P-36 Physics

Laws of Motion
Ist, Ind & IIIrd Laws of 1 § J · 1 § J ·
tan 1 ¨ V0 ¸ sin 1 ¨ V
TOPIC 1
Motion (a)
Jg ¨ ¸ (b)
Jg ¨ g 0 ¸¸
© g ¹ © ¹
1. A particle moving in the xy plane experiences a velocity 1 § J · 1 § 2J ·
(c) l n ¨1  V0 ¸ (d) tan 1 ¨ V0 ¸
G Jg © ¨ g ¹ ¸ 2J g ¨ ¸
dependent force F k ( v y i  v x j ) , where vx and vy are x © g ¹
G G 5. A ball is thrown vertically up (taken as + z-axis) from the
and y components of its velocity v . if a is the accelera-
ground. The correct momentum-height (p-h) diagram is:
tion of the particle, then which of the following statements
[9 April 2019 I]
is true for the particle? [Sep. 06, 2020 (II)]
G G
(a) quantity v u a is constant in time
G
(b) F arises due to a magnetic field (a) (b)
(c) kinetic energy of particle is constant in time
G G
(d) quantity v ˜ a is constant in time
2. A spaceship in space sweeps stationary interplanetary dust.
dM (t ) (c) (d)
As a result, its mass increases at a rate bv 2 (t ),
dt
where v (t) is its instantaneous velocity. The instantaneous 6. A particle of mass m is moving in a straight line with
acceleration of the satellite is : [Sep. 05, 2020 (II)]
momentum p. Starting at time t = 0, a force F = kt acts in the
bv 3 same direction on the moving particle during time interval
(a) bv3 (t ) (b) 
M (t ) T so that its momentum changes from p to 3p. Here k is a
2bv 3 bv 3 constant. The value of T is : [11 Jan. 2019 II]
(c)  (d) 
M (t ) 2 M (t ) k p
3. A small ball of mass m is thrown upward with velocity u (a) 2 (b) 2
p k
from the ground. The ball experiences a resistive force
mkv2 where v is its speed. The maximum height attained 2k 2p
by the ball is : [Sep. 04, 2020 (II)] (c) (d)
p k
1 ku 2 1 § ku 2 · 7. A particle of mass m is acted upon by a force F given by
(a) tan 1 (b) ln 1 
2k g k ¨© 2 g ¸¹ R
the empirical law F v(t). If this law is to be tested
1 ku 2
1 § ku 2· t2
(c) tan 1 (d) ln 1 
k 2g 2 k ¨© g ¸¹ experimentally by observing the motion starting from rest,
the best way is to plot : [Online April 10, 2016]
4. A ball is thrown upward with an initial velocity V0 from the
1
surface of the earth. The motion of the ball is affected by a (a) log v(t) against (b) v(t) against t2
drag force equal to mJv2 (where m is mass of the ball, v is t
its instantaneous velocity and Jis a constant). Time taken 1
(c) log v(t) against 2 (d) log v(t) against t
by the ball to rise to its zenith is : [10 April 2019 I] t
Laws of Motion P-37

8. A large number (n) of identical beads, each of mass m (a) Statement 1 is true, Statement 2 is true, Statement 2 is
and radius r are strung on a thin smooth rigid horizontal the correct explanation of Statement 1.
rod of length L (L >> r) and are at rest at random (b) Statement 1 is false, Statement 2 is true.
positions. The rod is mounted between two rigid (c) Statement 1 is true, Statement 2 is false.
supports (see figure). If one of the beads is now given (d) Statement 1 is true, Statement 2 is true, Statement 2 is
a speed v, the average force experienced by each support not the correct explanation of Statement 1.
after a long time is (assume all collisions are elastic): 12. Two fixed frictionless inclined planes making an angle 30°
[Online April 11, 2015] and 60° with the vertical are shown in the figure. Two
blocks A and B are placed on the two planes. What is the
relative vertical acceleration of A with respect to B ? [2010]
L A
B

30°
mv 2 mv 2 60°
(a) (b) (a) 4.9 ms–2 in horizontal direction
2(L  nr) L  2nr
2 (b) 9.8 ms–2 in vertical direction
mv (c) Zero
(c) (d) zero
L  nr (d) 4.9 ms–2 in vertical direction
9. A body of mass 5 kg under the action of constant force 13. A ball of mass 0.2 kg is thrown vertically upwards by applying
G G
F Fxˆi  Fy ˆj has velocity at t = 0 s as v 6iˆ  2ˆj m/s a force by hand. If the hand moves 0.2 m while applying the
force and the ball goes upto 2 m height further, find the
G G
and at t = 10s as v 6ˆj m / s . The force F is: magnitude of the force. (Consider g = 10 m/s2). [2006]
[Online April 11, 2014] (a) 4 N (b) 16 N (c) 20 N (d) 22 N
14. A player caught a cricket ball of mass 150 g moving at a
§ 3 ˆ 4 ˆ· rate of 20 m/s. If the catching process is completed in 0.1s,
(a) 3jˆ  4ˆj N (b) ¨  i  j ¸ N
© 5 5 ¹ the force of the blow exerted by the ball on the hand of the
§ 3ˆ 4 ˆ· player is equal to [2006]
(c) 3iˆ  4ˆj N (d) ¨ i  j ¸ N (a) 150 N (b) 3 N (c) 30 N (d) 300 N
©5 5 ¹
15. A particle of mass 0.3 kg subject to a force F = – kx with
10. A particle of mass m is at rest at the origin at time
k = 15 N/m . What will be its initial acceleration if it is
t = 0. It is subjected to a force F(t) = F0e–bt in the x direction.
Its speed v(t) is depicted by which of the following released from a point 20 cm away from the origin ?[2005]
curves? [2012] (a) 15 m/s2 (b) 3 m/s2 (c) 10 m/s2 (d) 5 m/s2
16. A block is kept on a frictionless inclined surface with angle
F0 F0 of inclination ‘D’. The incline is given an acceleration ‘a’
mb mb to keep the block stationary. Then a is equal to [2005]
(a) v (t ) (b) v (t )
t t
F0 F0
mb mb a
D
(c) v (t ) (d) v (t )
(a) g cosec D (b) g / tan D
t t (c) g tan D (d) g
11. This question has Statement 1 and Statement 2. Of the 17. A rocket with a lift-off mass 3.5 × 104 kg is blasted upwards
four choices given after the Statements, choose the one with an initial acceleration of 10m/s2. Then the initial thrust
that best describes the two Statements. of the blast is [2003]
Statement 1: If you push on a cart being pulled by a horse
(a) 3.5 u 10 5 N (b) 7.0 u 10 5 N
so that it does not move, the cart pushes you back with an
(c) 14.0 u 10 5 N (d) 1.75 u 10 5 N
equal and opposite force.
18. Three forces start acting simultaneously on a particle
Statement 2: The cart does not move because the force G
described in statement 1 cancel each other. moving with velocity, v . These forces are represented
[Online May 26, 2012] in magnitude and direction by the three sides of a triangle
ABC. The particle will now move with velocity [2003]
P-38 Physics

C 2g
(a)
3

g R
(b) m
2
A B
G 5g
(a) less than v (c)
G 6
(b) greater than v
G (d) g m
(c) v in the direction of the largest force BC
(d) vG , remaining unchanged 24. Two blocks of mass M1 = 20 kg and M2 = 12 kg are
connected by a metal rod of mass 8 kg. The system is
19. A solid sphere, a hollow sphere and a ring are released
pulled vertically up by applying a force of 480 N as shown.
from top of an inclined plane (frictionless) so that they
The tension at the mid-point of the rod is :
slide down the plane. Then maximum acceleration down
the plane is for (no rolling) [2002] [Online April 22, 2013]
480 N
(a) solid sphere (b) hollow sphere (a) 144 N
(c) ring (d) all same M1

(b) 96 N
Motion of Connected Bodies,
TOPIC 2 Pulley & Equilibrium of (c) 240 N
Forces
20. A mass of 10 kg is suspended by a rope of length 4 m, from (d) 192 N M2
the ceiling. A force F is applied horizontally at the mid-
point of the rope such that the top half of the rope makes 25. A uniform sphere of weight W and radius 5 cm is being
an angle of 45° with the vertical. Then F equals: held by a string as shown in the figure. The tension in the
(Take g = 10 ms–2 and the rope to be massless) string will be : [Online April 9, 2013]
[7 Jan. 2020 II]
(a) 100 N (b) 90 N
(c) 70 N (d) 75 N 8 cm
21. An elevator in a building can carry a maximum of 10
persons, with the average mass of each person being 68
kg. The mass of the elevator itself is 920 kg and it moves
with a constant speed of 3 m/s. The frictional force
opposing the motion is 6000 N. If the elevator is moving
up with its full capacity, the power delivered by the motor
to the elevator (g =10 m/s2) must be at least: W W W W
(a) 12 (b) 5 (c) 13 (d) 13
[7 Jan. 2020 II] 5 12 5 12
(a) 56300 W (b) 62360 W 26. A spring is compressed between two blocks of masses m1
(c) 48000 W (d) 66000 W and m2 placed on a horizontal frictionless surface as shown
22. A mass of 10 kg is suspended vertically by a rope from in the figure. When the blocks are released, they have
initial velocity of v1 and v2 as shown. The blocks travel
the roof. When a horizontal force is applied on the rope
distances x1 and x2 respectively before coming to rest.
at some point, the rope deviated at an angle of 45°at the
roof point. If the suspended mass is at equilibrium, the §x ·
The ratio ¨ 1 ¸ is [Online May 12, 2012]
magnitude of the force applied is (g = 10 ms–2) © x2 ¹
m1 m2
[9 Jan. 2019 II]
(a) 200 N (b) 140 N
v1 v2
(c) 70 N (d) 100 N
23. A mass ‘m’ is supported by a massless string wound around
a uniform hollow cylinder of mass m and radius R. If the
m2 m1 m2 m1
str ing does not slip on the cylinder, with what
(a) m1 (b) m2 (c) m1 (d) m2
acceleration will the mass fall or release? [2014]
Laws of Motion P-39

27. A block of mass m is connected to another block of mass 33. When forces F1, F2, F3 are acting on a particle of mass m
M by a spring (massless) of spring constant k. The block such that F2 and F3 are mutually perpendicular, then the
are kept on a smooth horizontal plane. Initially the blocks particle remains stationary. If the force F1 is now removed
are at rest and the spring is unstretched. Then a constant then the acceleration of the particle is [2002]
force F starts acting on the block of mass M to pull it. (a) F1/m (b) F2F3 /mF1
Find the force of the block of mass m. [2007]
(c) (F2 - F3)/m (d) F2 /m.
MF mF 34. Two forces are such that the sum of their magnitudes is
(a) (b)
(m  M ) M 18 N and their resultant is 12 N which is perpendicular
mF to the smaller force. Then the magnitudes of the forces
(c) ( M  m) F (d)
m (m  M ) are [2002]
28. Two masses m1 5g and m2 4.8 kg tied to a string (a) 12 N, 6 N (b) 13 N, 5 N
are hanging over a light frictionless pulley. What is the (c) 10 N, 8 N (d) 16N, 2N.
acceleration of the masses when left free to move ? 35. A light string passing over a smooth light pulley connects
( g 9.8m / s 2 ) [2004] two blocks of masses m1 and m2 (vertically). If the
acceleration of the system is g/8, then the ratio of the
masses is [2002]
(a) 8 : 1 (b) 9 : 7 (c) 4 : 3 (d) 5 : 3
36. Three identical blocks of masses m = 2 kg are drawn by a
force F = 10. 2 N with an acceleration of 0. 6 ms-2 on a
frictionless surface, then what is the tension (in N) in
the string between the blocks B and C? [2002]
C B A F
(a) 5 m/s2 (b) 9.8 m/s2
(c) 0.2 m/s 2 (d) 4.8 m/s2 (a) 9.2 (b) 3.4 (c) 4 (d) 9.8
29. A spring balance is attached to the ceiling of a lift. A man 37. One end of a massless rope, which passes over a massless
hangs his bag on the spring and the spring reads 49 N, and frictionless pulley P is tied to a hook C while the other
when the lift is stationary. If the lift moves downward with end is free. Maximum tension that the rope can bear is 360
an acceleration of 5 m/s2, the reading of the spring balance N. With what value of maximum safe acceleration (in ms-2)
will be [2003] can a man of 60 kg climb on the rope? [2002]
(a) 24 N (b) 74 N (c) 15 N (d) 49 N P
30. A block of mass M is pulled along a horizontal frictionless
C
surface by a rope of mass m. If a force P is applied at the
free end of the rope, the force exerted by the rope on the
block is [2003]
(a) 16 (b) 6 (c) 4 (d) 8
Pm Pm PM
(a) (b) (c) P (d)
M m M m M m
31. A light spring balance hangs from the hook of the other
TOPIC 3 Friction
light spring balance and a block of mass M kg hangs from
the former one. Then the true statement about the scale 38. An insect is at the bottom of a hemispherical ditch of
reading is [2003] radius 1 m. It crawls up the ditch but starts slipping
(a) both the scales read M kg each after it is at height h from the bottom. If the coefficient
(b) the scale of the lower one reads M kg and of the of friction between the ground and the insect is 0.75,
upper one zero then h is : (g = 10 ms–2) [Sep. 06, 2020 (I)]
(c) the reading of the two scales can be anything but the (a) 0.20 m (b) 0.45 m
sum of the reading will be M kg (c) 0.60 m (d) 0.80 m
(d) both the scales read M/2 kg each 39. A block starts moving up an inclined plane of inclination
32. A lift is moving down with acceleration a. A man in the lift 30° with an initial velocity of v0. It comes back to its
drops a ball inside the lift. The acceleration of the ball as v
observed by the man in the lift and a man standing initial position with velocity 0 . The value of the
2
stationary on the ground are respectively [2002] coefficient of kinetic friction between the block and the
(a) g, g (b) g – a, g – a I
(c) g – a, g (d) a, g inclined plane is close to . The nearest integer to I
1000
is _________. [NA Sep. 03, 2020 (II)]
P-40 Physics

40. A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled 44. Two masses m1 = 5 kg and m2 = 10 kg, connected by an
in case (B), by a force F=20 N, making an angle of 30 o with inextensible string over a frictionless pulley, are moving
the horizontal, as shown in the figures. The coefficient of as shown in the figure. The coefficient of friction of
friction between the block and floor is P = 0.2. The horizontal surface is 0.15. The minimum weight m that
difference between the accelerations of the block, in case should be put on top of m2 to stop the motion is: [2018]
m T
(B) and case (A) will be : (g =10 ms–2) (a) 18.3 kg m
m2
2
[12 April 2019 II]
(b) 27.3 kg
T
(c) 43.3 kg
m1
(d) 10.3 kg
m1g
45. A given object takes n times more time to slide down a 45°
rough inclined plane as it takes to slide down a perfectly
(a) 0.4 ms–2 (b) 3.2 ms–2 smooth 45° incline. The coefficient of kinetic friction
(c) 0.8 ms–2 (d) 0 ms–2
between the object and the incline is :
41. Two blocks A and B masses mA=1 kg and mB = 3 kg are kept
on the table as shown in figure. The coefficient of friction [Online April 15, 2018]
between A and B is 0.2 and between B and the surface of 1 1
the table is also 0.2. The maximum force F that can be (a) 1 (b) 1  2
n2 n
applied on B horizontally, so that the block A does not
slide over the block B is : [Take g = 10 m/s2] 1 1
[10 April 2019 II] (c) 2 (d)
2n 1  n2
46. A body of mass 2kg slides down with an acceleration of
3m/s2 on a rough inclined plane having a slope of 30°.
The external force required to take the same body up the
(a) 8 N (b) 16 N (c) 40 N (d) 12 N plane with the same acceleration will be: (g = 10m/s2)
42. A block kept on a rough inclined plane, as shown in the [Online April 15, 2018]
figure, remains at rest upto a maximum force 2 N down (a) 4N (b) 14N (c) 6N (d) 20N
the inclined plane. The maximum external force up the 47. A rocket is fired vertically from the earth with an acceleration
inclined plane that does not move the block is 10 N. The of 2g, where g is the gravitational acceleration. On an
coefficient of static friction between the block and the inclined plane inside the rocket, making an angle T with
plane is : [Take g = 10 m/s2] [12 Jan. 2019 II] the horizontal, a point object of mass m is kept. The
10
N minimum coefficient of friction Pmin between the mass and
the inclined surface such that the mass does not move is :
[Online April 9, 2016]
2N (a) tan2T (b) tanT
30°
(c) 3 tanT (d) 2 tan T
3 3 48. Given in the figure are two blocks A and B of weight 20 N
(a) (b)
2 4 and 100 N, respectively. These are being pressed against a
1 2 wall by a force F as shown. If the coefficient of friction
(c) (d) between the blocks is 0.1 and between block B and the
2 3
wall is 0.15, the frictional force applied by the wall on
43. A block of mass 10 kg is kept on a rough inclined plane as
block B is: [2015]
shown in the figure. A force of 3 N is applied on the block.
The coefficient of static friction between the plane and
the block is 0.6. What should be the minimum value of F
force P, such that the block doesnot move downward? A B
(take g = 10 ms–2) [9 Jan. 2019 I]
P

k g
10
N

45° (a) 120 N (b) 150 N

3

(a) 32 N (b) 18 N (c) 23 N (d) 25 N (c) 100 N (d) 80 N

Laws of Motion P-41

49. A block of mass m = 10 kg rests on a horizontal table. The A

coefficient of friction between the block and the table is
0.05. When hit by a bullet of mass 50 g moving with speed
h
Q, that gets embedded in it, the block moves and comes to
stop after moving a distance of 2 m on the table. If a freely
Q
falling object were to acquire speed after being dropped
10
B L C
from height H, then neglecting energy losses and taking g
= 10 ms–2, the value of H is close to: 2h vo2 h v2o
(a)  (b) 
[Online April 10, 2015] P 2Pg P 2Pg
(a) 0.05 km (b) 0.02 km
h v2o h v2
(c) 0.03 km (d) 0.04 km (c)  (d)  o
50. A block of mass m is placed on a surface with a vertical 2P Pg 2P 2Pg

x3 54. A block A of mass 4 kg is placed on another block B of

cross section given by y . If the coefficient of friction mass 5 kg, and the block B rests on a smooth horizontal
6
table. If the minimum force that can be applied on A so
is 0.5, the maximum height above the ground at which the
that both the blocks move together is 12 N, the maximum
block can be placed without slipping is: [2014]
force that can be applied to B for the blocks to move
1 2 1 1 together will be: [Online April 9, 2014]
(a) m (b) m (c) m (d) m
6 3 3 2 (a) 30 N (b) 25 N (c) 27 N (d) 48 N
51. Consider a cylinder of mass M resting on a rough horizontal 55. A block is placed on a rough horizontal plane. A time
rug that is pulled out from under it with acceleration ‘a’ dependent horizontal force F = kt acts on the block, where
perpendicular to the axis of the cylinder. What is Ffriction k is a positive constant. The acceleration - time graph of
at point P? It is assumed that the cylinder does not slip. the block is : [Online April 25, 2013]
[Online April 19, 2014] a a
w
v
O
(a) (b)
P
O t O t
a
a a

F friction
Ma Ma (c) (d)
(a) Mg (b) Ma (c) (d)
2 3
52. A heavy box is to dragged along a rough horizontal floor. O t O t
To do so, person A pushes it at an angle 30° from the 56. A body starts from rest on a long inclined plane of slope
horizontal and requires a minimum force FA, while person 45°. The coefficient of friction between the body and
B pulls the box at an angle 60° from the horizontal and the plane varies as P = 0.3 x, where x is distance travelled
needs minimum force FB. If the coefficient of friction down the plane. The body will have maximum speed
3 FA (for g = 10 m/s2) when x = [Online April 22, 2013]
between the box and the floor is , the ratio F is (a) 9.8 m (b) 27 m (c) 12 m (d) 3.33 m
5 B
[Online April 19, 2014] 57. A block of weight W rests on a horizontal floor with
coefficient of static friction P. It is desired to make the
5
(a) (b) block move by applying minimum amount of force. The
3 3 angle T from the horizontal at which the force should be
3 2 applied and magnitude of the force F are respectively.
(c) (d) [Online May 19, 2012]
2 3
53. A small ball of mass m starts at a point A with speed vo PW
(a) T tan 1 P , F
and moves along a frictionless track AB as shown. The
1  P2
track BC has coefficient of friction P. The ball comes to
stop at C after travelling a distance L which is: § 1· PW
[Online April 11, 2014] (b) T tan 1 ¨ ¸ , F
© P¹ 1  P2
P-42 Physics

(c) T 0, F PW 64. A block rests on a rough inclined plane making an angle of

30° with the horizontal. The coefficient of static friction
§ P · PW
(d) T tan 1 ¨ ,F between the block and the plane is 0.8. If the frictional
© 1  P¹¸ 1 P force on the block is 10 N, the mass of the block (in kg) is
58. An insect crawls up a hemispherical surface very slowly. (take g 10 m / s 2 ) [2004]
The coefficient of friction between the insect and the (a) 1.6 (b) 4.0 (c) 2.0 (d) 2.5
surface is 1/3. If the line joining the centre of the 65. A horizontal force of 10 N is necessary to just hold a block
hemispherical surface to the insect makes an angle D with stationary against a wall. The coefficient of friction between
the vertical, the maximum possible value of D so that the the block and the wall is 0.2. The weight of the block is
insect does not slip is given by [Online May 12, 2012] [2003]

D
10N
(a) cot D = 3 (b) sec D = 3
(c) cosec D = 3 (d) cos D = 3
59. The minimum force required to start pushing a body up (a) 20 N (b) 50 N (c) 100 N (d) 2 N
rough (frictional coefficient P) inclined plane is F1 while 66. A marble block of mass 2 kg lying on ice when given a
the minimum force needed to prevent it from sliding down velocity of 6 m/s is stopped by friction in 10 s. Then the
is F2. If the inclined plane makes an angle T from the coefficient of friction is [2003]
F1 (a) 0.02 (b) 0.03 (c) 0.04 (d) 0.06
horizontal such that tan T = 2P then the ratio is
F2
Circular Motion, Banking of
(a) 1 (b) 2 [2011 RS] TOPIC 4
(c) 3 (d) 4 Road
60. If a spring of stiffness ‘k’ is cut into parts ‘A’ and ‘B’ of 67. A disc rotates about its axis of symmetry in a hoizontal
length A A : A B 2 : 3, then the stiffness of spring ‘A’ is plane at a steady rate of 3.5 revolutions per second. A coin
given by [2011 RS] placed at a distance of 1.25cm from the axis of rotation
remains at rest on the disc. The coefficient of friction
3k 2k between the coin and the disc is (g = 10m/s2)
(a) (b)
5 5 [Online April 15, 2018]
5k (a) 0.5 (b) 0.7 (c) 0.3 (d) 0.6
(c) k (d) 68. A conical pendulum of length 1 m makes an angle T= 45°
2
w.r.t. Z-axis and moves in a circle in the XY plane.The
61. A smooth block is released at rest on a 45° incline and then
radius of the circle is 0.4 m and its centre is vertically be-
slides a distance ‘d’. The time taken to slide is ‘n’ times as
low O. The speed of the pendulum, in its circular path, will
much to slide on rough incline than on a smooth incline.
be :
The coefficient of friction is [2005]
(Take g = 10 ms–2) Z
[Online April 9, 2017]
1 1 (a) 0.4 m/s O
(a) Pk = 1 – 2 (b) Pk = 1 T
n 2
n (b) 4 m/s
1 1 (c) 0.2 m/s
(c) Ps = 1  (d) Ps = 1 C
n 2
n2
(d) 2 m/s
62. The upper half of an inclined plane with inclination I is 69. A particle is released on a vertical smooth semicircular
perfectly smooth while the lower half is rough. A body track from point X so that OX makes angle T from the
starting from rest at the top will again come to rest at the vertical (see figure). The normal reaction of the track on
bottom if the coefficient of friction for the lower half is the particle vanishes at point Y where OY makes angle I
given by [2005] with the horizontal. Then: [Online April 19, 2014]
(a) 2 cos I (b) 2 sin I (c) tan I (d) 2 tan I X
63. Consider a car moving on a straight road with a speed of Y
100 m/s . The distance at which car can be stopped is T
[ P k 0.5 ] [2005]
(a) 1000 m (b) 800 m (c) 400 m (d) 100 m I

O
Laws of Motion P-43

1 Q2 Q2
(a) sin I = cos I (b) sin I cos T (a)  cos T iˆ  sin T ˆj
2 R R
2 3
(c) sin I cos T (d) sin I cos T Q2 Q2
3 4 (b)  sin T iˆ  cos T ˆj
70. A body of mass ‘m’ is tied to one end of a spring and R R
whirled round in a horizontal plane with a constant angular Q2 Q2
(c)  cos T iˆ  sin T ˆj
velocity. The elongation in the spring is 1 cm. If the R R
angular velocity is doubled, the elongation in the spring Q2 ˆ Q2 ˆ
is 5 cm. The original length of the spring is : (d) i j
R R
[Online April 23, 2013] 73. An annular ring with inner and outer radii R1 and R2 is
(a) 15 cm (b) 12 cm rolling without slipping with a uniform angular speed. The
(c) 16 cm (d) 10 cm ratio of the forces experienced by the two particles situated
71. A point P moves in counter-clockwise direction on a F1
on the inner and outer parts of the ring , F is [2005]
circular path as shown in the figure. The movement of 'P' is 2
such that it sweeps out a length s = t3 + 5, where s is in 2 R2
§ R1 ·
metres and t is in seconds. The radius of the path is 20 m. (a) ¨© R ¸¹ (b) R1
The acceleration of 'P' when t = 2 s is nearly. [2010] 2
y R1
(c) R (d) 1
2
B 74. Which of the following statements is FALSE for a particle
P(x,y) moving in a circle with a constant angular speed ? [2004]
(a) The acceleration vector points to the centre of the
m circle
20
(b) The acceleration vector is tangent to the circle
x (c) The velocity vector is tangent to the circle
O A (d) The velocity and acceleration vectors are
(a) 13m/s2 (b) 12 m/s2 perpendicular to each other.
(c) 7.2 ms2 (d) 14m/s2 75. The minimum velocity (in ms-1) with which a car driver must
72. For a particle in uniform circular motion, the acceleration traverse a flat curve of radius 150 m and coefficient of friction
G 0.6 to avoid skidding is [2002]
a at a point P(R,T) on the circle of radius R is (Here T is (a) 60 (b) 30
measured from the x-axis) [2010] (c) 15 (d) 25
P-44 Physics

1. (a) Given
G 1 ª ku 2  g º
F k ( v y iˆ  vx ˆj ) Ÿ ln « » h
2k ¬ g ¼
? Fx kv y iˆ, Fy kv x ˆj 4. (a) Net acceleration
mdvx dv x k dv
kv y Ÿ vy = a = – (g + Jv2)
dt dt m dt
dv y Let time t required to rise to its zenith (v = 0) so,
k
Similarly, vx
dt m 0 t
dv
dv y vx
Ÿ ³ v y dvy
³ g  Jv2
³ dt [for Hmax, v = 0]
dvx vy ³ vx dvx v0 0

1 § J v0 ·
v 2y v x2  C ?t tan 1 ¨ ¸
Jg © g ¹
v 2y  v x2 constant
5. (d) v2 = u2 – 2gh
G G k
v ua (v x iˆ  v y ˆj ) u (v y iˆ  vx ˆj ) or v = u 2  2 gh
m
2ˆ 2ˆ k k
m u 2  2 gh
(v x k  v y k ) (vx2  v 2y ) kˆ constant Momentum, P = mv
m m
2. (b) From the Newton's second law, u2
At h = 0, P = mu and at h ,P 0
dp d (mv) § dm · f
F v¨ ¸ ...(i) upward direction is positive and downward direction
dt dt © dt ¹
is negative.
dM (t ) 6. (b) From Newton’s second law
We have given, bv 2 (t ) ...(ii)
dt
dp
Thrust on the satellite, F kt
dt
§ dm ·
F v ¨ ¸ v (bv 2 ) bv3 [Using (i) and (ii)] Integrating both sides we get,
© dt ¹ T
3p T ª t2 º
bv3 ³p dp ³0 kt dt Ÿ > p @
3p
k« »
ŸF M (t ) a bv3 Ÿ a p
M (t ) ¬« 2 ¼» 0
3. (d) v=0 kT 2 p
2
(g+kv ) = a (acceleration) Ÿ 2p ŸT 2
H 2 k
u R dv R
7. (a) From F  2 v(t) º m  2 v(t)
t dt t
G dv Rdt
F mkv 2  mg (' mg and mkv2 act Integrating both sides ¨
dt ¨ mt 2

opposite to each other)
G R
G F In v  
a [ kv 2  g ] mt
m
1
dv § dv · = ln v r
Ÿ v˜ [kv 2  g ] ¨©' a v ¸ t
dh dh ¹ 8. (b) Space between the supports for motion of beads is
0 h
v ˜ dv L–2nr
Ÿ³ 2 ³ dh Average force experienced by each support,
u kv  g 0
2mV mV 2
1 0 F
Ÿ ln ª¬kv 2  g º¼ h 2( L – 2nr ) L – 2 nr
2k u
V
Laws of Motion P-45

mv Ÿ 40  0 2 (a) 0.2 Ÿ a 100 m/s2

L–2nr
? F ma 0.2 u 100 20 N
mv Ÿ N  mg 20 Ÿ N 20  2 22 N
Note :
9. (a) From question, Whand + Wgravity = 'K
Mass of body, m = 5 kg Ÿ F (0.2)  (0.2)(10)(2.2) 0 Ÿ F 22 N
Velocity at t = 0, 14. (c) Given, mass of cricket ball, m = 150 g = 0.15 kg
u = (6iˆ  2 ˆj) m/s Initial velocity, u = 20 m/s
Velocity at t = 10s, Force,
m(v  u ) 0.15(0  20)
v = + 6 ĵ m/s F 30 N
t 0.1
Force, F = ? 15. (c) Mass (m) = 0.3 kg
v u Force, F = m.a = –kx
Acceleration, a =
t Ÿ ma = –15x
6 ˆj  (6iˆ  2 ˆj ) 3iˆ  4 ˆj Ÿ 0.3a = –15x
= = m/s2 15 150
10 5 x  50 x
Ÿ a= – x
Force, F = ma 0.3 3
( 3iˆ  4 ˆj ) a = –50 × 0.2 = 10m/s2
= 5u ( 3iˆ  4 ˆj) N 16. (c) When the incline is given an acceleration a towards
5
10. (c) Given that F(t) = F0e–bt the right, the block receives a reaction ma towards left.
ma
g cos
dv N
Ÿ m = F0e–bt
dt D a
dv F0 bt a D
= e mg cosD
dt m + ma sinD mg mg sin D
v t
F0 bt For block to remain stationary, Net force along the incline
³ dv = m³
e dt should be zero.
0 0
mg sin D = ma cos D Ÿ a = g tan D
t
F ª e  bt º F0 ª 17. (b) In the absence of air resistance, if Thrust (F)
v= 0« » =  e  bt  e 0 º the rocket moves up with an acceleration
m ¬« b ¼» mb ¬ ¼
0 a, then thrust
F bt
Ÿ v = 0 ª¬1  e º¼ F = mg + ma
mb
a
11. (a) According to newton third law of motion i.e. every ? F = m ( g + a) = 3.5 × 104 ( 10 + 10)
action is associated with equal and opposite reaction.
12. (d) mg sin T= ma = 7 × 105 N mg
?a = g sin T 18. (d) Resultant force is zero, as three forces are represented
? Vertical component of acceleration by the sides of a triangle taken in the same order. From
= g sin2 T G G
Newton’s second law, Fnet ma.
?Relative vertical acceleration of A with respect to B is
Therefore, acceleration is also zero i.e., velocity remains
g (sin 2 60  sin 2 30] unchanged.
§3 1· g 19. (d) This is a case of sliding (if plane is friction less) and
g¨ – ¸ 4.9 m/s2
©4 4¹ 2 therefore the acceleration of all the bodies is same.
in vertical direction 20. (a) From the free body diagram
13. (d) For the motion of ball, just after the throwing
v = 0, s = 2m, a = –g = –10ms–2
v2 – u2 = 2as for upward journey
Ÿ u 2 2( 10) u 2 Ÿ u 2 40
When the ball is in the hands of the thrower
u = 0, v = 40 ms–1
s = 0.2 m
v2 – u2 = 2as
P-46 Physics

T cos45° = 100 N ...(i) 480

T sin45° = F ...(ii) 12 ms 2
20  12  8
On dividing (i) by (ii) we get Tension at the mid point
T cos 45q 100
§ Mass of rod ·
T sin 45q F T ¨ M2  ¸a
© 2 ¹
Ÿ F = 100 N
= (12 + 4) × 12 = 192 N
21. (d) Net force on the elevator = force on elevator
25. (d) P
+ frictional force
Ÿ F = (10 m + M)g + f T
8 cm
where, m = mass of person, M = mass of elevator, T
f = frictional force
Ÿ F = (10 × 68 + 920) × 9.8 + 600 Q 5 cm O
Ÿ F = 22000 N
Ÿ P = FV = 22000 × 3 = 66000 W
w wcosT
22. (d)
PQ OP 2  OQ2
o
45
132  52 12
Tension in the string
13
F T w cos T W
45
o
12
26. (a)
100 N 27. (d) Writing free body-diagrams for m & M,
At equilibrium, M
m
mg 100 K
tan 45° = F
F F
N N
? F = 100 N
a
23. (b) From figure,
T T M
m F

D mg Mg
R we get T = ma and F – T = Ma
where T is force due to spring
T Ÿ F – ma = Ma or,, F = Ma + ma
? Acceleration of the system
T
F
m a a .
mg M m
Now, force acting on the block of mass m is
Acceleration a = RD …(i)
and mg – T = ma …(ii) § F · mF
ma = m ¨ .
From equation (i) and (ii) © M  m ¸¹ m  M
§ a· If a is the acceleration along the inclined plane, then
T × R = mR2D = mR2 ¨© ¸¹ 28. (c) Here, m1 = 5kg and m2 = 4.8 kg.
R
If a is the acceleration of the masses,
or T = ma m1a = m1g – T ...(i)
Ÿ mg – ma = ma m2a = T – m2g ...(ii)
g Solving (i) and (ii) we get
Ÿ a
2 § m1  m2 ·
24. (d) Acceleration produced in upward direction a ¨ ¸g
© m1  m2 ¹
F
a (5  4.8) u 9.8
M1  M 2  Mass of metal rod Ÿa m / s2 = 0.2 m/s2
(5  4.8)
Laws of Motion P-47

29. (a) When lift is stationary, W1 = mg ...(i) 34. (b) Let the two forces be F1 and F2 and let F2 < F1. R is
When the lift descends with acceleration, a the resultant force.
W2 = m(g – a) Given F1 + F2 = 18 ...(i)
49 From the figure F22  R 2 F12
W2 = (10 – 5) 24.5 N
10
F12  F22 R2
? F12  F22 144 ...(ii)
Only option (b) follows equation (i) and (ii).
T F1
a
mg R
F2
30. (d) Taking the rope and the block as a system
a F1
M m 35. (b) For mass m1
T P m1g – T = m1a ...(i)
For mass m2
we get P = (m + M)a T–m2g = m2a ...(ii)
P
? Acceleration produced, a
mM
Taking the block as a system,
Force on the block, F = Ma
MP T
? F T
mM a m2
31. (a) The Earth exerts a pulling force Mg. The block in turn m1 a
exerts a reaction force Mg on the spring of spring balance m2g
S1 which therefore shows a reading of M kgf. m1g
As both the springs are massless. Therefore, it exerts a Adding the equations we get
force of Mg on the spring of spring balance S2 which (m1  m2 ) g
a
shows the reading of M kgf. m1  m2
g
s2 Mkgf Here a
8
m1
1
s1 Mkgf 1 m2 m m m 9
? Ÿ 1 1 8 1  8 Ÿ 1
8 m1 m2 m2 m2 7
1
M m2
Mg
36. (b) Force = mass × acceleration
32. (c) Case - I: For the man standing in the lift, the ? F = (m + m + m) × a
acceleration of the ball F = 3m × a
G G G
abm ab  am Ÿ abm = g – a F
Case - II: The man standing on the ground, the acceleration a=
3m
of the ball
G G G 10.2
abm ab  am Ÿ abm = g – 0 = g ?a m / s2
6
33. (a) When forces F1, F2 and F3 are acting on the particle, 10.2
it remains in equilibrium. Force F2 and F3 are perpendicular ? T2 = ma 2 u 3.4N
to each other, 6
2 kg 2 kg 2 kg
F1 = F2 + F3 F
T2
C B A
? F1 = F22  F32 T2 T1 T1

The force F1 is now removed, so, resultant of F2 and F3 37. (c) Tension, T = 360 N
will now make the particle move with force equal to F1. Mass of a man m = 60 kg
F mg – T = ma
Then, acceleration, a = 1
m
P-48 Physics

T g Pg 3 §g P 3·
? a g Ÿ  4¨  ¸
m 2 2 ¨2 2 ¸¹
©
360
10  4m / s 2 Ÿ 5  5 3P 4(5  5 3P) (Substituting, g = 10 m/
60
s2)
38. (a) For balancing, mg sin T f Pmg cos T
Ÿ 5  5 3P 20  20 3P Ÿ 25 3P 15
3
Ÿ tan T P 0.75 3 346
4 ŸP 0.346
5 1000
f = mgcosT
RcosT T R I 346
So,
1000 1000
40. (c) A : N = 5g + 20 sin30°
h h T 1
= 50 + 20 × = 60 N
2
mgsinT N

T R 20 cos 30°
RcosT 5 f
3
h T 5g
4
20 sin 30°
§ 4· R
Ff 20 cos 30q  μN
h R  R cos T R  R¨ ¸
© 5¹ 5 Accelaration, a1 =
m 5
R
?h 0.2 m [' radius, R = 1m] ª º
5 3
39. (346) « 20 u  0.2 u 60 »
2
Acceleration of block while moving up an inclined plane, = « » = 1.06 m/s2
« 5 »
a1 g sin T  Pg cos T «¬ »¼
Ÿ a1 g sin 30q  Pg cos30q
N 20 sin 30°
g Pg 3
 ..(i) ('
T= 30o)
2 2
20 cos 30°
Using v 2  u 2 2a ( s ) f
Ÿ v02  02 2a1 ( s) ('
u = 0)
Ÿ v02  2a1 ( s ) 0 5g
v02
Ÿs ...(ii) B : N = 5g – 20 sin 30°
a1
Acceleration while moving down an inclined plane 1
= 50 – 20 × = 40 N
a2 g sin T  Pg cos T 2

Ÿ a2 g sin 30q  Pg cos 30q F  f ª 20cos 30q  0.2 u 40 º

a2 = « » = 1.86 m/s2
m ¬ 5 ¼
g P 3
Ÿ a2  g ...(iii) Now a2 – a1 = 1.86 – 1.06 = 0.8 m/s2
2 2
41. (b) Taking (A + B) as system
Using again v 2  u 2 2as for downward motion
F – P(M + m)g
2
§v · v02 = (M + m)a
Ÿ¨ 0¸ 2a2 ( s ) Ÿ s ...(iv)
© 2¹ 4a2
F – P(M  m) g
Equating equation (ii) and (iv) Ÿa= ( M  m)
v02 v02
Ÿ a1 4a2 F  (0.2)4 u10 § F 8·
a1 4a2 a ¨ ¸ ...(i)
4 © 4 ¹
Laws of Motion P-49

But, amax = Pg = 0.2 × 10 = 2 45. (b) The coefficients of kinetic friction between the object
and the incline
F 8
? 2
4 § 1· 1
P tan T ¨1  2 ¸ Ÿ P 1  2 (' T = 45°)
ŸF = 16 N © n ¹ n
46. (d) Equation of motion when the mass slides down
42. (a) From figure, 2 + mg sin 30° = Pmg cos 30° and Mg sin T – f = Ma
10 = mg sin 30° + P mg cos 30° Ÿ 10 – f = 6 (M = 2 kg, a = 3 m/s2, T = 30° given)
A
= 2Pmg cos 30° – 2 ? f = 4N f
Ÿ 6 = Pmg cos 30° and kg
4 = mg cos 30° Equation of motion when the block is 2
By dividing above two pushed up Ma
T 30°
3
Let the external force required to take C B
Ÿ Pu 3 the block up the plane with same
2 acceleration be F Ma A
f
3 F – Mg sin T – f = Ma g
?Coefficient of friction, P k
2
2
Ÿ F – 10 – 4 = 6 F 30°
T
F = 20 N C B
P
47. (b) Let P be the minimum coefficient of friction
P P3mgcosT
g

vmgcos?
k

v = 0.6
10

43. (a)
3 + mgsing?
N

45°
3

3PgsinT 3g

100
mg sin 45° = = 50 2
2
At equilibrium, mass does not move so,
['m = 10kg, g = 9.8 m s -2
] 3mg sinT = P3mg cosT
= min  tan
1
Pmg cosT = 0.6 × mg × = 0.6 ´ 50 2
2 f1 f2
3 + mg sinT= P + Pmg cosT
F N
 2 = P + 30 2 48. (a) A B
? P = 31.28 = 32 N
44. (b) Given : m1 = 5kg; m2 = 10kg; P = 0.15 f1
20N 100N
FBD for m1, m1g – T = m1a
Ÿ 50 – T = 5 × a Assuming both the blocks are stationary
and, T – 0.15 (m + 10)g = (10 + m)a N= F
For rest a = 0
f1 = 20 N
or, 50 = 0.15 (m + 10) 10
f2 = 100 + 20 = 120N
N
m f
m2 T
P(m+m2)g (m+m2)g T
m1

m1g = 50N

3
Ÿ 5 (m  10) 120N
20
Considering the two blocks as one system and due to
100 equilibrium f = 120N
m  10 ? m = 23.3kg; close to option (b)
3
P-50 Physics

49. (d) f = μ(M + m) g R

52. (d) F (Push)
f μ( M  m) g A T = 60°
a μg
M m ( M  m) 30°
= 0.05 × 10 = 0.5 ms–2
Initial momentum 0.05V f
V0
(M  m) 10.05 mg
m = 50g M = 10 kg Pmg
FA
sin T  P cos T
Q
V0 Similarly,
Pmg
FB
sin T  P cos T
Pmg
FA sin T  P cos T
?
FB Pmg
sin T  P cos T
v2 – u2 = 2as
0 – u2 = 2as Pmg ª 3 º
u2 = 2as = «P given »
3 ¬ 5 ¼
2 sin 60q  cos 60q
§ 0.05v ·
¨ ¸ 2 u 0.5 u 2 5
© 10.05 ¹ Pmg
Solving we get v 201 2 3
sin 30q  cos 30q
Object falling from height H. 5
V 
2 gH sin 30q  cos30q
10 5
201 2 =
2 u10 u H 3
sin 60q  cos 60q
10 5
H = 40 m = 0.04 km
50. (a) At limiting equilibrium, 1 3 3
 u
2 5 2
P = tan T =
3 3 1
m  u
y 2 5 2
T
1§ 3· 1 8
¨1  ¸ u
2© 5¹ 2 5
=
3§ 1· 3u4
¨1  ¸
dy x 2 5 © 5¹ 10
tanT = P= (from question)
dx 2 8
'
Coefficient of friction P = 0.5 10 8 2
=
x2 3u4 3u4 3
? 0.5 10
2
Ÿ x = ±1 53. (b) Initial speed at point A, u = v0
Speed at point B, v = ?
x3 1 v2 – u2 = 2gh
Now, y m
6 6 v2 = v20 + 2gh
51. (d) Force of friction at point P, Let ball travels distance ‘S’ before coming to rest
1 v2 v 2  2 gh
Ffriction Ma sin T S= = 0
3 2Pg 2Pg
1
= Ma sin 90q [ here T = 90°] v02 2 gh h v2
3 =  =  0
Ma 2Pg 2Pg P 2Pg
=
3
Laws of Motion P-51

54. (c) Minimum force on A 1

= frictional force between the surfaces cos T
= 12 N 1  P2
Hence, Fmin
A 4 kg
Pw Pw
5 kg =
B 1 P 2
1  P2

Smooth table 1  P2 1  P2
58. (a) O r

F1 D
R y Bowl
Therefore maximum acceleration
A h
12N
amax = 3m / s 2
4kg mg cos D
mg sin D
Hence maximum force, mg
Fmax = total mass × amax The insect crawls up the bowl upto a certain height h only
= 9 × 3 = 27 N till the component of its weight along the bowl is balanced
55. (b) Graph (b) correctly dipicts the acceleration-time by limiting frictional force.
graph of the block. For limiting condition at point A
56. (d) When the body has maximum speed then R = mg cosD ...(i)
P 0.3x tan 45q F1 = mg sinD ...(ii)
? x = 3.33 m Dividing eq. (ii) by (i)
57. (a) Let the force F is applied at an angle T with the 1 F1
horizontal. tan D P > As F1 PR @
cot D R
R
1ª 1 º
Ÿ tan D P 'P Given »
F sinT F 3 «¬ 3 ¼
F = μR
? cot D = 3
F cosT 59. (c) N1 F1

w mg sin T
f1 mg cos T
For horizontal equilibrium, T mg
F cos T = μR ...(i)
For vertical equilibrium, N2
2
F

R + F sin T= mg f2
or, R = mg – F sinT ...(ii)
Substituting this value of R in eq. (i), we get
F cosT = μ (mg – F sinT) mg sin T
mg cos T
= μ mg – μ FsinT T mg
or, F (cosT+ μsinT) = μmg
When the body slides up the inclined plane, then
μmg mg sin T + f1 = F1
or, F = ...(iii)
cosT  P sinT or, F1 = mg sin T + Pmg cos T
For F to be minimum, the denominator (cosT + μ sinT) When the body slides down the inclined plane, then
should be maximum. mg sin T – f 2 F2
d
? (cosT  P sinT ) = 0 or F2 = mg sin T– Pmg cos T
dT F1 sin T  P cos T
or, – sinT + μ cosT = 0 ? =
or, tanT = μ F2 sin T  P cos T
or, T= tan–1(μ) F1 tan T  P 2P  P 3P
Ÿ F 3
P 2 tan T  P 2P  P P
Then, sinT = and
1  P2
P-52 Physics

60. (d) It is given AA : AB = 2 : 3 g sin I ( g sin I  Pg cos I)

2A § 3A · Ÿ P = 2 tan I
AA , AB
5 ©¨ 5 ¹¸ NOTE
1 According to work-energy theorem, W = 'K = 0
? We know that k v (Since initial and final speeds are zero)
A
If initial spring constant is k, then ? Workdone by friction + Work done by gravity = 0
A
kA k AA A kBA B i.e., ( μ mg cos I )  mg A sin I 0
2
§ 2A · μ
kA kA ¨ ¸ or cos I sin I or μ 2 tan I
© 5¹ 2
5k 63. (a) Given, initial velocity, u = 100m/s.
kA Final velocity, v = 0.
2 Acceleration, a = Pkg = 0.5 × 10
61. (b) a = g sin T  Pg cos T
v2 – u2 = 2as or
d T Ÿ 02 – u2 = 2(–Pkg)s
g sin d 1
a = 45q 45q Ÿ 1002 2 u  u 10 u s
2
smooth rough Ÿ s = 1000 m
On smooth inclined plane, acceleration of the body = g
sin T. Let d be the distance travelled 64. (c) fs

1 N sT
? d = ( g sin T)t12 , m g co
2 3 0° mg
s in
2d mg
t1 , 30°
g sin T
Since the body is at rest on the inclined plane,
On rough inclined plane,
mg sin 30° = Force of friction
mg sin T – PR
a= Ÿ m u 10 u sin 30q 10
m Ÿ m u 5 10 Ÿ m 2.0 kg
mg sin T – Pmg cos T 65. (d) Horizontal force, N = 10 N.
Ÿ a=
m Coefficient of friction P = 0.2.
Ÿ a = g sin T – Pkg cos T f = PN
1 ˆ cos T) t 2
? d = ( g sin T  Pkg 2
2
2d 10N 10N 10N
t2
ˆ cos T
g sin T  Pkg
According to question, t2 = nt1 W
The block will be stationary so long as
2d 2d Force of friction = weight of block
n = ˆ cos T ? PN = W
g sin T g sin T  Pkg
Ÿ 0.2 × 10 = W
Here, P is coefficient of kinetic friction as the block Ÿ W = 2N
moves over the inclined plane. 66. (d) u = 6 m/s, v = 0, t = 10s, a = ?
? sin T = (sin T – Pk̂ cos T)n2 v–u
Acceleration a =
1 1 t
2
Ÿ n= Ÿ n 0–6
1  Pk 1  Pk Ÿ a=
10
1 6
Ÿ Pk 1
Ÿ a 0.6m / s2
n2 10
62. (d) For first half
mg
acceleration = g sin I;
For second half f = PN N
acceleration = – ( g sin I  Pg cos I) The retardation is due to the frictional force
For the block to come to rest at the bottom, acceleration ? f = ma Ÿ PN ma
in I half = retardation in II half.
Laws of Motion P-53

ma At t = 2s, at 6 u 2 12 m/s2
Ÿ Pmg ma Ÿ P = mg
9 u 16
a 0.6 ac 7.2 m/s2
ŸP 0.06 20
g 10 ? Resultant acceleration
mv 2 = at2  ac2
67. (d) Using, Pmg mrZ 2
r
Z = 2Sn = 2S × 3.5 = 7S rad/sec = (12) 2  (7.2) 2 = 144  51.84
Radius, r = 1.25 cm = 1.25 × 10–2 m
Coefficient of friction, μ = ?
= 195.84 = 14 m/s2
G
72. (c ) Clearly a ac cos T(iˆ) ac sin T( ˆj )
m( rZ ) 2
Pmg (' v = rZ)
r v 2 v2
= cos T iˆ  sin T ˆj
R R
Y

P( R, T)
O
1.25 cm
R
Disc T
X
O

μmg = mrZ2
2 73. (c)
§ 22 ·
2 1.25 u 10 2 u ¨ 7 u ¸ a2
R2
rZ © 7¹
Ÿ P v2 ZR 2
g 10 R1
a1 v1 ZR 1
1.25 u 102 u 222
0.6
10
68. (d) Given, T = 45°, r = 0.4 m, g = 10 m/s2
Let m is the mass of each particle and Z is the angular
mv 2
T sin T ...... (i) speed of the annular ring.
r
T cos T mg ...... (ii) v12 Z 2 R12
a1 Z 2 R1
From equation (i) & (ii) we have, R1 R1
v2 v22
tan T a2 Z 2 R2
rg R2
Taking particle masses equal
T T
F1 ma1 mR1Z2 R1
F2 ma2 mR2Z2 R2
NOTE :
The force experienced by any particle is only along radial
direction.
v2 = rg




' T = 45° Force experienced by the particle, F = mZ2R
Hence, speed of the pendulum in its circular path,
F1 R1
v rg 0.4 u 10 = 2 m/s ?
F2 R2
69. (c) 70. (a)
3 74. (b) Only option (b) is false since acceleration vector is
71. (d) s t 5 always radial (i.e. towards the center) for uniform circular
ds motion.
Ÿ velocity, v 3t 2
dt 75. (b) The maximum velocity of the car is
dv vmax = Prg
Tangential acceleration at = 6t
dt Here P = 0.6, r = 150 m, g = 9.8
v2 9t 4 vmax = 0.6 q150 q9.8  30m / s
Radial acceleration ac =
R R