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Single Correct Type Questions
1. As shown in the figure a block of mass 10 kg lying on a
horizontal surface is pulled by a force F acting at an angle
30 o , with horizontal. For m s = 0.25, the block will just start
to move for the value of F : [Given g = 10 ms
[1 Feb, 2023 (Shift-II)]
o
F
( a ) 33.3 N ( b ) 25.2 N
( c ) 20 N ( d ) 35.7 N
2. A body of mass 10 kg is moving with an initial speed of
20 m/s. The body stops after 5 s due to friction between
body and the floor. The value of the coefficient of friction is:
(Take acceleration due to gravity g = 10 ms
[31 Jan, 2023 (Shift-II)]
( a ) 0.2 ( b ) 0.
( c ) 0.5 ( d ) 0.
3. 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 block slides through a
distance of 50 m in an interval of time 10 s. Coefficient
of kinetic friction is (given, g = 10 ms
[1 Feb, 2023 (Shift-1)]
( a ) 0.60 ( b ) 0.
( c ) 0.50 ( d ) 0.
4. A bag is gently dropped on a conveyor belt moving at a
speed of 2 m/s. The coefficient of friction between the
conveyor belt and bag is 0.4. Initially, the bag slips on the
belt before it stops due to friction. The distance travelled
by the bag on the belt during slipping motion is:
[Take g = 10 m/s
- ] [27 July, 2022 (Shift-I)]
( a ) 2 m ( b ) 0.5 m
( c ) 3.2 m ( d ) 0.8 m
5. A block of mass 40 kg slides over a surface, when a mass
of 4 kg is suspended through an inextensible massless
string passing over frictionless pulley as shown below.
40 kg
4 kg
The coefficient of kinetic friction between the surface and
block is 0.02. The acceleration of block is.
(given g = 10 ms
- ) [29 June, 2022 (Shift-II)]
( a ) 1 ms
( c ) 4/5 ms
- ( d ) 8/11 ms - 6. A block of mass m slides along a floor while a force of
magnitude F is applied to it at an angle q as shown in
figure. The coefficient of kinetic is m K
. Then, the block’s
acceleration ‘a’ is given by: [16 March, 2021 (Shift-I)]
(g is acceleration due to gravity)
F
q
( a ) K
F F
cos g sin m m
θ + μ − θ
( b ) K
F F
cos g sin m m
θ + μ + θ
( c ) K
F F
cos g sin m m
θ − μ − θ
( d ) K
F F
cos g sin m m
− θ − μ − θ
Friction
7. An insect is at the bottom of a hemispherical ditch 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
- ) [6 Sep, 2020 (Shift-I)]
( a ) 0.45 m ( b ) 0.80 m ( c ) 0.20 m ( d ) 0.60 m
8. 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 30º with the horizontal, as shown in the figures. The
coefficient of friction between the block and floor is
m = 0.2. The difference between the accelerations of
the block, in case ( b ) and case ( a ) will be: ( g = 10 ms
[12 April, 2019 (Shift-II)]
( a )
o
F = 20 N
( b )
o
F = 20 N
( a ) 0 ms
- ( b ) 0.8 ms - ( c ) 0.4 ms - ( d ) 3.2 ms - 9. A system of two blocks of masses m = 2 kg and m = 8
kg is placed on a smooth table as shown in figure. The
coefficient of static friction between two blocks is 0.5.
The maximum horizontal force F that can be applied to
the block of mass M so that the blocks move together will
be:
[27 June, 2022 (Shift-I)]
M
F
m
( a ) 9.8 N ( b ) 39.2 N ( c ) 49 N ( d ) 78.4 N
10. Two blocks A and B of masses m A = 1 kg and m B = 3 kg
are kept on the table as shown in figure. The coefficient
of friction between A and B is 0.2 and between B and the
surface of the table is also 0.2. The maximum force F that
can be applied on horizontally, so that the block A does
not slide over the block B is: (Take g = 10 m/s 2 )
[10 April, 2019 (Shift-II)]
A
B
( a ) 16 N ( b ) 40 N ( c ) 12 N ( d ) 8 N
11. The time taken by an object to slide down a 45
o rough
inclined plane is n times as it takes to slide down a
perfectly smooth 45 o inclined plane. The coefficient of
kinetic friction between the object and the incline plane is
[29 Jan, 2023 (Shift-II)]
( a ) 2
1 − n
( b ) (^2)
n
( c ) 2
n
n
12. Consider a block kept on an inclined plane (inclined at
45º) as shown in the figure. If the force required to just
push it up the incline is 2 times the force required to just
prevent it from sliding down, the coefficient of friction
between the block and inclined plane (m) is equal to:
[25 Jan, 2023 (Shift-II)]
( a ) 0.33 ( b ) 0.
( c ) 0.25 ( d ) 0.
13. A block of mass m slides down an inclined plane inclined
at angle 30 o with an acceleration 4
g
. The value of
coefficient of kinetic friction will be:
[29 Jan, 2023 (Shift-I)]
( a )
( b )
( c )
( d )
14. 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 q. The magnitude of the contact force
will be: [27 July, 2022 (Shift-II)]
( a ) Mg
( b ) Mg cos q
( c ) (^) Mg sin θ + Mg cosθ
( d ) (^) Mg sin θ 1 + μ
15. A block ‘ A ’ takes 2 s to slide down a frictionless incline
of 30° and length ‘’, kept inside a lift going up with
uniform velocity ‘ v ’. If the incline is changed to 45°, the
time taken by the block, to slide down the incline, will be
approximately: [27 July, 2022 (Shift-II)]
( a ) 2.66 s ( b ) 0.83 s
( c ) 1.68 s ( d ) 0.70 s
24. The coefficient of static friction between two blocks is 0.
and the table is smooth. The maximum horizontal force that
can be applied to move the blocks together is _______ N.
(take g = 10 ms
- ) [26 Aug, 2021 (Shift-II)]
μ = 0.
2 kg
1 kg
F
Table
25. A boy of mass 4 kg is standing on a piece of wood having
mass 5 kg. If the coefficient of friction between the wood
and the floor is 0.5, what is the maximum force that the
boy can exert on the rope so that the piece of wood does
not move from its place is ______ (in N). (Round off to
the Nearest Integer) [Take g = 10 ms
[17 March, 2021 (Shift-II)]
T
R
F
T
T
T
26. Two blocks ( m = 0.5 kg and M = 4.5 kg ) are arranged
on a horizontal frictionless table as shown in figure.
The coefficient of static friction between the two blocks
is
. Then the maximum horizontal force that can be
applied on the larger block so that the blocks move
together is ______ N. (Round off to the Nearest Integer)
[Take g as 9.8 ms
- ] [17 March, 2021 (Shift-I)]
m
M
F
27. Two inclined planes are placed as shown in figure. A block
is projected from the point A of inclined plane AB along
its surface with velocity just sufficient to carry it to the
top Point B at a height 10 m. After reaching the point B
the block slide down on inclined plane BC. Time it takes
to reach to the point C from point A is t (^) ( 2 + (^1) ) s.
The value of t is _____ (use g = 10 m/s
2 )
[27 July, 2022 (Shift-II)]
A
B
C
30°
10 m
45°
28. When a body slides down from rest along a smooth
inclined plane making an angle of 30° with the 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 distance, it takes time a T , where
a is a constant greater than 1. The co-efficient of friction
between the body and the rough plane is
2
2
x
α − α where x = _________.
[1 Sep, 2021 (Shift-II)]
29. A body of mass ‘ m ’ is launched up on a rough inclined
plane making an angle of 30º with the horizontal. The
coefficient of friction between the body and plane is 5
x
if the time of ascent is half of the time of descent. The
value of x is __________. [20 July, 2021 (Shift-II)]
30. A block starts moving up an inclined plane of inclination
30° with an initial velocity of v 0
. It comes back to its initial
position with velocity
0
2
v
. The value of the coefficient of
kinetic friction between the block and the inclined plane is
close to 1000
I
. The nearest integer to I is ___________.
[3 Sep, 2020 (Shift-II)]
1. ( b ) 2. ( d ) 3. ( c ) 4. ( b ) 5. ( d ) 6. ( c ) 7. ( c ) 8. ( b ) 9. ( c ) 10. ( a ) 11. ( d ) 12. ( a ) 13. ( b ) 14. ( a ) 15. ( c ) 16. ( b ) 17. ( a ) 18. ( a ) 19. ( a ) 20. [2]
21. [5] 22. [25] 23. [3.3] 24. [ 15] 25. [30] 26. [21] 27. [2] 28. [ 3 ] 29. [3] 30. [346]
ANSWER KEY
EXPLANATIONS
1. ( b )
Equating the vertical forces,
N = Mg − F sin
F F
mg
F cos30 = μ N
F − F
= ×
⇒ F = = N
2. ( d ) Using formula, v = u + at
⇒ 0 = 20 + (–mg) (5) [ ⸪ v = 0 , a = – m g ]
⇒ m = 0.
3. ( c )
S = ut + at
= + ´ a ´
⇒ a = 1 m/s 2
F – m mg = ma
30 – m × 50 = 5 × 1
m=
4. ( b ) In frame of belt
a = m g = 4 m/s 2 , v = 2m/s, u = 0 m/s
v
2 = u
2
⇒ s = 0.5 m
5. ( d ) For 4 kg block
24 g – T = 4 a ...( i )
For 40 kg block
T – 40g × 0.02 = 40 a ...( ii )
Adding ( i ) and ( ii )
40 – 8 = 44 a
m/s 44 11
a = =
6. ( c ) From the free body diagram given below,
N
F sinq
F cosq
F
q
mg
F
K
N = mg – F sinq ...( i )
ma = F cosq – m K N ...( ii )
On solving ( i ) and ( ii ) we get
ma = F cosq – m K
(mg – F sinq)
cos (^) K sin
F F
a g m m
⇒ = θ − μ − θ
7. ( c ) R = 1m y q
q
h
m = tan θ
= tan θ
⇒ θ = 37º
∴ h = R – R cos θ = 1 – 1 ×
= 0.2 m
q
m
mg
sin
q
mg cosq mg
N
R
f
15. ( c ) Acceleration of the block down the inclined plane,
a = g sinq
From Kinematics equations,
sin 30 (2) 2
= g
... ( i )
sin 45 ( ) 2
= g t
... ( ii )
( i ) = ( ii )
( )
t t
16. ( b ) a = –m g = –0.5 × 9.8 = –4.9 m/s
2
2 9.8 9.
u d a
×
= 9.8 m
17. ( a ) Using FBD,
F = mg sinq – qE cosq– m N
Here N = mg cosq + qE sinq
3 9.8 5 10 200 3
F
− × × × ∴ = −
3 5 10 200 1 9.8 3
2 2
− (^) × × × × − +
F = 2.25 N ⇒ a = 2.25 m/s
2
S = ut + at
t 1.31sec a
qE
N
f r
1 m mg cosq
mg sinq
18. ( a ) mg sin θ + 2 = μ mg cos θ ... ( i )
10 – mg sin θ = μ mg cos θ ... ( ii )
On adding ( i ) + ( ii )
12 = 2 μ mg
; μ mg =
On ( a ) – ( b ):
8 = 2 mg ×
; mg = 8; μ =
19. ( a ) For equilibrium of the block net force be zero. Hence
we can write.
mg sin θ + 3 = P + friction
mg sin θ + 3 = P + m mg cos θ
After solving, we get, P = 32 N
20. [2] Mass per unit length = l
N = mg = l( L – x ) g
f s, max = m s
N
mg x
L x –
N
f s ,max = (0.5) (l) ( L – x ) g
And also f s ,max
= m x
g
0.5l ( L – x ) g = l xg
L x x
2 m 2 2 3 3
L x L = ⇒ x = = =
21. [5] Minimum possible force,
2 1
mg F
μ
min
F
×
×
= 5N
22. [25] f = W = 0.5 × 10 = 5 N
N = F
The magnitude of horizontal force
should be smaller man frictional
force to keep it adhere to the wall.
⇒ f ≤ m N
⇒ 5 ≤ 0.2 F
⇒ F ≥ 25 N
23. [3.3] Net force in the Vertical direction
is zero so.
N = mg + F sin 60°
f im = F cos 60° ⇒ μ N = F cos60°
⇒ m( mg + F sin 60°) = F cos 60°
⇒ m mg = F [cos 60° – m sin 60°]
cos 60 sin 60 1 1 1 1 3
2 6 2 3 3 2
mg F
× ×
μ = = = − μ − × −
N N
= = = ×
F sin 60°
F cos 60°
F
lim
mg
N
m
24. [ 15]
max max
F
= × ⇒ F = N
25. [30]
T N
f
R
F
T
T
T
T
T
90g
m=9kg
∴ f = T
⇒ m N = T
⇒ m (90 – T ) = T
⇒ 0.5 (90 – T ) = T
⇒ 90 – T = 2 T
⇒ 3 T = 90
⇒ T = 30 N
0.5 kg
mg
N
F
f
26. [21] (^) m
M
F
m
F
a M m
F
f ma m M m
F
m mg M m
≤ μ
(for no slipping)
⇒ F ≤ μ ( m + M g )
max
∴ F = + × = N
27. [2] Acceleration between A and B = – g sin45° = 2
− g
Acceleration between B and C = +gsin30° = 2
g
10 m
B
A C
Velocity at point A.
2 0 2 10 2 2
g = u − ×
u = 10 2 m/s
From A to B
v = u + at
AB
g = − t
t AB = 25 ...( i )
From B to C
s = ut + at
BC
g = + × t
tBC = (^2 2) ...( ii )
2 (^2 1 ) t (^) AB + tBC = +
⇒ t = 2
