as shown in the figure. The speed?
[2004]
movement of 'P' is such that it sweeps out a length
(1) The acceleration vector points to the centre of
s = t3 + 5, where s is in metres and t is in seconds.
the circle
The radius of the path is 20 m. The acceleration of
(2) The acceleration vector is tangent to the circle
'P' when t = 2 s is nearly. [2010]
(3) The velocity vector is tangent to the circle
(4) The velocity and acceleration vectors are
perpendicular to each other

(1) 13 m/s2 (2) 12 m/s2

(3) 7.2 m/s2 (4) 14 m/s2
 22

Answer Key
1. (4) 46. (3) 91. (3) 136. (4)
2. (2) 47. (3) 92. (1) 137. (2)
3. (3) 48. (2) 93. (2) 138. (1)
4. (3) 49. (4) 94. (3) 139. (2)
5. (5) 50. (4) 95. (1) 140. (3)
6. (1) 51. (2) 96. (3) 141. (2)
7. (2) 52. (2) 97. (1) 142. (4)
8. (1) 53. (3) 98. (2) 143. (3)
9. (4) 54. (1) 99. (1) 144. (4)
10. (4) 55. (4) 100. (3) 145. (1)
11. (2) 56. (1) 101. (2) 146. (3)
12. (1) 57. (240) 102. (2) 147. (4)
13. (2) 58. (4) 103. (2) 148. (3)
14. (2) 59. (1) 104. (1) 149. (2)
15. (1) 60. (3) 105. (2) 150. (2)
16. (2) 61. (1) 106. (4) 151. (2)
17. (3) 62. (2) 107. (4) 152. (2)
18. (4) 63. (2) 108. (1) 153. (3)
19. (2) 64. (2) 109. (3) 154. (2)
20. (2) 65. (3) 110. (4) 155. (4)
21. (2) 66. (4) 111. (2) 156. (1)
22. (2) 67. (3) 112. (4) 157. (4)
23. (3) 68. (1) 113. (1) 158. (3)
24. (6) 69. (2) 114. (2) 159. (40)
25. (3) 70. (3) 115. (1) 160. (3)
26. (12) 71. (1) 116. (4) 161. (1)
27. (2) 72. (3) 117. (4) 162. (2)
28. (2) 73. (3) 118. (6) 163. (3)
29. (4) 74. (2) 119. (3) 164. (24)
30. (3) 75. (1) 120. (2) 165. (3)
31. (3) 76. (36) 121. (2) 166. (2)
32. (500) 77. (12) 122. (3) 167. (1)
33. (1) 78. (4) 123. (15) 168. (2)
34. (2) 79. (30) 124. (3) 169. (3)
35. (4) 80. (4) 125. (30) 170. (5)
36. (1) 81. (2) 126. (5) 171. (4)
37. (4) 82. (82) 127. (3.33) 172. (4)
38. (2) 83. (492) 128. (25) 173. (3)
39. (1) 84. (1) 129. (25) 174. (1)
40. (1) 85. (4) 130. (1) 175. (4)
41. (3) 86. (4) 131. (346) 176. (3)
42. (3) 87. (3) 132. (3) 177. (3)
43. (4) 88. (1) 133. (2) 178. (2)
44. (4) 89. (4) 134. (1)
45. (3) 90. (1) 135. (2)
 23

Solution
1. (4) 6. (1)
Sol. F = M × a (Force acting on body) Sol. P = cos(kt )iˆ − sin(kt ) ˆj
Given: v = 4 x dP
Squaring on both side, v2 = 16x  F= = −k sin(kt )iˆ − k cos(kt ) ˆj
dt
Differentiating on both side w.r.t. x,
dv vdv 16 FP −k cos(kt )  sin(kt ) + k sin(kt )  cos(kt )
2v = 16  a = = = 8 m/s2 cos  = = =0
dx dx 2 | F || P | 1 k
 F = 0.5 × 8 = 4 N.

 =
2. (2) 2
Sol. Given mass m = 2 kg, force F = 6 N along positive
7. (2)
z-axis.
Sol. Here force applied on the block increases linearly with
5
 Acceleration a = 3kˆ, t = s time, so
3 F kt
F = ma  a = = or a  t
u = 3iˆ + 4 ˆj m m
5 Hence a vs t graph will be a straight line passing
 v = u + at = 3iˆ + 4 ˆj + 3kˆ  = 3iˆ + 4 ˆj + 5kˆ through origin.
3
8. (1)
3. (3) p mv
Sol. The problem involves the buoyancy force F acting on a Sol. Using Newton's IInd law, F = =
balloon and change in its motion when a small mass is t t
0.12  25
released. From Newton's second Law of Motion. F= = 30N
F − Mg = Ma 0.1
 F = Ma + Mg …(i) 9. (4)
F − (M − x) g = (M − x)3a Sol. Impulse, I = P = Pf − Pi = m(v f − vi ) ( v = 2gh )
 Ma + Mg − Mg + xg = 3Ma − 3xa [using (i)] Mass of body, m = 0.1 kg
 x=
2Ma
g + 3a
I = P = 0.1 ( (
2  9.8  5 − − 2  9.8 10 ))
−1
= 0.1(14 + 7 2)  2.39 kg ms
4. (3)
Sol. Given, 10. (4)
Mass of cricket ball, m = 150g Sol. From principle of momentum conservation,
m1v1 = m2v2
Speed of cricket ball, V = 20 m/s
or, 1000 × 6 = 1200 × v = 5 m/s
P (mv − 0) 150  20
Force, F = = = = 30 N
t 0.1 1000  0.1 11. (2)
Sol. By momentum conservation,
5. (5) m1u1 + m2u2 = m1v1 + m2v2
Sol. 0 = 3(−v) + 0.01(600 − v)
v  2m/s

12. (1)
Sol. Given,
Mass of the particle, m = 5 kg
As the particle is at rest, So resultant of F2 and F3
should be opposite to F1
Fnet = F22 + F32 = 62 + 82 = 10 N
F 10
By Newton’s 2nd law, Fnet = ma  Acceleration, a = net = = 2m/s2
 294 – R = 30 (0.1)  R = 291 N m 5
 24

13. (2) 19. (2)

Sol. Given, mass of each bullets, m = 10g Sol.
Speed of each bullets, v = 250 m/s
Force F = n m v
Here n = number of bullets fired per second
F 125
 n= = = 50 Assume surface to be frictionless
mv 10 10−3  250
Then, 50 = V × 10  V = 5 m/s
As v = u + at V = 0 + a  20
14. (2)
1
Sol. Given,  5 = a  20  a = m/s2
500 4
Mass of body, m = = 0.5kg 1
1000 So, F = ma = 20  = 5N
4
Velocity, v = 10 x  v 2 = 100 x
dv vdv
 2v = 100  a = = 50 m/s2 20. (2)
dx dx Sol. Impulse, I = change in momentum, P
 Force, F = ma = 0.5  50 = 25 N P
Favg = P1 = P2 I1 = I 2
t
15. (1) Given t1 = 3s and t2 = 5s
Sol. Mass of particle,
Hence, Favg in case (i), when t2 = 3s is more than (ii)
m = 500 g = 0.5kg
when t2 = 5s
Velocity of a particle,
v = 2tiˆ + 3t 2 ˆj 21. (2)
dv dm 10
a= = 2iˆ + 6tjˆ Sol. Force, F = v =  4.5 = 9 dyne
dt dt 5
at t = 1s, a = 2iˆ + 6 ˆj
22. (2)
Force acting on the particle,
F = ma = 0.5(2iˆ + 6 ˆj) = iˆ + 3 ˆj ()
Sol. Initial momentum Pi = 0.15 12 iˆ

F = iˆ + xjˆ Hence x = 3 Final momentum Pf = 0.15 12 ( −iˆ )

P = 3.6 kg m/s or, 3.6 = Ft
16. (2)
Sol. We have, Impulse = Area under F-t curve, for fig (D), 3.6 = 100 t  t = 0.036 sec
area is highest. So, impulse is maximum,
23. (3)
17. (3) dp dm
Sol. F = =v = 10 1 = 10N
dp dt dt
Sol. =| F |= Slop of curve
dt F 10
a = = = 5 m/s2
As, shown in figure maximum slope represent (c) and m 2
minimum slope represent (b).
24. (6)
18. (4) Sol. By law of conservation of linear momentum Pi = Pf
Sol. Given, mass of machine gun, M = 10 kg
 60  V = (120 + 60)  2  60V = 360  V = 6 m/s
mass of bullet, m = 20g = 20 × 10–3 kg
velocity of bullet V = 100 ms–1, let V be the recoil
velocity of gun, using conservation of momentum 25. (3)
vdv 1.5 kx v
nmv = MV Sol. a = ; adx = vdx;  − dx =  vdv
180 dx 0.5 m 4
 20 10−3  100 = 10V
60 k v − 42
2
12 v2 −16
 − [1.52 − 0.52 ] = − [2] =
 n = 180 bullets per minute 2m 2 2 2 2
 v = 0.6 m/s  −3  4 = v2 − 16  v 2 = 4  v = 2 m/s

26. (12)
 25

Sol. = Pf − Pi = mv − (−mv) = 2mv = 2  0.4 15 = 12 Ns Sol. From the Newton’s second law of motion,
F = ma
F   T −t  
2
27. (2) F
a=  a = 0 1 −  
Sol. F = 10iˆ + 5 ˆj; m = 0.1 kg M M   T  

F 10iˆ + 5 ˆj dv F0   T − t  
2
a= = = 100iˆ + 50 ˆj  a = = 1 −  
m 0.1 dt M   T  
1 1
S = ut + at 2 = 0 + (100iˆ + 50 ˆj )  22 = 200iˆ + 100 ˆj v F0 2T   T − t  
2
  dv =
M 0   T  
2 2 1 −  dt
0

28. (2) 2T
F0  1 
Sol. While going upward, a1 = –(g + a) v=  t + 2 (T − t )3 
M  3T 0
 10 
= − 10 +  = −12 m/s2
 5 F0 
 1 3  T3  

v = 2T + 2 (T − 2T)  − 0 + 2  
M  3T   3T   
F0  4T  4F0T
v = =
M  3  3M

Now, V = u + a1t1  u = – a1t1 31. (3)

1
and, S = ut1 + at12 F 40iˆ + 10 ˆj
Sol. Acceleration, a = = = 8iˆ + 2 ˆj m/s2
2 m 5
1 Using,
S = ut1 − 12  t12
2 1 1
 s = ut + at 2  s = (8iˆ + 2 ˆj ) 100
= −at12 − 6t12 = 12t12 − 6t12 = 6t12 2 2
10  s = (400iˆ + 100 ˆj) m
a2 = g − a = 10 − = 8 m/s2
5
Here, u = 0 32. (500)
1 1 F
So, S = a2t22 =  8  t22 = 4t22 Sol. a = = 10iˆ + 5 ˆj
2 2 m
t1 4 2 Displacement of the box along x-axis,
So, S = 6t12 = 4t22  = = 1 1
t2 6 3 x = axt 2 = 10 100 = 500 m
2 2
29. (4)
33. (1)
Sol. Thrust force on rocket is given by
Sol. Given
 dm 
Fthrust = Vrel  F = k (vyiˆ + vx ˆj )
 dt 
 dm  Fx = kvy , Fy = kvx
 Vrel  − mg  = ma
 dt  m
dvx dv k
= kv y  x = v y
 dm  dt dt m
 500   − 10 10 = 10  20
3 3
dvy k
 dt  Similarly, = vx
dm dt m
 = 60 kg/s dv y vx
dt =   v y dv y =  vx dvx
dvx v y
30. (3)
v2y = vx2 + C
 26

 v2y − vx2 = constant C0 (let) 1 0 −1

 
 v0  g 2
dv = t
k
v  a = (vxiˆ + v y ˆj )  (v yiˆ + vx ˆj )  +v
2

m   
k k Ck 0
= (vx2 kˆ − v2y kˆ) = (vx2 − v2y ) kˆ = 0 kˆ = constant   
m m m   
1 1  −1  v  
  tan = −t
34. (2)  g   g 
   
Sol. From the Newton's second law,       v0
dp d (mv)  dm 
F= = = v  …(i)
−1   
dt dt  dt   tan−1  v = −t
g  g 0 
We have given,
dM(t )
= bv 2 (t ) …(ii)  
dt  v0 
1
t = tan −1 
 g 
Thrust on the satellite,
 dm   dM  g  
F = vrel   = (0 − v)  = −v(bv2 ) = −bv3
 dt   dt 
[Using (i) and (ii)] 37. (4)
−bv 3
Sol. v2 = u2 – 2gh or v = u 2 − gh
 F = M(t )a = −bv3  a =
M(t )
Momentum, P = mv = m u 2 − 2 gh
35. (4) u2
Sol. At h = 0, P =  mu and at h = ,P=0
2g
upward direction is positive and downward direction is
negative.

38. (2)
F = –mkv2 – mg
Sol. From Newton's second law
( mg and mkv2 act in downward direction)
dp
F = F = kt
a= = −[kv2 + g ] dt
m Integrating both sides we get,
dv  dv 
 v  = −[kv 2 + g ]  a=v  3p T
T
t2 
dh  dh  p d p =  kt dt  [ p]3pp = k  
0
 2 0
0 v  dv h 1
 2 =  dh  ln[kv2 + g ]u0 = −h
u kv + g 0 2k kT2 p
 2p = T = 2
1  ku + g 
2
1  ku 2  2 k
 ln   = h  h = ln  + 1
2k  g  2k  g 
39. (1)
36. (1) R dv R
Sol. From F = 2
v(t )  m = 2 v(t )
Sol. Net acceleration t dt t
dv dv Rdt
= a = −( g + v 2 ) Integrating both sides  = 2
dt v(t ) mt
Let time t required to rise to its zenith (v = 0) so, R
0 −dv t In v(t ) = −
v0 g + v2 = 0 dt [for Hmax, v = 0] mt
1
 ln ln v(t ) 
t
 27

40. (1)  Relative vertical acceleration of A with respect to B

Sol. From question, is
Mass of body, m = 5 kg
3 1 g
Velocity at t = 0, g (sin 2 60 − sin 2 30) = g  −  = = 4.9 m/s2
 4 4 2
u = (6iˆ − 2 ˆj) m/s
In vertical direction
Velocity at t = 10 s,
v = +6 ˆj m/s 44. (4)
Force, F = ? Sol. For the motion of ball, just after the throwing
v − u 6 ˆj − (6iˆ − 2 ˆj) v = 0 , s = 2m, a = – g = – 10 ms–2
Acceleration, a = =
t 10  −u2 = 2 ( −10)  2u2 = 40
ˆ ˆ ˆ
−3i + 4 j −3i + 4 j ˆ
= = m/s2 When the ball is in the hands of the thrower
5 5 u = 0, v = 40 ms–1
(−3iˆ + 4 ˆj)
Force, F = ma = 5  = (−3iˆ + 4 ˆj) N s = 0.2 m
5 v2 – u2 = 2as
 40 – 0 = 2 (a) 0.2  a = 100 m/s2
41. (3)
 F = ma = 0.2 × 100 = 20 N
dv
Sol. Given that F(t ) = F0e−bt  m = F0e−bt  N – mg = 20  N = 20 + 2 = 22N
dt
Note:
dv F0 −bt
= e Whand + Wgravity = K
dt m
 F(0.2) – (0.2)(10)(2.2) = 0  F = 22 N
v F t −bt
0 dv = m0 0 e dt
45. (3)
−bt  t
F0  e F0  −bt −0  Sol. Given, mass of cricket ball, m = 150 g = 0.15 kg
v=   = −(e − e ) 
m  −b  0 mb  Initial velocity, u = 20m / s
Force,
F m(u − v) 0.15(20 − 0)
 v = 0 [1 − e−bt ] F= = = 30N
mb t 0.1
42. (3)
46. (3)
Sol. If you push a cart with some force then according to
Sol. Mass (m) = 0.2 kg
Newton third law the cart will exert an equal and
opposite force on you. So the cart push you with same  ma = –15x  0.3a = – 15x
amount of force in opposite direction. So Statement 1 15 −150
 a=− x= x = −50 x
is correct. The action reaction pairs mentioned in 0.3 3
statement one cannot cancel each other because action At x = 0.2m, a = – 50 × 0.2 = – 10 m/s2 (towards origin)
and reaction forces act on two different bodies. So they  a = 10 m/s2 (away from origin)
cannot cancel each other. So Statement 2 is wrong.
47. (3)
43. (4)
Sol. When the incline is given an acceleration a towards the
Sol. mg sin  = ma
right, the block experience a pseudo force ma towards
 a = g sin 
left.
 Vertical component of acceleration = a sin 
= g sin2 

For block to remain stationary, Net force along the

incline should be zero.
mg sin  = ma cos    = g tan 
 28

48. (2) T1 = mgcos30°

Sol. In the absence of air resistance, if the rocket moves up 3
with an acceleration a, then thrust = 110  =5 3
2
1
T2 = mg sin 30 = 110  =5
2

53. (3)
Sol. For massless spring, F = kx
 kx1 = 5N
 kx2 = 7N
 k(5x1 – 2x2) 5kx1 – 2kx2
F = mg + ma = 5 × 5 – 2 × 7 = 11N
 F = m(g + a) = 3.5 × 104 (10 + 10)
= 7 × 105 N 54. (1)
Sol. Acceleration of the system is given by
49. (4)  m − m1  g
a= 2 g =
Sol. Resultant force is zero, as three forces are represented  m1 + m2  8
by the sides of a triangle taken in the same order. From
 8(m2 − m1 ) = m1 + m2  7m2 = 9m1
Newton's second law, Fnet = ma .
m2 9
Therefore, acceleration is also zero i.e., velocity  =
remains unchanged. m1 7

50. (4) 55. (4)

Sol. This is a case of sliding (if plane is frictionless) and Sol. From the free body diagram
therefore the acceleration of all the bodies is same. T1 sin 45° = F
T1 cos 45° = T2 = mg
51. (2) F
 tan45° =  F = mg
Sol. Given: T1 = 2 T2 mg

T1 sin 1 + T2 sin 2 = mg and T1 = 3T2

 T2  3 sin 1 + sin 2  = mg  F = 1 × 10 = 10 N ( m = 1 kg)
From the option,
For 1 = 60° and 2 = 30° 56. (1)
mg Sol. Acceleration is given as:
 T2 =
2  m − m1 
a = 2 g
 m1 + m2 
52. (2)
Sol. Using free body diagram g  m2 − m1 
 = g
2  m1 + m2 
 2(m2 − m1) = m1 + m2
m1  2 − 1 
 = 
m2  2 + 1 
 29

57. (240) 3g 3 3g
Sol. Taking masses M1, M2 and M3 as system,  Fnet = ma cos  =  = 30  =
T 2 T
 T1 – (4 + 6 + 10) × 10 = (4 + 6 + 10) × (2)
 T = 20 N
 T1 = 20(10 + 2) = 240 N

58. (4) 62. (2)

Sol. For 2 kg block Sol. From FBD of lift,
T – 2g sin 30 = 2a …(i) T1 = T2 + mg, m = mass of lift
 T = g + 2a  T1 – T2 = mg
For 4 kg block  Tnet = mg.
4g – 2T = 4a
 4g – 2(g + 2a) = 4a
g
a =
4

59. (1)
Sol. Horizontal force, F = 80 N So, (I) is true
F 80 From FBD of person,
aA = aB = aC = = = 8m/s2 N + ma = mg
5 + 3 + 2 10
N = mg – ma  N < mg
So, (II) is false.

T1 = 5 × 8 = 40 N 63. (2)
Sol. For 4 kg block
4g sin 30° – T = 4a …(i)
For 1 kg block
T2 – T1 = 3 × 8  T2 = 64 N T – 1g sin 30° = 1a
60. (3) 4(T – g sin 30°) = 4a …(ii)
Sol. From (i) from (ii), we get
4(T – g sin 30°) = 4g sin 60° – T
5T = 20 3 +20
T = 4( 3 +1)N

64. (2)
Sol. For equilibrium condition, m2g = m1g sin

From FBD of roller

1
N = mg + F sin 30° = 700 + 200 × = 800 N
2

61. (1)
Sol. m2 3
sin  = =
m1 5
4
cos  =
5
Normal force (N) on m1 = 5g cos 
4
= 5 10  = 40N
5
From the free body diagram shown above,
 30

65. (3)
Sol. Acceleration of block on smooth inclined plane,
a = g sin 
1
Using, s = ut + at 2
2
1
s = g sin 30(2)2
2
FBD of 20 kg block w.r.t. 100 kg
T – 20 g = 20(2)  T = 40 + 200
 T = 240 …(ii)
N1 = 20a1 …(iii)

When the incline is changed to 45°

1
s = g sin 45t 2
2
As distance travelled ins same
1 1 2
  (4) = t  t = 2 2  1.68
 2 2
FBD of 10 kg block w.r.t. 100 kg
66. (4)
Sol. Acceleration in such system is given as

10a1 − 240 = 10(2)  a1 = 26 m/s2

F = 240 + 20(26) + 100  26  F = 3360 N

(m2 − m1 ) 69. (2)

a= g
(m2 + m1 ) M2 g − M1g
Sol. a =
g ((L − ) −  ) g L L M1 + M2
 =  = =
2 L 4 x 2M1g − M1g g
So, x = 4 When, M2 = 2M1  a1 = =
3M1 3
67. (3) 3M1g − M1g g
When M2 = 3M1  a2 = =
Sol. Given that mass of monkey, m = 50kg 4M1 2
Acceleration due to gravity, g = 10 m/s2
g
Tension (T) = 350N
a 2
Given monkey climbs downward, acceleration of The ratio 1 = 3 =
monkey, a = 4 m/s2 a2 g 3
When monkey climbs upward, acceleration of monkey, 2
a = 5 m/s2 70. (3)
(For upward) Sol. Along horizontal
T – mg = ma  T = mg + ma = 50(10 + 5) = 750N F1 + 1cos 45 = 2sin 45
Rope will break while climbing upward
2 1 1
(For downward) F1 = − =
T = m(g – a) = 50(10 – 4) = 300N 2 2 2
Rope will not break while climbing downward Along vertical
F2 = 1sin 45 + 2sin 45
68. (1)
3 F 1 1
Sol. Let a1 be the acceleration of 100 kg block F2 = 3sin 45 = so, 1 = = so, x = 3
FBD of 100 kg block w.r.t. ground 2 F2 3 x
F – T – N1 = 100 a1 ….(i)
 31

71. (1)  –4Ta1 – 2Ta2 – Ta3 – Ta4

Sol. Let addition force required be = F =0
F + 5iˆ − 6iˆ + 7 ˆj − 8 ˆj = 0  4a1 + 2a2 + a3 + a4
=0
 F = iˆ + ˆj,| F|= 12 + 12 = 2N
y component 1 76. (36)
Angle with x-axis: tan  = = Sol. We have, acceleration of system as
x component 1 6mg 3g
So,  = tan–1(1) = 45° a= =
10m 5
72. (3)
Sol. For observer in box

N + Ma = Mg
 N = M(g – a)
Mg 3g Taking 8, 9, 10 together
 = M( g − a)  a = 3g 3  2  3 10
4 4 T = 3ma = 3m  = = 36N
5 5
73. (3)
Sol. FBD of middle point is as shown below. 77. (12)
From FBD Sol. Let draw FBD of block clearly for equilibrium
T sin  = 30
T cos  = 100

So, tan  = 0.3

= 3 × 10–1 F cos 60° = mg sin 60°
So,  = tan–1(3 × 10–1) F
 = tan60
mg
74. (2)
Sol. When elevator moves downward, then a pseudo force x
 = 3
acts on object in upward direction, due to which 0.2 10
effective weight of object decreases.  x = 2 3  x = 12
75. (1) 78. (4)
Sol. Sol. Let aw be the acceleration of wedge and ab be the
acceleration of block w.r.t wedge
For the wedge w.r.t. ground
 32

1 From diagram,  =  + 30°

Ncos60 = Maw  N  = 16aw  60° =  + 30°   = 30°
2
 N = 32 aw …(i)
For block w.r.t. wedge 80. (4)
Sol. From free body diagram
80 – 2T = 8a …(i)
T – 20 = 4a …(ii)

Balancing vertical forces

N + maw sin 30° = mg cos 30°
 N = 8g cos30° – 8aw sin 30° Multiply equation (ii) by 2 and adding with equation (i)
 32aw = 4 3g − 4aw (Using (i)) we get
40 10
3 (8 + 8)a = 40  a = = m/s2
 aw = g 16 4
9 1 2S
Along incline plane Using S = at 2  t 2 =
 mg sin 30 + maw cos30 = mab = 8ab
2 a
0.2  2  4 2
1 3 3  = t  t = 0.4 sec
8 g  + 8 g 10
 ab = 2 9 2
8 81. (2)
1 3 3 2g Sol. Total weight acting downward,
 ab = g  + g = W = (20 × 3 + 10)g = 70 g
2 9 2 3

79. (30)
Sol. Let draw FBD of Bob

Now from FBD,

W – N = 70a = 70 × 0.2
 N = 70 × 10 – 70 × 0.2 = 686 N

82. (82)
Sol. P makes angle of 35° with AC
So, component along AC = 100 cos 35 = 81.9 N 82 N
We have 83. (492)
T sin  = ma + mg sin 30° ….(i) Sol. Mass of the person, M = 60kg
T cos  = mg cos 30° …(ii) Tension in the rope of the lift when it moves downward
Dividing (i) by (ii), we get with acceleration a,
a + g sin30 10 + 5 T = M(g – a) = 60(10 – 1.8) = 492 N
tan  = = = 3
g cos30 5 3 84. (1)
So,  = 60° Sol. The free body diagram is
Let ‘’ be angle made by string with vertical By Lami’s theorem
 33

mg F 88. (1)
 = …(i)
sin135 sin135 Sol. When lift is stationary, W1 = mg
 F = mg = 100 N

85. (4)
480
Sol. a = = 12 m/s2
20 + 12 + 8

When the lift descends with acceleration, a

W2 = m(g – a)
49
W2 = ( 9.8 − 5) = 24N
9.8
480 – T = 24a
480 – T = 24 × 12 89. (4)
480 – T = 288 Sol. Taking the rope and the block as a system
T = 192 N

86. (4)
Sol. Writing free body-diagrams for m and M, we get P = (m + M)a
P
 Acceleration produced, a =
m+M
Taking the block as a system,
Force on the block, F = Ma
MP
 F=
m+M

90. (1)
we get T = ma and F–T= Ma Sol. As springs are massless
where T is force due to spring
 F–ma–Ma or, F = Ma+ma
 Acceleration of the system
F
a=
M+m
Now, force acting on the block of mass m is
 F  mF
ma = m   = m+M
 M + m 
So, tension in spring 1
87. (3) = Tension in spring 2
Sol. Here, m1 = 5kg and m2 = 4.8 kg. = Mg
If a is the acceleration of the masses, And, reading of spring balance is equal to tension of
spring so, both will show same reading.
m1a = m1g – T …(i)
m2a = T – m2g …(ii) 91. (3)
Solving (i) and (ii) we get Sol. Case-I: For the man standing in the lift, the
 m − m2  (5 − 4.8)  9.8 acceleration of the ball
a = 1 g a = m/s2 = 0.2 m/s2 abm = ab − am  abm = g − a
 1
m + m2 (5 + 4.8)
Case-II: The man standing on the ground, the
acceleration of the ball
abm = ab − am  abm = g − 0 = g
 34

92. (1) 96. (3)

Sol. When forces F1, F2 and F3 are acting on the particle, it Sol. Given: m = 25 Kg, µ = 0.25, S = 5 m
remains in equilibrium. Force F2 and F3 are
perpendicular to each other,
 F1 = F22 + F32
The force F1 is now removed, so, resultant of F2 and F3
will now make the particle move with force equal to F1. Block moves with uniform velocity
F F – f = ma
Then, acceleration, a = 1
m So, a = 0  F cos 45° = f
F=µ×N
93. (2) F  F 
 =  mg − 
Sol. Force = mass × acceleration 2  2
F  F 
 = 0.25 25  9.8 − 
2  2
F
 1.25 = 61.25
 F = (m + m + m) × a 2
F = 3m × a 61.25  2
F= = 49 2
F 10.2 1.25
a= a= m/s2 1
3m 6 Wext = FScos45 = 49 2  5  = 245 J
10.2 2
 T2 = ma = 2  = 3.4N
6
97. (1)
Sol. Accerlation if plane is smooth,
94. (3)
a = g sin
Sol. Tension, T = 360 N Displacement of object
mg – T = ma 2
1
T 360 = ( g sin ) t12 t1 =
 a = g − = 10 − = 4 m/s2 2 g sin 
m 60
Accelration of plane is rough,
a = g sin - g cos
95. (1) Displacement of object
Sol. The FBD of the block is 1
= ( g sin  − g cos )t22
2
2 2
t2 = = nt1n
g sin  − g cos  g sin 
on comparing,
1
 =1− 2
n

 Fnet = ma 98. (2)

 mgsin60° – µmgcos60° = ma Sol.
g  g
 g sin 60 − g cos60 =  a =
2 2 
3 1 1
 − =
2 2 2
 = 3 −1
Fk = µ kN = 0.3 × 50 × 9.8 = 147 N [ N = mg]
 35

99. (1) 103. (2)

Sol. From FBD, Sol. Maximum static friction
50  3
( f s )max = mg cos30 = 0.1 = 2.5 3N
2

For brick begins to slide,

Frictional force fL = mg sin 45°
Normal reaction N = mg cos 45° = N
 Coefficient of static friction
fL
S = = mg sin 45°/mg cos 45° = tan 45° = 1
N

100. (3) F1 − F2 = 25 + 2.5 3 − 25 + 2.5 3 = 5 3N

Sol. FBD for 6 kg block
6a = 6g – T …(i)
104. (1)
FBD for 20 kg block
Sol. Coefficient of friction, µ = 0.5
1 dy x
tan  =  = = =
2 dx 2
x2 (1)2 1
So x = 1m, y = = = m
4 4 4
T – fk = 20 a
T – µ(20)g = 20a …(ii)
105. (2)
comparing (i) and (ii)
Sol. Co-efficient of friction (µ) is independent of the area of
 6g – 6a = 20a + 20 µg
 6g – 20a × 0.04g = 26a contact and depends on nature of surface in contact, so,
10(6 − 0.8) it depends on material of object.
a= = 2m/s2
26
106. (4)
101. (2) Sol. Initially,
Sol. N = mg Friction, f = µ mg = m 2 R …(i)
For no slipping F  µN Finally,
g g  2 
 m2r  mg  2 = =  mg = m   R ' …(ii)
r r
 4 
102. (2) From (i) and (ii)
Sol. 4R
1=
R'
 R' = 4R = 4 × 1 = 4 cm ( R = 10 m given)
So distance from the center will be 4 cm.

107. (4)
For M block Sol.
Mg sin53° – µ (Mg) cos 53° - T = Ma
 T = 80 – 15 – 20  T = 45 N
For m block
T – mg sin 37° – µmg cos37° = 2m
Apply the conservation of momentum
 45 = 10 m  m = 4.5 kg
Pi = Pf (Collision)
 36

108. (1)
Sol. From equation of motion

1
S = ut + at 2
2
1 g 3 g 1 3 1
 50 = 0 +  a 100  − g =  − =
2 2 2 4 2 2 4
a = 1 m/s ;F −mg = ma
2
3 1 1
  =  =
1 2 4 2 3
30 −   50 = 5 1  50 = 25;  = = 0.50
2
112. (4)
109. (3)
Sol. Let a1 be the acceleration when it slide down smooth
Sol. Block will just start to move when F cos 30° = fs
incline plane.
Then, a1 = g sin45 = g / 2
Let a2 be the acceleration when it slide down rough
inclined plane
g k g
Then, a2 = g sin 45 − k g cos45 = −
2 2
Let ‘t1’ be the taken when it slide down smooth surface
3F
 = N and ‘t2’ be the time taken when is slide down rough
2
surface.
3F
 = (mg − Fsin30) 1 1
2 t2 = nt1 and a1t12 = a2t22
2 2
3F F 3F 0.25
 = 0.25(100 − )  = (200 − F) 1 g 2 1  g k g  2 2 1
2 2 2 2  t1 =  −  n t1  k = 1 − 2
2 2 2 2 2 n
 3F = 50 − 0.25F  F( 3 + 0.25) = 50
50
F= = 25.2 N 113. (1)
0.25 + 3 Sol. When block is just about to move up
F1 = mg sin 45° + µmg cos 45°
110. (4)
mg
Sol. We have = (1 + )
f mg 2
a= = = g
m m
| V − u | | 20 − 0 |
Also, | a |= = = 4 m/s2
t 5
4
So, g = 4   = = 0.4
10

111. (2) When block is just about to move down

Sol. From the free body diagram shown. F2 mg sin 45° – µmg cos 45°
mg sin 30° – µmg cos 300 = ma
mg
1 3  = (1 − )
 a = g  −  2
2 2 
As, F1 – 2F2
 37

117. (4)
Sol. For 4 kg block
4g – T = 4a ….(i)
For 40 kg block
T – 40g × 0.02 = 40a [ fk = µ mg]
T – 8 = 40a ….(ii)
Adding (i) and (ii),we get
40 – 8 = 44a
mg mg
 (1 + ) = 2 (1 − ) 32 8
2 2 a= = m/s2
44 11
 1 +  = 2(1 − )  1 +  = 2 − 2
1 118. (6)
 3 = 1   = = 0.33
3 Sol. For 4M
2Mg – T = 4 Ma …(i)
114. (2) For M
Mg
Sol. Retardation due to friction n == g  a = −g
M
v2 − u 2 02 − 22 2
Now, s = = = = 0.5m T – Mg = Ma …(ii)
2a −2g 0.4 10 Adding (i) and (ii), we get
g
115. (1) Mg = 5Ma  a =
5
Sol. From the free diagram shown
Mg 6
N = Mg cos  So, T = Ma + Mg = + Mg = Mg
f = Mg sin  5 5

119. (3)
Sol.

Contact force, R = N2 + f 2
Clearly, from (i) ‘a’ will be maximum when f = flim
 R = (Mgcos ) + (Mgsin )
2 2
f   2g
So, amax = lim = = g
2 2
= (Mg)2 (cos2  + sin 2 )  R = Mg From (ii), Fmax = 8amax + flim = 8µg + 2µg = 10µg = 49N

116. (4) 120. (2)

Sol. By FBD of block, we have Sol. For beaker to move with disc
N = 50 N fs = m2R
And F  mg + N So, R will be maximum, when fs = flim

i.e.  2g + 0.5 × 50
 20 + 25  45 N
So, Maximum force that can be applied is 45N.
 38

Therefore, 124. (3)

flim = m2Rmax Sol. From question,
µmg = m2Rmax 1
ta = td
g 2
Rmax = 2 2s 1 2s
 =
g aa 2 ad
So, R  2
 Or, aa = 4ad …(i)
g sin  + g cos  = 4( g sin  − g cos )
121. (2) 3tan 
 5g cos  = 3g sin    =
f mg 5
Sol. a = = = g
m m 3
2  = [  = 30°]
u 5
So, S = [ v = 0]
2a So, x = 3
(9.8)2 (9.8)2
S= = = 9.8 m 125. (30)
2 g 2  0.5  9.8 Sol. From FBD shown, T = mg – N
R = Mg + N = (M + m)g – T
122. (3)
Sol. Acceleration on smooth inclined plane
a = g sin 30° = g/2
1
Using S = ut + at2
2
1 2 g 2
 S = gT = T …(i) ( u = 0) For no movement of block,
2 4 T  R  T  [(M + m) g − T]
Acceleration on rough inclined plane (M + m) g (0.5)(5 + 4) 10
T  T 
g g 3 1+  1 + 0.5
a = g sin30 − g cos30 = −
2 2  T  30 N
g  Tmax = 30 N
a= (1 −  3)
2
1 2 126. (5)
Using again S = ut + at Sol. As block is at rest
2
So, F cos  = f = N
1
S= g (1 − 3)(T)2 …(ii)
4
By (i) and (ii)
1 1 1
= gT 2 = g (1 − 3)2T 2  1 − 3 = 2
4 4 
 2 − 1  1
  =  2   x = 3.00 F cos  =  (mg – F sin)
   3 F(cos  +  sin ) = mg
mg
F=
123. (15) cos  +  sin 
Sol. For block of mass 1 kg For Fmin
F = fs,max (cos  +  sin) = 0  tan  = 
 1 kg × a = µ N = µ × 1 × g ( N = mg) i.e.
d
(cos  +  sin ) = 0  tan  = 
d
 a = 0.5 × 1 × 10 = 5 m/s2
 1
 Maximum horizontal force, So, sin  = and cos  =
Fmax = ma = (1 + 2) × 5 = 15 N 1+  2
1 + 2
 39

mg mg
Thus, Fmin = =
1  1 + 2
+
1+  2
1+  2

127. (3.33)
Sol.

In limiting case,
fr = N = F …(iii)
Using equation (i) and (iii),
mg 0.5 10
F= F= = 25N
 0.2

From the condition of equilibrium 130. (1)

N = mg + F sin 60° Sol. For balancing, mg sin = f = mg cos
For no movement of the block 3
 tan  =  = = 0.75
F cos 60°  f 4
 F cos 60°   (mg + F sin 60°) ( f = µN)
mg
 F  Fcritical = 10 N
cos60 −  sin60
As Fcritical = 3x
So, 3x = 10  x = 3.33

128. (25)
Sol.

 4 R
h = R − R cos  = R − R   =
5 5
R
 h = = 0.2m [ radius, R = 1m]
5
Block will fall down when  = angle of repose i.e.,
tan  = µ 131. (346)
Sol. Sv = Sd
dy d  x 2  x
 tan  = =   = and at time of maximum v02 v02 / 4
dx dx  4  2 =  au = 4ad
2au 2ad
height tan  =  = 0.5
 g sin  + g cos  = 4( g sin  − g cos )
x2 (1)2
 x = 1 and therefore y = = = 0.25 m = 25 cm 3 3 1 3 346
4 4   = tan    =  = = 0.346 =
5 5 3 5 1000
So, I = 346
129. (25)
Sol. F.B.D. of the block is shown in the diagram
132. (3)
Since, block is at rest, Sol. A : N = 5g + 20 sin 30°
 fr – mg = 0 …(i) 1
F–N=0 …(ii) = 50 + 20  = 60N
2
Also, fr  N
 40

135. (2)
Sol. Given: m1 = 5kg; m2 = 10kg;  = 0.15
FBD for m1, m1g – T = m1a
= (10 + m)a

F− f
Accelaration, a1 =
m
20cos30 − N
=
5
For rest a = 0
 3 
20  − 0.2  60 or, 50 = 0.15(m + 10)10
= 
2 3
 5 = ( m + 10 )
5 20
= 1.06 m/s2 100
= m + 10
3
 m = 23.3 kg; close to option (2)

136. (4)
Sol. Equation of motion when the mass slides down
Mg sin  - f = Ma
 10 – f = 6 (M = 2 kg, a = 3 m/s2,  = 30° given)
 f = 4N
B : N = 5g – 20 sin 30°
1
= 50 − 20  = 40 N
2
F − f 20cos30 − 0.2  40
a2 = = = 1.86 m/s2
m 5
Now a2 – a1 = 1.86 – 1.06 = 0.8 m/s2
Equation of motion when the block is pushed up
133. (2)
Let the external force required to take the block up the
Sol. Taking (A + B) as system
plane with same acceleration be F
F – µ(M + m)g = (M + m)a
F − (M + m) g
a =
(M + m)
F − (0.2)4 10  F − 8 
a= = 
4  4 
But, amax = g = 0.2  10 = 2
F−8 F – Mg sin  – f = Ma
 = 2  F = 16N
4  F – 10 – 4 = 6
F = 20 N
134. (1)
Sol. From figure, 2 + mg sin 30° = µmg cos30° and 137. (2)
10 = mg sin 30° + µmg cos 30° = 2µmg cos 30° – 2 Sol. Let be the minimum coeffcient of friction
 6 = µmg cos 30° and 4 = mg sin 30°
3
By dividing above two  =   3
2
3
 Coefficient of friction,  =
2
 41

At equilibrium, mass does not move so, kt

a= − g
3 mg sin  = µ3 mg cos  m
 µmin = tan

138. (1)
Sol. So, a-t graph will be as shown

142. (4)
Sol. a = g sin 45° – µg cos 45°
a = g sin 45° – 0.3 xg cos 45°
g 0.3gx
= − = 5 2 − 0.3(5 2) x = 5 2 −1.5 2x
2 2
Velocity will increase until a = 0 and when v = vmax,
then a = 0
Along vertical direction 0 = 5 2 − 1.5 2x
A → f = 20 N
B → f ' = f + 100 = 20 + 100 = 120 N
5 2
x=
1.5 2
139. (2) x = 3.33 m
Sol. Initial speed at point A, u = v0
Speed at point B, v = ? 143. (3)
v2 – u2 = 2gh Sol.
v2 = v02 + 2gh
Let ball travels distance ‘S’ before coming to rest
v2 v02 + 2 gh v02 2 gh h v02
S= = = + = +
2g 2g 2g 2g  2g

140. (3)
Sol. Minimum force on A
= frictional force between the surfaces = 12 N or, F1 = mg sin  + µmg cos
When the body slides the inclined plane, then
mg sin –f2 = F2
or F2 = mg sin  – mg cos 
F sin  +  cos 
 1=
F2 sin  −  cos 
F tan  +  2 +  3
 1= = = =3
F2 tan  −  2 −  

Therefore maximum acceleration 144. (4)

amax =
12N
= 3m/s2 Sol. For first half acceleration = g sin ;
4kg For second half
Hence maximum force, acceleration = – (g sin  – g cos )
Fmax = total mass × amax For the block to come to rest at the bottom, acceleration
= 9 × 3 = 27 N in I half-retardation in II half.
g sin  = – (g sin  – g cos )  µ = 2 tan 
141. (2) NOTE:
Sol. According to work-energy theorem, W = K = 0
(Since initial and final speeds are zero)
 Workdone by friction + Work done by gravity = 0
kt − mg
a= i.e., −( mg cos ) + mg sin  = 0
m 2

or cos  = sin  or  = 2tan
2
 42

145. (1) From eq. (i) and (ii),

Sol. Given, initial velocity, u = 100 m/s. sin  +  cos  v02
Final velocity, v = 0 = ; rg tan  + rg = v02 − v02 tan 
cos  −  sin  rg
Acceleration, a = µkg = 0.5 × 10
v2 – u2 = 2as or v02 − rg tan 
=
 02 – µ2 = 2(–µkg)s rg + v02 tan 
 s = 100 m
149. (2)
146. (3) Sol. Centripetal accerlation is,
Sol. Since the body is at rest on the inclined plane, ac = 2x
mg sin 30° = Force of friction vdv  dv 
= 2 x  ac = v dx 
dx
On intigration by applying limits,
v 3 2
0 vdv = 1  xdx
v 3
 v2  2 x 
2
  =   
 m × 10 × sin 30° = 10  2 0  2 1
 m × 5 = 10  m = 2kg v2 2 2 2
= [3 − 1 ]  v = 2 2
2 2
147. (4) x=2
Sol. u = 6 m/s, v = 0, t = 10s, a = ?
v−u
Acceleration a = 150. (2)
t Sol. Maximum velocity is given by:
0−6  tan  +    tan 30 + 0.2 
a= V = Rg 
10 −    = 300  g  1 − 0.2  tan 30 
 1 tan   
−6
a= = −0.6m/s2
 0.57 + 0.2 
10 = 300 10   = 51.4 m/s
1 − 0.2  0.57 

151. (2)
The retardation force is due to the frictional force Sol. Given: R = 9m,
 f = – ma 120
 µN = – ma N= rpm = 40 rpm
3
 µmg = – ma 2N 2 40 4
ma = = = rad/s
 =− 60 60 3
mg 2
 4 
−a 0.6 acentripetal = 2 R =    9 = 162 m/s2
 =− = = 0.06  3 
g 10
152. (2)
148. (3) Sol. Maximum Tension, T = m2
mv02
Sol. Nsin  + f cos  = …(i) T 400
r = = = 40rad/s
N cos – f sin  = mg …(ii) m 0.5  0.5

153. (3)
mv2
Sol. Centripetal force, F =  v2  r
r
(For same m and F)
v1 r 3
 = 1 = [ r1 : r2 = 3: 4]
v2 r2 2
 43

154. (2) mV2

v2
12 12  = mg
Sol. To negotiate curve, tan  = = r
Rg 10  400 Vmax = rg = 0.34  50 10  13m/s

159. (40)
h h 144 dv dv v2 v dv 1 x
tan  =  = Sol. v = a v =  =  dx
1.5 1.5 4000 dx dx R 15 v R 0
Therefore, outer rail raised with height, h = 5.4 cm v x v
 ln =  e x /R =  v = 15e x /R
155. (4) 15 R 15
R
Sol. Given that dx t0
 = 15ex/R   2 e− x/R dx = 15 dt
900 9 dt 0 0
m = 900 g = kg = kg
1000 10  t0 = 40(1 − e−/2 )
r = 1 m, N = 10 rpm
2N 2(10) 
= = = rad/s 160. (3)
60 60 3
Sol. Given that speed, v = 20 m/s
Radius of circular track, R = 40 m
T cos  = mg …(i)
2
mv
T sin  = ...(ii)
R

For circular motion, at lowest point

Fnet = mr2
 T – mg = mr2
 T = mg + mr2 Divide equation (ii) by (i), we have
2
9 9   v2
 9.8 + 1  = 8.82 + 0.98 = 9.80N tan  =
10 10  3  Rg
202
156. (1) tan  =
mv2 40 10
Sol. Centripetal force Fc = and v = r ×  
r tan  = 1   =
 Fc = m r = 200 × (0.2) × 70 = 560 N
2 2 4

157. (4) 161. (1)

Sol. Given, mass of child, m = 5kg Sol. For observer on block,
Radius of merry-go-round, R = 2m mv2
2 N=  N  v2
Angular velocity,  = = 2 rad/s r
3.14
The centrifugal force on the child will be
F = m2 R = 5 × 22 × 2 = 40N

158. (3)
mv2
Sol. Centripetal force, fc =
r
So, curve is parabola,
Frictional force = mg
Here, centripetal force for motion is being provided by symmetric about N – axis
the friction.
 44

162. (2) 165. (3)

Sol. V =| V − u |= V2 + u 2  V ⊥ u  Sol.

Now, K1 + P1 = K 2 + P2
1 1
 mu 2 + 0 = mV 2 + mgl  V2 = u 2 − 2 gl
2 2
 V = u 2 − 2 gl Tmax = m2maxR
k 2 k 
2
 k  
80 = 0.1    2
 30    =   60 = 30 
302  80
 k2 =  k 2 = 360000  k = 600
2  0.1
So, V = V2 + u 2 = u 2 − 2 gl + u 2

= 2(u 2 − gl )  x = 2 166. (2)

Sol. Let us take a general point 'P'
163. (3)
Sol. By law of conservation of mechanical energy,
KP + PP = KQ + PQ
1
 0 + mgR sin  = mv2  v2 = 2Rg sin 
2

mv 2
At point ‘Q’, N = + mg sin 
R
N Rg Rg sin  3 mv2 mv2
 = 1 + 2 sin  = 1 + = T + mg cos  = T = − mg cos 
 mv 
2
v 2Rg sin  2 R R
 
 R  So, T will be minimum when, mg cos  is maximum
 A = constant i.e., when cos  is maximum
So, graph between A and a will be as shown below. i.e. when  = 0
and  is zero when string is at highest point.

167. (1)
Sol. From figure

164. (24)
2
mVmax
Sol. f s, max =
R
2
mVmax
 mg =
R r L 1
 Vmax = Rg sin  = = =   = 45
L L 2 2
V R2
So, 2,max = mv2
V1,max R1 Tsin  = …(i)
r
Tcos = mg …(ii)
2
v
tan  =  v = rg
rg
R2 48 4
 V2,max = V1,max  = 30  = 30  = 24 m/s.
R1 75 5
 45

168. (2)
Sol. Given, mass of the block, m = 200g = 200 × 10–3 Kg
Radius of the circular groove r = 20
Time taken to complete one round, T= 40 s
Here, normal force will provide the necessary
centripetal force.
2
 2 
N = m2 r = 200 10−3     0.2 = 9.859 10−4 N.
 40   22 
2
1.25 10−2   7  
r2  7 
169. (3)  = =
Sol. For statement-I g 10
The maximum speed by which cyclist can take a turn 1.25 10−2  222
on a circular path. = = 0.6
10
 v  rg  0.2  2  9.8
 v  3.92  vmax = 1.97m/s 172. (4)
Sol. Given,  = 45°, r = 0.4 m, g = 10 m/s2
5
Speed of cyclist, v = 7 kmh −1 = 7  = 1.94 m/s mv 2
18 T sin  = …(i)
As, v < vmax  cyclist will not slip r
The maximum safe speed on a banked frictional road T cos  = mg …(ii)
( + tan )
vallowable = rg
1 −  tan 
2  9.8(0.2 + tan 45)
v =
1 − 0.2  tan 45
2  9.8 1.2
= = 5.42 m/s
0.8
As, v < vallowable  cyclist cross the curve without From equation (i) & (ii) we have,
slipping v2
So, both the statements are true. tan  =
rg
170. (5) v2 = rg   = 45°
Sol. At the highest position, Hence, speed of the pendulum in its circular path,
mv2 v rg = 0.4 10 = 2m/s
Tmin = min − mg
l
At the lowest position, 173. (3)
mv2 Sol. K.EX + P.EX = K.EY + P.E Y
Tmax = + mg (Given)
l 1
And we know,  mgR(cos  − sin ) = mV2
2
Tmax − Tmin = 6mg  5Tmin − Tmin = 6mg
 V2 = 2gR(cos  − sin )
2
3 mvmin
 Tmin = mg = − mg
2 l
5 5
 vmin = gl = 10 1 = 5 m/s .
2 2

171. (4)
mv2 mV2
Sol. Using, mg = = mr2 at Y, = mg sin  [ N Y = 0]
r R
 = 2n = 2 × 3.5 7  rad/sec  2mg (cos  − sin ) = mg sin 
Radius, r = 1.25 cm = 1.25 × 10–2 m 2
 sin  = cos 
Coefficient of friction, µ = ? 3
 46

174. (1) 177. (3)

Sol. Initially, k × 1 = m2R …(i) Sol. Let m is the mass of each particle and  is the angular
Finally, k × 5 = m(2)2 (R + 5) …(ii) speed of the annular ring.
From (i) and (ii), we get: R = 15 cm v2 2 R12
So, original length, R = 15 cm. a1 = 1 = = 2 R1
R1 R1
175. (4) v22
a2 = = 2 R2
Sol. s = t 3 + 5 R2
ds
 velocity, v = = 3t 2
dt
dv
Tangential acceleration, at = = 6t
dt
v2 9t 4
Radial acceleration, ac = =
R R
At, t = 2s, at = 6  2 = 12 m/s2 F1 ma1 mR12 R1
= = =
9 16 F2 ma2 mR22 R2
ac = = 7.2 m/s2
20 Note:
 Resultant acceleration = at2 + ac2 The force experienced by any particle is only along
radial direction.
= (12)2 + (7.2)2 = 144 + 51.84 = 195.84 = 14 m/s2 Force experienced by the particle, F = m2R
F R
 1= 1
176. (3) F2 R2
Sol. a = ac cos (−iˆ) + ac sin (− ˆj)
178. (2)
Sol. Only option (2) is filse since accelaation vector is
always radial (i.e. towards the center) for uniform
circular motion.

−v2 v2
= cos iˆ − sin ˆj
R R

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