datavorous/entrance-exam-dataset huggingface dataset:
A bag is gently dropped on a conveyor belt moving at a speed of 2 m s-1. The coefficient of friction between the conveyor belt and bag is 0.4 Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is : [Take g=10 m s-2]

Options:
A 2 m
B 0.5 m
C 3.2 m
D 0.8 ms

The correct answer is:
0.5 m

Solution:
When the bag is dropped on the conveyor belt, the acceleration of the bag will be a=μg=4 m s-2

The bag will stop moving on the conveyor when the velocity of the bag and the velocity of the conveyor becomes equal.

u=0, v=2 m s-1

Using the equation of motion for constant acceleration,

v2=u2+2as

⇒s=0.5 m

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (27 Jul Shift 1)
A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed 10 ms-1 gets embedded in it, then loss of kinetic energy will be

Options:
A 4.9 J
B 9.8 J
C 14.7
D 19.6 J

The correct answer is:
9.8 J

Solution:
If the final velocity of the bullet and the bag is v, applying conservation of momentum,

mu=m+Mv⇒0.2×10=0.2+9.8v⇒v=0.2 m s-1

Loss in kinetic energy will be,

Ki-Kf=12×0.2×102-12×10×0.22=9.8 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (25 Jul Shift 2)
A bag $X$ contains 2 white and 3 black balls and another bag $Y$ contains 4 white and 2 black balls. One bag is selected at random and a ball is drawn from it. Then the probability for the ball chosen be white is

Options:
A $\frac{2}{15}$
B $\frac{7}{\frac{7}{15}}$
C $\frac{8}{15}$
D $\frac{14}{15}$

The correct answer is:
$\frac{8}{15}$

Solution:
Let $A$ be the event of selecting bag $X, B$ be the event of selecting bag $Y$ and $E$ be the event of drawing a white ball, then $P(A)=1 / 2, P(B)=1 / 2 \quad P(E / A)=2 / 5 \quad P(E / B)=4 / 6=2 / 3$
$P(E)=P(A) P(E / A)+P(B) P(E / B)=\frac{1}{2} \cdot \frac{2}{5}+\frac{1}{2} \cdot \frac{2}{3}=\frac{8}{15}$

Topic: Mathematics, Probability, JEE Main
A bakelite beaker has volume capacity of 500 cc at 30°C. When it is partially filled with Vm volume (at 30°C) of mercury, it is found that the unfilled volume of the beaker remains constant as temperature is varied. If γbester =6×10-60°C-1, where γ is the coefficient of volume expansion, then Vm (in cc ) is close to …

Options:

The correct answer is:
20

Solution:
∆Vm=∆Vc

Vmγm∆T=Vcγc∆T

Vm=Vcγcγm=500×6×10-61.5.×10-4=20cc

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2020 (03 Sep Shift 1)
A ball $A$ of mass $m$ moving along positive $x$-direction with kinetic energy $K$ and momentum $p$ undergoes elastic head on collision with a stationary ball $B$ of mass $M$ after collision the ball $A$ moves along negative $x$-direction with kinetic energy $\frac{K}{9}$, final momentum of $B$ is

Options:
A $p$
B $\frac{p}{3}$
C $\frac{4 p}{3}$
D $4 p$

The correct answer is:
$\frac{4 p}{3}$

Solution:
Kinetic energy
$\begin{gathered}K=\frac{1}{2} m u_1^2 \\ u_1=\sqrt{\frac{2 K}{m}} \\ \frac{1}{2} m v_1^2=\frac{K}{9} \\ v_1^2=\frac{2 K}{9 m} \\ \text { Velocity, } v_1=\sqrt{\frac{2 K}{9 m}}\end{gathered}$
Kinetic energy $K=\frac{p^2}{2 m}$
From conservation of the momentum
$\begin{aligned} m u_1 & =-m v_1+p_B \\ m\left(u_1+v_1\right) & =p_B\end{aligned}$
Final momentum of $B$
$\begin{aligned} p_B & =m\left[\sqrt{\frac{2 K}{m}}+\sqrt{\frac{2 K}{9 m}}\right] \\ & =\sqrt{2 m K}+\sqrt{\frac{2 m K}{3}}=p+\frac{p}{3}=\frac{4 p}{3}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball balanced on a vertical rod is an example of

Options:
A stable equilibrium
B unstable equilibrium
C neutral equilibrium
D perfect equilibrium

The correct answer is:
unstable equilibrium

Solution:
The body is said to be in the state of stable equilibrium when the centre of gravity of a body lies below the point of suspension or support. When the centre of gravity of a body lies above the point of suspension or support, it is said to be in the state of unstable equilibrium. Now, as the ball is at the top of the rod, is slightly disturbed, the centre of gravity is lowered down and line of action of its weight lies outside the base of the rod due to which a torque start acting on the ball. As a result the ball fell down. When the centre of gravity of a body lies on the point of suspension or support, it is said to be in state of neutral equilibrium.
Hence, the correct answer is "unstable equilibrium".

Topic: Physics, Work Power Energy, JEE Main
A ball falls freely from a height of 45 m. When the ball is at a height of 25 m, it explodes into two equal pieces. One of them acquires an additional horizontal component of velocity equal to 10 m s-1, while its vertical component remains the same. The distance between the two pieces when both strike the ground is

Options:
A 10 m

B 20 m

C 15 m

D 30 m

The correct answer is:
20 m

Solution:
Let at the time explosion velocity of one piece of mass is (10i^). If the velocity of other be v→2, then from conservation law of momentum (since there is no force in the horizontal direction), the horizontal component of v→2, must be -10 i^.
∴ The relative velocity of two parts in the horizontal direction =20 m s-1
Time taken by the ball to fall through 45 m,
=t=2hg=2×4510=3 s and time taken by the ball to fall through first 20m, t′=2h′g=2×2010=2 s. Hence time taken by ball pieces to fall from 25 m height to ground =t-t′=3-2=1 s.
∴ The horizontal distance between the two pieces at the time of striking on the ground
=20×1=20 m

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball floats on the surface of water in a container exposed to the atmosphere. When the container is covered and the air is partially removed, then the ball

Options:
A rises
B gets immersed more in water
C remains immersed at its former depth
D may rise or sink

The correct answer is:
gets immersed more in water

Solution:
In the given situation, the weight of the ball is supported by the buoyant force of water and buoyant force due to air. As the density of air becomes less, the buoyant force due to the air becomes less. Therefore, to compensate a greater buoyant force due to water is required which is achieved by a greater immersion of ball into the water.

Topic: Physics, Mechanical Properties of Fluids, COMEDK, COMEDK 2020
A ball hits the floor and rebounds after an inelastic collision. In this case

Options:
A the momentum of the ball just after the collision is the same as that just before the collision
B the mechanical energy of the ball remains the same in the collision
C the total momentum of the ball is conserved
D the total energy of the ball and the earth is conserved.

The correct answers are:
the total energy of the ball and the earth is conserved.

Solution:
The correct option is (4)
Momentum of the system changes only due to external force (or impulse). When a ball hits the floor and gets rebound, then no external force (Impulse) acts on the ball as well as the floor, thus total momentum of the "ball + earth" system remains conserved. Also, some of the mechanical energy of the ball is always lost in an inelastic collision due to its deformation, hence it cannot remain conserved in this collision.
Final Answer: (4).

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball (initially at rest) falls vertically for 2 s and hits a smooth plane inclined at 30Â° to the horizontal. The coefficient of restitution is 5 8. The distance along the plane between the first and second impact of the ball is

Options:
A 40.63 m
B 20.63 m
C 30.63 m
D 50.63 m

The correct answer is:
40.63 m

Solution:
Let the ball strike the plane at A. We assign a co-ordinate system with origin at A. The velocity of the ball while striking the 30o plane = gt = 2g m/s. The components of u along and perpendicular to the plane are u sin 30o and u cos 30o respectively. After impact, the velocity component u sin 30o along the plane remains unaffected while the component u cos 30o perpendicular to the plane becomes e(u cos 30o) on rebound.
Thus velocity of rebound = eâ¡ . 2 g . 3 2 = 3 eg
The acceleration perpendicular to the plane = - g cos 3 0 â = - g 3 / 2
If t be the time of striking the plane again, then
0 = 3 egt - 1 2 3 2 g t 2
â t = 4 eâ¡ = 4 Ã 5 8 = 2 . 5 sec
The distance AB= uâ¡ sin 3 0 Î¿ t + 1 2 g sin 3 0 Î¿ t 2
= 2 g Ã 2 . 5 Ã 1 2 + 1 2 Ã 1 0 Ã 1 2 5 2 2
= 2 5 + 2 5 0 1 6 = 4 0 . 6 3 m

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball is bouncing elastically with a speed $1 \mathrm{~m} / \mathrm{s}$ betwen walls of a railway compartment of size $10 \mathrm{~m}$ in a direction perpendicular to walls. The train is moving at a constant velocity of $10 \mathrm{~m} / \mathrm{s}$ parallel to the direction of motion of the ball. As seen from the ground.

Options:
A
The direction of motion of the ball changes every 10 seconds.

B
Speed of ball changes every 10 seconds.

C
Average speed of ball over any 20 seconds interval is fixed.

D
The acceleration of ball is the same as from the train

The correct answers are:
Speed of ball changes every 10 seconds.
,
Average speed of ball over any 20 seconds interval is fixed.
,
The acceleration of ball is the same as from the train

Solution:
As the motion is observed from ground, time to strike the ball with wall will after every 10 sec and compare to velocity of trains $(10 \mathrm{~m} / \mathrm{s})$, speed of ball is less $(1 \mathrm{~m} / \mathrm{s})$. The speed of ball is uniform, so average speed at any time remain same or $1 \mathrm{~m} / \mathrm{sec}$. with respect totrain or ground. The speed of the ball before collision with side of train is $10+1=11 \mathrm{~m} / \mathrm{s}$.
Speed after collision with side of train $=10-1=9 \mathrm{~m} / \mathrm{s}$. (Opposite to the direction of train) As speed is changing after travelling $10 \mathrm{~m}$ and speed is $1 \mathrm{~m} / \mathrm{s}$ hence, time duration of the changing speed is 10 . Change in velocity on collision will be in magnitude $=(11-9)=2 \mathrm{~m} / \mathrm{sec}$.
So magnitude of acceleration on both walls is same but direction will be opposite.
Thus, the train is moving with constant velocity hence, it will act as inertial frame of reference as that of Earth and acceleration will be same in both frames.

Topic: Physics, Motion In One Dimension
A ball is dropped from a building of height $45 \mathrm{~m}$. Simultaneously another ball is thrown up with a speed 40 $\mathrm{m} / \mathrm{s}$. Calculate the relative speed of the balls as a function of time.

Options:

None

Solution:
For the first ball falling from the top of building, $u_1=0, u_2=40 \mathrm{~m} / \mathrm{s}, h=45 \mathrm{~m}, a_1=g, t_1=t, v=v_1$ Velocity of the dropped ball after time $t$,
$$
\begin{aligned}
&v_1=u_1+a_1 t \\
&v_1=u_1+g t \\
&\Rightarrow v_1=g t
\end{aligned}
$$
(downward)
For the second ball thrown upward,
$$
u_2=40 \mathrm{~m} / \mathrm{s}, a_2=-g, v=v_2, t_2=t
$$
Velocity of the ball after time $t$,
$$
\begin{aligned}
&v_2=u_2+a_2 t \\
&v_2=u_2-g t \\
&v_2=(40-g t) \\
&\text { (upward) } \\
&v_2=-(40-g t) \quad \text { (downward) } \\
&\therefore \text { Relative velocity of Ist ball w.r.t. IInd ball } \\
&v_2=v_1-v_2 \\
&=g t-[-(40-g t)]=40 \mathrm{~m} / \mathrm{s} \quad \text { (downward) } \\
&
\end{aligned}
$$
So their relative speed remain $=(40-0)=40 \mathrm{~m} / \mathrm{s}$.

Topic: Physics, Motion In One Dimension
A ball is dropped from a building of height \( 45 \mathrm{~m} \). Simultaneously another ball is thrown up with a speed \( 40 \mathrm{~m} / \mathrm{s} \). Calculate the relative speed of the balls as a function of time.

Options:
A \( 40 \mathrm{~m} / \mathrm{s} \)
B \( 30 \mathrm{~m} / \mathrm{s} \)
C \( 20 \mathrm{~m} / \mathrm{s} \)
D \( 10 \mathrm{~m} / \mathrm{s} \)

The correct answer is:
\( 40 \mathrm{~m} / \mathrm{s} \)

Solution:
For the ball dropped from building, u1 = 0
Velocity of the dropped ball after time t
v1 = u1 + gt
v1 = gt (downward)
For the ball thrown up, u2 = 40 m/s
Velocity of the ball after time t
v2 = u2 – gt
= (40 – gt) (upward)
&therefore; Relative velocity of one ball w.r.t another ball
= v1 – v2
= gt – [– (40 – gt) ] = 40 m/s

Topic: Physics, Motion In One Dimension, JEE Main
A ball is dropped from a height h on a floor. The coefficient of restitution for the collision between the ball and the floor is e. The total distance covered by the ball before it comes to the rest.

Options:
A h1-2e2
B h1+e21-e2
C h1-e21+e2
D he2

The correct answer is:
h1+e21-e2

Solution:
Height achieved in nth collision is given by

hn=e2nh
Distance covered in first collision = h,

Distance covered in 2nd collision = 2e2h, and so on.

Total distance covered by the ball before it comes to rest:

s=h+2e2h+2e4h+......

â s=h1+2e2+2e4+2e6+......

from the sum of infinite geometric progression:

â s=h1+2e21-e2

âs=h1+e21-e2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball is dropped from a height of 20 m. If the coefficient of restitution for the collision between ball and floor is 0.5, after hitting the floor, the ball rebounds to a height of ______ m.

Options:

The correct answer is:
5

Solution:
As we know h'=e2h, where h'=height after the collision, h=height before the collision and e=coefficient of restitution.

Therefore,

h'=0.52×20 m=5 m

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2023 (31 Jan Shift 2)
A ball is dropped from height 5 m. What is the time (in seconds) after which ball stops rebounding if the coefficient of restitution between ball and ground e=12?

Options:

The correct answer is:
3

Solution:
T=vg+2evg+2e2vg+………∞
T=vg1+2e+2e2+………∞
T=vg2+2e+2e2+………∞-1
T=vg21+e+e2+………∞-1
T=vg2×11-e-1 ⇒ T=2ghg2×11-12-1 ⇒ T=3 s

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball is dropped from height H on to a horizontal surface. If the coefficient of restitution is e then the total time after which it comes to rest is

Options:
A 2Hg1- e1+ e
B 2Hg1+e1-e
C 2Hg1+e21- e2
D 2Hg1- e21+e2

The correct answer is:
2Hg1+e1-e

Solution:
The time it takes for the ball to fall from a height is given by

t=2Hg

The height up to which the ball bounces after the nth collision from the floor is

Hn=e2nH

t=2Hg+22H1g+22H2g+…∞
t=2Hg1+2e+2e2+ ……..∞
t=2Hg1+e1-e

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball is dropped from the top of a 100 m high tower on a planet. In the last 12s before hitting the ground, it covers a distance of 19 m. Acceleration due to gravity (in m s-2 ) near the surface on that planet is______

Options:

The correct answer is:
8

Solution:
gt-1222×12=81 ... (1)
12gt2=100 …. (2)
devide equation 1 & 2
tt-12=109

t=5 s

put t in equation 2

g=8 ms2

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2020 (08 Jan Shift 2)
A ball is dropped on a smooth inclined plane and is observed to move horizontally after the impact. The coefficient of restitution between the plane and ball is e. The inclination of the plane is

Options:
A 45o
B tan-1⁡e
C tan-1e
D tan-1e2

The correct answer is:
tan-1e

Solution:
Along the inclined plane
mu sinθ=mv cosθ
⇒tanθ=vu
Now, e=-v2-v1u2-u1
=-0--v sinθ0-u cosθ
tan2⁡θ=e

θ=tan-1e

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball is dropped on to a horizontal plate from a height h=9 m above it. If the coefficient of restitution is e=12, the total distance travelled before the ball comes to rest is

Options:
A 10 m

B 15 m

C 20 m

D 25 m

The correct answer is:
15 m

Solution:
The impact between the ball and the plate is direct impact. If u is the velocity of the ball before impact, its velocity after is eu, where e is the coefficient of restitution.
But uâ¡ = 2 gh since the ball falls from a height h. So the velocity of the ball at the first rebound
= v 1 = eâ¡ uâ¡ = eâ¡ 2 gh
The height h1 to which it rises after first rebound
hâ¡ 1 = v 1 2 2 g = eâ¡ 2 Â· 2 gh 2 g = eâ¡ 2 hâ¡
The velocity v2 after second rebound v2 = ev1 = e2u
The height h2 it rises after second rebound
hâ¡ 2 = v 2 2 2 g = eâ¡ 4 Â· uâ¡ 2 2 g = eâ¡ 4 Â· 2 gh 2 g = eâ¡ 4 hâ¡
In general, the height hn to which the ball rises after nth rebound is given by hn = e2nh.
Total distance travelled
= h + 2h1 + 2h2 + 2h3 + ................+2hn+............to â
= h + 2 (e2h + e4h + e6 h +.......)
= h + 2he2(1 + e2 + e4+...........)
= hâ¡ + 2 hâ¡ eâ¡ 2 1 - eâ¡ 2 = hâ¡ 1 + 2 eâ¡ 2 1 - eâ¡ 2 = hâ¡ 1 + eâ¡ 2 1 - eâ¡ 2
= 9 1 + 1 2 2 1 - 1 2 2 = 1 5 m

Topic: Physics, Work Power Energy, JEE Main
A ball is dropped vertically downwards from a height $h$ above the ground. It hits the ground inelastically and bounces up vertically. Neglecting subsequent motion and air resistance, which of the following graph represents variation between speed $(v)$ and height $(h)$ correctly?

Options:
A

B

C

D

The correct answer is:

Solution:
When a ball hits the ground vertically then some of its energy is lost due to which it bounces back to lesser height with less velocity. The energy of the ball keep on decreasing with time and finally it become zero. Due to this variation of velocity with height can be represented as follows :

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2012 (26 May Online)
A ball is dropped vertically from a height of $h$ onto a hard surface. If the ball rebounds from the surface with a fraction $r$ of the speed with which it strikes the latter on each impact, what is the net distance traveled by the ball up to the $10^{\text {th }}$ impact ?

Options:
A $2 \mathrm{~h} \frac{1-\mathrm{r}^{10}}{1-\mathrm{r}}$
B $\mathrm{h} \frac{1-\mathrm{r}^{20}}{1-\mathrm{r}^{2}}$
C $2 \mathrm{~h} \frac{1-\mathrm{r}^{22}}{1-\mathrm{r}^{2}}-\mathrm{h}$
D $2 \mathrm{~h} \frac{1-\mathrm{r}^{20}}{1-\mathrm{r}^{2}}-\mathrm{h}$

The correct answer is:
$2 \mathrm{~h} \frac{1-\mathrm{r}^{20}}{1-\mathrm{r}^{2}}-\mathrm{h}$

Solution:
$\quad$ Total distance $=\left(\frac{\mathrm{v}_{0}^{2}}{g}+\mathrm{r}^{2} \frac{\mathrm{v}_{0}^{2}}{\mathrm{~g}}+\mathrm{r}^{4} \frac{\mathrm{v}_{0}^{2}}{\mathrm{~g}}+\ldots\right.$ upto $10^{\text {th }}$ terms $)-\mathrm{h}=\frac{\mathrm{v}_{0}^{2}}{\mathrm{~g}}\left(1+\mathrm{r}^{2}+\mathrm{r}^{4}+\ldots .+10^{\text {th }}\right.$ term $)-\mathrm{h}$
also $v_{0}=\sqrt{2 g h}$
$$
\begin{array}{l}
\therefore \text { Total distance }=2 \mathrm{~h}\left(\frac{1-\left(\mathrm{r}^{2}\right)^{10}}{1-\mathrm{r}^{2}}\right)-\mathrm{h} \\
\text { or } \text { total distance }=\frac{2 \mathrm{~h}\left(1-\mathrm{r}^{20}\right)}{\left(1-\mathrm{r}^{2}\right)}-\mathrm{h}
\end{array}
$$

Topic: Physics, Motion In One Dimension, JEE Main
A ball is held at rest at position A by two light strings. The horizontal string is cut and the ball starts swinging like a pendulum. Point B is the farthest to the right the ball goes as it swings back and forth. What is the ratio of the tension in the supporting string in position B to its value at A before the horizontal string was cut?

Options:
A sin2⁡β
B cos2⁡β
C tan2⁡β
D cos⁡β

The correct answer is:
cos2⁡β

Solution:
At A:

Tâ²cosâ¡Î²=mg
Tâ²=mgcosâ¡Î²
At: B
Tâ²â²- mg cosâ¡Î²= mv21
at B ; v=0
Tâ²â²=mg cosâ¡Î²
Tâ²â²Tâ²= mg cosâ¡Î²mg/cosâ¡Î²
Tension at BTension at A=Tâ²â²Tâ²=cos2Î²

Topic: Physics, Laws of Motion, JEE Main
A ball is let fall from a height $h_0$. It makes $n$ collisions with the earth. After $n$ collisions it rebounds with a velocity $v_n$ and the ball rises to a height $h_n$, then coefficient of restitution is given by

Options:
A $e=\left[\frac{h_n}{h_0}\right]^{1 / 2 n}$
B $e=\left[\frac{h_0}{h_n}\right]^{1 / 2 n}$
C $e=\frac{1}{n} \sqrt{\frac{h_n}{h_0}}$
D $e=\frac{1}{n} \sqrt{\frac{h_0}{h_n}}$

The correct answer is:
$e=\left[\frac{h_n}{h_0}\right]^{1 / 2 n}$

Solution:
The velocity of ball after $n$th rebound will be
$v_n=e^n v_0 \quad\left(\text { where } v_0=\sqrt{2 g h_0}\right)$
Therefore, the height after $n$th rebound will be
$\begin{aligned}
h_n & =\frac{v_n^2}{2 g}=\frac{e^{2 n} v_0^2}{2 g}=\frac{e^{2 n} \times 2 g h_0}{2 g} \\
\Rightarrow \quad e^{2 n} & =\frac{h_n}{h_0} \text { or } e=\left[\frac{h_n}{h_0}\right]^{1 / 2 n}
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball is made of a material of density $\rho$ where $\rho_{\text {oil }} < \rho < \rho_{\text {water }}$ with $\rho_{\text {oil }}$ and $\rho_{\text {water }}$ representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position?

Options:
A

B

C

D

The correct answer is:

Solution:
$$
\rho_{\text {oil }} < \rho < \rho_{\text {water }}
$$
Oil is the least dense of them so it should settle at the top with water at the base. Now the ball is denser than oil but less denser than water. So, it will sink through oil but will not sink in water. So it will stay at the oil-water interface.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2010
A ball is moving uniformly in a circular path of radius 1 m with a time period of 1.5 s. If the ball is suddenly stopped at t=8.3 s, the magnitude of the displacement of the ball with respect to its position at t=0 s is closest to

Options:
A 1 m
B 33 m
C 3 m
D 2 m

The correct answer is:
2 m

Solution:
Time period of rotation of ball =1.5 s

So, in time interval of 7.5 s(=1.5×5)s ball completes 5 revolutions. Also, ball covers one-fourth of circular path in time 1.54=0.375 s
So, in remaining 0.8 s(=8.3-7.5 s) ball is very near to other end of diameter as shown in the figure.

Cleary, displacement of ball is nearly equals to diameter (=2 m) of circular path.

Topic: Physics, Motion In Two Dimensions, KVPY, KVPY 2019 (SA)
A ball is projected at an angle 60Â° with the horizontal with speed 30 m/s. What will be the speed of the ball when it makes an angle 45Â° with the horizontal?

Options:
A 30 m/s
B 152 m/s
C 152 m/s
D 302 m/s

The correct answer is:
152 m/s

Solution:
Horizontal velocity remains same 30Â°cos60Â°=vcos45Â°

âv=152m/s

Topic: Physics, Motion In Two Dimensions, NEET
A ball is projected at an angle of 30Â° above with the horizontal from the top of a tower and strikes the ground in 5 s at an angle of 45Â° with the horizontal. Find the speed with which it was projected.

Options:
A 503-1 m s-1
B 502-1 m s-1
C 703-1 m s-1
D 803-1 m s-1

The correct answer is:
503-1 m s-1

Solution:
vx = - vy
u cos 30Â° = - (u sin 30Â° - gt)
u32=-u2-10Ã5
u3+12=50
âu=10013+1â u=503-1 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main
A ball is projected from point A with a velocity, 10 m s-1 perpendicular to the inclined plane as shown in the figure. Range of the ball on the inclined plane is :

Options:
A 4 0 3 m
B 2 0 13 m
C 13 20 m
D 13 40 m

The correct answer is:
4 0 3 m

Solution:
T = 2u sin 90 o g cos 30 o = 2 u g cos 3 0 o
R = 1 2 Ã g sin 30 o Ã 4 u 2 g 2 cos 2 30 o
R = 2 u 2 Ã 4 2 Ã 1 0 Ã 3 = 2 Ã 1 0 Ã 1 0 Ã 4 2 Ã 1 0 Ã 3 = 4 0 3 m

Topic: Physics, Motion In Two Dimensions, JEE Main
A ball is projected from the ground at an angle of 45o with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30o with the horizontal surface. The maximum height it reaches after the bounce (in meters) is _____________.

Options:

The correct answer is:
30.00

Solution:
H1=u2sin2â¡ 45 o 2g=120
â u24g=120 .....(i)
when half of kinetic energy is lost v=u2
H2=u22sin2â¡ 30 o 2g=u216g ......(ii)
From (i) and (ii)
H2=H14=30 m on 30.00

Topic: Physics, Motion In Two Dimensions, JEE Main
A ball is projected from the ground at an angle Î¸ with the horizontal. After 1 s it is moving at an angle 45Â° with the horizontal and after 2 s, it is moving horizontally. What is the velocity of projection of the ball?

Options:
A 103 m s-1
B 203 m s-1
C 105 m s-1
D 202 m s-1

The correct answer is:
105 m s-1

Solution:
The ball will be vertical at the highest point so
tup= usinθg=2 ⇒usinθ=20 .. …….(1)
∵ tanα=usinθ-gtu cosθ
⇒1=20-g(1)ucosθ⇒ucosθ=10… . .(2)
Squaring and adding equation (1) & equation (2), we get
usinθ2+ucosθ2=202+102
⇒ u2=500
⇒ u=105 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main