Exercise - 1 Objective Problems | JEE Main

Section A - Distance, Displacement, Velocity 7. A particle, after starting from rest , experiences,
and Acceleration, Equation of constant acceleration for 20 seconds. If it covers a
Motion distance of S1, in first 10 seconds and distance S2 in
next 10 sec, then
1. A particle covers half of the circle of radius r. Then (A) S2 = S1/2 (B) S2 = S1
the displacement and distance of the particle are (C) S2 = 2S1 (D) S2 = 3S1
respectively -
(A) 2r, 0 (B) 2r, r
8. A body sliding on a smooth inclined plane requires
r 4sec to reach the bottom after starting from rest at
(C) , 2r (D) r, r
2 the top. How much time does it take to cover one
fourth the distance starting from the top
2. A hall has the dimensions 10m × 10m × 10 m. A fly (A) 1sec (B) 2 sec
starting at one corner ends up at a diagonally opposite (C) 0.4sec (D) 1.6 sec
corner. The magnitude of its displacement is nearly
(A) 5 3 m (B) 10 3 m 9. The initial velocity of a particle is 10 m/sec and its
retardation is 2 m/sec2. The distance covered in the
(C) 20 3 m (D) 30 3 m fifth second of the motion will be
(A) 1m (B) 19m
3. A car travels from A to B at a speed of 20 km h–1 (C) 50m (D) 75m
and returns at a speed of 30 km h–1. The average
speed of the car for the whole journey is 10. A body starts from rest, the ratio of distances
(A) 5 km h–1 (B) 24 km h–1 travelled by the body during 3rd and 4th seconds is :
–1
(C) 25 km h (D) 50 km h–1
(A) 7/5 (B) 5/7
(C) 7/3 (D) 3/7
4. A car travels a distance of 2000 m. If the first half
distance is covered at 40 km/hour and the second
half at velocity v and if the average velocity is 48
km/hour, then the value of v is - Section B - Motion under Gravity
(A) 56 km/hour (B) 60 km/hour
11. A body is dropped from a height h under acceleration
(C) 50 km/hour (D) 48 km/hour
due to gravity g. If t1 and t2 are time intervals for its
fall for first half and the second half distance, the
5. A car runs at constant speed on a circular track of radius
relation between them is
100 m taking 62.8 s on each lap. What is the average
speed and average velocity on each complete lap ? (A) t1 = t2 (B) t1 = 2t2
(A) velocity 10 m/s speed 10 m/s (C) t1 = 2.414 t2 (D) t1 = 4t2
(B) velocity zero, speed 10 m/s
(C) velocity zero, speed zero 12. Two bodies of different masses ma and mb are
(D) velocity 10 m/s, speed zero dropped from two different heights, viz a and b.
The ratio of times taken by the two to drop through
6. A body starts from rest and is uniformly accelerated these distances is
for 30 s. The distance travelled in the first 10s is x1,
ma b
next 10 s is x2 and the last 10 s is x3. Then x1 : x2 : (A) a : b (B) m : a
x3 is the same as b

(A) 1 : 2 : 4 (B) 1 : 2 : 5 (C) (D) a2 : b2

a: b
(C) 1 : 3 : 5 (D) 1 : 3 : 9
13. A body is thrown upward and reaches its maximum 19. A ball of mass m is thrown upward with a velocity
height. At that position- v. If air exerts an average resisting force F, the
(A) its velocity is zero and its acceleration is also zero velocity with which the ball returns back to the
(B) its velocity is zero but its acceleration is thrower is :
maximum
mg F
(C) its acceleration is minimum (A) v (B) v
mg  F mg  F
(D) its velocity is zero and its acceleration is the
acceleration due to gravity
mg  F
(C) v (D) None of these
mg  F
14. A ball is thrown upwards from the foot of a tower.
The ball crosses the top of the tower twice after an
interval of 4 second and the ball reaches ground 20. A bullet is fired vertically upwards with an initial
after 8seconds then the height of tower in meters is. velocity of 50 ms–1. If g = 10 ms–2, what is the ratio
(A) 20 m (B) 30 m of the distances travelled by the bullet during the
(C) 60 m (D) 50 m first and the last second of its upward motion ?
(A) 9 : 1 (B) 9 : 2
15. A ball is thrown upward from the edge of a cliff (C) 3 : 1 (D) 9 : 4
with an initial velocity of 6 m/s How fast is it moving
1/2 s later? (g = 10 m/s2)
(A) 2 (B) 3 Section C - Graphs
(C) 4 (D) 1
21. The displacement-time graph of a moving particle is
16. A particle is thrown upward from ground. It shown below. The instantaneous velocity of the
experiences a constant resistive force which can produce particle is negative at the point
x
retardation of 6 m/sec2. What will be the ratio of time (A) C
D
of descent to time of ascent (g = 10 m/sec2) (B) D
E F
(A) 2 (B) 3 (C) E
C
(C) 4 (D) 1 (D) F t

17. A particle is thrown upwards from ground. It 22. The variation of velocity of a particle moving along
experiences a consistant air resistance force which straight line is shown in the figure. The distance
can produce retardation 2 m/s2. The ratio of time of travelled by the particle in 4 s is
ascent to the time of descent is : [g = 10 m/s2] (A) 25m v(m/s)

2 (B) 30m
(A) 1 : 1 (B) (C) 55m
20
3 10
(D) 60m
t(s)
2 3 1 2 3 4
(C) (D)
3 2 23. The displacement time graphs of two particles A
and B are straight lines making angles of respectively
18. A ball is thrown vertically upward from a height of 30º and 60º with the time axis. If the velocity of A is
40 m and hits the ground with a speed that is three vA
times its initial speed. What is the time taken (in vA and that of B is vB then the value of v is
B
sec) for the fall?
(A) 2 sec (B) 3 sec (A) 1/2 (B) 1/ 3
(C) 4 sec (D) 8 sec (C) (D) 1/3
3
24. The v-t graph of a linear motion is shown in adjoining 27. If position time graph of a particle is sine curve as
figure. The distance from origin after 8 seconds is - shown, what will be its velocity-time graph

v v

(A) (B)
(A) 18 meters (B) 16 meters t t
(C) 8 meters (D) 6 meters

v
v
25. The adjoining curve represents the velocity-time
graph of a particle, its acceleration values along OA, (C) (D)
t
AB and BC in metre/sec2 are respectively- t

1
28. The diagram shows variation of with respect to
v
time (where v is in m/s).
1s 
 
v m

45º
t(s)
3
what is the instantaneous acceleration of body at

(A) 1, 0, -0.5 (B) 1, 0, 0.5  m

t = 3 sec.  in 2  .
(C) 1, 1, 0.5 (D) 1, 0.5, 0  s 

1
(A) 2 (B)
26. In the following velocity-time graph of a body, the 3
distance and displacement travelled by the body in (C) 1 (D) None of these
5 second in meters will be -
29. The particle moves with rectilinear motion given
the acceleration-displacement (a-S) curve is shown
in figure, determine the velocity after the particle
has traveled 30 m.
If the initial a(m/s2 )
velocity is
10 m/s. 10
(A) 10m/s
(A) 75,115 (B) 105, 75 (B) 40m/s.
(C) 45, 75 (D) 95, 55 (C) 20m/s. S(m)
15 30
(D) 60m/s.
30. Figure shows the graph of the x-co-ordinate of a 33. Velocity-time graph of a particle moving in a straight
particle going along the x-axis as function of time. line is shown in figure. In the time interval from
Find the instantaneous speed of particle at t = 12.5 s t = 0 to t = 14 s, find:
(in m/s)
v(m/s)
x
20

10
A 10 12 14
8m 0 t(s)
2 4 6
4m –10

O 4s 8s 12s 16s B (a) average velocity and

 50   25 
(A)   m/s (B)   m/s
(A) 2m/s (B) 8m/s.  7   7 
(C) 4m/s. (D) 6m/s.
 20   15 
(C)   m/s (D)   m/s
 7  7
31. Figure shows the graph of velocity versus time for a
particle going along x axis. Initially at t = 0, particle is (b) average speed of the particle
at x = 3m. Find (A) 20 m/s (B) 40 m/s
position of v(m/s) (C) 10 m/s (D) 30 m/s
particle at 10
t = 2s. (in m) 34. A body initially at rest moving along x-axis in such
(A) 9m 2 a way so that its acceleraation Vs displacement
(B) 3m plot is as shown in figure. What will be the maximum
(C) 12m
O 8 t velocity of particle in m/sec.
(A) 1 a
(D) 6m
(B) 2 1m/s2

32. Displacement-time graph of a particle moving in a (C) 3

straight line is as shown in figure. (D) 5
0.5 1m S
s c
a b d Section D - Variable Acceleration
o t
35. At an instant t , the co-ordinates of a particle are
x = at2, y = bt2 and z = 0 , then its velocity at the
instant t will be
(a) Find the sign of velocity in regions oa, ab, bc and cd
(A) negative , positive, positive, negative (A) t a 2  b2 (B) 2t a 2  b 2
(B) positive , positive, positive, negative (C) (D) 2t2
a 2  b2 a 2  b2
(C) positive , negative, positive, positive
(D) positive , positive, negative, negative
36. The displacement of a body is given by 2s = gt2
(b) Find the sign of acceleration in the above region
where g is a constant. The velocity of the body at
(A) zero, postive, negative, negative any time t is
(B) positive , zero, negative, positive (A) gt (B) gt/2
(C) positive , zero, positive, negative (C) gt2/2 (D) gt3/3
(D) positive , zero, negative, negative
37. A particle is moving so that its displacement s is 43. A particle moving in a straight line has velocity
given as s = t3- 6t2 + 3t + 4 meter. Its velocity at the displacement equation as v  5 1  x . Here v is in
instant when its acceleration is zero will be - m/sec and x in metre. Select the correct alternative:-
(A) 3 m/s (B) -12 m/s (A) Particle is initially at rest.
(C) 42 m/s (D) -9 m/s (B) Initial velocity of the particle is 5 m/sec and the
particle has a constant acceleration of 12.5 m/sec2.
38. A particle starting from rest undergoes acceleration (C) Particle moves with uniform velocity.
given by a = | t – 2| m/s2 where t is time in sec. (D) None of these.
Velocity of particle after 4 sec is -
(A) 1 m/s (B) 2 m/s 44. The velocity of a body depends on time according
to the equation v = 20 + 0.1 t 2. The body is
(C) 8 m/s (D) 4 m/s
undergoing :
(A) uniform acceleration
39. The acceleration a in ms–2 of a particle is given by a (B) uniform retardation
= 3t2 + 2t + 2, where t is the time. If the particle (C) non–uniform acceleration
starts out with a velocity u = 2 ms–1 at t = 0, then (D) zero acceleration
find the velocity at the end of 2s.
(A) 16 m/s (B) 36 m/s 45. Starting from rest a particle moves in a straight line
(C) 18 m/s (D) 9 m/s with acceleration
1/ 2

a  25  t 2  m/s2 for 0  t  5s
40. The acceleration of a particle is given by
a = t3 – 3t2 + 5, where a is in ms–2 and t in sec. At t 3
a m / s 2 for t > 5s
= 1s, the displacement and velocity are 8.30 m and 8
6.25 ms–1, respectively. Calculate the displacement The velocity of particle at t = 7s is :
and velocity at t = 2 sec. (A) 11 m/s (B) 22 m/s
(A) 15.6 m, 4 m/s (B) 5.3 m, 4 m/s (C) 33 m/s (D) 44 m/s
(C) 15.6 m, 8 m/s (D) 15.3 m, 8 m/s
46. A particle moves along positive branch of the curve
Y = X2/2 with X governed by X = t2/2, where X
41. A particle starts moving along x-axis from t = 0, its
and Y are measured in metres and time t is seconds.
position varying with time as x = 2t3 – 3t2 + 1.
At t = 2s, the velocity of the particle is :
(a) At which time instants is its velocity zero ?
(A) 2iˆ  4ˆj (B) 4iˆ  2ˆj
(A) t = 0, 2s (B) t = 0, 3s
(C) t = 0, 4s (D) t = 0, 1s (C) 4iˆ  2ˆj (D) 2iˆ  4ˆj
(b) What is the velocity when it pass through origin ?
(A) v = 0 (B) v = 1 47. The motion of a body falling from rest in a viscous
(C) v = 5 (D) v = 3 dv
medium is described by  A  Bv , where A and
dt
42. A particle moves in the x-y plane with the velocity B are constants.
 The velocity at time t is given by :
v  aiˆ  bt ˆj . At the instant t  a 3 / b the
A
magnitude of total acceleration is : (A)
B

1  e  Bt  
(B) A 1  e
B t2


(A) b (B) a
(C) ABe–t (D) AB2 (1 – t)
(C) b/a (D) 2
a b 2
48. A street car moves rectilinearly from station A to 52. A body is thrown with a velocity of 9.8 m/s making
the next stop B with an acceleration varying an angle of 30º with the horizontal. It will hit the
according to the law f = a – bx, where a and b are ground after a time-
positie constants and x is its distance from station (A) 3 s (B) 2 s
A. The distance AB is then equal to : (C) 1.5 s (D) 1 s
2a a
(A) (B) 53. A projectile thrown with a speed v at an angle  has
b b
a range R on the surface of the earth. For same v
a and , its range on the surface of moon will be-
(C) (D) 2a – b
2b (A) R/6 (B) 6R
(C) R/36 (D) 36 R
49. The speed of a body moving on a straight track
varies according to v = 2t + 13 for 0  t  5s, 54. In a projectile motion the velocity -
v = 3t + 8 for 5 < t  7 s and v = 4t + 1 for (A) is always perpendicular to the acceleration
t < 7 s. The distances are measured in metre. The
(B) is never perpendicular to the acceleration
distance in metres moved by the particle at the end
(C) is perpendicular to the acceleration for one instant only
of 10 second is :
(D) is perpendicular to the acceleration for two instant
(A) 127 (B) 247
(C) 186 (D) 313
55. Two projectile A and B are projected with angle of
projection 15º for the projectile A and 45º for the
50. A particle moving along x–axis has acceleration 'f' at
projectile B. If RA and RB be the horizontal range
time t, given by
for the two projectiles, then -
 t (A) RA < RB (B) RA = RB
f  f0 1  
 T (C) RA > RB
where f0 and T are constants. The particle at t = 0 (D) The information is insufficient to decide the
has zero velocity. In the time interval between t = 0 relation of RA with RB
and the instant when f = 0, the particle's velocity
(vx) is : 56. A ball is thrown upwards. It returns to ground
describing a parabolic path. Which of the following
1 remains constant ?
(A) f0 T 2 (B) f0T2
2 (A) speed of the ball
1 (B) kinetic energy of the ball
(C) f 0T (D) f0T (C) vertical component of velocity
2
(D) horizontal component of velocity.

Section E - Ground-to Ground projectile 57. The angle of projection of a body is 15º . The other
Motion, Equation of Trajectory, angle for which the range is the same as the first
Projectile from tower, Projectile one is equal to-
Motion in Inclined plane (A) 30º (B) 45º
51. The horizontal and vertical distances travelled by a (C) 60º (D) 75º
particle in time t are given by x = 6t and
y = 8t – 5t2. If g = 10 m/sec2, then the initial velocity 58. A ball is thrown at an angle of 45º with the horizontal
of the particle is- with kinetic energy E. The kinetic energy at the
(A) 8 m/sec (B) 10 m/sec highest point during the flight is-
(C) 5 m/sec (D) zero (A) Zero (B) E/2
(C) E (D) (2)1/2E
59. The maximum range of a projectile is 22 m. When 65. A ball is hit by a batsman at an angle of 37º as
it is thrown at an angle of 15º with the horizontal, its shown in figure. The man standing at P should run
range will be- at what minimum velocity so that he catches the
(A) 22 m (B) 6 m ball before it strikes the ground. Assume that height
(C) 15 m (D) 11 m of man is negligible in comparison to maximum height
of projectile.
(A) 3 ms–1
5x 2 (B) 5 ms–1
60. The equation of projectile is y = 16x – . The
4 (C) 9 ms–1
horizontal range is- (D) 12 ms–1
(A) 16 m (B) 8 m
(C) 3.2 m (D) 12.8 m 66. Trajectories of two projectiles are shown in the
figure. Let T1 and T2 be the time periods and u1 and
61. If four balls A, B, C, D are projected with same u2 be their speeds of projection.
speed at angles of 15º, 30º, 45º and 60º with the Then -
(A) T2 > T1 Y
horizontal respectively, the two balls which will fall
at the same place will be- (B) T1 > T2
(A) A and B (B) A and D (C) u1 > u2
1 2
(C) B and D (D) A and C (D) u1 < u2
X

62. The velocity at the maximum height of a projectile 67. A hollow vertical cylinder of radius r and height h
is half of its initial velocity u. Its range on the has a smooth internal surface. A small particle is
horizontal plane is placed in contact with the inner side of the upper
rim, at point A, and given a horizontal speed u,
2u 2 3 u2 tangential to the rim. It leaves the lower rim at
(A) (B) point B, vertically below A. If n is an integer then-
3g 2g
u
u2 u2 (A) 2h / g = n A u
(C) (D) 2 r
3g 2g
h
(B) =n
2 r h
63. A marble A is dropped vertically, another identical
B
marble B is projected horizontally from the same 2 r
(C) =n
point at the same instant h r
(A) A will reach the ground earlier than B
u
(B) B will reach the ground earlier than A (D) 2gh = n
(C) both Aand B will reach the ground at the same instant
(D) none of the above
68. Time taken by the projectile to reach A to B is t.
64. An aeroplane is flying at a height of 1960 m in Then the distance AB is equal to –
horizontal direction with a velocity of 360 km/hr. ut
When it is vertically above the point. A on the ground, (A)
3
it drops a bomb. The bomb strikes a point B on the u
ground, then the time taken by the bomb to reach
3ut B
the ground is- (B)
2
60°
(A) 20 2 sec (B) 20 sec 30°
(C) 3 ut A
(C) 10 2 sec (D) 10 sec (D) 2 ut
69. Rank the launch angles for the five paths in the 73. An object A is moving with 10 m/s and B is moving with
figure below with respect to time of flight, from the 5 m/s in the same direction of positive x-axis. A is 100 m
shortest time of flight to the longest– behind B as shown. Find time taken by A to Meet B
(A) 18 sec.
10m/s 5m/s
(B) 16 sec.
y(m ) A B
(C) 20 sec.
150 vi = 50 m/ s
(D) 17 sec. 100m

75° 74. A police jeep is chasing a culprit going on a motor

100
60°
45°
bike. The motor bike crosses a turning at a speed of
50
30° 72 km/h. The jeep follows it a speed of 90 km/h
15°
x(m)
crossing the turning ten seconds later than the bike.
100 150 200 250
50 Assuming that they travel at constant speeds, how
far from the turning will the jeep catch up with the
(A) 15°, 30°, 45°, 60°, 75° bike ? (in km)
(A) 3 km (B) 5 km
(B) 75°, 60°, 45°, 30°, 15°
(C) 1 km (D) 7 km
(C) 15°, 75°, 30°, 60°, 45°
(D) 30°, 60°, 15°, 45°, 75° 75. A wooden block of mass 20g is dropped from the
top of the cliff 50 m high. Simultaneously a bullet of
70. A particle is projected with a velocity of 20 m/s at mass 20 g is fired from the foot of the cliff upwards
an angle of 30º to an inclined plane of inclination with a velocity 25 ms-1. The bullet and the wooden
30º to the horizontal. The particle hits the inclined block will meet each other after time :
plane at an angle 30º, during its journey. The time (A) 2 s (B) 1 s
of flight is - (C) 0.5 s (D) 4 s

4 2 76. A cart is moving horizontally along a straight line

(A) (B)
3 3 with constant speed 30m/sec. A particle is to be fired
vertically upwards from the moving cart in such a
3 way that it returns to the cart at the same point from
(C) 3 (D)
2 where it was projected after the cart has moved 80
m. At what speed (relative to cart) must the projectile
be fired? (g = 10m/sec2)
(A) 10 m/sec (B) 10 8 m / sec
Section F - Relative Motion
40
71. Two trains each of length 50 m are approaching (C) m / sec (D) None of these
3
each other on parallel rails. Their velocities are 10
m/sec and 15 m/sec. They will cross each other in -
77. A passenger in a train drops a ball from the window
(A) 2 sec (B) 4 sec of the train running at an acceleration a. A pedestrian,
(C) 10 sec (D) 6 sec on the ground, by the side of the rails, observes the
ball falling along
72. A car A is going north-east at 80 km/hr and another (A) a vertical with an acceleration g2  a 2
car B is going south-east at 60 km/hr. Then the
direction of the velocity of A relative to B makes (B) a vertical acceleration g2  a 2
with the north an angle such that tan  is -
(A) 1/7 (B) 3/4 (C) a parabola with an acceleration g2  a 2
(C) 4/3 (D) 3/5 (D) a parabola with an acceleration g
78. A monkey is climbing up a tree at a speed of 3 m/s. A 83. A boat moves relative to water a velocity which is
dog runs towards the tree with a speed of 4 m/s. What n times the river flow velocity. At what angle to the
is the relative speed of the dog as seen by the monkey? stream direction must be boat move to minimize
(A) > 7 m/s drifting ?
(B) Between 5 m/s and 7 m/s (A) /2 (B) sin–1 (1/n)
(C) 5 m/s (D) < 5 m/s  
(C) + sin–1 (1/n) (D) – sin–1 (1/n)
2 2
79. A car is going eastwards with a velocity of 8 m s–1.
To the passengers in the car, a train appears to be
84. A swimmer’s speed in the direction of flow of river
moving north wards with a velocity 15 m s–1. What
is 16 km h–1. Against the direction of flow of river,
is the actual velocity of the train?
the swimmer’s speed is 8 km h–1. Calculate the
(A) 7 ms–1 (B) 17 ms–1
–1
swimmer’s speed in still water and the velocity of
(C) 23 ms (D) None of the above
flow of the river.
(A) 12 km/h, 4 km/h (B) 10 km/h, 3 km/h
80. A particle is moving in x-y plane. At certain instant,
(C) 10 km/h, 4 km/h (D) 12 km/h, 2 km/h
the components of its velocity and acceleration are
as follows Vx = 3 m/s, Vy = 4 m/s, ax = 2 m/s2 and
85. A man wishes to cross a rive in a boat. If he crosses
ay = 1 m/s2. The rate of change of speed at this
the river in minimum time, then he takes 10 min
moment is
with a drift of 120 m. If he crosses the river taking
(A) 4 m/s2 (B) 2 m/s2
shortest route, he takes 12.5 min, find the velocity
2 2
(C) 3 m/s (D) 5 m/s of the boat with respect to water.
1 1
(A) (B)
3 5
Section G - River Boat Problems
1 1
(C) (D)
81. A river has width 0.5 km and flows from West to 6 2
East with a speed 30 km/hr. If a boatman starts
sailing his boat at a speed 40 km/hr normal to 86. A man crosses the river perpendicular to river flow
bank, the boat shall cross the river in time – in time t seconds and travels an equal distance down
(A) 0.6 minute (B) 0.75 minute the stream in T seconds. The ratio of man’s speed
(C) 0.45 minute (D) 3 minute in still water to the speed of river water will be:

t2  T2 T2  t 2
82. A boat man could row his boat with a speed (A) (B)
t 2  T2 T2  t2
10 m/sec. He wants to take his boat from P to a
point Q just opposite on the other bank of the river
flowing at a speed 4 m/sec. He should row his boat– t 2  T2 T2  t2
(C) (D)
t2  T2 T2  t 2
Q

vs = 4 m/s 87. A swimmer crosses the river along the line making
an angle of 45° with the direction of flow. Velocity
P of the river is 5 m/s. Swimmer takes 6 seconds to
cross the river of width 60 m. The velocity of the
(A) at right angle to the stream swimmer with respect to water will be :
(B) at an angle of sin–1 (2/5) with PQ up the stream (A) 10 m/s (B) 12 m/s
(C) at an angle of sin–1 (2/5) with PQ down the stream (C) 5 5 m/s (D) 10 2 m/s
(D) at an angle cos–1 (2/5) with PQ down the stream
88. A swimmer crosses a river with minimum possible 92. An aeroplane flying at a constant velocity releases a
time 10 second. And when he reaches the other end bomb. As the bomb drops down from the aeroplane.
starts swimming in the direction towards the point
(A) it will always be vertically below the aeroplane
from where he started swimming. Keeping the
direction fixed the swimmer crosses the river in 15 (B) it will always be vertically below the aeroplane
sec. The ratio of speed of swimmer with respect to only if the aeroplane is flying horizontally
water and the speed of river flow is (Assume (C) it will always be vertically below the aeroplane
constant speed of river & swimmer) : only if the aeroplane is flying at an angle of 45° to
3 9 the horizontal.
(A) (B)
2 4 (D) it will gradually fall behind the aeroplane if the
aeroplane is flying horizontally
2 5
(C) (D)
5 2
93. A helicopter is flying south with a speed of 50 kmh–1.
89. STATEMENT 1 : The magnitude of velocity of two A train is moving with the same speed towards east.
boats relative to river is same. Both boats start The relative velocity of the helicopter as seen by
simultaneously from same point on one bank may the passengers in the train will be towards.
reach opposite bank simultaneously moving along (A) north east (B) south east
different paths.
(C) north west (D) south west
STATEMENT 2 : For boats to cross the river in
same time. The component of their velocity relative
to river in direction normal to flow should be same. 94. Two particles are moving with velocities v1 and v2.
(A) Statement-1 is True, Statement-2 is True; Their relative velocity is the maximum, when the
Statement-2 is a correct explanation for Statement-1. angle between their velocities is
(B) Statement-1 is True, Statement-2 is True; Statement-
(A) zero (B) /4
2 is NOT a correct explanation for Statement-1.
(C) Statement-1 is True, Statement-2 is False. (C) /2 (D) 
(D) Statement-1 is False, Statement-2 is True.
95. A man in a balloon, throws a stone downwards with
90. A swimmer jumps from a bridge over a canal and swims a speed of 5 m/s with respect to balloon. The balloon
1 km upstream. After that first km, he passes a floating
is moving upwards with a constant acceleration of 5
cork. He continues swimming for half an hour and then
m/s2. Then velocity of the stone relative to the man
turns around and swims back to the bridge. The swimmer
after 2 second is :
and the cork reach the bridge at the same time. The
swimmer has been swimming at a constant speed. How (A) 10 m/s
fast does the water in the canal flow ? (B) 30 m/s
(A) 2 km/h (B) 3 km/h
(C) 15 m/s
(C) 1 km/h (D) 4 km/h
(D) 35 m/s

Section H - Rain Problems, Aircraft wind

96. Three stones A, B and C are simultaneously
problems, Relative Motion
between two projectiles projected from same point with same speed. A is
thrown upwards, B is thrown horizontally and C is
91. A man is walking on a road with a velocity
thrown downwards from a building. When the
3 km/hr. Suddenly rain starts falling. The velocity
distance between stone A and C becomes 10 m,
of rain is 10 km/hr in vertically downward direction.
The relative velocity of the rain is - then distance between A and B will be :

(A) (A) 10 m (B) 5 m

13 km/hr (B) 7 km/hr
(C) (C) 5 2 m/s (D) 10 2 m/s
109 km/hr (D) 13 km/hr
97. Two aeroplanes fly from their respective position 99. STATEMENT 1 : Three projectiles are moving in
‘A’ and ‘B’ starting at the same time and reach of differnt paths in the air. Vertical component of
point ‘C’ (along straight line) simultaneously when relative velocity between any of the pair does not
wind was not blowing. On a windy day they head change with time as long as they are in air. Neglect
towards ‘C’ but both reach the point ‘D’ the effect of air friction.
simultaneously in the same time which they took to STATEMENT 2 : Relative acceleration between any
reach ‘C’. Then the wind is blowing in : of the pair of projectiles is zero.
(A) Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct explanation for
Statement-1.
(C) Statement-1 is True, Statement-2 is False.
(A) North-East direction
(D) Statement-1 is False, Statement-2 is True.
(B) North-West direction
(C) Direction making an angle 0 <  < 90 with North
100. STATEMENT 1 : Two stones are projected with
towards West.
different velocities from ground from same point
(D) North direction
and at same instant of time. Then these stones cannot
collide in mid air. (Neglect air friction)
98. A particle is thrown up inside a stationary lift of STATEMENT 2 : If relative acceleration of two
sufficient height. The time of flight is T. Now it is particles initially at same position is always zero,
thrown again with same initial speed v0 with respect then the distance between the particle either remains
to lift. At the time of second throw, lift is moving up constant or increases continuously wiht time.
with speed v0 and uniform acceleration g upward
(A) Statement-1 is True, Statement-2 is True;
(the acceleration due to gravity). The new time of
Statement-2 is a correct explanation for Statement-1.
flight is:
(B) Statement-1 is True, Statement-2 is True;
T T Statement-2 is NOT a correct explanation for
(A) (B)
4 2 Statement-1.

(C) T (D) 2T (C) Statement-1 is True, Statement-2 is False.

(D) Statement-1 is False, Statement-2 is True.