Q.1.

When a bird of weight W sits on a stretched wire, the tension T in the wire is
Option 1 W
>
2
Option 2 =W
Option 3 <W
Option 4 None of these
Correct Answer 1
Explanation

Considering the equilibrium of bird

W
2T cosθ = W ⇒ T =
2cos θ
W
∴ cos θ < 1 ⇒ T >
2
So option (1) is correct.

Q.2. In the figure, the blocks A, B and C of mass m each have acceleration a1, a2 and a3
respectively. F1 and F2 are external forces of magnitudes 2 mg and mg respectively.

Option 1 a1 = a2 = a3
Option 2 a1 > a3 > a2
Option 3 a1 = a2 , a2 > a3
Option 4 a1 > a2 , a2 = a3
Correct Answer 2
Explanation

Case A Case B Case C

Fnet = F1 - mg Fnet = 2mg – mg Fnet = 2mg - mg
= 2 mg - mg = mg = mg
mg mg mg
⇒ a1 = =g ⇒ a2 = ⇒ a3 =
m 3m 2m
g g
⇒ a1 = g ⇒ a2 = ⇒ a3 =
3 2
∴ a1 > a3 > a2
So option (2) is correct.

Q.3. For the arrangement shown in the figure the tension in the string is given by

Option 1 mg
2
Option 2 mg
Option 3 3
mg
2
Option 4 2 mg
Correct Answer 2
Explanation

Acceleration of system
m g - 2m g sin30 0
a=
3m
mg -mg
=
3m
=0
∵ acceleration of system = 0

So option (2) is correct.

Q.4. A person swimming in a fresh water pool is obeying:

Option 1 Newton’s second law
Option 2 Gravitational law
Option 3 Newton’s third law
Option 4 Newton’s first law
Correct Answer 3
Explanation A person swimming moves forward due to reactionary force of water on person. It is
based on Newton’s third law.
So option (3) is correct.

Q.5. The passenger move forward when train stops, due to:
Option 1 Inertia of passenger
Option 2 Inertia train
Option 3 Gravitation pull by earth
Option 4 None of the above
Correct Answer 1
Explanation The passenger move forward when train stops due to inertia of motion of passenger.
So option (1) is correct.

Q.6. A mass of 3 kg descending vertically downward supports a mass of 2 kg by means of a

light string passing over a pulley. At the end of 5 s the string breaks. How much high
from now the 2 kg mass will go? (g = 9.8 m/s2)
Option 1 4.9 m
Option 2 9.8 m
Option 3 16.9 m
Option 4 2.45 m
Correct Answer 1
Explanation

Initially acceleration ‘a’ of the system is

3g - 2g
a=
3+2
g
⇒ a=
5
At 5s. V = u + at
g
⇒ v = 0 + × 5 ⇒ V = g = 9.8m / s
5
When string breaks at t = 5s, the mass 2 kg mores under gravity. So height realted can
be calculated as
2
V 2 ( 9.8 ) 9.8
h= = = = 4.9m
2g 2 × 9.8 2
h = 4.9 m
so option (1) is correct

Q.7. A man of mass 60 kg is standing on a horizontal conveyor belt. When the belt is given
an acceleration of 1 ms-2, the man remains stationary with respect to the moving belt.
 If g = 10 ms-2, the net force acting on the man in:

Option 1 Zero
Option 2 120 N
Option 3 60 N
Option 4 600 N
Correct Answer 3
Explanation ∵ man is stationary w.r.t. conveyor belt
∵ acceleration of man = /m/s2.
So net force acting on the man is
Fnet = mass × acceleration = 60 × 1 = 60N.

Q.8. In the arrangement shown, if the surface is smooth, the acceleration of the block m2
will be

Option 1 m2g
4m 1 + m 2
Option 2 2m 2 g
4m 1 + m 2
Option 3 2m 2 g
m 1 + 4m 2
Option 4 2m 1 g
m1 + m2
Correct Answer 1
Explanation

→ →
∵ T⋅ a =O
⇒ Ta1 - 2Ta2 = 0
⇒ a1 = 2a2 –-(1)
For m1 block
T = m1a1 -–--(2)
For m2 block
 m2g - 2T = m2a2 ----(3)
put (1) and (2) in (3)
m2g - 2m1 (2a2) = m2a2
⇒ m2g = (m2 + 4m1) a2
m2g
⇒ a2 =
m 2 + 4m 1

Q.9. Two masses are connected by a string which passes over a pulley accelerating upward
at a rate as shown. If a1 and a2 be the accelerations of bodies 1 and 2 respectively,
then:

Option 1 A = a1 - a2
Option 2 A = a1 + a2
Option 3 a -a
A= 1 2
2
Option 4 a1 + a2
A=
2
Correct Answer 3
Explanation

→ →
a 1 /pulley = - a 2 /pulley
→ → → →
a1 -ap =  a2 -ap 
 
→ → → →
a1 -ap =a2 +ap
→ → →
⇒ a1 + a2 = -2ap
→ →
→a +a
⇒ ap = 1 2
2
a1 -a2
⇒ A=
2
upward direction = +ve
downward = -ve
so option (3) is correct.
Q.10. A block of mass m is resting on a wedge of angle θ as shown in the figure. The wedge
is given an acceleration a. What is the value of a so that the mass m falls freely

Option 1 g
Option 2 g cos θ
Option 3 g cot θ
Option 4 g tan θ
Correct Answer 3
Explanation Mass m falls freely when normal reaction N = O

When block loose contact

N = O ⇒ ma sin θ =mg cos θ
⇒ a = gcot θ
So option (3) is corret

Q.11. A light string going over a clamped pulley of mass m supports a block of mass as shown
in the figure. The force on the pulley by the clamp is given by :

Option 1 2Mg
Option 2 2mg
Option 3 g ( M + m )2 + m 2
Option 4 g ( M + m )2 + M 2
Correct Answer 4
Explanation

For equilibrium of M block

T = Mg ---(1)
For equilibrium of pulley

→ 2
R = ( Mg )2 + {( M+m) g}
Net force of clamp will balance R’
2
R = R ' = M 2 + (M + m ) g
So option (4) is correct.

Q.12. Tension in the cable supporting an elevator, is equal to the weight of the elevator.
From this, we can conclude that the elevator is going up or down with a
Option 1 Uniform speed
Option 2 Uniform acceleration
Option 3 Variable acceleration
Option 4 Either (b) and (c).
Correct Answer 1
Explanation

∵ T = mg ⇒a = O
∴ Elevator is moving up or down with uniform speed.
So option (a) is correct.
Q.13. A body of mass m is suspended by two string making angles α and β with the
horizontal. Tensions in the two strings are

Option 1 m g cos β
T1 = = T2
sin ( α + β )
Option 2 m g sin β
T1 = = T2
sin ( α + β )
Option 3 mg cos β mg cos α
T1 = , T2 =
sin ( α + β ) sin ( α + β )
Option 4 None of these
Correct Answer 1
Explanation

From equilibrium of block

T1 sin β + T1 sin α = mg − − (1)
T1 cos α = T2 cos β − − (2)
T cos α
T2 = 1
cos β
cos α
T1 = sin β + T1 sin α = mg
cos β
m g cos β
⇒ T1 =
sin ( α + β )
Similarly
mg cos α
T2 =
sin ( α + β )

Q.14. An automobile enters a turn whose radius is R. The road is banked at angle θ . Friction
is negligible between the wheels of the automobile and road. Mass of the automobile
 is m and speed is u. Select the correct alternative:
Option 1 Net force on the automobile is zero
Option 2 Normal reaction on the automobile is mg cos θ
Option 3 Normal reaction on the automobile is mg sec θ
Option 4 None of the above
Correct Answer 3
Explanation

Net force = ma
mu2
=
R
Towards the center of the circle
Also Ncos θ = mg
mg
⇒ N=
cos θ
⇒ N = mg secθ
So option (2) is correct.

Q.15. A ball is suspended by a thread from the ceiling of a car. The brakes are applied and
the speed of the car changes uniformly from 10 m/s to zero in 5s. The angle by which
the ball deviates from the vertical (g = 10 m/s2) is:
Option 1 1
tan -1  
3
Option 2 1
sin -1  
 5
Option 3 1
tan -1  
 5
Option 4 1
cot -1  
 3
Correct Answer 3
Explanation Applying v = u + at
V = 0, u = 10m/s, t = 5s
O=10 × a × 5 ⇒ a=-2m/s2

For equilibrium of ball

T cos θ = mg
and Tsin θ =ma
 T sin θ m a a
⇒ = ⇒ tan θ =
T cos θ m g g
2 1 1
tan θ = = ⇒ θ = tan -1  
10 5 5

Q.17. A sphere of mass m is held between two smooth inclined walls. For sin 370 = 3/5, the
normal reaction of the wall (2) is equal to:

Option 1 16 mg
25
Option 2 25 mg
21
Option 3 39 mg
25
Option 4 Mg
Correct Answer 4
Explanation

From equilibrium of sphere.

From equilibrium of sphere.

N1cos370 = N2 cos 740 + mg –(1)

N1 sin 370 = N2 sin 740 ---(2)
2 2
 4   3
cos740 = cos 2 × 370 = cos2 370 - sin2 370 =   -  
 5  5
16 16 7
⇒ cos 74 0 = - =
25 25 25
3 4 24
sin74 0 = 2sin370 cos370 = 2 × × =
5 5 25
 4 7
N1 × = N2 × + mg in eqn − − (1)
5 25
3 24 8
N1 × = N2 × ⇒ N1 = N2
5 25 5
8 4 7  32 7 
∴ N2 × = N2 × + mg ⇒  -  N2 = m g
5 5 25  25 25 
8
⇒ N2 = mg ∴ N1 = mg
5
So option (4) is correct.

Q.18. Three masses of 1 kg, 6 kg and 3 kg are connected to each other with threads and
placed on a table as shown in figure. If g = 10 ms-2, the acceleration with which the
system is moving is

Option 1 Zero
Option 2 1 ms-2
Option 3 2 ms-2
Option 4 3 ms-2
Correct Answer 3
Explanation

Fnet on system
= 3g - 1g
= 2g
Mass of system = (1 + 6 + 3)kg
F 2g
Acceleration of system = net = = 2m / s2
M 10
Option (3) is correct.

Q.19. The engine of a car produces acceleration of 4 m/s2 in car. If this car pulls another car
of same mass, what will be the acceleration produced?
Option 1 8 m/s2
Option 2 2 m/s2
Option 3 4 m/s2
Option 4 1
m / s2
2
Correct Answer 2
Explanation Force of engine = µngt = F = m1a1 = m2a2
⇒ m × 4 = ( 2m) × a ⇒ a= 2m/ s2
So option (2) is correct.
Q.20. All surfaces shown in figure are smooth. System is released with the spring
unstretched. In equilibrium, compression in the spring will be:

Option 1 mg
2k
Option 2 2 mg
k
Option 3 (M+m) g
2k
Option 4 mg
k
Correct Answer 4
Explanation

Considering equilibrium of rod

N1
= mg
2
N1
= N2
2
Considering equilibrium of wedge.

N1
= kx
2
N1 mg
⇒ x= =
2k k
So option (4) is correct.

Q.21. Three identical blocks are suspended on two identical springs one below the other as
shown in figure. If thread is cut that supports block 1, then initially (choose one
alternative only):
Option 1 The second block falls with zero acceleration
Option 2 The first block falls with maximum acceleration
Option 3 Both (a) and (b) are wrong
Option 4 Both (a) and (b) are correct
Correct Answer 2
Explanation

From F.B.D. of blocks (1), (2) and (3) in equilibrium

T1 = mg + kx3
kx3 = mg + kx2 = 2mg
kx2 = mg
when string is cut T1 = 0 but spring force will not change instantaneously.
∴ block (2) and (3) will have acceleration = 0
and block (1) will passes maximum acceleration

mg + mg
⇒ a1 = ⇒ a1 = 2g
m
∴ So option (2) is correct.

Q.22. A uniform rope of mass m hangs freely from a ceiling. A bird of mass M climbs up the
rope with an acceleration a. The force exerted by the rope on the ceiling is:

Option 1 Ma + mg
Option 2 M (a + g) + mg
Option 3 M (a + g)
Option 4 Dependent on the position of bird on the rope
Correct Answer 2
Explanation

When bird climbs up the rope with acceleration a then

T - Mg = Ma ⇒ T = M(g + a)
The force exerted by force on the ceiling is = T + mg

= M(g + a) + mg

Q.23. Two masses m and M are attached with strings as shown. For the system to be in
equilibrium we have:

Option 1 2M
tan θ = 1 +
m
Option 2 2m
tan θ = 1 +
M
Option 3 M
tan θ = 1 +
2m
Option 4 m
tan θ = 1 +
2M
Correct Answer 1
Explanation

For equilibrium at A
T3 = mg --(1)
T1 cos 450 = T2 cos 450 ⇒ T1 = T2 ---(2)
T1 sin 450 + T2 sin 450 = T3 ---(3)
 By using (1) and (2) in (3)
T1 T2 2T1 mg
+ = mg ⇒ = mg ⇒ T1 =
2 2 2 2
mg
∴ T2 =
2
Consider F.B.D. of junction B in equilibrium

T2 mg
Tu sin θ = Mg + = Mg +
2 2
T2 mg
Ta cos θ = =
2 2
 m
M+ g
Tacosθ  2
∴ dividing =
Tacosθ g
m
2
2M + m
⇒ tan θ =
m
2M
⇒ tan θ = 1 +
m
So option (1) is correct.

Q.24. In order to raise a mass of 100 kg a man of mass 60 kg fastens a rope to it and passes
the rope over a smooth pulley. He climbs the rope with an acceleration 5g/4 relative to
rope. The tension in the rope is
Option 1 1432 N
Option 2 928 N
Option 3 1218 N
Option 4 642 N
Correct Answer 3
Explanation 1218 N

Q.25. Two blocks of masses 5 kg and 3 kg are attached to the ends of a string passing over a
smooth pulley fixed to the ceiling of an elevator. A man inside the elevator accelerated
9
upwards, finds the acceleration of the blocks to be g. The acceleration of the
32
elevator is:
Option 1 g
3
Option 2 g
4
Option 3 g
8
Option 4 g
6
Correct Answer 3
Explanation

If acceleration of elevator is a upwards then w.r.t. man in lift the F.B.D. of blocks is as
shown below.

So relative acceleration of the system is

5( g+a) -3( g+a)
arel =
( 5+3)
2 ( g + a)
=
8
9 9g 2 ( g + a )
∴ arel = g ⇒ =
32 32 8
9g 9g g
⇒ = g+a ⇒ a= -g =
8 8 8
g
⇒ a=
8
So option (3) is correct.

Q.26. A man has weight 80 N. He stands on a weighing scale in a lift which is moving
upwards with a uniform acceleration of 5 m/s2. What would be the reading on the
scale? (g = 10 m/s2)
Option 1 800 N
Option 2 120 N
Option 3 Zero
Option 4 400 N
Correct Answer 2
Explanation

The reading of scale is due to normal reaction between man and machine. So from
F.B.D. of man N - mg = ma ⇒ N = m (g + a)
80
N = mg + ma = 80 + × 5 = 120N
10
⇒ reading on scale = 120 N

Q.27. A trolley is accelerating down an incline of angle θ with acceleration g sin θ . Which of
the following is not correct? ( α is angle made by the string with vertical).

Option 1 α=θ
Option 2 α = 00
Option 3 Tension in the string T = mg cos θ
Option 4 All of the above.
Correct Answer 2
Explanation

w.r.t. point O i.e. the point of suspatation F.B.D. of mass ‘m’ is

so in equilibrium condition tension T balances mg cos θ component and string

becomes perpendicular to the incline

so, T = mg cos θ
α=θ
So option (2) is incorrect.
Q.28. A trolley car slides down a smooth inclined plane of angle of inclination θ . If a body is
suspended form the roof or the trolley car by an inextensible string, the corresponding
tension in the string will be:
Option 1 mg
Option 2 mg cos θ
Option 3 mg sinθθ
Option 4 Zero
Correct Answer 2
Explanation

In equilibrium T = mg cos θ
So option (2) is correct.

Q.29. A mortor cycle and a car are moving on a horizontal road with the same velocity. If
they are brought to rest by the application of brakes, which provided equal
retardation, then
Option 1 A motor cycle will stop at shorter distance
Option 2 Car will stop at a shorter distance
Option 3 Both will stop at the same distance
Option 4 Nothing can be predicted.
Correct Answer 3
Explanation As slopping distance due to retardation ‘a’ is calenlated V2 = u2 + 2a s
As V = 0, u = +u; retardation = -a
O2 = u2 - 2as
a2
⇒ S=
2a
As u and a are same so S is same.
So option (3) is correct.

Q.30. A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10
m/s. A plumb bob is suspended from the roof of the car by a light rigid rod. The angle
made by the bob with the vertical is (g = 10 m/s2)
Option 1 Zero
Option 2 300
Option 3 450
Option 4 600
Correct Answer 3
Explanation

For equilibrium of bob of mass ‘m’

mv 2
T sin θ = and T cos θ = mg
R
Tsinθ mv2 v2
= ⇒ tanθ =
Tcos θ R mg Rg

tan θ =
( 10 ) 2 = 1 ⇒ θ = 450
10 × 10
So option (3) is correct.

Q.31. A block slides down an inclined plane of slope of angle θ with a constant velocity. It is
then projected up the plane with an initial velocity u. The distance up to which it will
rise before coming to rest is:
Option 1 u2
θ
4 gsinθ
Option 2 u2
2g
Option 3 u2 sin2 θ
2g
Option 4 usin θ
2g
Correct Answer 1
Explanation

When block slides down the incline with constant velocity ⇒ acceleration = 0
Then mg sin θ component is balanced by kinetic friction fk
∴ mg sin θ = fk

When block is projected up the incline then retardation

mg sin θ + fk
a=
m
mg sin θ + mg sin θ
=
m
= 2g sin θ
 u2 u2
∴ slopping distance S = =
2a 4gsinθ
So option (1) is correct.

Q.32. The time taken by a body to slide down a rough 450 incline plane is twice that required
to slide down a smooth 450 incline plane. The coefficient of kinetic friction between
the object and rough plane is given by
Option 1 1
3
Option 2 3
4
Option 3 3
4
Option 4 4
3
Correct Answer 2
Explanation

g µg
a1 = -
2 2

While sliding from rese time taken to over distance down the incline is
2
t=
a
t1 a2 g/ 2
∴ = =
t2 a1 g 2 (1 - µ )
2t2 1 1 3
⇒ = ⇒ 1- µ = ⇒ µ =
t2 1- µ 4 4
So option (2) is correct.

Q.33. Two blocks are connected over a massless pulley as shown in figure. The mass of block
A is 10 kg and the coefficient of kinetic friction is 0.2. Block A slides down the incline at
constant speed. The mass of block B in Kg is

Option 1 3.5
Option 2 >2.5
Option 3 3.3
Option 4 3.0
Correct Answer 3
Explanation

∵ System moves constant speed ⇒ acceleration = 0

0
(
⇒ 10g sin 30 µ 10g cos 30 0
) = Mg
1 3
⇒ 10 × - 0.2 × 10 × =M
2 2
⇒5- 3 =M
⇒ M = 5 - 1.732 = 3.3 kg
⇒ M = 3.3 kg
So option 3 is correct.

Q.34. A block of mass m is kept on an inclined plane of a lift moving down with acceleration
of 2 m/s-2. What should be the coefficient of friction to let the block move down with
constant velocity relative to lift:

Option 1 1
µ=
3
Option 2 µ = 0.4
Option 3 µ = 0.8
Option 4 3
µ=
2
Correct Answer 1
Explanation

w.r.t the lift

 Considering equilibrium of block w.r.t edge.
3 3
N + ma =mg
2 2
3
N = ( mg - ma )
2
∵ block slides with uniform velocity

⇒ mg sin300=masin 300 + µN
1 1 3
⇒ mg = ma + µ ( mg - ma )
2 2 2
10 2 3
∵a= 2 ⇒ = + µ ( 10 - 2 )
2 2 2
1
⇒ 5=1+µ ×4 3 ⇒ µ =
3
So option (a) is correct.

Q.35. Block A of mass m rests on the plank B of mass 3m which is free to slide on a
frictionless horizontal surface. The coefficient of friction between the block and plank
is 0.2. If a horizontal force of magnitude 2 mg is applied to the plank B, the
acceleration of A relative to the plank and relative to the ground respectively, are:

Option 1 g
0,
2
Option 2 2g
0,
3
Option 3 3g g
,
5 5
Option 4 g 2g
,
5 5
Correct Answer 4
Explanation
 Normal reaction between A and B is
N = mg ⇒ flim = µ N
⇒ flim = (0.2) mg ---(1)
To find the force required at which the block starts slipping
F
F = ( 4m ) a ⇒ a = − − − (2)
4m

for stock A fs = ma
F F
⇒ fs = m ⇒ fs = − − − (2)
4m 4
F
∵ fs ≤ flim ⇒ ≤ ( 0.2 ) mg
4
⇒ F ≤ 0.8 mg
∴ 2 mg is greater than this value so slipping will start between the blocks A and B.
So now F.B.D. of block A and block B

fk g
acceleration of a A = = 0.2g =
m 2
2mg - fk
acceleration of aB =
3m
g
2mg -
= 5 = 9g = 3g
3 15 5
g 3g -2g
relative acceleration of aw.r.t.b = - =
5 5 5

Q.36. A body is moving down along an inclined plane of angle of inclination θ . The coefficient
of friction between the body and the plane varies as µ= 0.5x, where x is the distance
moved down the plane. The body will have the maximum velocity when it has
travelled a distance x given by:
Option 1 x = 2 tan θ
Option 2 2
x=
tan θ
Option 3 x = 2 cot θ
Option 4 2
x=
cot θ
Correct Answer 1
Explanation

Normal reaction
N = mg cos θ
∴ kinetic friction
fk= µN= ( 0.5x) mgcosθ
∴ acceleration
mg sin θ - fk
a=
m
mg sin θ - ( 0.5x ) mg cos θ
=
m
xcos θ g
a = g sin θ -
2
dN
∵a = ν
dx

dN
For maximum velocity =0 ⇒ a = 0
dx
x cos θ g sin θ
∴ g sin θ = ⇒ x=2
2 cos θ
⇒ x = 2tanθ
So option (a) is correct.

Q.37. In the shown arrangement mass of A = 1 kg, mass of B = 2 kg. Coefficient of friction
between A and B = 0.2. There is no friction between B and ground. The frictional force
exerted by A on B equals:

Option 1 2N
Option 2 3N
Option 3 4N
Option 4 5N
Correct Answer 1
Explanation

When the system accelerates with common acceleration ‘a’

 F
F = 3a ⇒ a =
3
Fo r 1 kg F
⇒ fs =
fs = 1 × a 3

∵ fs ≤ flim
Common acceleration = a
flim = 0.2 N = 0.2 mg
F
∵ ≤ 0.2 × 1 × 10 ⇒ F ≤ 6N ∵ maximum
3
Force F possible without slipping is 6 N.
So slipping wire take place and fk = µN = 2N

Q.38. A body of mass 60 kg is dragged along a horizontal surface by a horizontal force which
is just sufficient to start the motion of the body from rest. If the coefficients of static
and kinetic friction are 0.5 and 0.4 respectively, the acceleration of the body is
Option 1 0.98 m/s2
Option 2 9.8 m/s2
Option 3 0.54 m/s2
Option 4 5.292 m/s2
Correct Answer 1
Explanation ∴ F = flim = µ s N = 0.5 × 60 × g
⇒ F = 300 N
Also kinetic friction is fk = µ kN
⇒ fk = 0.4 × 60 × g = 240 N

F-fk
acceleration of block is a =
m
300 - 240
=
60
60
=
60
= 1 m/s2
If g = 9.8 m/s2 then a = 0.98 m/s2
so option (a) is correct.

Q.39. Two blocks of mass 4 kg and 2 kg are connected by a heavy string and placed on rough
horizontal plane. The 2 kg block is pulled with a constant force F. The coefficient of
friction between the blocks and the ground is 0.5. What is the value of F so that
tension in the string is constant throughout during the motion of the blocks? (g = 10
m/s2)

Option 1 40 N
Option 2 30 n
Option 3 50 N
Option 4 60 N
Correct Answer 2
Explanation

From F.B.D. of system N1 = 4g and N2 = 2g

Tension is constant when system moves with uniform velocity.
∴ F = µ N1 + µ N 2
for system of blocks
F = µ( 4g) + µ( 2g) = µ × 6g =0.5 × 6 × 10 =30N
F = 30N
So, option (2) is correct.

Q.40. In the figure, MA = 2 kg, For what minimum value of F, A starts slipping over B? (g = 10
m/s2)

Option 1 24 N
Option 2 36 N
Option 3 12 N
Option 4 20 N
Correct Answer 2
Explanation Let the system moves with common acceleration a then

∴F - fk = 6a
F - 24 = 6a
F
⇒ a= - 4
6
∴ A moves due to static friction fs m
 ∴ For A
fs = 2a
F 
⇒ fs = 2  - 4 
6 
F
⇒ fs = -8
5
F
∵ fs ≤ ( flim )1 ⇒ -8 ≤ 4
3
F
⇒ ≤ 12 ⇒ F ≤ 36N
3
∴ slipping will not occur uptp 36 N.
∴ minimum force required for A slipping over B is 36 N.
So option (2) is correct.

Q.41. A block of mass m is placed on another block of mass M which itself is lying on a
horizontal surface. The coefficient of friction between two blocks is µ 1 and that
between the block of mass M and horizontal surface is µ2 . What maximum horizontal
force can be applied to the lower block, so that the two blocks move without
separation?

Option 1 (M+m) ( µ2 -µ1 ) g

Option 2 (M-m) ( µ2 -µ1 ) g
Option 3 (M-m) ( µ2 +µ1 ) g
Option 4 (M+m) ( µ2 +µ1 ) g
Correct Answer 4
Explanation

∴ Normal reaction on upper and lower block is N1 = mg and N2 (m + M)g respectively.

∴ limiting friction on upper and lower blocks are µ 1 N1 and µ 2 N2
 when system moves with common acceleration ‘a’ then

fk= µ2 ( m+M) g
F - µ2 ( m + M) g
a=
( m + M)
So static friction will accelerate m

fs = ma
 F -µ 2 (m + M) g 
⇒ fs = m 
 ( m + M ) 
 F - µ 2 (m + M) g 
∵ fs ≤ flim ⇒ m  ≤µ mg
 ( m + M )  1
F - µ2 ( m+M) g ≤µ1 ( m+M) g
F ≤ ( µ1 + µ2 ) ( m+M) g
∴maximumforceF =( µ1 + µ2 ) ( m+M) g
So option (4) is correct

Q.42. In the arrangement shown in figure, there is a friction force between the blocks of
masses m and 2m. The mass of the suspended block is m. The block of mass m is
stationary with respect to block of mass 2 m. The minimum value of coefficient of
friction between m and 2 m is:

Option 1 1
2
Option 2 1
2
Option 3 1
4
Option 4 1
3
Correct Answer 3
Explanation
 From F.B.D. of ‘m’ mg - 7 = ma ---(1)
From F.B.D. of system of (m + M)
T = 3ma ---(2)
By (1) + (2) ⇒ mg - T + T = 3ma + ma
⇒ mg = 4 ma
g
⇒ a=
4

For system of m and 2m moving with common acceleration ‘a’ static friction fs will
provide necessary force on m ∴ fs = ma
mg
⇒ fs =
4
and N = mg
∵ fs ≤ µ s N ⇒ fs ≤ µ s mg
mg 1
⇒ ≤ µ s mg ⇒ µ s ≥
4 4
1
∴ minimum µ s =
4

Q.43. The rear side of a truck is open and a box of mass 20 kg is placed on the truck 4 m
away from the open end. µ =0.15 and g = 10 m/s2. The truck starts from rest with an
acceleration of 2 m/s2 on a straight road. The box will fall off the truck when it is at a
distance from the starting point equal to:
Option 1 4m
Option 2 8m
Option 3 16 m
Option 4 32 m
Correct Answer 3
Explanation Let the box falls from the rear of truck at t s. with respect to truck D.B.D. is as shown
∴ relative acceleration of box in

20a- µ mg
arel =
20
20 × 2 - 0.15 × 20 × 10
=
20
= 0.5
So distance travelled by truck in
1 1
t = 4s = at 2 = × 2 × 4 2 = 16m
2 2
1 2 1
By s = at+ at ⇒ 4 = × 0.5 t 2
2 2

Q.44. two blocks A and B are pushed against the wall with the force F. The wall is smooth but
 the surface in contact of A and B are rough. Which of the following is true for the
system of the blocks?

Option 1 F should be equal to weight of A and B.

Option 2 F should be less than the weight of A and B.
Option 3 F should be more than the weight of A and B
Option 4 The system cannot be in equilibrium (at rest).
Correct Answer 4
Explanation

As wall is smooth no force is present to balance the weight in vertical direction.

So system cannot be in equilibrium.

Q.45. A block of mass M lies on a rough surface of coefficient of friction µ A force is applied
on it an angle θ to the horizontal as shown, and the block is at rest. The frictional force
acting on the block will be

Option 1 F cos θ
Option 2 µ ( mg +F sinθ)
Option 3 µ( mg-Fsinθ)
Option 4 µmg
Correct Answer 1
Explanation

∵ block at rest ⇒ force are balanced

 ∴ fs = Fcosθ
is the fictional force.

Q.46. A block of mass 3 kg is at rest on a rough inclined plane as shown in the figure. The
magnitude of net force exerted by surface on the block will be

Option 1 26 N
Option 2 19.5 N
Option 3 10 N
Option 4 30 N
Correct Answer 4
Explanation As 3 kg block is at rest on the incline
∴ net force exerted by surface on the block will balance weight mg.
Then

  
Fnet =N+fs

Fnet = 3g
=3 × 10 = 30N

Q.47. For the arrangement shown in the figure the tension in the string is

Option 1 6N
Option 2 6.4 N
Option 3 0.4 N
Option 4 zero
Correct Answer 4
Explanation

∴ limiting firction
flim = µ s N
=0.8 × 1g × cos370
4
= 0.8 × 1 × 10 ×
5
32
= = 6.4N
5
∴ Component of weight down the incline
3
= 1g sin 37 0 = 10 × = 6N
5
∴ flim = 6.4 N ∴ static friction can along balance the weight component. So tension in
the string will remain zero.

Q.48. The force required to just move a body up the inclined plane is double the force
required to just prevent the body from sliding down the plane. The coefficient of
friction is µ. The inclination θ of the plane is
Option 1 tan-1 µ
Option 2 µ
tan-1  
2
Option 3 tan-1 ( 2µ
µ)
Option 4 tan-1 ( 3µ
µ)
Correct Answer 4
Explanation To just move a body up the inclined plane

∵flim = µN= µ ( mg cosθ)

F1 = mg sin θ + flim
F1 = mg sin θ + µ mg cos θ − − − (1)
To just prevent the body from sliding down.

F2 + flim = mg sin θ
F2 + µ mg cos θ = mg sin θ
 F2 = mg sin θ - µ mg cos θ − − − (2)
Given F1 = 2 F2
⇒ mg sinθ + µ mg cosθ =2( mg sinθ - µ mg cosθ)
sin θ + µ cos θ = 2 sin θ - µ cos θ
⇒ sin θ =2µ cos θ ⇒ tan θ = µ
⇒ θ =tan-1 ( 2µ )

Q.49. A body of mass m rests on horizontal surface. The coefficient of friction between the
body and the surface is µ. If the mass is pulled by a force P as shown in the figure, the
limiting friction between body and surface will be:

Option 1 µ mg
Option 2   P 
µ mg +   
  2 
Option 3   P 
µ mg -   
  2 
Option 4   3P  
µ mg -  
  2  
Correct Answer 3
Explanation

∵ flim = µ N
P P
or N + = mg ⇒ N = mg -
2 2
So limiting friction
flim = µ N

Q.50. A rough vertical board has an acceleration a along the horizontal, so that a block of
mass M pressing against it does not fall. The coefficient of friction between block and
the board is:
Option 1 a
>
g
Option 2 g
<
a
Option 3 a
<
g
Option 4 g
>
a
Correct Answer 4
Explanation

∵ M is atrest w.r.e. vertical board.

⇒ N = Ma or fs = Mg
∵ fs ≤ flim ⇒ fs ≤ µ sN
g
⇒ Mg ≤ µ s Ma ⇒ µ s ≥
a

Q.51. A block of weight 5N is pushed against a vertical wall by a force of 12 N. The coefficient
of friction between the wall and the block is 0.6. the magnitude of total force exerted
by the wall on the block is

Option 1 7.2 N
Option 2 5N
Option 3 12 N
Option 4 13 N
Correct Answer 4
Explanation

According to F.B.D. of block

N = F = 12 N
∴ flim = µ N = 0.6 × 12 =7.2N

As the weight of block is 5 N

∴ Static friction will balance this out.
Hence total force exerted by wall.
→ → →
R = fs + N ⇒ R = (12) 2 + ( 5) 2
→ → →
R = fs + N ⇒ R = (12) 2 + ( 5) 2
Q.52. A uniform chain of length l is placed on a rough table with length l/n ( n > 1) hanging
over the edge. If the chain just begins to slide to off the table by itself from this
position the coefficient of friction between the chain and the table is
Option 1 1
n
Option 2 1
n -1
Option 3 1
n+1
Option 4 n-1
n+1
Correct Answer 2
Explanation

let the mass per unit length of chain = λ

∴ Mass kept on table
 1
m 1 = λ (  )  1- 
 n

mass surpended = m2 = λ
n
∵ system in limiting equilibrium, so it is equivalent to selor system.

∴ m2g = flim
m2 g = µ N
m 2 g = µ m1 g
m2
µ=
m1
λ /n 1×n
⇒ µ= ⇒ µ=
 1 n(n - 1 )
λ   1- 
 n 
1
⇒ µ=
n-1

Q.53. Two masses A and B of 7 kg and 3 kg respectively are connected with a string passing
over a frictionless pulley fixed at the corner of table as shown in the figure. The
coefficient of friction between A and horizontal surface is 0.3. The minimum mass of C
that may be placed on A to prevent it from moving is equal to
Option 1 15 kg
Option 2 10 kg
Option 3 5 kg
Option 4 3 kg
Correct Answer 4
Explanation

Considering the equilibrium of system

3g = fs ---(1)
Also normal reaction on A from horizontal surface as 3g = T = fs
N = (7 + m)g ---(2)
∵ fs ≤ flim ⇒ fs ≤ µ s N
⇒3g ≤µs ( 7+m) g
3
⇒ µs ≥
7+m
3
⇒ 0.3 ≥ ⇒ 7 + m ≥ 10 ⇒ m ≥ 3
7+m
⇒ m ≥3
∴ minimum mass of C = 3 kg.

Q.54. If µ is coefficient of friction between the tyres and road, then the minimum stopping
distance for a car of mass m moving with velocity V is
Option 1 µ Vg
Option 2 V2
2µ g
Option 3 V2µ g
Option 4 V
µ
2g
Correct Answer 2
Explanation

The retardation of car is due to kinetic friction fk

 fk µ mg
⇒ a= = ⇒ a = µg
m m
∵V2= µ2+ 2as ⇒ 02=- 2 µg × s
µ2
⇒ S= is the stopping distance.
µ 2g

Q.55. 2
A parabolic bowl with its bottom at origin has the shape y= x . Here, x and y are in
20
metres. The maximum height at which a small mass m can be placed on the bowl
without slipping (coefficient of static friction is 0.5) is:

Option 1 2.5 m
Option 2 1.25 m
Option 3 1.0 m
Option 4 4.0 m
Correct Answer 2
Explanation 1.25 m

Q.56. the system is pushed by a force F as shown in figure All surfaces are smooth except
between B and C. Friction coefficient between B and C is µ. Minimum value of F to
prevent block B from downward slipping is

Option 1  3 
  mg
 2µ 
Option 2  5
 2µ  mg
 
Option 3 5
 2  µ mg
 
Option 4 3
 2  µ mg
 
Correct Answer 2
Explanation  5
 2µ  mg
 

Q.57. A homogeneous chain of length L lies on a table. The coefficient of friction between
the chain and the table is µ. The maximum length which can hang over the table in
equilibrium is
Option 1  µ 
 L
 µ +1 
Option 2  1- µ 
 L
 µ 
Option 3  1- µ 
 L
 1+ µ 
Option 4  2µ 
 L
 2µ +1 
Correct Answer 1
Explanation

Let the maximum length which can hang over the table in equilibrium is x
Let mass per unit length = λ

For limiting equilibrium of chain

( λ x) g = µN
( λx) g= µλ(L - x) g
x = µL - µ x
⇒ x ( 1 + µ) = µL
µL
⇒ x=
1+µ

Q.58. A wedge of mass 2m and a cube of mass m are shown in figure. Between cube and
wedge, there is no friction. The minimum coefficient of friction between wedge and
ground so that wedge does not move is

Option 1 0.10
Option 2 0.20
Option 3 0.25
Option 4 0.50
Correct Answer 2
Explanation

So cube of mass m will slide down the incline

So N = mg cos θ ---(1)

From equilibrium of wedge

N' = Ncos θ + 2mg
= ( mgcosθ) cosθ + 2mg
=mgcos2θ + 2mg
= mg cos2 450 + 2mg
2
 1 
= mg   + 2mg
 2
mg 5mg
= + 2mg =
2 2
5mg
⇒ N' = − − (2)
2
Also,

fs = N sin θ
= = mg cos θ sin θ
1 1 mg
fs = mg × = ---(3)
2 2 2
∵ fs ≤ flim ⇒ fs ≤ µ N
mg 5mg 1
⇒ ≤µ ⇒µ≥
2 2 5
⇒ µ m in = 0.2

Q.59. During paddling of a bicyle, the force of friction exerted by the ground on the two
wheels is such that it acts.
Option 1 In the backward direction on the front wheel and in the forward direction on the rear
wheel
Option 2 in the forward direction on the front wheel and in the backward direction on the rear
wheel
Option 3 in the backward direction on both the front and the rear wheel
Option 4 in the forward direction on both the front and the rear wheels
Correct Answer 1
Explanation In a bicycle, the rear wheel moves with muscle force. In forward motion of the cycle,
the wheel moves such that any point on the wheel goes towards backward direction.
 In the case of real wheel, the direction of force of friction will be opposite to the
direction of motion of any point M the wheel subbing the ground i.e. direction of force
of friction will be in forward direction. The foreard wheel moves because of friction
itself so direction of force of friction is same as that of motion of any point on the
wheel subbing the froung i.e. backward direction.

Q.60. A block of mass 0.1 kg is held atainst a wall by applying a horizontal force of 5 N on the
block. If the coefficient of friction between the block and the wall is 0.5, the magnitude
of the frictional force acting on the block is
Option 1 2.5 N
Option 2 0.98 N
Option 3 4.9 N
Option 4 0.49 N
Correct Answer 2
Explanation

Normal reaction from wall = 5N and static friction fs = mg

∵ flim = µN= ( 0.5) × 5 =2.5N
∵ mg = 0.1 × 9.8 = 0.98 = < flim
∴ static friction will act with fs = mg = 0.98 N.

Q.61. What is the maximum value of the force F such that the block shown in the
arrangement, does not move?

Option 1 20 N
Option 2 10 N
Option 3 12 N
Option 4 15 N
Correct Answer 1
Explanation

From F.B.D of block

 fs = F cos 600 = F/2---(1)
or N = mg + F sin 600
F 3
⇒ N = mg + − − − (2)
2
 F 3 1  F 3
∴ flim = µ N ⇒ flim = µ  mg + =  mg+  − − − (3)
 2  2 3 2 
F mg f f mg 2mg
∴ fs ≤ flim ⇒ ≤ + ⇒ ≤ ⇒F≤
2 2 3 4 4 2 3 3
2× 3 ×g
⇒F ≤ ⇒ F ≤ 2g ⇒ F ≤ 20
3

Q.62. Which of the following are correct?

Option 1 A parachute of weight W strikes the ground with his legs and comes to rest ☒
with an upwar acceleration of magnitude 3g. Force exerted on him by
ground during landing is 4 W.
Option 2 Two massless spring balances are hung vertically in series from a fixed point ☒
and a mass M kg is attached to the lower end of the lower spring balance.
Each spring balance reads M kg.
Option 3 A rough vertical board has an acceleration a along the horizontal direction ☒
so that a block of mass m passing against it does not fall. The coefficient of
friction between the block and the board is greater than g/a.
Option 4 A man is standing at a spring platform. If man jumps away from the ☒
platform the reading of the spring balance first increases and then
decreases to zero.
Explanation (a) From F.B.D. of parachulist
R = reaction from ground

⇒ R - W = ma
W
⇒ R-W= × 3g
g
(b) In series combination of masseess spring common tension will be developed
i.e. T = K1x1 = K2 x2 = Mg
So each spring balance reads Mg incorrect
 (c) For equilibrium of m

N = ma
fs = mg
∵ fs ≤ flim ⇒ mg ≤ µ ma
D) When man jumps spring compressed and man accelerates up so from F.B.D. of man

N - mg = ma ⇒ N = m(g + a)
∴ reading of balance in caasess but when than jumps completely N = O in correct.

Q.63. Two block of masses m1 and m2 are connected through a massless inextensible string.
Block of mass m1 is placed at the fixed rigid inclined surface while the block of mass m2
hanging at the other end of the string. Which is passing through a fixed massless
frictionless pulley shown in the figure. The coefficient of static friction between the
block and the inclined plane is 0.8. The system of masses m1 and m2 is released from
rest.

Option 1 The tension in the string is 20 N after releasing the system. ☒

Option 2 the contact force by the inclined surface on the block is along normal to the ☒
inclined surface.
Option 3 the magnitude of contact force by the inclined surface on the block m1 is ☒
20 3 N.
Option 4 None of these ☐
Explanation

As there is no tendency of slipping as net force is balanced ⇒a = O. So no

requirement of friction on 4 kg block. So normal reactin will be only contact force.
Hnece,
3 4 × 10
N = 4g cos30 = 4g = 3 ⇒ N = 20 3
2 2

Q.64. A body of mass 5 kg is suspended by the strings making angles 600 and 300 with the
horizontal as shown in the figure (g = 10 m/s2)

Option 1 T1 = 25 N ☒
Option 2 T2 = 25 N ☐
Option 3 T1 = 25 3N ☐
Option 4 T2 = 25 3N ☒
Explanation

From F.B.D. of 5 kg block

T1 cos 300 = T2 cos 600
3 T
⇒ T1 = 2 ⇒ T2 = 3 T1 ---(1)
2 2
0 0
And T2 sin 60 + T1 sin 30 = 5g
3 T
⇒ T2 + 1 = 5g
2 2
⇒ T1 + 3 T2 = 10g ---(2)
Put (1) in (2) T1 + 3T1 = 10g
5
⇒ 4T1 = 10g ⇒ T1 = g = 25N
2
5g 50
and T2 = 3 T1 = 3 ⇒ T2 = 3 ⇒ T2 = 25 3 N
2 2
So options a, d are correct.

Q.65. The string shown in the figure is passing over small smooth pulley rigidly attached to
trolley A. If speed of trolley is constant and equal to VA. Speed and magnitude of
acceleration of block B at the instant shown in figure is
Option 1 VB = VA , a B = 0 ☐
Option 2 aB = 0 ☐
Option 3 3
VB = VA
☒
5
Option 4
aB =
16VA2 ☒
125
Explanation

Since the length of rope is constant so

l1 + l2 = constant
d 1 d 2
⇒ + =0
dt dt
d 2 d
⇒ =- 1
dt dt
d 1
⇒ VB = - ---(1)
dt

Also,
l2 = x2 + h2; now both sides
d dx
⇒ 2 1 1 = 2x +0
dt dt
d dx
⇒ 1 1 = x ---(2)
dt dt
d 1 x dx
⇒ =
dt  1 dt
d1 x dx
⇒ = ---(3)
dt x2 + h2 dt
Putting (3) in (1)
x dx
⇒ VB = -
x + h dt
2 2

Q.66. In the figure, if F = 4N, m = 2kg, M = 4 kg then

Option 1 2
The acceleration of m w.r.t . ground is m / s 2 .
☐
3
Option 2 The acceleration of m w.r.t. ground id 1.2 m/s2. ☒
Option 3 the acceleration of M is 0.4 m/s2. ☒
Option 4 3
The acceleration of m w.r.t . ground is m / s 2 .
☐
2
Explanation

m = 2 kg, m = 4 kg
flim between m and
N = µ sN
⇒ flim = 0.1 × 2 × 10 = 2
To check the condition for slipping let us assume that two blocks move with same
acceleration a
⇒ F = (m + M) a ⇒ 4 = (2 + 4)a
4 2
⇒ a= = m/s 2
6 3
∴ friction on the loner block is fs = Ma

2 8
fs = ( 4 ) × = = 2.67 N
3 3
as fs > flim
Two blocks will slip over each other due to insufficient friction and kinetic friction acts.
From F.B.D. of blocks.

F - µ kN 4 - 0.08 × 2 × 10
a1 = =
m 2
⇒ a1 = 2 – 0.8 = 1.2 m/s2
f µ N 0.08 × 2 × 10
and a2 = k = k = = 0.4 m/s2
M M 4
So acceleration of m w.r.t ground = 1.2 m/s2
And acceleration of M w.r.t. ground = 0.4 m/s2
So sption 2, 3 are correct
Q.67. A 20 kg block is placed on top of 50 kg block as shown in the figure. A horizontal force
F acting on A causes an acceleration of 3 m/s2 to A and 2 m/s2 to B. For this situation
mark out the correct statement(s).

Option 1 The friction force between A and B is 40 N. ☒

Option 2 The net force acting on A is 150 N. ☒
Option 3 The value of F is 190 N ☒
Option 4 the value of F is 150 N. ☐
Explanation

Relative acceleration of B w.r.t. A is

aB/A = aB - aA
= 2 - 3 = -1m/s2
So relative acceleration is in left word direction.
←
-ve
 →
+ve
So kinetic friction act on B in rightword direction.
From F.B.D of B
fk = 20 aB = 20 × 2
⇒ fk = 40 newton
∵ fnet = mass × acceleration
∴ net force on A is = 50 × 3 = 150 N
Also from F.B.D. of A
F - fk = 50 aA
⇒ F - 40 = 50 × 3 ⇒ F = 150 + 40
⇒ F = 190 N

Q.68. Two block A and B masses mA and mB velocity v and 2v, respectively, at a given instant
(see figure). A horizontal force F acts on the block A. There is no friction between
ground and block B and there is coefficient of friction between A and B is µ. The
friction

Option 1 On A oppose its motion. ☐

Option 2 On B oppose its motion relative to A. ☒
Option 3 On A and B is µ m A g. ☒
Option 4
On the block B is
µmB
F.
☐
( mA + mB )
Explanation On B oppose its motion relative to A, On A and B is µ m A g.

Q.69. → →
Two rough blocks A and B, A placed over B, move with acceleration a A and aB ,
→ → → →
velocity v A and v B by the action of horizontal force FA and FB , respectely (figure).
 When no friction exists between the blocks A and B.

Option 1 VA = VB and aA = AB ☒
Option 2 aA = aB ☒
Option 3 both (1) and (2) ☒
Option 4 FA F
= B and v A = v B
☒
mA mA
Explanation VA = VB and aA = AB, aA = aB, both (1) and (2), FA = FB and v A = v B
mA mA

Q.70. Some statements are given below, which are in one way or other can be explained by
Newton’s law of motion. Mark the correct statement(s).
Option 1 In a tug of war, the team that pushes the ground harder, wins. ☐
Option 2 In a tug of war, the team that pushes the ground harder (horizontally), wins ☒
Option 3 Observers win two different inertial frames will measure the same ☐
acceleration of a moving object then the velocity of the object w.r.t. two
observers would be also same.
Option 4 A horizontal force acts on a body that is free to move. Can it produce an ☒
acceleration if this force is equal to half of the weight of that body?
Explanation In a tug of war, the team that pushes the ground harder (horizontally), wins, A
horizontal force acts on a body that is free to move. Can it produce an acceleration if
this force is equal to half of the weight of that body?

Q.71. Mark the correct statement(s) regarding friction.

Option 1 Frictin force can be zero, even through the contact surface is rough. ☒
Option 2 Even through there is no relative motion between surfaces, frictional force ☒
may exist between them.
Option 3 The expressions fL = µ SN or fk = µ KN are approsimate expressins. ☒
Option 4 Does the expression fL = µ SN tells that direction of fL and N are the same. ☐
Explanation Frictin force can be zero, even through the contact surface is rough, Even through
there is no relative motion between surfaces, frictional force may exist between them,
The expressions fL = µ SN or fk = µ KN are approsimate expressins

Q.72. A block is resting over a rough horizontal floor. At t = 0, a time varying force starts
acting on it, the force is described by equation F = kt, where k is constant and t is in
seconds. Mark the correct statement(s) for this situation.

Option 1 Curve 1 shows acceleration-time graph. ☐

Option 2 Curve 2 shows acceleration-time graph. ☒
Option 3 Curve 3 shows velocity-time graph. ☒
Option 4 Curve 4 shows displacement-time graph. ☒
Explanation Curve 2 shows acceleration-time graph, Curve 3 shows velocity-time graph, Curve 4
shows displacement-time graph

Passage Text In the system shown in the figure, m1 > m 2 . System is held at rest by thread BC. Now
thread BC is burnt. Answer the following:
Q.78. Before burning the thread, what are the tensions is spring and thread BC, respectively?

Option 1 m1g, m2g

Option 2 m1g, m1g + m2g
Option 3 m2g, m1g
Option 4 m1g, m1g + m2g
Correct Answer 2
Explanation

Before burning the thread Tension in spring is T1 = m1g

Tension in rope is attained from F.B.D. of m2
T1 = m2g + T2 ⇒ T2 = T1 - m2g
⇒ T2 = m1g - m2g in tension in rope BC.

Q.79. Just after burning the thread, what is the tension in the spring?
Option 1 m1 g
Option 2 m2 g
Option 3 0
Option 4 can’t say
Correct Answer 1
Explanation Just after burning the thread alongation in spring remains unchanged so tension T1
remains same ∴ T1 = m1g

Q.80. Just after burning the thread, what is the acceleration of m2?
Option 1  m2 - m1 
 g
 m2 
Option 2  m1 - m2 
 g
 m1 + m2 
Option 3 0
Option 4  m1 - m2 
 g
 m2 
Correct Answer 4
Explanation Just after burning the thread from F.B.D. of m2

T1 - m2g = m2a
T2 - m 2 g
⇒ acceleration a =
m2
m1 g - m 2 g
⇒a=
m2
 m - m2 
⇒a =  1 g
 m2 

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

Q.81. Maximum force F0 exerted by the bat on the ball is

Option 1 4,000 N
Option 2 5,000 N
Option 3 3,000 N
Option 4 2,500 N
Correct Answer 1
Explanation

←
-ve
+ve
→
Pi = Initial momentium of ball = - mu
Pf = final momentium of ball = + mv
∴ Change in linear momentum
→ → →
= ∆ P = Pf - Pi
= mv - (-mu) = m (v + u)
 ∵ area of
force - line graph = change in linear momentum
∵ Area of
F - t graph
1 1
= F0 × 6ms = F0 × 6 × 10 -3
2 2
→ 200 60
Also ∆P = ( 40 + 20 ) = = 12
1000 50
1
∴ F0 × 6 × 10 -3 = 12 ⇒ F0 = 4000 N
2

Q.82. Average force exerted by the bat on the ball is

Option 1 5,000 N
Option 2 2, 000 N
Option 3 2,500 N
Option 4 6,000 N
Correct Answer 2
Explanation Average force exerted by bat on the ball
→
∆P 12
= = = 2000 N
∆t 6 × 10 -3

Q.83. What is the speed of the ball at the istant the force acting on it is maximum?
Option 1 40 m/s
Option 2 30 m/s
Option 3 20 m/s
Option 4 10 m/s
Correct Answer 3
Explanation

∵ Area of F - t graph = mv' - ( -m × 20)

upto 4 m/s = mv’ + 20m
1
∵ Area = × 4 × 10 -3 × F0 = 2 × 10 -3 F0
2
200
⇒ 2 × 10 -3 × 4000 = ( V'+ 20 )
1000
⇒ 40 = V’ + 20 ⇒ V = 20.

Passage Text In the following figure the weight w is 60.0 N.

Q.84. The tension in the diagonal string is
Option 1 60 N
Option 2 90 N
Option 3 85 N
Option 4 100 N
Correct Answer 3
Explanation

From F.B.D. at Junction P

From equilibrium at P
T2
T1 = − − (1)
2
T
and F1 = 2 − −(2)
2
From F.B.D. at Junction S

From equilibrium at S
T2
= T3 − − (3)
2
T
and F2 = 2 − − (4)
2
T3 = W = 60 N --(5)
Put equation (5) in (3)
T2 = 2 T3 ⇒ T2 = 60 2 N
Put in equation (1) we get
T2 60 2
T1 = = ⇒ T1 = 60 N
2 2
 T2 60 2
Also, F1 = F2 = = ⇒ F1 = F2 = 60 N
2 2
∴ tension in diagonal string in 60 2 ≈ 85N

Q.85. Find the magnitudes of the horizontal forces F1 and F2 that must be applied to hold the
system in the position shown.
Option 1 75 N, 90 N, respectively
Option 2 60 N, 60 N, respectively
Option 3 90 N 90 N, respectively
Option 4 45 N, 90 N, respectively
Correct Answer 2
Explanation As, F1 = F2 = 60 N
as calculated above.

Passage Text Block A has a mass of 40 kg and block B has a mass of 15 kg, and F of 500 N is applied
parallel to smooth inclined plane (figure). The system is moving together.
Q.86. The acceleration of the system is

Option 1 45
m / s2
11
Option 2 23
m / s2
11
Option 3 13
m / s2
7
Option 4 8
m / s2
3
Correct Answer 1
Explanation

Mass of system = 40 + 15 = 55 kg

According to F.B.D. of system

Fnet = (Msystem) a
0
⇒ F - 55 g sin 30 = 55a ⇒ 500 - 275 = 55 a
225 45
⇒a= ⇒a= m / s2
55 11
Q.87. The least coefficient of friction between A and B is
Option 1 5 2
12
Option 2 9 3
53
Option 3 9 2
28
Option 4 5 3
18
Correct Answer 2
Explanation On drawing F.B.D. of w.r.t A.

Resolving force on B along vertical and horizontal

B at rest w.r.t A

ma
⇒N = + mg
2
∴ Flim = µN
a 
⇒ Flim = µ m =  + g  ---(1)
2 
Also, fs = ma cos300
3
⇒ fs = m a ---(2)
2
3 a  a 3 /2
∵ fs ≤ Flim ⇒ ma ≤ µ m + g  ⇒ µ ≥
2 2  ( a + 2g ) / 2
45 45
3 3 3
⇒µ ≥ 11 ⇒ ma ≤µ 11
 45  2 45 + 220
 11 + 2 × 10 
  11
45 3 9 3
⇒µ≥ ⇒ µ min =
265 53

Passage Text Three blocks A, B and C of mass 3M, 2M and M, respectively, are suspended vertically
with the help of springs PQ and TU and a string RS as shown. If acceleration of block A,
B and C are a1, a2 and a3, respectively.
Q.88. The value of acceleration a3 at the moment spring PQ is cut is
Option 1 g downward
Option 2 g upward
Option 3 More than g downward
Option 4 Zero
Correct Answer 4
Explanation

From F.B.D. of system of blocks in equilibrium initiall,

T1 = 6 Mg is the tension in spring PQ
T2 = 3 Mg is the tension in spring RS
T3 = 3 Mg is the tension in spring TU
As spring PQ is cut tension in PQ suddenly becines zero. But tension T3 will remains
unchanged. So just after cut for block C
T3 = Mg and acceleration a3 = 0

Q.89. The value of acceleration a1 at the moment string RS is cut is

Option 1 g downward
Option 2 g upward
Option 3 More than g downward
Option 4 Zero
Correct Answer 2
Explanation As RS is cut tension in Rs Suddenly becomes zero but spring tension T1 remains
unchanged.
From F.B.D. pf block A

T1 - 3 Mg = 3 Ma1
 ⇒ 6 Mg - 3Mg = 3 Ma1 ⇒ a1 = g upward

Q.90. The value of acceleration a2 at the moment spring TU is cut is

Option 1 g/5 upward
Option 2 g/5 downward
Option 3 g/3 upward
Option 4 Zero
Correct Answer 1
Explanation As To spring is cut tension T3 becomes 0. So for the system of 3M and 2M there is a
common acceleration a1 = a2. But spring tension T1 is PQ remains unchanged as T1 = 6
Mg

T1 - 5mg 6Mg - 5mg

∴ for system a1 = =
5M 5M
g
⇒ a1 = upwards = a2
5

Passage Text In the following figure both the pulley and the string are massless and all the surfaces
are frictionless.

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

Q.91. The tension in the string is
Option 1 120
N
7
Option 2 240
N
7
Option 3 130
N
7
Option 4 None of these
Correct Answer 2
Explanation

By constraint equation of pulley attached to wedge

let acceleration of masses m1, m2 and m3 are respectively a1, a2 and a3
By string constraint a2 = a3 - a1 in the acceleration of m2 w.r.t. pulley in vertical
direction
From F.B.D. of m1, m2 and m3 we have

m2 acceleration horizontally as well as vertically

For m1 T - N = m1a1 ---(1)
for m2 T - m2g = m2a2 ---(2)
N = m2a1 ------(3)
For m3 m3g - T = m3a3 ---(4)
and a2 = a3 - a1 ---(5)
Put (3) in (1) ⇒ T - m2a1 = m1a1 ⇒ T = (m1 + m2)a1 ---(6)
Put (5) in (2) ⇒ T - m2g = m2 (a3 - a1) ----(7)
From (4)
T
a3 = g -
m3
and from (6)
T
a1 =
m1 + m 2
Put in (7
 T T 
T - m2g = m2  g- - 
 m3 m1 + m2 
m2 m2
⇒ T+ T+ T = 2m 2 g
m3 m1 + m 2
 m m2 
⇒ T 1 + 2 +  = 2m2g
 m3 m1 + m2 
 2 2   3+2+2  12g
⇒ T 1 + +  = 2 × 2g ⇒ T   = 4g ⇒ T =
 3 1+2   3  7
12 × 10 120
∴ Tension in string = T = = N
7 7
By equation (6) acceleration a1 = T
m1 + m2
120
⇒ a1 =
7( 1+2)
40
⇒ a1 = m/ s2
7
Putting in equation (4)
T
a3 = g -
m3
120 40
⇒ a 3 = 10 - =10 -
7×3 7
 30
⇒ a3 = m / s2
7
Also acceleration of m2 is
= a22 +a12
30 40 -10
as a2 = a3 - a1 = - = m / s2
7 7 7
2 2
 10   40 
⇒ net acceleration of m 2 =   +  
 7   7 
17
= 10 m / s2
7

Passage Text A student performs two exeriments to determine the coefficient of static and kinetic
friction between a block of mass 100 kg and the horizontal floor.
Ist Experiment: He applies a gradual increasing force on the block and is just able to
slide the block when force is 450 N
IIst Experiment: He applies contant force of different magnitudes for the duration of 2
s and determine the distance travelled by block in this duration.
Set Force Distance
1. 300 N 0.5 m
2. 600 N 2.2 m
3. 750 N 3.0 m
Assume all the forces have been applied horizontally.
Q.94. The coefficient of static friction between the block and the floor is
Option 1 0.45
Option 2 0.5
Option 3 0.3
Option 4 1.45
Correct Answer 1
Explanation As the block begins to slide when
Fapplied = 450 N ⇒ flim = 450 N

as normal reaction
N = 100g
flim = µ N
⇒ 450 = µ × 100 × 10
45
⇒ µ = = 0.45
100

Q.95. Which set of the readings of Experiment II is absolutely wrong?

Option 1 1
Option 2 2
Option 3 3
Option 4 None of these
Correct Answer 1
Explanation As limiting friction to 450 N so at forces lesser than flim block will not slide. Hence at
300 N block remains at rere. So set 1 is absolutely wrong.
Q.96. The speed of the block after 3s (beginning from the starting of application of force) in
set 2 for IInd experiment is
Option 1 6 m/s
Option 2 2 m/s
Option 3 3 m/s
Option 4 Information is insufficient
Correct Answer 3
Explanation For 0 -25 acceleration of block is
F - fk 600 - fk
a= =
m 100
1 2 1
∵ S = at + at ⇒ a starting from rest S= at 2
2 2
1 600 - fk
⇒ 2= × 22 ⇒ 600 - fk = 100 ⇒ fk = 500 N
2 100
Under F speed of block at 3s is
600 - 500
V1 =0+at ⇒ V1 = ×2
100
⇒ V1 = 3 m/s

Passage Text When two masses are connected by a string (which we assume to be weighless and
does not stretch), they both have same acceleration of the same magnitude. Two
unequal masses are attached by a lightweight string that passes over a pulley of
neglibile mass as shown in figure. The block of mass m2 lies on a frictionless incline of
angle θ.

With the help of the comprechensin given above, choose the most appropriate
alternative to each of the following questions.
Q.97 The acceleration of the two masses is
Option 1 m 1 g + m 2 g sin θ
m1 + m 2
Option 2 m 1 g - m 2 g sin θ
m1 + m 2
Option 3 m 2 sin θ - m 1 g
m1 + m 2
Option 4 m1m2g( 1 + sin θ)
m1 + m2
Correct Answer 3
Explanation

Case (a) If m2 sin θ > m1 then m2 slides down the incline and m1 moves vertically up
with same acceleration a
 Case (b) If m1 m2 sin θ then m2 slides up the inclines
∴ according to case (a) acceleration of system is

a=
( m2 sinθ - m1 ) g
m1 + m2
and according to case (b) acceleration of system is

a=
( m1 - m2 sinθ - m1 ) g
m1 + m2

Q.98. The tension in the string is given by

Option 1 m 2 m 2 sin θ
m1 + m2
Option 2 m1m2g ( 1 + sin θ)
m1 + m2
Option 3 m1m2g (1 - sin θ)
m1 + m2
Option 4 m 1m 2 g sin θ
m1 + m2
Correct Answer 2
Explanation So according to case (a) T - m2g sin θ = m2a

⇒ T = m2gsinθ + m2
( m1 - m2 sinθ) g
m1 + m2
( m1 + m2 ) ( m2 gsin θ ) +m1 m2 g - m 22 sin θ g
T=
m1 + m2
m1 m2 g + m 2 sin θ g + m1 m2 g - m 2 sin θ g
2 2
T=
m1 + m 2
m1 m2g( 1+ sinθ)
T=
m1 + m2

Q.101. If m1 = 10.0kg, m2 = 5.00kg and θ = 450, the acceleration is

Option 1 4.22 m/s2
Option 2 -4.22 m/s2
Option 3 8.44 m/s2
Option 4 -4.44 m/s2
Correct Answer 1
Explanation In case (b) when m1 = 10 kg, m2 = 5 kg and θ = 450
5
 10 - 5 × sin450  10 -
a =  2 × 10 20 - 5 2
10 =
 10 + 5  15 3
a ≈ 4.22 m/s2 is the magnitude of acceleration.

Q.117. Match the enentries of coloumn I with the entries of Coloumn II.
No. Column A Column B Column C Id of Additional
Answer
1 (A) Fritctional force (s) Kinetic friction
is less than applied
force
2 (B) Frictional force (p) m = tan θ (q) Limiting friction,
is equal to the (r) Static
 applied force
3 (C) An object is on (p) m = tan θ (q) Limiting friction,
the verge of motion (r) Static
4 (D) An object is (p) m = tan θ (q) Limiting friction,
about to slide doen (r) Static
when placed on an
inclined plane, θ
being the angle of
inclination with the
horizontal
Explanation (A)When frictional force is less than applied force then body slides w.r.t. surface and
kinetic friction acts so A → S
(B) When frictional force is equal to the applied force then it is the condition of limiting
friction than
µ = tan (angle of repose) and static friction acts.
So B → p, q, r
(C) An object is on the verge of motion condition of limiting friction.
So C → p, q, r
(D) An object is about to slide down at angle of repose θ which is the condition of
limiting friction.
So D → p, q, e

Q.118. For the situation shown in the figure, in Coloumn I, the statements regarding friction
forces are mentioned, while in Coloumn II some information related to friction forces
are given. Match the entries of Coloumn I with the entries of Coloumn II.

No. Column A Column B Column C Id of Additional

Answer
1 (A) Total friction (q) towards left
force on 3 kg block
is
2 (B) Total friction (r) zero
force on 5 kg block
is
3 (C) Friction force on (p) towards right (r) (s) non-zero
2 kg block due to 3 zero
kg block is
4 (D) Friction force (q) towards left (s) non-zero
on 3 kg block due
to 5 kg block is
Explanation

When force F = 100 N is applied on the middle block then 5 kg can only slide due to
 frictional force f2

∵ ( flim)2 < ( flim) 3

∴ 5 kg block can never slide and f2 = 5 N
If we assume that upper 2 blocks move as a system then

Common acceleration of 2kg and 3 kg in

100 - 5 95
a' = = = 19
5 5
2
⇒ a’ = 19 m/s
But to slide with such acceleration friction required on 2 kg = f1 = 2 × a' = 38 N
Which is not possible as (flim)1 = 4 N.
Hence sliding will take place between 2 kg 4 3kg block
So final situation is

So total frictional force on 3 kg block = 9 N towards left.

So total friction on 5 kg block = 0
So friction on 2 kg = 4 N right
Friction on 3 kg due to 5 kg = 5 N left
So A → q, s, B → r, C → p, a, D → q, s

Q.119. The system shown in the figure is initially in equilibrium.

No. Column A Column B Column C Id of Additional

Answer
1 (A) Just after the (p) Accelerates up
Spring 2 is cut, the
block D
2 (B) Just after the (r) Momentarily at
 Spring 2 is cut, the rest
block C
3 (C) Just after the (r) Momentarily at
Spring 2 is cut, the rest
block A
4 (D) Just after the (r) Momentarily at
Spring connecting A rest
and B is cut, the
block D
Explanation Just after the spring (2) is cut the tension in spring (1) remains unchanged

when spring (2) is cut then block will momentarily be at rest but the to decrease in the
down ward forces on D block D will accelerate up momentarily.
Also then spring (2) in cut then along action in spring (1) remain unchanged
momentarily. So block A momentarily at rest.
When spring connecting A and B is cut then alongation in spring (1) and (2) remains
momentarily unchanged
∴ block D in momentarily at rest
So, A → p, B → r, C → r, D → r

Q.120. A light string fixed at one end to a clamp on ground passes over a fixed pulley and
hangs at the other side. It makes an angle of 300 with the ground. A monkey of mass 5
kg climbs up the rope. The clamp can tolerates a vertical force of 40 N only. The
maximum acceleration (in m/s2) in upward direction with which the monkey can climb
safely is
Correct Answer 6
Is Integer Type ☒
Explanation

∵ maximum vertical force on clamp in

T sin 300 = 40
T
⇒ = 40 ⇒ T = 80 N
2
When monkey climbs with acceleration a
T - 5g = 5a
30
⇒ 80 - 50 = 5a ⇒ a =
5
2
⇒ a = 6 m/s

Q.121. An inclined plane makes an angle of 300 with the horizontal. A groove OA = 5 m cuts in
 the plane makes an angle of 300 with OX. A short smooth cylinder is free to slide down
under the ingluence of gravity. The time taken by the cylinder to reach from A to O is
W sec. Find W.

Correct Answer 2
Is Integer Type ☒
Explanation

Component of g along the incline is g sin 300. The component of g sin 300 along the
groove = (g sin 300) cos 600
g 1 g
= × =
2 2 4
1
∴ appling S = at + at 2 along the grove
2
1 2
OA = at as u = 0
2
1 g 10
⇒ 5= × × t2 ⇒ 5 = × t 2 ⇒ t 2 = 4s
2 4 8
⇒ t = 2s

Q.122. In the arrangement shown in figure, m1 = 1 kg and m2 = 2 kg. The pulleys are massless
and strings are light. The value of M for which the mass m1 moves with constant
velocity (neglecting frictin) is W kg. Find W.

Correct Answer 8
Is Integer Type ☒
Explanation

∵ m1 moves with uniform velocity ⇒ a1 = 0

 By constraint equation
a2 = 2a ------(1)
From F.B. D of m1
T = 1g = 10 N----(2)
From F.B.D. of m2 2g - T = m2 a2
20 - 10 = 2 × a 2
⇒ a2 = 5 m/s2
2
∴ acceleration of pulley and block of Mass M = a = a = 5
2 2
From F.B.D. of M
5
2T = Ma ⇒ 2T = M ×
2
5
⇒ 2 × 10 = M ×
2
⇒ M = 8 kg
∴W=8

Q.123. A car is going at a speed of 6 m/sec when it encounters a slope of angle 370. The length
of the sloping slide is 7 m. The friction coefficient between the road and the tyre is 0.5.
The driver applied brakes. The minimum speed of the car with which it can reach the
bottom is W m/sec. Find W.

Correct Answer 8
Is Integer Type ☒
Explanation When car slides doen under gravity and friction then acceleration of car down the
incline is
mgsin θ - µ mgcos θ
a=
m
⇒ a = gsin θ - µ gcos θ
 1 
⇒ a = 10  sin37 0 - cos37 0 
 2 
3 1 4 1
⇒ a = 10  - ×  = 10 × = 2m/s 2
5 2 5 5
On applying V2 = u2 + 2a S down the incline
2
V2 = ( 6 ) + 2 × ( 2) × 7
V2 = 36 + 28 ⇒ V2 = 74
⇒ V = 74 = 8.6m/s
So minimum speed on reaching 8 m/s

Q.124. A monkey of mass m = 1 kg clings to a rope slung over a light frictionless pulley. The
opposite end of the rope is tied to a weight of mass M = 2 m lying on a smooth
horizontal plane. The tension of the rope for the monkey moves upwards with an
acceleration a = 2 m/sec2 relative to the rope is T = W N (neglect the mass of pulley
and rope). Find W.
Correct Answer 8
Is Integer Type ☒
Explanation Form F.B.D. of M
T = Ma = T = 2a ---(1)
Form F.B.D. of m
acceleration of monkey w.r.t. ground is
→ → →
aM/G = aM/Rope + aRope/G
⇒ aM = (arel - a) = (2 - a) ---(2)
T - 1g = 1 am ⇒ T - 10 = 1 (2 - a) ⇒ T = 12 - a ---(3)
Put (1) in (3) 2a = 12 - a ⇒ a = 4 m/s2
∴ Tension T = 2a = 2 × 4 ⇒ T = 8 N