NEET / JEE Main 11th PHYSICS KOSH EDUTECH [2]

4.1 Uniform and Non-uniform

Circular Motion
Single Correct Answer Type
1. A particle of mass 𝑚 is moving in a horizontal circle of radius 𝑟, under a centripetal force
𝑘
= 2 where 𝑘 is a constant.
𝑟 ′
a) The potential energy of the particle is zero 3. A body just being revolved in a vertical circle of
𝑘 radius 𝑅 with a uniform speed. The string
b) The potential energy of the particle is breaks when the body is at the highest point.
𝑟
𝑘 The horizontal distance covered by the body
c) The total energy of the particle is −
2𝑟 after the string breaks is
𝑘 a) 2 𝑅 b) 𝑅 c) 𝑅√2 d) 4 𝑅
d) The Kinetic energy of the particle is −
𝑟 Ans. a
𝑘
e) The potential energy of the particle is − Sol.: As the body just completes the circular path,
2𝑟 hence critical speed at the highest point.
Ans. a
𝑣𝐻 = √𝑔𝑅
Sol.: For horizontal planes potential energy remains
constant equal to zero, if we assumes surface to which is totally horizontal.
be the zero level. As the string breaks at the highest point, hence
2. A particle is moving in a circle of radius 𝑅 in form here onwards the body will follow
such a way that at any instant the normal and parabolic path. Time taken by the body to reach
tangential components of its acceleration are the ground
equal. If its speed at 𝑡 = 0 is 𝑣0 , the time taken to 2ℎ 2 × 2𝑅
complete the first revolution is [UP SEE 2005] 𝑡=√ =√
𝑔 𝑔
𝑅 𝑅
a) b) (1 − 𝑒 −2𝜋 ) Hence, horizontal distance covered by the body
𝑣0 𝑣0
𝑅 −2𝜋 2𝜋𝑅 = 𝑣𝐻 × 𝑡
c) 𝑒 d)
𝑣0 𝑣0 4𝑅
= √𝑔𝑅 × √ = 2𝑅
Ans. b 𝑔
Sol.: When a particle moves in a circular motion, it is 4. A ball of mass(𝑚)0.5 kg is attached to the end of
acted upon by centripetal force directed towards a string having length (𝐿) 0.5 m. The ball is
the centre. Hence, centripetal acceleration is rotated on a horizontal circular path about
𝑑𝑣 𝑣 2 vertical axis. The maximum tension that the
𝑎𝑁 = =
𝑑𝑡 𝑅 string can bear is 324 N. The maximum possible
𝑡 𝑣
𝑑𝑡 𝑑𝑣
or ∫ =∫ 2 value of angular velocity of ball (in rad/s) is
0 𝑅 𝑣0 𝑣
1𝑣
or 𝑡 = −𝑅 [ ]
𝑣 𝑣0
𝑣0 𝑅
𝑣=
𝑅 − 𝑣0 𝑡
𝑑𝑟 𝑣0 𝑅
Also =
𝑑𝑡 (𝑅 − 𝑣0 𝑡)
2𝜋 𝑅 𝑇
𝑑𝑡
∫ 𝑑𝑟 = 𝑣0 𝑅 ∫
0 0 𝑅 − 𝑣0 𝑡 a) 9 b) 18 c) 27 d) 36
𝑅
⟹ 𝑇 = (1 − 1 − 𝑒 −2𝜋 ) Ans. d
𝑣0 Sol.: 𝑇 cos θ component will cancel 𝑚𝑔.
NEET / JEE Main 11th PHYSICS KOSH EDUTECH [3]

c) 5 ms −1 d) 2√5 ms −1
Ans. c
Sol.: For water not to spill out of the bucket,
𝑣min = √5𝑔𝑅 (at the lowest point)
= √5 × 10 × 0.5 = 5 ms −1
7. A particle of mass 𝑚 is rotating in a horizontal
circle of radius 𝑅 and is attached to a hanging
mass 𝑀 as shown in the figure. The speed of
rotation required by the mass 𝑚 keep 𝑀 steady
is
𝑇 sin θ Component will provide necessary
centripetal force the ball towards center C.
∴ 𝑇 sin θ = 𝑚𝑟𝜔2 = 𝑚(𝑙 sin θ)ω2
𝑇
or 𝑇 = 𝑚𝑙ω2 ⟹ ω = √ml rad/s

𝑇max 324
or 𝜔max = √ =√ = 36 rad/s
𝑚𝑙 0.5 × 0.5
𝑚𝑔𝑅 𝑚𝑔𝑅 𝑚𝑔 𝑚𝑅
5. A wheel of radius 1m rolls forward half a a) √ b) √ c) √ d) √
𝑀 𝑚 𝑀𝑅 𝑀𝑔
revolution on a horizontal ground. The
magnitude to the displacement of the point of Ans. b
the wheel initially in contact with the ground is Sol.: To keep the mass M steady, let T is the tension in
[BCECE 2005] the string joining the two. Then for particle 𝑚,
𝑚𝑣 2
a) 2𝜋 b) √2𝜋 𝑇= … (i)
𝑅
c) √𝜋 + 4
2 d) 𝜋 For mass 𝑀,
Ans. c 𝑇 = 𝑀𝑔 … (ii)
Sol.: When wheel rolls half a revolution, the point (P ) From Eqs. (i) and (ii)
of the wheel which is in contact with the ground
initially, moves at the top of the wheel as shown. 𝑚𝑣 2 𝑀𝑔𝑅
= 𝑀𝑔 ⟹ 𝑣 = √
𝑅 𝑚

Horizontal displacement of point P,

𝑦 = 2𝑟
Net displacement = √𝑥 2 + 𝑦 2
= √(𝜋𝑟)2 + (2𝑟)2 8. A stone is swinging in a horizontal circle 0.8 m in
= 𝑟√𝜋 2 + 4 diameter, at 30rev/min. A distant light causes a
= √𝜋 + 4
2 shadow of the stone to be formed on a nearby
6. A bucket filled with water is tied to a rope of wall. What is the amplitude of the motion of the
length 0.5 m and is rotated in a circular path in shadow? What is the frequency? [BVP 2010]
vertical pane. The least velocity it should have at a) 0.4 m, 1.5 Hz b) 0.4 m, 0.5 Hz
the lowest point of circle so that water dose not c) 0.8 m, 0.5 Hz d) 0.2 m, 0.5 Hz
spill is, (𝑔 = 10 ms−2 ) [EAMCET 2007] Ans. b
a) √5 ms −1 b) √10 ms −1 Sol.: The amplitude is the radius of the circle
NEET / JEE Main 11th PHYSICS KOSH EDUTECH [4]

0.8 Equating Eqs. (iii) and (iv), we get

𝑅= = 0.4 m
2 𝑣2
The frequency of the shadow is the same as that ℎ=
𝑔
of the circular motion, so Given, 𝑣 = 0.5 ms−1 and 𝑔 = 10 ms−2
𝜔 = 30 rev/min (0.5)2
= 0.5rev/s = π rads −1 ∴ ℎ = 0.025 m = 2.5 cm
10
𝜔 𝜋
and v = = = 0.5 Hz.
2𝜋 2𝜋
9. The centripetal acceleration of a body moving in
11. The minimum velocity (in ms −1 ) with which a
a circle of radius 100 m with a time period of 2 s
car driver must traverse a flat curve of radius of
will be [UP SEE 2009]
150 m and coefficient of friction 0.6 to avoid
a) 98.5 ms −2 b) 198.5 ms −2
skidding is
c) 49.29 ms −2 d) 985.9 ms −2
a) 60 b) 30 c) 15 d) 25
Ans. d
Ans. b
Sol.: Centripetal acceleration, 𝑎 = 𝜔2 𝑟
2 Sol.: Using the relation
1
= 4𝜋 2 𝑛2 𝑟 = 4𝜋 2 ( ) × 100 𝑚𝑣 2
2 = 𝜇𝑅, 𝑅 = 𝑚𝑔
𝑟
𝜋 × 100ms = 985.9ms −2
2 −2
𝑚𝑣 2
= 𝜇𝑚𝑔 ⟹ 𝑣 2 = 𝜇𝑟𝑔
𝑟
10. A particle describes a horizontal circle in a 𝑣 2 = 0.6 × 150 × 10
conical funnel whose inner surface is smooth 𝑣 = 30 ms−1
−1
with speed of 0.5 ms . What is the height of the 12. A body is moving with a certain velocity in a
plane of circle from vertex of the funnel? circular path. Now, the body reverses its
a) 0.25 cm b) 2 cm c) 4 cm d) 2.5 cm direction, then [OJEE 2005]
Ans. d a) the magnitude of centripetal force remains
Sol.: The various forces acting on the particle, are its same
weight mg acting vertically downwards, normal b) the direction of centripetal force remains
reaction N. Equating the vertical forces, we have same
c) the direction of centripetal acceleration
remains same
d) the of centripetal force does not change
Ans. a
13. If the length of the second’s hand in a stop-clock
is 3 cm, the angular velocity and linear velocity
of the tip is [WB-JEE 2007]
a) 0.2047 rads −1 , 0.0314 ms−1
b) 0.2547 rads −1 , 0.0314 ms−1
c) 0.1472 rads −1 , 0.06314 ms −1
d) 0.1047 rads −1 , 0.00314 ms −1
Ans. d
𝑁 sin θ = 𝑚𝑔 … (i) Sol.: Angular velocity of second’s hand
Also, centripetal force, 2𝜋 𝜋 3.14
𝑚𝑣 2 = = =
60 30 30
= 𝑁 cos θ … . (ii)
𝑅 = 0.1047 rad s −1
From Eqs. (i) and (ii), we get Linear velocity, 𝑣 = 𝑟𝜔
𝑅𝑔 = 3 × 10−2 × 0.1047 = 0.00314 ms −1
tan θ = 2 … . (iii)
𝑣 14. For a particle in uniform circular motion the
Also, from triangle 𝑂𝐴𝐵, acceleration a at a point 𝑃(𝑅, 𝜃) on the circle of
𝑅
tan θ = … . . (iv) the radius R is (here 𝜃 is measured from the
ℎ 𝑥 −axis) [AIEEE 2010]
NEET / JEE Main 11th PHYSICS KOSH EDUTECH [5]

𝑣2 𝑣2 16. A cyclist is travelling on a circular section of

a) − cos θ 𝐢̇ + sin θ ̂𝐣̇
̂
𝑅 𝑅 highway of radius 2500 ft at the speed of 60 mile
𝑣2 𝑣2 h−1. The cyclist suddenly applies the brakes
b) − sin θ 𝐢̇̂ + cos θ ̂𝐣̇
𝑅 𝑅 causing the bicycle to slown sown at constant
2
𝑣 𝑣2 rate. Knowing that after 8 s the speed has been
c) − cos θ 𝐢̇̂ − sin θ ̂𝐣̇
𝑅 𝑅 reduced to 45 mileh−1. The acceleration of the
𝑣2 𝑣2 bicycle immediately after the brakes have been
d) − 𝐢̇̂ + ̂𝐣̇
𝑅 𝑅 applied is [MP PET 2007]
Ans. c
a) 2 ft/s 2 b) 4.14 ft/s2
Sol.: For a particle in uniform circular motion
c) 3.10 ft/s2 d) 2.75 ft/s2
Ans. b
Sol.: Tangential acceleration, 𝑣𝑓 − 𝑣𝑖
𝑎𝑡 =
𝑡
45 − 60 22 −1
=( ) fts
8 15
11 −2
=− fts
4
22 2
𝑣 2 𝑣 2 (60 × 15)
𝐚= = towards center of circle Radial acceleration, 𝑎𝑟 = =
𝑅 𝑟 2500
𝑣2 = 3.1 fts −2
𝐚= (− cos θ 𝐢̇̂ − sin θ ̂𝐣̇)
𝑅 Then, 𝑎 = √𝑎𝑟2 + 𝑎𝑡2 = 4.14 fts −2
𝑣2 𝑣2
or 𝐚 = cos θ 𝐢̇̂ − sin θ ̂𝐣̇ 17. A small block is shot into each of the four tracks
𝑅 𝑅
15. A bob of mass M is suspended by a massless as shown below. Each of the frictionless track
string of length L. The horizontal velocity 𝑣 at rises to the same height. The speed, which the
position A is just sufficient to make it reach the block enters the tracks, is same in all cases. At
point B. The angle 𝜃 at which the speed of the the highest point of the track, normal reaction is
bob is half of that at A, satisfies maximum in [WB-JEE 2006]

a) b)

c) d)
[IIT JEE 2008]
𝜋 𝜋 𝜋
a) 𝜃 = b) < 𝜃 <
4 4 2 Ans. a
𝜋 3𝜋 3𝜋
c) < 𝜃 < d) <𝜃<𝜋 Sol.: Since the block rises to the same heights in all
2 4 4
the four cases, from conversation of energy,
Ans. d
speed of the block at highest point will be same
Sol.: Velocity of the bob at the point A
in all four cases. Say it is 𝑣0 . Equation of motion
𝑣 = √5𝑔𝐿 …..(i)
will be
𝑣 2
( ) = 𝑣 2 − 2𝑔ℎ … . (ii)
2
ℎ = 𝐿(1 − cos θ) … (iii)
Solving Eqs. (i), (ii)and (iii), we get
7
cos θ = −
8
7
or θ = cos−1 (− ) = 151°
8
NEET / JEE Main 11th PHYSICS KOSH EDUTECH [6]

𝑚𝑣02 motion towards the centre, hence acceleration is

𝑁 + 𝑚𝑔 =
𝑅 perpendicular to velocity.
2
𝑚𝑣0
or 𝑁 = − 𝑚𝑔
𝑅
R (the radius of curvature) in first case
minimum. Therefore, normal reaction N will be
maximum in first case.
18. A point P moves in counter-clockwise direction
on a circular path as shown in the figure. The
movement of P is such that it sweeps out length 20. A cyclist moves in such a way that he track 60°
𝑠 = 𝑡 3 + 5, where 𝑠 is in metre and t is in turn after 100 m. What is the displacement
second. The radius of the path is 20 m. The when to takes seventh turn?
acceleration of P when t =2s is nearly a) 100 m b) 200 m c) 100√3 m d) 100√3 m
Ans. a
Sol.: In 6 turns each of 60°, the cyclist traversed a
regular hexagon path having each side 100 m.
So, at 7th turn, he will be again at

a) 13 ms −2 b) 12 ms −2 c) 7.2 ms −2 d) 14 ms −2
Ans. d
Sol.: Given, s = 𝑡3 + 5
𝑑𝑠
Speed, v= = 3𝑡 2 Point B (as shown) which is at distance 100 m
𝑑𝑡
𝑑𝑣 from starting point A. Hence, net displacement
and rate of change of speed, 𝑎𝑡 = = 6𝑡
𝑑𝑡 of cyclist is 100 m.
∴ Tangential acceleration at 𝑡 = 2 𝑠, 21. A bridge is in the form of a semi-circle of radius
𝑎𝑡 = 6 × 2 = 12 ms−2 40 m. The greatest speed with which a motor
and at 𝑡 = 2𝑠, 𝑣 = 3(2)2 = 12ms−1 cycle can cross the bridge without leaving the
v 2 144 −2 ground at the highest point is (𝑔 =
∴ Centripetal acceleration, 𝑎𝑐 = = ms
𝑅 20 10ms−2 ) (frictional force is negligibly small)
∴ Net acceleration = 𝑎𝑡2 + 𝑎𝑖2 ≈ 14ms −2 a) 40 ms−1 b) 20 ms−1 c) 30 ms−1 d) 15 ms−1
19. A body executing uniform circular motion has at e) 25 ms−1
any instant its velocity vector and acceleration Ans. b
vector [DCE 2003] Sol.: Given, 𝑟 = 40 m and g = 10m/s2
a) along the same direction we have 𝑣 = √𝑔𝑟
b) in opposite direction
= 10 × 40 = √400
c) normal to each other
= 20 m/s
d) not related to each other
22. A can filled with water is revolved in a vertical
Ans. c
of radius 4 m and the water does not fall down.
Sol.: An object moving in uniform circular motion is
The time period for a revaluation is about
moving around the perimeter of the circle with a
a) 2 s b) 4 s c) 8 s d) 10 s
constant speed. While the speed of object is
Ans. b
constant, its velocity is changing. Velocity being
Sol.: When a body is revolving in circular motion it is
a vector quantity has a constant magnitude but a
acted upon by a centripetal force directed
changing direction. The direction is always
towards the center. Water will not fall if weight
directed tangent line is always pointing in a new
is balanced by centripetal force. Therefore
direction. Also when it is moving in circular
NEET / JEE Main 11th PHYSICS KOSH EDUTECH [7]

a) 1 cm b) 2 cm c) 4 cm d) 8 cm
Ans. a
Sol.: The object will slip if centripetal force acting on
it is more than friction force.
So, 𝑚𝑟𝜔2 > 𝜇𝑚𝑔
𝑚𝑣 2 𝑟𝜔2 ≥ μ𝑔
𝑚𝑔 =
𝑟 𝑟𝜔2 = constant
⟹ 𝑣 2 = 𝑟𝑔 … (i) 𝑟1 𝜔2 2
Circumference of a circle is 2𝜋𝑟. =( )
𝑟2 𝜔1
2π𝑟
Time of revoluation = 4 2𝜔 2
𝑣 =( )
𝑟2 𝜔
Putting the value of 𝑣 from Eq. (i), we get
𝑟2 = 1 cm
2𝜋𝑟 𝑟 26. One end of a string of length 𝑙 is connected to a
𝑇= = 2𝜋√
√𝑔𝑟 𝑔 particle of mass 𝑚 and other to a small peg on a
m2 smooth horizontal table. If the particle moves in
Given, 𝑟 = 4 m, 𝑔 = 9.8 a circle with speed 𝑣, the net force on the
s
particle (directed towards the centre) is [BCECE
4
∴ 𝑇 = 2𝜋√ 2006, RPET 2006]
9.8
𝑚𝑣 2
4𝜋 a) 𝑇 b) 𝑇 −
⟹𝑇= = 4s 𝑙
√9.8 𝑚𝑣 2 d) zero
23. A particle moves in a circular path with c) 𝑇 +
𝑙
decreasing speed. Choose the correct statement. Ans. a
[IIT JEE 2005] Sol.: When particle moves in circle, then the resultant
a) Angular momentum remains constant force must act towards the center and its
b) Acceleration (a) is towards the center magnitude F must satisfy
c) Particle moves in a spiral path with 𝑚𝑣 2
decreasing radius 𝐹=
𝑙
d) The direction of angular momentum remains This resultant force is directed towards the
constant center and it is called centripetal force. This
Ans. d force originates form tension T.
Sol.: 𝐋 = 𝑚(𝐫 × 𝐯) 𝑚𝑣 2
∴𝐹= =𝑇
Direction of (𝐫 × 𝐯), hence the direction of 𝑙
angular momentum remains the same. 27. A car runs at a constant speed on a circular track
24. A pallet of mass 1 g is moving with an angular of radius 100 m, taking 62.8 s for every circular
velocity of 1 rads −1 along a circle of radius 1 m lap. The average velocity and average speed for
the centrifugal force is each circular lap respectively is [AMU 2007]
a) 0.1 dyne b) 12 dyne c) 10 dyne d) 100 dyne a) 0,0 b) 0, 10 ms−1
Ans. d c) 10 ms−1 , 10 ms−1 d) 10 ms−1 , 0
Sol.: Given, 𝑚 = 1 × 10−3 kg, ω = 1 rad s −1 and 𝑟 Ans. b
= 1m Sol.: On a circular path in completing one turn, the
Hence, centrifugal force = 𝑚𝜔2 𝑟 = 10−3 N distance traveled is 2𝜋𝑟 while displacement is
1N = 105 dyne zero.
∴ Centrifugal force = 100 dyne displacement 0
Hence, average velocity = =
25. A small object placed on a rotating horizontal time interval 𝑡
turn table just slips when it is placed at a =0
distance of 4 cm from the axis of rotation, if the distance
Average speed =
angular velocity of the turn table is doubled the time interval
2π𝑟 2 × 3.14 × 100
object slips when its distance from the axis of = = = 10 ms−1
𝑡 62.8
rotation is
NEET / JEE Main 11th PHYSICS KOSH EDUTECH [8]

28. A cyclist is travelling with velocity 𝑣 on a banked conserved? [IIT JEE 2003]
curved road of radius R. The angle 𝜃 through a) center of the circle
which the cyclist leans inwards is given by [J & b) on the circumference of the circle
K CET 2008] c) inside the circle
𝑅𝑔 d) outside the circle
a) tan θ = 2 b) tan θ = 𝑣 2 𝑅𝑔
𝑣 Ans. a
𝑣 2𝑔 𝑣2 Sol.: In uniform circular motion the only force acting
c) tan θ = d) tan θ =
𝑅 𝑅𝑔 on the particle is centripetal (towards centre).
Ans. d Torque of this force about the centre is zero.
Sol.: 𝑅 cos θ = 𝑚𝑔 … (i) Hence angular momentum about centre remains
𝑚𝑣 2 conserved.
𝑅 sin θ = … (ii)
𝑅 32. What is the angular velocity of earth? [OJEE
From, Eqs. (i) and (ii) we get 2005]
𝑣2 2𝜋 2𝜋
∴ tan θ = a) rad s −1 b) rad s −1
𝑅𝑔 86400 3600
29. A cyclist goes round a circular path of 2𝜋 2𝜋
c) rad s −1 l d) rad s −1
circumference 34.3 m in √22 s, the angle made 24 6400
by him with the vertical will be Ans. a
a) 45° b) 40° c) 42° d) 48° Sol.: Time period of earth on its own axis
Ans. a 𝑇 = 24 h
Sol.: 3403 2𝜋𝑟 = 24 × 60 × 60 s
Here, 2𝜋𝑟 = 34.3 ⟹ 𝑟 = and 𝑣 = 2𝜋
2𝜋 𝑇 ∴ Angular velocity ω =
2𝜋𝑟 𝑇
= 2𝜋
√22 =
24 × 60 × 60
−1
𝑣2 2𝜋
Angle of banking θ = tan ( ) = 45° = rads −1
𝑟𝑔 86400
30. A toy cyclist completes one round of a square 33. A roller coaster is designed such that riders
track of side 2 m in 40 s. What will be the experience ‘weightlessness’ as they go round the
displacement at the end of 3 min? top of a hill whose radius of curvature is 20 m.
a) 52 m b) Zero c) 16 m d) 2√2 m The speed of the car at the top the hill is
e) 4√2 m between [MHT-CET 2008]
Ans. d a) 14 ms−1 and 15 ms −1 b) 15 ms−1 and 16 ms −1
Sol.: Displacement is distance from initial to final c) 16 ms−1 and 17 ms −1 d) 13 ms−1 and 14 ms −1
position In 40s cyclist completes =1 round Ans. a
∴ In 3 min(180 s) cyclist will complete Sol.: Balancing the force, we get
1 𝑣2
= 4 round Displacement for 4 round is zero. 𝑀𝑔 − 𝑁 = 𝑀
2 𝑅
𝑙 For weightlessness, 𝑁 = 0
Displacement for round = length of diagonal
2 𝑀𝑣 2
= 2√2m. ∴ = 𝑀𝑔
𝑅
or 𝑣 = √𝑅𝑔
Putting the values, 𝑅 = 20 m, 𝑔 = 10.0 ms−2
So, 𝑣 = √20 × 10.0 = 14.14 ms−1
Thus, the speed of the car at the top of the hill is
between 14 ms−1 and 10 ms−1

31. A particle undergoes uniform circular motion.

About which point on the plane of the circle, will
the angular momentum of the particle remain
NEET / JEE Main 11th PHYSICS KOSH EDUTECH [9]

with a constant speed of 31.4 ms−1. What is its

average speed for one complete revolution?
[DCE 2004]
a) Zero b) 31.4 ms−1
c) 3.14 ms−1 d) √2 × 31.4 ms−1
Ans. b
Sol.: The time taken by the particle for one complete
34. A particle originally at a rest at the highest point revolution.
of a smooth circle in a vertical plane, is gently 2𝜋𝑟
𝑡=
pushed and starts sliding along the circle. It will speed
leave the circle at a vertical distance ℎ below the 2 × 3.14 × 100
= = 20s
highest point such that 31.4
Hence, averge speed is
2 × 3.14 × 100
𝑣av = = 31.4 ms −1
20

𝑅 𝑅
a) ℎ = 2𝑅 b) ℎ = c) ℎ = 𝑅 d) ℎ =
2 3
Ans. d 36. A stone tied to one end of rope and rotated in a
Sol.: From law of conservation of energy, potential circular motion. If the string suddenly breaks,
energy of fall gets converted to kinetic energy. then the stone travels [J & K CET 2007]
a) in perpendicular direction
b) in direction of centrifugal force
c) towards centripetal force
d) in tangential direction
Ans. d
Sol.: If the string suddenly breaks, the centripetal
∴ PE = KE force will be zero only tangential force will be
1 present, then the stone travels in tangential
𝑚𝑔ℎ = 𝑚𝑣 2
2 direction.
or 𝑣 = √2𝑔ℎ … . (i) 37. If KE of the particle of mass 𝑚 performing UCM
Also, the horizontal component of force is equal in a circle of radius 𝑟 is E. Find the acceleration
centrifugal force. of the particale
𝑚𝑣 2 2𝐸 2𝐸 2 4𝐸
∴ 𝑚𝑔 cos θ = … (ii) a) b) ( ) c) 2 𝐸𝑚𝑟 d)
𝑅 𝑚𝑟 𝑚𝑟 𝑚𝑟
From Eq. (i) Ans. a
2𝑚𝑔ℎ Sol.: Kinetic energy
∴ 𝑚𝑔 cos θ = … (iii)
𝑅 1
From ∆ 𝐴𝑂𝐵, 𝐸 = 𝑚𝑣 2
2
(𝑅 − ℎ 1 𝑣2
cos θ = or 𝑚𝑟 =𝐸
𝑅 2 𝑟
(𝑅 − ℎ) 2𝑚𝑔ℎ 1
⟹ 𝑚𝑔 ( )= or 𝑚𝑟𝑎 = 𝐸
𝑅 𝑅 2
⟹ 3ℎ = 𝑅 2𝐸
or 𝑎 =
𝑅 𝑚𝑟
⟹ℎ= 38. Which of the following statements is false for a
3
particle moving in a circle with a constant
angular speed? [AIEEE 2004]
35. An object is moving in a circle of radius 100 m a) The velocity vector is tangent to the circle
NEET / JEE Main 11th PHYSICS KOSH EDUTECH [10]

b) The acceleration vector is tangent to the Substituting the value of 𝑣 from Eq. (i) in Eq. (ii),
circle we get
c) The acceleration vector point to the center of 𝑚 𝐿 2 𝐿2
the circle 𝐹𝑐 = [ ] =
𝑟 𝑚𝑟 𝑚𝑟 3
d) The velocity and acceleration vectors are
perpendicular to each other 42. A particle is moving in a circle with uniform
Ans. b speed𝑣. In moving from a point to another
Sol.: For a particle moving in a circle with constant diametrically opposite point [OJEE 2003]
angular speed, velocity vector is always tangent a) the momentum changes by 𝑚𝑣
to the circle and the acceleration vector always b) the momentum changes by 2 𝑚𝑣
points towards the center of circle or is always c) the kinetic energy changes by (1/2) 𝑚𝑣 2
along radius of the circle. Since, tangential d) the kinetic energy changes by 𝑚𝑣 2
vector is perpendicular to the acceleration Ans. b
vector. But in no case acceleration vector is Sol.: Initial velocity
tangent to the circle. 𝑣1 = 𝑣
39. Given that centripetal force𝐹 = −𝑘/𝑟 2 . The Final velocity 𝑣2 = −𝑣
total energy is Initial momentum 𝑝1 = 𝑚𝑣
a) −𝑘/𝑟 2 b) 𝑘/𝑟 c) −𝑘/2𝑟 2 d) −𝑘/2𝑟 Final momentum 𝑝1 = 𝑚(−𝑣) = −𝑚𝑣
Ans. d Change in momentum ∆𝑝 = 𝑝1 − 𝑝2
Sol.: 𝑘 = 𝑚𝑣 − (−𝑚𝑣)
Centripetal force 𝐹 = − 2
𝑟 = 2𝑚𝑣j
𝑚𝑣 2 𝑘 𝑘
= 2 ⟹ 𝑚𝑣 2 =
𝑟 𝑟 𝑟
1 𝑘
Kinetic energy = 𝑚𝑣 2 =
2 2𝑟
Since the centripetal force is a conservative
force, and for a conservative force,
𝑑𝑈
𝐹= ⟹ 𝑈 = − ∫ 𝐹 ∙ 𝑑𝑟
𝑑𝑟 43. A weightless thread can bear tension upto 37 N.
𝑘 𝑘 A. stone of mass 500 g is tied to it and revolved
𝑈 = ∫ 2 𝑑𝑟 = −
𝑟 𝑟 in a circular path of radius 4 m in a vertical
𝑘 𝑘 𝑘
Toatal energy = 𝐾 + 𝑈 = − =− plane. If g = 10 ms−2 , then the maximum
2𝑟 𝑟 2𝑟 angular velocity of the stone will be [WB-JEE
40. A car moving with the speed of 10 m/s takes a
2006]
circular turn of radius 20 m. The magnitude of
a) 2 rad s −1 b) 4 rad s −1
the acceleration of the car is [MP PET 2010]
c) 8 rad s −1 d) 16 rad s −1
a) 5.0 ms −2 b) 50.0 ms −2
Ans. b
c) 0.25 ms −2 d) 0.5 ms −2
Sol.: Maximum tension in the thread is given by
Ans. a
𝑚𝑣 2
Sol.: 𝑣 2 (10)2 𝑇max = 𝑚𝑔 +
Centripetal acceleration = = = 5m/s2 𝑟
𝑟 20 or 𝑇max = 𝑚𝑔 + 𝑚𝑟𝑤 2 (∵ 𝑣 = 𝑟𝜔)
41. A particle of mass 𝑚 is circulating on a circle of
𝑇max − 𝑚𝑔
radius 𝑟 having angular momentum 𝐿, then the or 𝜔2 =
𝑚𝑟
centripetal force will be Given, 𝑇max = 37 N, m = 500g = 0.5 kg, 𝑔
a) 𝐿2 /𝑚𝑟 b) 𝐿2 𝑚/𝑟 c) 𝐿2 /𝑚𝑟 3 d) 𝐿2 /𝑚𝑟 2 = mg −2 ,
Ans. c 𝑟 = 4m
Sol.: Angular momentum 𝐿 = 𝑟 × 𝑝 = 𝑟 × 𝑚 × 𝑣 37 − 0.5 × 10 37 − 5
𝐿 ∴ 𝜔2 = =
𝑣= … . (i) 0.5 × 4 2
𝑚𝑟 or 𝜔2 = 16
𝑚𝑣 2 or ω = 4 rad s −1
Now, as centripetal force, 𝐹𝑐 = … . (ii)
𝑟
NEET / JEE Main 11th PHYSICS KOSH EDUTECH [11]

Ans. d
44. The time period of the second’s hand of a watch Sol.: Angle made by the cyclist with vertical is given
is by
a) 1 h b) 1 s c) 12 h d) 1 min 𝑣2
tan θ =
e) 0.1 h rg
Ans. d 10 × 10 5
∴ θ = tan−1 ( ) (∴ 𝑣 = 36 ×
Sol.: Second’s hand of a watch completes its one 80 × 10 18
rotation in 1 min. So, its time period is 1 min. = 10 ms−1 )
45. A wheel completes 2000 revolutions to cover
1
the 9.5 km distance, then the diameter of the = tan−1 ( )
8
wheel is
49. A particle moves in circle of radius 25 cm at the
a) 1.5 m b) 1.5 cm c) 7.5 cm d) 7.5 m
rate of two revolutions per second. The
Ans. a
acceleration of particle is
Sol.: Distance covered in n revoulation=n2𝜋𝑟 = 𝑛𝜋𝐷
a) 2𝜋 2 ms−2 b) 4𝜋 2 ms−2 c) 8𝜋 2 ms−2 d) 𝜋 2 ms−2
⟹ 2000𝜋𝐷 = 9500
Ans. b
Sol.: Acceleration of the particle is
9500
⟹𝐷= = 1.5m 𝑎 = 𝑟𝜔2 = 𝑟(2𝜋𝑛)2
2000 × 𝜋
46. If 𝛼 is angular acceleration, 𝜔 is angular velocity = 0.25 × (2𝜋 × 2)2
and 𝑎 is the centripetal acceleration then, which = 16𝜋 2 × 0.25
of the following is true? = 4𝜋 2 ms−2
𝜔𝑎 𝑣 𝑣𝑎 𝑎 50. A body moves along a circular path of radius 10
a) 𝛼 = b) 𝛼 = c) 𝛼 = d) 𝛼 = m and the coefficient of friction is 0.5. What
𝑣 𝜔𝑎 𝜔 𝜔𝑣
Ans. a should be its angular speed in rad s −1 , if it is not
Sol.: Centripetal acceleration to slip from the surface? (𝑔 = 9.8 ms −2 )
𝑣 a) 5 b) 10 c) 0.1 d) 0.7
𝑎=
𝑡 Ans. d
ω ω𝑣
Angular acceleration ∝= = Sol.: For body to move on circular path. Frictional
t 𝑣𝑡
𝜔𝑎 force provides the necessary centripetal force,
∴ ∝=
𝑣 𝑖𝑒, frictional force = centripetal force
47. A 500 kg car takes a round of radius 50 m with a 𝑚𝑣02
velocity of 36 kmh−1 . The centripetal force is or 𝜇𝑚𝑔 = = 𝑚𝑟𝜔2
𝑟
a) 250 N b) 750 N c) 1000 N d) 1200 N or μ𝑔 = 𝑟𝜔2
Ans. c ∴ 0.5 × 9.8 = 10 𝜔2
Sol.: Given, 𝑚 = 500 kg, or ω = 0.7 rad s −1
5
𝑣 = 36 kmh−1 = 36 × = 10 ms−1 and 𝑟
18 51. What will be the maximum speed of a car on a
= 50 m road-turn of radius 30m if the coefficient of
𝑚𝑣 2 friction between the tyres and the road is 0.4?
Centripetal force 𝐹 =
𝑟 [OJEE 2008]
500 × (10)2 a) 10.84 ms −1 b) 9.84 ms−1
Hence, 𝐹 = = 1000 N
50 c) 8.84 ms−1 d) 6.84 ms−1
48. A cyclist is moving on a circular track of radius
Ans. a
80 m with a velocity𝑣 = 36kmh−1. He has to
Sol.: Maximum speed 𝑣 = √𝜇𝑟𝑔 = √0.4 × 30 × 9.8
lean from the vertical approximately through an
angle = 10.84ms−1
(take 𝑔 = 10 ms−2 [WB-JEE 2007] 52. The angle of banking is independent of [MHT-
1 CET 2005]
a) tan−1 (4) b) tan−1 ( ) a) speed of vehicle
3
1 1 b) radius of curvature of road
c) tan−1 ( ) d) tan−1 ( )
4 8 c) height of inclination
NEET / JEE Main 11th PHYSICS KOSH EDUTECH [12]

d) None of the above Ans. a

Ans. d Sol.: Given, 𝑣 = 400 ms −1 , 𝑟 = 160 m, 𝑎 =?
53. The maximum and minimum tensions in the 𝑚𝑣 2
Centripetal force, 𝐹=
string whirling in a circle of radius 2.5 m are in 𝑟
the ratio 5:3, then its velocity is [Jamia Millia 𝑚𝑣 2
𝑚𝑎 =
Islamia 2004] 𝑟
a) √98 ms −1 b) 7 ms −1 𝑣2
or 𝑎 =
c) √490 ms−1 d) √4.9 ms−1 𝑟
(400)2 16 × 104
Ans. a So, 𝑎 = =
160 160
Sol.: 𝑚𝑣 2
Minimum tension 𝑇1 = − 𝑚𝑔 = 103 ms −2 = 1 kms−2
𝑟 56. Find the maximum speed at which a car can
𝑚𝑣 2
Maximum tension 𝑇2 = + 𝑚𝑔 turn round a curve of 30 m radius on a level
𝑟
road if the coefficient of friction between the
𝑚𝑣 2
Let =𝑥 tyres and the road is 0.4
𝑟
So, 𝑇1 = 𝑥 − 𝑚𝑔 …(i) (Acceleration due to gravity = 10 ms−2 )
𝑇2 = 𝑥 + 𝑚𝑔 …(ii) a) 12 ms −2 b) 10 ms −2 c) 11 ms −2 d) 15 ms −2
Diving Eq. (i) by Eq. (ii) Ans. c
𝑇1 𝑥 − 𝑚𝑔 𝑇1 3 Sol.: Centripetal force is provided by friction, so
= (∵ Given = ) 𝑚𝑣 2 𝑚𝑣 2
𝑇2 𝑥 + 𝑚𝑔 𝑇2 5 < 𝑓𝐿 𝑖𝑒, < 𝜇 𝑚𝑔
3 𝑥 − 𝑚𝑔 𝑟 𝑟
∴ = 𝑖. 𝑒, 𝑣 < √𝜇𝑔𝑟 so that, 𝑣max = √μ𝑔𝑟
5 𝑥 + 𝑚𝑔
or 3𝑥 + 3𝑚𝑔 = 5𝑥 − 5𝑚𝑔 Here, μ = 0.4, 𝑟 = 30m and 𝑔 = 10ms −2
or 𝑥 = 4 𝑚𝑔 ∴ 𝑣max = √0.4 × 30 × 10 = 11m/s
𝑚𝑣 2 57. When a body moves in a circular path, no work
𝑖𝑒, = 4𝑚𝑔
𝑟 is done by the force since [K CET 2004]
∴ 𝑣 2 = 4 𝑟𝑔 a) force and displacement are perpendicular
or 𝑣 = √4𝑟𝑔 other
or 𝑣 = √4 × 2.5 × 9.8 b) the force is always away from the center
𝑣 = √98 ms−1 c) there is no displacement
d) there is no net force
54. A car is circulating on the path of radius 𝑟 and at Ans. a
any time its velocity is 𝑣 and rate of increases of Sol.: When a body moves on a circular path then
velocity is 𝑎. The resultant acceleration of the force and distance are perpendicular to each
car will be other. Therefore, work done by the force is
𝑊 = 𝐹 ∙ 𝑑 cos θ
𝑣2 𝑣2 𝑣4 𝑣2 = 𝐹 ∙ 𝑑 cos 90° (∵ θ = 90°)
a) √ + 𝑟 b) √ + 𝑎 c) √ + 𝑎 d) ( + 𝑎)
2 2
𝑎2 𝑟 𝑟2 𝑟 = 0 (∵ cos 90° = 0)
Ans. c 58. A body is tied to one end of the string and
Sol.: Resultant acceleration whirled in a vertical circle, the physical quantity
which remains constant is [J & K CET 2007]
tangential 2 centripetal 2 a) Momentum b) Speed
= √( ) +( )
acceleration acceleration c) Kinetic energy d) Total energy
𝑣2
2
𝑣4 Ans. d
= √𝑎2 + ( ) = √ 2 + 𝑎2 Sol.: Momentum, speed and kinetic energy change
𝑟 𝑟
continuously in a vertical circular motion. The
55. The acceleration of a vehicle travelling with
physical quantity which remains constant is the
speed of 400ms−1 as it goes round a curve of total energy.
radius 160 m, is 59. A particle of mass 𝑚 is tied to one end of a string
a) 1 kms −2 b) 100 ms−2c) 10 ms−2 d) 1 ms −2 of length 𝑙 and rotated through the other end
NEET / JEE Main 11th PHYSICS KOSH EDUTECH [13]

along a horizontal circular path with speed 𝑣. = 𝜔2 𝑟 = (2𝜋𝑓)2 𝑟

The work done in half horizontal circle is [MP 22 1200 2 30
PET 2008] = (2 × × ) ×
7 60 100
a) Zero 𝑚𝑣 2 = 4740 ms −2
b) ( ) 2𝜋𝑙
𝑙 62. A particle is moving in a circle of radius 𝑅 with
𝑚𝑣 2 𝑚𝑣 2 constant speed𝑣. If radius is doubled, then its
c) ( ) 𝜋𝑙 d) ( )𝑙 centripetal force to keep the same speed gets
𝑙 𝑙
Ans. a [BCECE 2005]
Sol.: Here, centripetal force, a) twice as great as before
𝑚𝑣 2 b) half
𝐹= c) one-fourth
𝑙
But the angle between force and displacement is d) remains constant
90° because the direction of centripetal force is Ans. b
always towards the center and direction of Sol.: Centripetal force is given by
displacement is always tangential. 𝑚𝑣 2
𝐹=
Then work done 𝑅
1
𝑊 = 𝐅 ∙ 𝐬 = 𝐹𝑠 cos 90° ⟹𝐹∝
⟹ 𝑊=0 𝑅
𝐹2 𝑅1
or =
𝐹1 𝑅2
Given, 𝑟2 = 2𝑟1
𝐹2 𝑅1 1
∴ = =
𝐹1 2𝑅1 2
𝐹1
or 𝐹2 =
2
60. A car is moving along a straight horizontal road therefore, centripetal force will become half.
with a speed𝑣0 . If the coefficient of friction 63. The figure shows a circular path of a moving
between tyres and the road is 𝜇, the shortest particle. If the velocity of the particle at same
distance in which the car can be stopped is instant is 𝐯 = −3𝐢̇̂ − 4𝐣̇̂, through which quadrant
𝑣2 𝑣0 𝑣0 2 𝑣0 is the particle moving when clockwise and anti-
a) 0 b) c) ( ) d)
2𝜇𝑔 𝜇𝑔 𝜇𝑔 𝜇 clockwise respectively
Ans. a
Sol.: Retarding force 𝐹 = 𝑚𝑎 = 𝜇𝑅 = 𝜇 𝑚𝑔
𝑎 = 𝜇𝑔
Now, from equation of motion, 𝑣 2 = 𝑢2 − 2𝑎𝑠
∴ 0 = 𝑢2 − 2𝑎𝑠
𝑢2 𝑢2 𝑣02
∴ 𝑠= = =
2𝑎 2𝜇𝑔 2𝜇𝑔
61. An electric fan has blades of length 30 cm as [AMU 2005]
measured from the axis of rotation. If the fan is a) 1st and 4 th b) 2nd and 4th
rotating at 1200 rpm, the acceleration of a point c) 2 nd and 3rd d) 3 rd and 4 th
on the tip of the blade is about [MHT-CET 2006] Ans. b
a) 1600 ms −2 b) 4740 ms −2 Sol.: The figure shows a circular path of moving
c) 2370 ms −2 d) 5055 ms −2 particle. At any instant velocity of particle.
Ans. b
Sol.: Given, 𝑓 = 1200 rpm,
30
𝑟 = 30 cm = m
100
Acceleration of a point at the tip of the blade
= centripetal acceleration
NEET / JEE Main 11th PHYSICS KOSH EDUTECH [14]

v = −3𝐢̇̂ − 4𝐣̇̂ = (−3, −4) (in coordinate from). a) 𝑣𝑟 = 2𝑣𝑓 b) 𝑣𝑓 = 2𝑣𝑟 c) 𝑣𝑓 = 𝑣𝑟 d) 𝑣𝑓 > 𝑣𝑟
The coordinates of velocity show that particle is Ans. c
in 3rd quadrant at that instant. While moving Sol.: Speeds at the top point of each wheel will equal
clockwise particle will enter into 4th quadrant and is equal to the speed of centre of mass.
and these into 3rd and while moving 68. A cyclist moves in such a way that he track 60°
anticlockwise particle will enter into 2nd turn after 100 m. What is the displacement
quadrant and then into 3rd quadrant. when to takes seventh turn? [OJEE 2007]
∴ 4th and 2nd quadrants. a) 100 m b) 200 m
64. A particle is moving with a constant speed 𝑣 in a c) 100√3 m d) 100√3 m
circle. What is the magnitude of average after Ans. a
half rotation? Sol.: In 6 turns each of 60°, the cyclist traversed a
𝑣 𝑣 𝑣
a) 2𝑣 b) 2 c) d) regular hexagon path having each side 100 m.
𝜋 2 2𝜋 So, at 7th turn, he will be again at
Ans. b
Sol.: 2𝜋𝑟
Time 𝑇 =
𝑣

Point B (as shown) which is at distance 100 m

from starting point A. Hence, net displacement
𝑇 𝜋𝑟 of cyclist is 100 m.
and 𝑡0 = =
2 v 69. A stone of mass 1 kg tied to the end of a string of
2𝑟 2v
∴ v𝑎v = = length 1 m, is whirled in a horizontal circle with
𝜋 𝑟/v π
a uniform angular velocity 2 rads −1. The tension
65. A particle is acted upon by a force of constant
of the string is (in newton)
magnitude which is always perpendicular to the 1 1
velocity of the particle. The motion of the a) 2 b) c) 4 d)
3 4
particle takes place in a plane it follows that Ans. c
[Jamia Millia Islamia 2007] Sol.: The tension of the string,
a) Its velocity is constant 𝑇 = 𝑚𝑟𝜔2
b) Its acceleration is constant = 1 × 1 × (2)2 = 4N
c) Its kinetic energy is constant 70. The centripetal acceleration of a particle of mass
d) It moves in a straight line 𝑚 moving with a velocity 𝑣 in a circular orbit of
Ans. c radius 𝑟 is [Kerala CEE 2003, DCE 2003]
Sol.: When a force of constant magnitude acts on a) 𝑣 2 / 𝑟 along the radius, towards the center
velocity of particle perpendicularly, then there is b) 𝑣 2 / 𝑟 along the radius, away from the center
no change in the kinetic energy of particle. 𝑚𝑣 2 /𝑟 along the redius, away from the
Hence, kinetic energy remains constant. c)
center
66. A body moving with constant speed in a circular d) 𝑚𝑣 2 / 𝑟 along the radius, towards the center
path has [OJEE 2004] Ans. a
a) angular momentum b) constant acceleration 71. A body moving along a circular path of radius 𝑅
c) constant velocity d) no work done with velocity 𝑣, has centripetal acceleration𝑎. If
Ans. d its velocity is made equal to 2𝑣, then its
67. In a bicycle the radius of rear wheel is twice the centripetal acceleration is
radius of front wheel. If 𝑟𝑓 and 𝑟𝑟 are the radius, 𝑎 𝑎
a) 4𝑎 b) 2𝑎 c) d)
𝑣𝑓 and 𝑣𝑟 are the speed of top most points of 4 2
wheel, then Ans. a
NEET / JEE Main 11th PHYSICS KOSH EDUTECH [15]

Sol.: Centripetal force 𝑣

At highest point 𝑣𝑦 = 0, 𝑣𝑥 =
𝑚𝑣 2 √2
= 𝑚𝑎 Maximum height achieved,
𝑅
𝑣2 𝑣 2 sin2 45° 𝑣 2
or 𝑎 = 𝐻= =
𝑅 2𝑔 4𝑔
𝑎1 𝑣12 Now, angular momentum about O
∴ =
𝑎2 𝑣22 𝑚𝑣 𝑣 2 𝑚𝑣 3
= ∙ =
Here, 𝑣1 = 𝑣, 𝑣2 = 2𝑣, 𝑎1 = 𝑎 √2 4𝑔 4√2𝑔
2
𝑎 𝑣 1 74. A bullet is to be fired with a speed of 20 ms −1 to
∴ = 2
=
𝑎2 (2𝑣) 4 hit a target 200m away on a level ground. If 𝑔 =
or 𝑎2 = 4𝑎 10 ms−2 , the gun should be aimed [Kerala CEE
72. A plumb line is suspended from a ceiling of a car 2009]
moving with horizontal acceleration of𝑎. What a) directly at the target
will be the angle of inclination with vertical? b) 5 cm below the target
[OJEE 2003] c) 5 cm above the target
𝑎 𝑔
a) tan−1 ( ) b) tan−1 ( ) d) 2 cm above the target
𝑔 𝑎 e) 2 cm below the target
𝑎 𝑔
c) cas−1 ( ) d) cas−1 ( ) Ans. c
𝑔 𝑎
Sol.: Horizontal distance of target is 200 m.
Ans. a
Speed of bullet = 2000 ms−1
Sol.: Let the angle from the vertical be𝜃. The diagram
Time taken by bullet to cover the horizontal
showing the different forces is given
distance
200 1
𝑡= = s
2000 10
1
During s, the bullet will fall down vertically
10
due to gravitational acceleration.
Therefore, height above the target, so that the
bullet hits the target is
1 1 1
ℎ = 𝑢 + 𝑔𝑡 2 = (0 × ) + × 10 × (0.1)2
2 10 2
𝑎 = 0.05m = 5cm
Form the figure, tan θ =
𝑔 75. A particle moves along a parabolic path 𝑦 = 9𝑥 2
𝑎 in such a way that the 𝑥-componentes of
θ = tan−1
𝑔 velocity remains constant and has a
1
value3 ms −1. The acceleration of the projectile is
4.2 Projectile Motion 1 2
a) 𝐣̇̂ ms−2 b) 3𝐣̇̂ ms−2 c) 𝐣̇̂ ms−2 d) 2𝐣̇̂ ms−2
73. A particle of mass 𝑚 is projected with a velocity 2 3
𝑣 making an angle of 45° with the horizontal. Ans. d
The magnitude of the angular momentum of the Sol.: Given, equation is
particle about the point of projection when the 𝑦 = 9𝑥 2 … . (i)
particle is at its maximum height, is Since, 𝑥-component of velocity remains
𝑚𝑣 3 𝑚𝑣 3 constant, we have
a) 𝑚√2𝑔ℎ3 b) c) d) Zero 𝑑𝑥 1 −1
√2𝑔 4√2𝑔 = ms … (ii)
Ans. c 𝑑𝑡 3
Sol.: The speed of projectile is 𝑣 and angle of From Eq. (i), we have 𝑦-component of velocity.
projection is45°. 𝑑𝑦 𝑑𝑥 2
= 18𝑥 ∙ ( )
𝑣 𝑑𝑡 𝑑𝑡
𝑣𝑥 = 𝑣 cos 45° =
√2 𝑑𝑦 𝑑𝑥 2 1 2
𝑣 = 18 ( ) = 18 × ( ) = 2ms−2
𝑣𝑦 = 𝑣 sin 45° − 𝑔𝑡 = − 𝑔𝑡 𝑑𝑡 𝑑𝑡 3
√2