AIR VP PRACTICE SHEET

NEET CLASS-11th
PHYSICS
Laws of Motion
Topic-I: Inertia, First Law of Motion, Second Law of 5. A force of 100 dynes acts on mass of 5 gm for 10 sec.
Motion (Impulse, Change in Momentum, Force), Third The change in velocity is:
Law, Rocket propulsion (1) 2 cm/sec
1. The velocity-time graph of a body of 25 kg is shown (2) 20 cm/sec
in figure. The net force acting on the body at t = 1.5 s (3) 200 cm/sec
is; (4) 2000 cm/sec

6. From Newton’s second law of motion, it can be inferred

that;
(1) No force is required to move a body uniformly
(1) 25 N (2) 50 N along straight line
(3) 12.5 N (4) 100 N (2) Accelerated motion is always due to an
external force
2. In a rocket, fuel burns at the rate of 1 kg/s. This fuel (3) Inertial mass of a body is equal to force
is ejected from the rocket with a velocity of 60 km/s. required per unit acceleration in the body
This exerts a force on the rocket equal to:
(4) All of these
(1) 6000 N (2) 60000 N
(3) 60 N (4) 600 N
7. A 10 g bullet moving at 200 m/s stops after penetrating
3. A particle of mass 2 kg is increasing its velocity 5 cm of wooden plank. The average force exerted on the

as, v = 4t iˆ m s −1 where t = time in seconds. Its bullet will be:

linear momentum at an instant will be: (1) 2000 N

(1) (4t ˆj )kg ms −1 (2) –2000 N

(3) +4000 N
(2) (8t kˆ) kg ms −1
(4) –4000 N
(3) (4t iˆ + 4t ˆj )kg ms −1

(4) (8t iˆ) kg ms −1 8. In accordance with Newton's third law of motion

(1) Action and reaction never balance each other
4. A particle is moving with a constant speed along a (2) For appearance of action and reaction, physical
straight-line path. A force is not required to
contact is not necessary
(1) Increase its speed
(3) This law is applicable whether the bodies are at
(2) Decrease the momentum
(3) Change the direction rest or they are in motion
(4) Keep it moving with uniform velocity (4) All of these
[1]
9. In which of the following graphs, the total change in 13. When a horse pulls a wagon, the force that causes the
momentum is zero? horse to move forward is the force
(1) The ground exerts on it
(2) It exerts on the ground
(1) (3) The wagon exerts on it
(4) It exerts on the wagon

14. A ball of mass 150 g starts moving with an

acceleration of 20m/s2 . When hit by a force, which
(2)
acts on it for 0.1 sec. The impulse of force is:
(1) 0.5 N-s (2) 0.1 N-s
(3) 0.3 N-s (4) 1.2 N-s

15. Figures I, II, III and IV depict variation of force with

(3) time

(I)
(4)

10. Assertion (A): No force is required by the body to

change its state of rest or motion.
Reason (R): In motion with uniform velocity, (II)
acceleration will be zero.
(1) Both Assertion (A) and Reason (R) are true
and Reason (R) is a correct explanation of
Assertion (A).
(2) Both Assertion (A) and Reason (R) are true
but Reason (R) is not a correct explanation of
Assertion (A). (III)
(3) Assertion (A) is true and Reason (R) is false.
(4) Assertion (A) is false and Reason (R) is true.

11. A machine gun fires 20 bullets per second into a target.

Each bullet has a mass 150 gm and has a speed of 800
m/sec. Find the force necessary to hold the gun in
position (IV)
(1) 800 N (2) 1000 N
(3) 1200 N (4) 2400 N

12. The linear momentum ‘p’ of a body moving in one The impulse is highest in the case of situations
dimension varies with time according to the equation depicted. Figure
p = a + bt 2 , where a and b are positive constants. The (1) I and II
net force acting on the body is (2) III and I
(1) proportional to t2 (2) a constant (3) III and IV
(3) proportional to t (4) inversely proportional to t (4) IV only
[2]
16. Statement-I: Linear momentum of a body changes 21. A mass of 1 kg is thrown up with a velocity of 100 m/s.
even when is moving uniformly in a circle. After 5 seconds, it explodes into two parts. One part
Statement-II: Force required to move a body of mass 400 gram comes down with a velocity of 25
uniformly along a straight line is zero. m/s. Calculate the velocity of other part.
(1) Statement I and Statement II both are correct. (1) 40 m/s upward
(2) Statement I is correct but Statement II is (2) 40 m/s downward
incorrect. (3) 100 m/s upward
(3) Statement I is incorrect but Statement II is (4) 60 m/s downward
correct.
(4) Statement I and Statement II both are incorrect. 22. A rigid ball of mass m strikes a rigid wall at 60° with
normal and gets reflected without loss of speed as
17. A body of mass 3kg initially at rest is acted on by a shown in the figure below. The magnitude of
force which varies with time t as shown in the impulse imparted by the wall on the ball will be:
graph below. The final momentum of the body at
t = 6 s is:

(1) Zero (2) 5 Ns

(3) 30 Ns (4) 50 Ns
mv mv
(1) (2)
18. In an explosion a body breaks up into two pieces of 2 3
unequal masses. In this situation; (3) mv (4) 2mv
(1) both parts will have equal magnitude of
momentum. 23. Assertion A: Real forces are always present in
(2) the lighter part will have greater momentum. pairs.
(3) the heavier part will have greater momentum. Reason R: For every action there is an equal and
(4) both parts will have equal kinetic energy. opposite reaction.
(1) A is true but R is false
19. If a ball of mass m strikes the wall with speed u at (2) A is false but R is true
an angle  with the vertical and gets reflected with (3) Both A and R are true and R is the correct
the same speed as shown in the figure, then the explanation of A.
magnitude of impulse imparted on the ball is: (4) Both A and R are true but R is NOT the correct
explanation of A.

24. Figure shown an estimated force-time graph for a

base ball struck by a bat. The magnitude of impulse
delivered to the ball is:

(1) mu cos  (2) 2mu cos 

(3) mu sin  (4) 2mu sin 

20. A constant force (F) is applied on a stationary particle

of mass ‘m’. The velocity attained by the particle in a
certain displacement will be proportional to;
1
(1) m (2)
m
1 (1) 30000 Ns (2) 45000 Ns
(3) m (4)
m (3) 13500 Ns (4) 10000 Ns

[3]
25. A block of metal weighting 2 kg is resting on a 29. A particle is on a smooth horizontal plane. A force F
frictionless plane (as shown in figure). It is struck is applied, whose F-t graph is given where t is time.
by a jet releasing water at a rate of 1 kg/s and at a
speed of 10 m s–1. The initial acceleration of the
block will be: (neglect any dissipative force)

A. In region AB acceleration is constant.

B. Initially the particle must be at rest.
C. At time t2, acceleration is constant.
(1) 3 m/s2
D. The initial acceleration is zero.
(2) 6 m/s2 Choose the most appropriate answer from the option
(3) 5 m/s2 given below which contains correct statements.
(4) 4 m/s2 (1) A and C only (2) A, B and D only
(3) C and D only (4) B and C only
26. A balloon of mass M is descending with a constant
30. A disc of mass 2 kg is kept floating horizontally in
g air by firing bullets of mass 0.1 kg each vertically
acceleration . When a mass m is released from the
3 upward at the rate of 5 per second. If the bullets
g rebound with the same speed, the speed with which
balloon it starts rising with the same acceleration these are fired will be: (Take g = 9.8 m/s2)
3
. The value if m is: (Assuming that air resistance is
constant)
M M
(1) (2)
2 4
(3) 4M (4) 2M
(1) 1.96 m/s (2) 0.98 m/s
27. A man fires a bullet of mass 200 g at a speed of (3) 9.8 m/s (4) 19.6 m/s
5 m/s. The gun is of one kg mass. What is the recoil
31. A body of mass M at rest explodes into three pieces,
speed of the gun?
M
(1) 0.1 m/s two of which of mass each are thrown off in
4
(2) 10 m/s perpendicular directions with speed of 3 m/s and
(3) 1 m/s 4 m/s respectively. The third piece will be thrown
(4) 0.01 m/s off with a speed of :
(1) 1.5 m/s (2) 2.0 m/s
(3) 2.5 m/s (4) 3.0 m/s
28. A cricket ball of mass 100 g has an initial velocity
u = (2iˆ + 3 ˆj ) m/s and a final velocity v = −(2iˆ + 3 ˆj ) m/s 32. Assertion A: Property of inertia applies only to the
bodies that are at rest.
after being hit by bat. The change in momentum (in
Reason R: The bodies do not change their state
kg m/s) of the ball is: unless acted upon by an unbalanced external force.
(1) −(0.9iˆ + 1.2 ˆj ) (1) Both A and R are true and R is the correct
explanation of A.
(2) −(0.4iˆ + 0.6 ˆj ) (2) Both A and R are true but R is NOT the correct
explanation of A.
(3) −(4iˆ + 6 ˆj ) (3) A is true but R is false.
(4) zero (4) A is false but R is true.

[4]
33. An object of mass 3 kg is at rest at t = 0. If a force Topic-II: FBD, Motion of Bodies Connected by String,
F = (9t 2iˆ + 6tjˆ) N is applied on the object (where t Spring, Pulley System, Unstraint Motion
38. A mass of 1 kg is suspended by a string A. Another
is time in second), then the velocity of the object at
string C is connected to its lower end (see figure). If
t = 2 s is:
a sudden jerk is given to C, then;
(1) ( 8iˆ + 3 ˆj ) m/s
(2) ( 8iˆ + 6 ˆj ) m/s
(3) ( 3iˆ + 8 ˆj ) m/s
(4) ( 8iˆ + 4 ˆj ) m/s
(1) the portion AB of the string will break.
34. A bomb of mass 4 kg initially at rest, explodes and (2) the portion BC of the string will break.
breaks into three fragments of masses in the ratio (3) none of the strings will break.
1 : 1 : 2. The two pieces of equal mass fly off (4) the mass will start rotating.
perpendicular to each other with a speed 10 m/s
39. A block of mass 10 kg is suspended through two
each. The speed of heavier fragment is:
light spring balance as shown in figure.
(1) 5 m/s (2) 15 m/s
(3) 12 m/s (4) 5 2 m/s

35. For a variable mass system, the correct form of

Newton’s second law will be;
→ dv dm
(1) F = m +v
dt dt
→ dv dm
(2) F = m −v (1) Both the scales will read 10 kg.
dt dt
→ dm dv (2) Both the scales will read 5 kg.
(3) F = v −m (3) The upper scale will read 10 kg and the lower
dt dt
→ zero.
mdv
(4) F= (4) The readings may be anything but their sum
dt
will be 10 kg.

36. The maximum force exerted on a passenger by the 40. What is the acceleration of 5 kg mass? All surface
floor of the elevator is not to exceed 1.6 times the smooth)
actual weight of the passenger. The elevator
accelerates upwards with an acceleration of a. What
can be the maximum acceleration of the elevator? (1) 2 m/s2 (2) 4 m/s2
(g = 10 m/s2) (3) 1 m/s2 (4) 3 m/s2
(1) 4 m s−2 (2) 5 m s−2
41. In the arrangement shown in two figures, the
(3) 6 m s−2 (4) 9 m s−2 mass m will ascend with acceleration respectively.

37. A variable force (F) is acting on a body which

displaces the body from x = 1 m to x = 2 m. If
F = (3x2) N, then the average force acting between
x = 1 m to x = 2 m is;
(1) 8 N (2) 4 N (1) g/2, g/2 (2) g, g/2
(3) 7 N (4) 10 N (3) 3g, g/2 (4) 2g, g/2
[5]
42. Two blocks of mass 8 kg and 5 kg are connected by 46. The spring balance inside a lift suspends an object. As
a heavy rope of mass 3 kg. An upward force of 180 the lift begins to ascent, the reading indicated by the
N is applied as shown in the figure. The tension in
spring balance will
the string at point P will be:
(1) Increase
180 N (2) Decrease
8 kg (3) Remain unchanged
P (4) Depend on the speed of ascend
3kg

5 kg 47. Two masses of 10 kg and 20 kg respectively are

(1) 60 N (2) 90 N connected by a massless spring as shown in fig. A
(3) 120 N (4) 150 N force of 200 N acts on the 20 kg mass. At the instant
shown in the figure 10 kg mass has acceleration 12
43. Initially the spring is relaxed. If the mass m is slowly
released, then the elongation produced in the spring m/s² towards right. The acceleration of 20 kg mass
in equilibrium. at this instant is:

(1) mg/k (2) mg/2k (1) 12 m/s2 (2) 4 m/s2

(3) 2mg/k (4) mg/4k (3) 10 m/s2 (4) Zero

44. A system consists of three masses m1, m2 and m3 48. Match List-I with List-II to find out the correct
connected by a string passing over a pulley P. The
option.
mass m1 hangs freely and m2 and m3 are on a rough
horizontal table (The coefficient of friction = µ). A man of mass m is on the floor of a lift then match the
The pulley is frictionless and of negligible mass. following.
The downward acceleration of mass m1 is: (Assume List-I List-II
m1 = m2 = m3 = m)
(A) Lift is moving (I) Apparent
up with weight is greater
acceleration a than true weight
(B) Lift is moving (II) Apparent
down with weight is zero
g (1 − g ) 2 g acceleration a
(1) (2)
9 3 (C) Lift is moving (III) Apparent
g (1 − 2) g (1 − 2) with uniform weight is equal
(3) (4)
3 2 velocity to true weight
(D) Lift is freely (IV) Apparent
45. The tension in the string as shown in the figure is:
[g = 10 m/s2] (All surfaces are smooth) falling weight is less
than true weight
A B C D
(1) II III IV I
(2) I IV III II
(1) 23.3 N (2) 50 N (3) II IV III I
(3) 30 N (4) 46.67 N (4) III II IV I
[6]
49. With what minimum acceleration can a fireman 53. A uniform rope of mass M and length L is hanging
slides down a rope while breaking strength of the from a fixed ceiling. If the tension at point A located
2 L
rope is of his weight? at distance from the free end is T, the tension at
3 4
2 3L
(1) g (2) g point B, located at distance from the free end
3 4
1 would be:
(3) g (4) Zero
3 (1) T (2) 2T

50. A monkey of mass 20 kg is holding a vertical rope. T

(3) (4) 3T
The rope will not break when a mass of 25 kg is 2
suspended from it but will break if the mass exceeds
25 kg. What is the maximum acceleration with 54. The spring-mass system shown in the figure is in
which the monkey can climb up along the rope equilibrium. The extension in spring A is:
(Take g = 10 m/s2)
( g = 10 m/s 2 )

(1) 10 m/s 2 (2) 25 m/s 2

(3) 2.5 m/s 2 (4) 5m/s2

51. In the figure, the blocks A, B and C of mass m each have

accelerations a1, a2 and a3 respectively. F1 and F2 are
external forces of magnitude 2mg and mg respectively.

(1) 0.4 m (2) 0.3 m

(3) 0.2 m (4) 0.1 m

55. The figure shows a horizontal force F acting on a

(1) a1 = a2 = a3
block of mass m on a smooth inclined plane of angle
(2) a1 > a3 > a2
. The normal reaction on the block is;
(3) a1 = a2, a2 > a3
(4) a1 > a2, a2 = a3

52. A block is dragged on a smooth plane with the help

of a rope whose free end is moved with a velocity v
as shown in figure. The horizontal velocity of the
block at the given instant is: (1) mgsin + Fcos (2) mgsin – Fcos
(3) mgcos – Fsin (4) mgcos + Fsin

56. A mass of 1 kg is suspended by a thread. It is lifted up

with an acceleration 4.9 m/s 2 and then lowered with

v an acceleration 4.9 m/s 2 . The ratio of the tensions in

(1) v (2)
sin  the two cases respectively is; (g = 9.8 m/s2)
v (1) 3 : 1 (2) 1 : 3
(3) v sin  (4)
cos  (3) 1 : 2 (4) 2 : 1
[7]
57. The acceleration of the block of mass 2M shown in 60. A monkey of mass 40 kg climbs on a rope (Fig.)
the figure is: (All contact surfaces are frictionless) Which can stand a maximum tension of 600 N. In
which of the following cases will the rope break: The
monkey; (ignore the mass of the rope) (g = 10 m s–2)

(1) 0
g sin 
(2)
2
2 g sin  (1) climbs up with an acceleration of 6 ms–2.
(3) (2) climbs down with an acceleration of 4 ms–2.
3
(3) climbs up with a uniform speed of 5 ms–1.
(4) 2 g sin 
(4) falls down the rope nearly free under gravity.

58. One end of string of length l is connected to a 61. The pulleys and strings shown in the figure are
particle of mass m and the other end is connected to smooth and are of negligible mass. For the system
a fixed small peg on a smooth horizontal table. If the of blocks to remain in static equilibrium, the angle
particle moves in horizontal circle with speed v with  should be:
tension T in the string, the net force acting on the
particle will be:
mv 2
(1) T +
l
mv 2
(2) T −
l
(3) Zero (1) 0° (2) 30°
(4) T (3) 45° (4) 60°

62. A block of mass M is pulled along a horizontal

59. Three blocks of masses m, 2 m and 3 m are connected
frictionless surface by a rope of mass m as shown in
by strings, as shown in the figure. After an upward force
figure. A horizontal force F is applied to one end of
F is applied on block m, the system of masses move the rope. The tension at the mid-point of the rope is:
upward at constant speed v. What is the net force on the
block of mass 2 m (g is the acceleration due to gravity)
F
v
m 2(m + M ) F
(1)
(2m + M )
(2m + M ) F
2m (2)
2(m + M )
(m + M ) F
3m (3)
(2m + M )
(1) 2 mg (2) 3 mg (m + 2M ) F
(4)
(3) 6 mg (4) zero 2(m + M )

[8]
63. Four blocks are moving on smooth surface under a 66. A 70 kg man standing on a weighing machine in a 50 kg
constant force F = 28 N, as shown in the figure. T1, lift pulls on the rope, which supports the lift as shown in
T2 and T3 are tensions in massless strings connecting the figure. The force with which the man should pull the
the blocks. Acceleration of each block is a. Match rope to keep the lift stationary is: (g = 10 m s–2)
List-I with List-II. (All units are in SI units.)

List -I List -II

(Numeric value)
A. T1 I. 2
B. T2 II. 4
C. T3 III. 18
D. a IV. 12
(1) A-I; B-II; C-IV; D-III
(2) A-II; B-I; C-IV; D-III
(3) A-IV; B-I; C-II; D-III (1) 300 N (2) 200 N
(4) A-II; B-IV; C-III; D-I (3) 400 N (4) 150 N
64. Tension T in the block of mass M having uniformly
67. In the arrangement shown below force F is just
distributed mass, at a distance ‘x’ from left end is
sufficient to keep in equilibrium the 100 N block,
given by:
T1, T2 and T3 are tension, in strings AB, CD and EF
and T4 is total sum of all tensions T1, T2 and T3 on
block 100 N.

Fx FL
(1) T = (2) T =
L x
F ( L − x) F ( L − x)
(3) T = (4) T =
L x
65. Which figure represents the correct F.B.D. of rod
of mass m as shown in figure: (Where T is the
tension in the string and R is the reaction force)
Match List-I with List-II
List-I List-II
A. T1 I. 400
N
7
B. T2 II. 100
N
7
(1) (2) C. T3 III. 200
N
7
D. T4 IV. 100 N
(1) A-II, B-III, C-I, D-IV
(2) A-I, B-III, C-II, D-IV
(3) (4) (3) A-I, B-II, C-IV, D-III
(4) A-I, B-IV, C-II, D-III

[9]
68. Two masses M1 = 4 kg, M2 = 12 kg are connected at 71. Two blocks of mass 4 kg and 6 kg are attached to
the ends of an inextensible string passing over a the ends of a string passing over a pulley. The 4 kg
frictionless fixed pulley as shown. When masses are mass is attached to the table top by other string. The
tension in this string T1 is equal to: (g = 10 m/s2)
released, then magnitude of acceleration of masses
will be: (g is acceleration due to gravity)

(1) 10 N (2) 10.6 N

g
(1) g (2) (3) 25 N (4) 20 N
2
g g 72. The total lengths of a spring are l1 and l2 when
(3) (4)
3 4 stretched with a force of 4 N and 5 N respectively.
The natural length of spring is:
69. A person of mass 60 kg stands on a weighing scale (1) l2 + l1 (2) 2(l2 – l1)
inside a lift of mass 940 kg. Suddenly the cable (3) 5l1 – 4l2 (4) 5 l2 – 4l1

snaps and lift starts falling freely. What is the

73. A mass of 2 kg is tied with a massless string and it
reading shown by the weighing scale? (g = 10 m/s2) is pulled by a force in upward direction as shown in
(1) 600 N (2) 104 N the diagram given below. If acceleration is 2 m/s2
(3) 9400 N (4) 0 then tension in the string will be; (take g = 10 m/s2 )

70. Find the magnitude of acceleration of the blocks for

the given system. The inclined plane is fixed and
pulley, strings are massless. (g in acceleration due
to gravity)

(1) 20 N (2) 12 N
(3) 24 N (4) 16 N

74. Consider the situation shown in the diagram given

below. The net force on the pulley due to tension T
in the string will be;
(1) Zero
g
(2)
5
g
(3)
10
3g (1) T (2) T 2
(4) (3) 2T (4) 0
10
[10]
75. A heavy block of mass M hangs in equilibrium from 79. An object is subjected to a force in the north-east
the end of the rope of mass m and length which is direction. To balance this force, a second force
connected to the ceiling. The tension in the rope at should be applied in the direction.
distance x from the ceiling is; (neglect size of block) (1) North-East (2) South
 −x
(1) Mg + mg  
(3) South-West (4) West
 x 
 −x 80. Three spring balances are attached to the ring as
(2) Mg + mg  
  shown in the figure. There is an angle of 90°
 x between the balance A and balance B. There is a
(3) Mg + mg  
  reading of 5 N on balance A and 12 N on the balance
 1− x  B.
(4) mg + Mg  
 

Topic-III: Non Inertial Frame, Pseudo Force, B

Equilibrium of Body
76. In translatory equilibrium; 90º

(1) the net external force acting on particle is zero. A
(2) the net external force acting on particle is
C
constant (non-zero).
(3) particle is always at rest.
(4) velocity of particle changes linearly with time. (1) Reading in the balance C is 13 N and angle  is
77. A block of weight W is supported by three strings as 67.4°
shown in figure. Which of the following relations is (2) Reading in the balance C is 13 N and angle  is
true for tension in the strings? 22.6°
(Here T1, T2 and T3 are the tension in the strings A, (3) Reading in the balance C is 5 N and angle  is
B and C respectively) 67.4°
(4) Reading in the balance C is 5 N and angle  is
22.6°

81. Three forces start acting simultaneously on a particle

(1) T1 = T2 (2) T1 = T3 moving with velocity v . These forces are represented
(3) T2 = T3 (4) T1 = T2 = T3 in magnitude and direction by the three sides of a
triangle ABC (as shown). The particle will now move
78. A body of weight 2 kg is suspended as shown in the
with velocity.
figure. The tension T1 in the horizontal string (in kg
wt) is:

(1) v remaining unchanged

(2) Less than v
2 3
(1) (2) (3) Greater than v
3 2
(4) in the direction of the largest force BC
(3) 2 3 (4) 2
[11]
82. A uniform rope lies on a table such that a part of it 86. Select the incorrect statements (s) from the following.
hangs down the table. When the length of the A. All Newton’s laws of motion hold good for
hanging part is one-fourth of the entire length, the both inertial and non-inertial frame of
rope just begins to slide. The coefficient of friction reference.
between the rope and the table is: (g = 10 m/s2) B. During explosion, linear momentum of the system
remains conserved.
2 1
(1) (2) C. Force of static friction is zero when no sliding
3 2
tendency.
1 1 (1) A only (2) B only
(3) (4)
3 6 (3) A and B (4) B and C

83. A small sphere is suspended by a string from the 87. A ball is suspended by a thread from the ceiling of
ceiling of a car. If the car begin to move with a a tram car. The brakes are applied and the speed of
g the car changes from 36 km h −1 to zero uniformly in
constant acceleration on a horizontal road, the
2 5 seconds. The angle by which ball deviates from
inclination of the string with the vertical is: the vertical is ( g = 10 ms−2 ) ;

(1) tan −1   in the direction of motion 1

1
(1) tan −1  
2  3

(2) tan −1   opposite to the direction of motion 1

1
(2) tan −1  
2 4
(3) tan–1 (2) in the direction of motion 1
(3) tan −1  
(4) tan–1 (2) opposite to the direction of motion 5
1
(4) tan −1  
84. A car is moving in a circular horizontal track of radius 2
10 m with a constant speed of 10 m/s. A bob is
suspended from the roof of the car by a light string of Topic-IV: Friction and its type, Dynamics of Circular
length 1.0 m. The angle made by the string with the Motion (Banking of Road, Conical Pendulum etc.)
vertical is: (Take g = 10 m/s2) 88. A block B is pushed momentarily along a horizontal
(1) 0 surface with an initial velocity v. If µ is the
 coefficient of sliding friction between B and the
(2) radian surface then, block B will come to rest after a time.
3

(3) radian
6

(4) radian v2 v
4 (1) (2)
g g

85. A block of mass m is placed on a smooth wedge of g g

(3) (4)
inclination . The whole system is accelerated v v
horizontally so that the block does not slip on the
89. The value of frictional force on block in the given
wedge. The normal force exerted by the wedge on
diagram is: (Take g = 10 m/s2)
the block (g is acceleration due to gravity) will be:
(1) mg sin  (2) mg

mg
(3) (4) mg cos  (1) 4 N (2) 5 N
cos  (3) 6 N (4) 9 N
[12]
90. If the coefficient of friction between the block of 94. In the figure shown, the coefficient of static friction
mass 5 kg and wall is 0.5, then minimum force F between the block A of mass 20 kg and horizontal table
required to hold the block with the wall is: is 0.2. What should be the minimum mass of hanging
2
(g = 10 m/s )
block just beyond which blocks start moving?

(1) 10 N (2) 100 N

(1) 2 kg (2) 3 kg
(3) 40 N (4) 50 N
(3) 4 kg (4) 5 kg

91. The maximum speed of car with which it can go

95. A bus turns a slippery road having coefficient of friction
around a level road of radius 10 m is: (coefficient of
friction between the road and tyre is 0.5). of 0.5 with speed of 10 m/s. The minimum radius of the
(g = 9.8 m/s2) arc in which bus turns is

(1) 19 m/s (2) 29 m/s [Take g = 10 m/s2]

(1) 4 m (2) 10 m
(3) 39 m/s (4) 49 m/s
(3) 15 m (4) 20 m

92. Assertion (A): On a rainy day, it is difficult to drive a

car or bus at high speed. 96. The time period of a conical pendulum having string
Reason (R): The value of coefficient of friction is length L and making an angle  with the vertical as
lowered due to wetting of the surface. shown in figure is: (g = acceleration due to gravity)
(1) Both Assertion (A) and Reason (R) are true and
Reason (R) is a correct explanation of
Assertion (A).
(2) Both Assertion (A) and Reason (R) are true but
Reason (R) is not a correct explanation of
Assertion (A).
(3) Assertion (A) is true and Reason (R) is false.
L
(4) Assertion (A) is false and Reason (R) is true. (1) T = 2  
g

93. A body is sliding down an inclined plane having  L cos  

(2) T = 2  
coefficient of friction 0.5. If the normal reaction is  g 
twice that of the resultant downward force along the
 L sin  
incline, the angle between the inclined plane and the (3) T = 2  
 g 
horizontal is:
(1) 15º (2) 30º  L tan  
(4) T = 2  
(3) 45º (4) 60º  g 
[13]
97. In a two block system, the lower block of mass 4 kg 101. A particle moves in a circular orbit under the action of
kept on a smooth surface is pulled horizontally by a a central attractive force which provide necessary
force of magnitude 5 N as shown. The friction force K
centripetal force. The attractive force is F =
(where
acting on the lower block will be: (coefficient of r
friction between the blocks (µ) = 0.5, g = 10 m/s2) K is constant and r is radius of circular orbit). The
speed of the particle is
(1) proportional to r2
(2) independent of r
(3) proportional to r
1
(1) 4 N (2) 5 N (4) proportional to
r
(3) 1 N (4) 2.5 N
102. A stone of mass 2 kg is tied to a string of length 0.5 m.
98. Assertion (A): It is easier to pull a moving body as If the breaking tension of the string is 900 N then the
compared to a stationary body. maximum angular velocity, the stone can have in
Reason(R): The kinetic friction acting on body is uniform circular motion is
less than limiting static friction. (1) 30 rad s-1 (2) 20 rad s-1
(1) Both Assertion (A) and Reason (R) are true and (3) 10 rad s-1 (4) 25 rad s-1

Reason (R) is a correct explanation of

Assertion (A). 103. A force F = iˆ + 4 ˆj acts on the block of mass 1 kg

(2) Both Assertion (A) and Reason (R) are true but placed on rough horizontal surface as shown in the
Reason (R) is not a correct explanation of figure. The force of friction (in newton) acting on the
block is: (Take g = 10 m/s2)
Assertion (A).
(3) Assertion (A) is true and Reason (R) is false.
(4) Assertion (A) is false and Reason (R) is true.

99. A biker is performing stunt in a cylindrical death

well of radius 16 m. If the speed of biker is constant
(1) – i (2) 18 i
and coefficient of friction of well’s surface is 0.2,
(3) –2.4 i (4) –3 i
find the minimum speed required to perform the
stunt safely. (Take g = 9.8 m/s2)
104. A block of mass 10 kg is in contact against the inner
(1) 24 m/s wall of a vertical hollow cylindrical drum of radius
(2) 28 m/s 1 m. The coefficient of friction between the block
(3) 16 m/s and the inner wall of the cylindrical drum is 0.1. The
(4) 49 m/s minimum angular velocity for the drum about its
axis to keep the block stationary will be:
100. The coefficient of static friction between two surfaces (g = 10 m/s2)
depends on (1) 10 rad/s
(1) The nature of both the surfaces in contact 10
(2) rad/s
(2) The shape of the surface in contact 2
(3) The area of contact between the surfaces (3) 10 rad/s
(4) All of above (4) 10  rad/s
[14]
105. A rough vertical board has an acceleration ‘a’ so that a 108. A block is placed on an inclined plane as shown in
2 kg block pressing against it does not fall. The the figure. For what value of  will the block slide,
coefficient of friction between the block and the board if the coefficient of static friction is 0.6?
 −1  3  
 tan   = 31
should be:
a  5 

2 kg

(1) 20° (2) 30°

g g (3) 45° (4) 15°
(1)  (2) 
a a
109. A force acting parallel to a rough incline plane is
a a
(3) = (4)  required to just move a body up the incline and it is
g 2g
double the force required to just prevent the body
from sliding down the plane. The coefficient of
106. In a conical pendulum, string makes an angle  = 45° friction is µ. The inclination  of the plane is:

with the vertical as shown in the figure. Mass m (1) tan–1(µ) (2) tan −1  
2
attached to the end of string is rotating in horizontal
(3) tan −1 (2) (4) tan −1 (3)
circle of radius 0.4 m with constant speed v. The
value of v is: (g = 10 m s–2) 110. In the figure shown mA = 10 kg, mB = 15 kg. The
maximum value of F, for which the blocks move
together is: (g = 10 m s–2)

(1) 0.4 m/s (2) 2 m/s

100 250
(3) 0.2 m/s (4) 4 m/s (1) N (2) N
3 3
25 125
(3) N (4) N
107. A particle describes a horizontal circle of radius r on 3 3
the smooth surface of an inverted cone as shown.
111. A block B is pushed momentarily along a horizontal
The height of plane of circle above vertex is h. The
surface with an initial velocity v. if  is the
speed of the particle should be: coefficient of kinetic friction between block B and
the surface, the block will come to rest after
covering a distance: (g is acceleration due to
gravity)

v2 g
(1) (2)
2 g v2
(1) rg (2) 2rg
v2 v 2
(3) (4)
(3) gh (4) 2gh g 2g

[15]
112. A cubical block rests on a horizontal plane of 116. A particle of mass m is moving with speed v = kt , in
 = 3. The angle through which the plane be horizontal circle of radius r (where k is constant and
inclined to the horizontal so that the block just t is time). The net force acting on particle at any time
begins to slide down will be: t will be:
(1) 30° (2) 45° k 2t 4
(3) 60° (4) 75° (1) mkrt (2) mk 1 +
r2
113. A gramophone record is revolving with an angular mk 2t 2 mkt
(3) (4)
velocity . A coin is placed at a distance r from the r2 r
centre of the record. The static coefficient of friction
is 0.5. The coin will revolve with the record without 117. A vehicle is moving on a track with constant speed
slipping if: (g is acceleration due to gravity) as shown in figure. The apparent weight of the
g 2 22 vehicle is: (Radius of curvature at B & C is same)
(1) r = (2) r 
2 g
g g
(3) r  (4) r 
2 2
22
(1) Maximum at A
114. A sports car is rounding a flat unbanked curve with (2) Maximum at B
radius of 230 m. The coefficient of friction between (3) Maximum at C
road and tyre is 0.96, the maximum speed at (4) Same at A, B and C
which driver can take the turn without sliding is
nearly; (g = 10 m/s2) 118. Two particles each of mass m are moving in
(1) 32 m/s (2) 42 m/s horizontal circle with same angular speed. If both
(3) 47 m/s (4) 52 m/s string 1 and 2 are of same length then the ratio of
T 
115. Two blocks A and B having masses 5 kg and 10 kg tension in string  1  is;
 T2 
respectively are placed one over other on a smooth
horizontal surface as shown in the diagram. The co-
efficient of friction between the blocks is 0.2. The
maximum force that should be applied on upper
block so that there is no relative motion between the
blocks will be; (g = 10 m/s2)


(1) (2) 3
2
(1) 20 N (2) 15 N 1
(3) 2 (4)
(3) 30 N (4) 60 N 3

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