datavorous/entrance-exam-dataset huggingface dataset:
A ball of mass $2 \mathrm{~kg}$ and another of mass $4 \mathrm{~kg}$ are dropped together from a 60 feet tall building. After a fall of 30 feet each towards earth, their respective kinetic energies will be in the ratio of:

Options:
A $\sqrt{2}+1$
B $1: 4$
C $1: 2$
D $1: \sqrt{2}$

The correct answer is:
$1: 2$

Solution:
Let $E_1$ and $E_2$ be the K.E. of two bodies.
Then $\quad E_1=\frac{1}{2} m_1 v_1^2$
and $E_2=\frac{1}{2} m_2 v_2{ }^2$
$$
\therefore \quad \frac{E_1}{E_2}=\frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_2 v^2}=\frac{m_1 v_1^2}{m_2 v_2^2}
$$
The initial velocity of both the bodies are zero.
$$
\begin{aligned}
& \therefore \quad v^2=2 g h \text { is same for both } \\
& \text { bodies } \\
& \therefore \quad \frac{E_1}{E_2}=\frac{m_1}{m_2}=\frac{2}{4}=\frac{1}{2}
\end{aligned}
$$

Topic: Physics, Work Power Energy, JEE Main
A ball of mass $0.2 \mathrm{~kg}$ is thrown vertically down from a height of $10 \mathrm{~m}$. It collides with the floor and loses $50 \%$ of its energy and then rises back to the same height. The value of its initial velocity is

Options:
A Zero
B $14 \mathrm{~ms}^{-1}$
C $196 \mathrm{~ms}^{-1}$
D $20 \mathrm{~ms}^{-1}$

The correct answer is:
$14 \mathrm{~ms}^{-1}$

Solution:
Given, mass of the ball, $m=0.2 \mathrm{~kg}$
Height from the surface of floor, $h=10 \mathrm{~m}$
Let initial velocity be $u$.
Total energy as initial point $\mathrm{TE}=\frac{1}{2} m v^2+m g h$
After collision, remaining energy
$=\frac{\frac{1}{2} m u^2+m g h}{2}$
With this remaining energy the ball bounces upto height $h$
Therefore, $\frac{\frac{1}{2} m u^2+m g h}{2}=m g h$
$\begin{aligned} & \Rightarrow \frac{1}{2} m u^2+m g h=2 m g h \\ & \Rightarrow \begin{aligned} u^2 & =\sqrt{2 g h} \\ & =\sqrt{2 \times 10 \times 10}=14 \mathrm{~m} / \mathrm{s}\end{aligned}\end{aligned}$

Topic: Physics, Work Power Energy, KCET, KCET 2023
A ball of mass $0.2 \mathrm{~kg}$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2 \mathrm{~m}$ which applying the force and the ball goes upto $2 \mathrm{~m}$ height further, find the magnitude of the force. Consider $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$

Options:
A
$22 \mathrm{~N}$

B
$4 \mathrm{~N}$

C
$16 \mathrm{~N}$

D
$20 \mathrm{~N}$

The correct answer is:
$20 \mathrm{~N}$

Solution:
$m g h=F s$
$F=20 N$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2006
A ball of mass 3 kg, moving with a speed of 100 m s-1, strikes a wall at an angle 60° (as shown in figure). The ball rebounds at the same speed and remains in contact with the wall for 0.2 s; the force exerted by the ball on the wall is:

Options:
A 15003 N
B 1500 N
C 3003 N
D 300 N

The correct answer is:
15003 N

Solution:
Change in momentum, ∆p=m∆v

⇒∆p=3100cos30°

⇒∆p=3×2×100×32

⇒∆p=3003

∆F=∆p∆t

⇒∆F=30030.2=15003 N

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2021 (20 Aug Shift 2)
A ball of mass 4 kg, moving with a velocity of 10 m s-1, collides with a spring of length 8 m and force constant 100 N m-1. The length of the compressed spring is x m. The value of x, to the nearest integer, is ___ .

Options:

The correct answer is:
6

Solution:
Let's say the compression in the spring by:y. So, by work-energy theorem we have

⇒12mv2=12ky2⇒y=mk·v⇒y=4100×10⇒y=2 m

⇒ Final length of spring,=8-2=6 m.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (18 Mar Shift 2)
A ball of mass 10 kg moving with a velocity 103 m s-1 along X-axis, hits another ball of mass 20 kg which is at rest. After the collision, the first ball comes to rest and the second one disintegrates into two equal pieces. One of the pieces starts moving along Y-axis at a speed of 10 m s-1. The second piece starts moving at a speed of 20 m s-1 at an angle θ (degree) with respect to the X-axis. The configuration of pieces after the collision is shown in the figure. The value of $\theta$ to the nearest integer is _________.

Options:

The correct answer is:
30

Solution:
Before Collision

After Collision

From conservation of momentum along x axis;

P→i=P→f

10×103=200cosθ

cosθ=32

θ=30°

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (16 Mar Shift 1)
A ball of mass 160 g is thrown up at an angle of 60° to the horizontal at a speed of 10 m s-1. The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly g=10 m s-2

Options:
A 1.73 kg m2 s-1
B 3.46 kg m2 s-1
C 3.0 kg m2 s-1
D 6.0 kg m2 s-1

The correct answer is:
3.0 kg m2 s-1

Solution:
The angular momentum L→=r→×p→

At maximum height angular momentum about O,

L=m vcosθ H

=mvcosθ×u2sin2θ2g for horizontal v=u

=mu3sin260°cos60°2g

=3 kg m2 s-1

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2014 (19 Apr Online)
A ball of mass 200 g rests on a vertical post of height 20 m. A bullet of mass 10 g, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance 30 m and the bullet at a distance of 120 m from the foot of the post. The value of initial velocity of the bullet will be (if g=10 m s-2) :

Options:
A 120 m s-1
B 60 m s-1
C 400 m s-1
D 360 m s-1

The correct answer is:
360 m s-1

Solution:
Let initial velocity of bullet be u.

Given, mass of bullet m=10 g and mass of ball M=200 g,

Initially ball is at height 5 m and at rest, the only acceleration acting is due to gravity.

Time of flight of each ball and bullet is t=2hg.

Now, velocity of ball and bullet as v1=302hg and v2=1202hg

As the collision is elastic, so applying conservation of linear momentum.

mu+M0=Mv1+mv2

⇒0.01u=0.230g2h+0.01120g2h

⇒u=300+60=360 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2023 (30 Jan Shift 1)
A ball of mass m1 is moving with velocity 3v. It collides head on elastically with a stationary ball of mass m2. The velocity of both the balls become v after collision. Then the value of the ratio m2m1 is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
3m1v=m1v+m2v
3m1v-m1v=m2v
2m1v=m2v
â´ m2m1=2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball of mass $m$ hits the floor with a speed $v$ making an angle of incidence $\theta$ with the normal. The coefficient of restitution is $e$.
The speed of reflected ball and the angle of reflection of the ball will be

Options:
A $v^{\prime}=v, \theta=\theta^{\prime}$
B $v^{\prime}=\frac{v}{2}, \theta=2 \theta^{\prime}$
C $v^{\prime}=2 v, \theta=2 \theta^{\prime}$
D $v^{\prime}=\frac{3 v}{2}, \theta=\frac{2 \theta^{\prime}}{3}$

The correct answer is:
$v^{\prime}=v, \theta=\theta^{\prime}$

Solution:
The parallel component of velocity of the ball remains unchanged. This gives
$v^{\prime} \sin \theta^{\prime}=v \sin \theta$ ...(i)
For the components normal to the floor, the velocity of separation $=v^{\prime} \cos \theta^{\prime}$
and the velocity of approach $=v \cos \theta$
Hence, $\quad v^{\prime} \cos \theta^{\prime}=e v \cos \theta$ ... (ii)
From Eqs. (i) and (ii)
$\begin{aligned} v^{\prime} & =v \sqrt{\sin ^2 \theta+e^2 \cos ^2 \theta} \\ \text { and } \quad \tan \theta^{\prime} & =\frac{\tan \theta}{e}\end{aligned}$
For elastic collision, $e=1$, so that $\theta^{\prime}=\theta$ and $v^{\prime}=v$

Topic: Physics, Center of Mass Momentum and Collision, JIPMER, JIPMER 2015
A ball of mass ' $\mathrm{m}$ ' is attached to the free end of a string of length ' $l$ '. The ball is moving in horizontal circular path about the vertical axis as shown in the diagram.
The angular velocity ' $\omega$ ' of the ball will be [ $\mathrm{T}=$ Tension in the string. $]$

Options:
A $\sqrt{\frac{\mathrm{T} l}{\mathrm{~m}}}$
B $\sqrt{\frac{\mathrm{Tm}}{l}}$
C $\sqrt{\frac{\mathrm{m} l}{\mathrm{~T}}}$
D $\sqrt{\frac{\mathrm{T}}{\mathrm{m} l}}$

The correct answer is:
$\sqrt{\frac{\mathrm{T}}{\mathrm{m} l}}$

Solution:
The tension in the string can be resolved in two components along the perpendicular axis. The gravitational force is acting downwards and the centrifugal force is acting in $-\mathrm{x}$ direction
$$
\begin{aligned}
& \mathrm{T} \sin \theta=\mathrm{mr} \omega^2 \\
\therefore \quad & \omega^2=\frac{\mathrm{T} \sin \theta}{\mathrm{mr}}
\end{aligned}
$$
$$
\therefore \quad \omega=\sqrt{\frac{\mathrm{T} \sin \theta}{\mathrm{mr}}}
$$
From figure, $\sin \theta=\frac{\mathrm{r}}{l}$
$$
\begin{aligned}
& \therefore \quad \omega=\sqrt{\frac{\mathrm{Tr}}{\mathrm{mr} l}} \\
& \therefore \quad \omega=\sqrt{\frac{\mathrm{T}}{\mathrm{m} l}} \\
&
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main
A ball of mass m is dropped from a height h in a tunnel, made across the earth (mass = M, radius = R) passing through its centre. If hâªR such that the motion of the particle through h can be considered uniformly accelerated at g, then the time period of the particle is

Options:
A 2πRg+2hg
B 2πRg+2hg
C 2πRg+42hg
D 2π2Rg+hg

The correct answer is:
2πRg+42hg

Solution:
The time, the particle takes to go through h is

t=2hg and in one complete oscillation, the particle travels through h four times.

For the motion of the particle through the tunnel, the force acting on the particle is towards the centre of the earth and it is proportional to the distance of the particle from the centre of the earth. Hence the motion of the particle through the tunnel will be like an SHM.

F=-GMmR3x

a=-GMR3x

TSHM=2πR3GM=2πRg

So the time period of the particle is

T=TSHM+42hg

T=2πRg+42hg

Topic: Physics, Gravitation, JEE Main
A ball of mass m is thrown vertically upward. Another ball of mass 2 m is thrown an angle θ with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is 1x. The value of x is _____ .

Options:

The correct answer is:
1

Solution:
When the first ball of mass m is projected vertically upwards, the second ball of mass 2m makes an angle θ with the vertical.

The time of flight is the same for both the balls.

Then, T1=T2 ⇒2u1g=2u2sinθg

⇒u1=u2sinθ ...1

Maximum height of projectile is given by H=u2sin2θ2g

Ratio of the heights is H1H2=u12u22sin2θ ...2

From equation 1 and 2, we have

⇒H1H2=1.

Thus, x=1.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2022 (27 Jul Shift 1)
A ball of mass m moving with a velocity v strikes the bob of a pendulum at rest. The mass of the bob is also m. If the collision is perfectly inelastic, the height to which the bob will rise is given by h=v2xÂ·g, then the value of x is

Options:

The correct answer is:
8

Solution:
Because the collision is perfectly inelastic, the two blocks stick together. By conservation of linear momentum,
2mV=mv or V=v2
By conservation of energy,
2mgh=12Ã2mV2 or h=v28Â·g

âx=8

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball of mass m moving with velocity v strikes the bob of a pendulum at rest. The mass of the bob is also m. If the collision is perfectly inelastic, the height to which the bob will rise is given by

Options:
A v28g
B v22g
C 2v2g
D v2g

The correct answer is:
v28g

Solution:
Because the collision is perfectly inelastic, hence the two blocks stick together. By conservation of linear momentum,
2mV=mv or V=v2
By conservation of energy,
2mgh=122mV2 or h=v2/8g

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball of mass m moving with velocity v strikes the bob of a pendulum at rest. The mass of the bob is also m . If the collision is perfectly inelastic, the height to which the bob will rise is given by h = v 2 x.g , then x=?

Options:

The correct answer is:
8

Solution:
Because the collision is perfectly inelastic, hence the two blocks stick together. By conservation of linear momentum,
2 mV = mv or V = v 2
By conservation of energy,
2mgh= 1 2 2m V 2 2 or h = v 2 / 8 g
x=8

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball of radius $r$ and density $\rho$ falls freely under gravity through a distance $h$ before entering water. Velocity of ball does not change even on entering water. If viscosity of water is $\eta$, the value of $h$ is given by

Options:
A $\frac{2}{9} r^2\left(\frac{1-\rho}{\eta}\right) g$
B $\frac{2}{81} r^2\left(\frac{\rho-1}{\eta}\right) g$
C $\frac{2}{81} r^4\left(\frac{\rho-1}{\eta}\right)^2 g$
D $\frac{2}{9} r^4\left(\frac{\rho-1}{\eta}\right)^2 g$

The correct answer is:
$\frac{2}{81} r^4\left(\frac{\rho-1}{\eta}\right)^2 g$

Solution:
Velocity of ball when it strikes the water surface $v=\sqrt{2 g h}$...(i)
Terminal velocity of ball inside the water $v=\frac{2}{9} r^2 g \frac{(\rho-1)}{\eta}$ ....(ii)
Equating (i) and (ii) we get $\sqrt{2 g h}=\frac{2}{9} \frac{r^2 g}{\eta}(\rho-1)$
$\Rightarrow \quad h=\frac{2}{81} r^4\left(\frac{\rho-1}{\eta}\right)^2 g$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A ball $P$ is dropped vertically and another ball $Q$ is thrown horizontally with the same velocities from the same height and at the same time. If air resistance is neglected, then

Options:
A Ball $P$ reaches the ground first
B Ball $Q$ reaches the ground first
C Both reach the ground at the same time
D The respective masses of the two balls will decide the time

The correct answer is:
Both reach the ground at the same time

Solution:
Vertical component of velocities of both the balls are same and equal to zero.
So $t=\sqrt{\frac{2 h}{g}}$
\(\Rightarrow\) Both reach the ground at the same time

Topic: Physics, Motion In One Dimension, JEE Main
A ball $P$ moving with a speed of $v \mathrm{~ms}^{-1}$ collides directly with another identical ball $Q$ moving with a speed $10 \mathrm{~ms}^{-1}$ in the opposite direction. $P$ comes to rest after the collision. If the coefficient of restitution is 0.6 , the value of $v$ is

Options:
A $30\ ms^{-1}$
B $40\ ms^{-1}$
C $50\ ms^{-1}$
D $60\ ms^{-1}$

The correct answer is:
$40\ ms^{-1}$

Solution:
From conservation of moment,
$\begin{aligned}
& m(v-10)=m v_2 \\
& v_2=(v-10) \\
& \longrightarrow \mathrm{V} \stackrel{10}{\bullet} \mathrm{m} / \mathrm{s} \\
& \mathrm{V}=0 \quad \mathrm{~V}_2 \quad \text { Berore }
\end{aligned}$
The coefficient of restitution,
$\begin{aligned}
\mathrm{e} & =\frac{\text { Velocity of separation }}{\text { Velocity of approach }} \\
& =\frac{v_2-v_1}{u_1+u_2}=\frac{(v-10)-0}{(v+10)} \\
\Rightarrow & 0.6=\frac{v-10}{v+10} \\
\Rightarrow & 0.6 v+6=v-10 \\
\Rightarrow & 0.4 v=16 \Rightarrow v=40 \mathrm{~ms}^{-1}
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball projected from ground at an angle of $45^{\circ}$ just clears a wall in front. If point of projection is $4 \mathrm{~m}$ from the foot of wall and ball strikes the ground at a distance of $6 \mathrm{~m}$ on the other side of the wall, the height of the wall is :

Options:
A
$4.4 \mathrm{~m}$

B
$2.4 \mathrm{~m}$

C
$3.6 \mathrm{~m}$

D
$1.6 \mathrm{~m}$

The correct answer is:
$2.4 \mathrm{~m}$

Solution:
As ball is projected at an angle $45^{\circ}$ to the horizontal therefore Range $=4 \mathrm{H}$
or $10=4 \mathrm{H} \Rightarrow \mathrm{H}=\frac{10}{4}=2.5 \mathrm{~m}$
$(\because$ Range $=4 \mathrm{~m}+6 \mathrm{~m}=10 \mathrm{~m})$
Maximum height, $\mathrm{H}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}$
$\therefore \mathrm{u}^2=\frac{\mathrm{H} \times 2 \mathrm{~g}}{\sin ^2 \theta}=\frac{2.5 \times 2 \times 10}{\left(\frac{1}{\sqrt{2}}\right)^2}=100$
or, $\mathrm{u}=\sqrt{100}=10 \mathrm{~ms}^{-1}$
Height of wall PA
$$
\begin{aligned}
& =\mathrm{OA} \tan \theta-\frac{1}{2} \frac{\mathrm{g}(\mathrm{OA})^2}{\mathrm{u}^2 \cos ^2 \theta} \\
& =4-\frac{1}{2} \times \frac{10 \times 16}{10 \times 10 \times \frac{1}{\sqrt{n}} \times \frac{1}{\sqrt{0}}}=2.4 \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2013 (22 Apr Online)
A ball rests upon a flat piece of paper on a table top. The paper is pulled horizontally but quickly towards right as shown relative to its initial position with respect to the table, the ball
(1) remains stationary if there is no friction between the paper and the ball.
(2) moves to the left and starts rolling backwards, ie, to the left if there is a friction between the paper and the ball.
(3) moves forward, ie, in the direction in which the paper is pulled.
Here, the correct statement/s is/are

Options:
A Both (1) and (2)
B only (3)
C only (1)
D only (2)

The correct answer is:
Both (1) and (2)

Solution:
The ball remains there as such due to the inertia of rest if there is no friction but will move in opposite direction if friction is present.

Topic: Physics, Laws of Motion, KCET, KCET 2010
A ball rolls off the top of a stairway with horizontal velocity u. The steps are 0.1 m high and 0.1 m wide. The minimum velocity u with which that ball just hits the step 5 of the stairway will be x m s-1, where x=_______ [use g=10 m s-2].

Options:

The correct answer is:
2

Solution:
The ball needs to just cross 4 steps to just hit 5th step

Therefore, horizontal range

R=4×0.1 m= 0.4 m

The formula to calculate the horizontal range is given by

R=ut ...1

Similarly, the vertical height covered by the ball is given by

h=12gt2 ...2

From equations (1) and (2), it follows that

4×0.1=12gt2⇒0.4=12 g0.4u2⇒u2=2⇒u=2 m s-1

Therefore, x=2.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2024 (29 Jan Shift 1)
A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$. If radius of the ball be $R$, then the fraction of total energy associated with its rotational energy will be:

Options:
A $\frac{K^2+R^2}{R^2}$
B $\frac{K^2}{R^2}$
C $\frac{K^2}{K^2+R^2}$
D $\frac{R^2}{K^2+R^2}$

The correct answer is:
$\frac{K^2}{K^2+R^2}$

Solution:
Total energy $=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2$
$=\frac{1}{2} m v^2\left(1+\frac{K^2}{R^2}\right)$
$\begin{aligned}
& \text {Rotational energy }=\frac{1}{2} I \omega^2 \\
& =\frac{K^2+R^2}{1+\frac{K^2}{R^2}}
\end{aligned}$
$\begin{aligned}
& \text {Required fraction }=\frac{K^2 / R^2}{1+K^2 / R^2} \\
& =\frac{K^2}{R^2+K^2}
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A ball starts falling under the effect of gravitational force from a height of 45 m. When it reaches a height of 25 m it explodes into two pieces of mass ratio 1:2. There is no change in the vertical motion of the pieces after the explosion but they acquire horizontal velocity. If the heavier piece gains a horizontal velocity of 10 m s-1, then the distance between the two pieces when both of them strike the ground is

Options:
A 30 m
B 10 m
C 20 m
D 15 m

The correct answer is:
30 m

Solution:
Let us assume the mass of the ball is 3m and the velocity of the lighter piece in horizontal direction after the explosion is v. We know that just after the collision, the horizontal velocity of the heavier piece is 10 m s-1. Using conservation of momentum in the horizontal direction we get

mv=2m×10

⇒v=20 m s-1

Since there is no change in vertical motion, the time taken by the pieces to reach the ground can be calculated in the following way.

Total time to fall through 45 m is

T45=2×4510=3 s

The time taken to fall through first 20 m is

T20=2×2010=2 s.

Hence time taken by the pieces to fall from 25 m height to ground is

T40-T20=3-2=1 s

The relative velocity of the pieces in the horizontal direction is

vrel=v--10=30 m s-1

The horizontal distance between the two pieces at the time of striking the ground is

x=30×1=30 m

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball strikes a horizontal floor at an angle of Î¸=45Â°. The coefficient of restitution between the ball and the floor is 0.5. The fraction of its kinetic energy lost in collision is

Options:
A 58
B 38
C 34
D 14

The correct answer is:
38

Solution:
Let ball strikes at a speed u the K1=12mu2
Due to collision tangential component of velocity remains unchanged at u sin 45Â°, but the normal component of velocity change to
u sin45Â°=12ucosâ¡45Â°
â´ Final velocity of ball after collision
v=usinâ¡45Â°2+12ucosâ¡45Â°2
=u22+u222=53u.
Hence final kinetic energy K2=12mv2=516mu2.
â´ Fractional loss in KE
=K1-K2K1=12mu2-516mu212mu2=38

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball strikes a smooth horizontal ground at an angle of 45°with the vertical. What cannot be the possible angle of its velocity with the vertical after the collision ?

Options:
A 45°
B 30°
C 53°
D 60°

The correct answer is:
30°

Solution:
Since surface is smooth so force is normal so horizontal component will remain unchanged while vertical component decreases or remain same (e≤1) that's why angle from vertical will increases or remains same.

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball suspended by a thread, swings in a vertical plane so that its acceleration in the extreme position and lowest position are equal in magnitude. Angle Î¸ of thread deflection in the extreme position will be:

Options:
A 2tan-1⁡12
B tan-1⁡12
C tan-1⁡2
D tan-1⁡2

The correct answer is:
2tan-1⁡12

Solution:
aA=g sinÎ¸ (only tangential)
aB=v2l (only radial)
K.E. + P.E. = 12m02+mgl 1-cosâ¡Î¸=12mv2
v2=2gl 1-cosâ¡Î¸ ...........(i)
â´aB=v2l=2g (1-cosÎ¸)
Since, aA=aB
â´g sinÎ¸=2g 1-cosÎ¸â¹2sinâ¡Î¸2 cosâ¡Î¸2=2 Ã2sin2â¡Î¸2
â¹tanâ¡Î¸2=12 â¹ Î¸=2tan-1â¡12

Topic: Physics, Work Power Energy, JEE Main
A ball weighing 10 g is moving with a velocity of 90 m s-1. If the uncertainty in its velocity is 5%, then the uncertainty in its position is _ ×10-33 m. (Rounded off to the nearest integer)

[Given: h=6.63×10-34 Js]

Options:

The correct answer is:
1

Solution:
Δv=90×5100

=4.5 m s-1

Δv.Δx=h4πm

Δx=h4πm·Δv

=6.63×10-344×3.14×0.01×4.5

=1.17×10-33

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2021 (26 Feb Shift 2)
A ball whose density is $0.4 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$ falls into water from a height of $9 \mathrm{~cm}$. To what depth does the ball sink

Options:
A $9 \mathrm{~cm}$
B $6 \mathrm{~cm}$
C $4.5 \mathrm{~cm}$
D $2.25 \mathrm{~cm}$

The correct answer is:
$6 \mathrm{~cm}$

Solution:
The velocity of ball before entering the water surface
$v=\sqrt{2 g h}=\sqrt{2 g \times 9}$
When ball enters into water, due to upthrust of water the velocity of ball decreases (or retarded)
The retardation, $a=\frac{\text{ apparent weight }}{\text { mass of ball }}$
$\frac{=V(\rho-\sigma) g}{V \rho}=\left(\frac{\rho-\sigma}{\rho}\right) g=\left(\frac{0.4-1}{0.4}\right) \times g=-\frac{3}{2} g$
If $h$ be the depth upto, which ball sink, then,
$0-v^2=2 \times\left(-\frac{3}{2} g\right) \times h \Rightarrow 2 g \times 9=3 g h \therefore h=6 \mathrm{~cm}$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A ball whose kinetic energy is E, is projected at an angle of $45^{\circ}$ to the horizontal. The kinetic energy of the ball at the highest point of its flight will be

Options:
A
$\mathrm{E}$

B
$\mathrm{E} / \sqrt{2}$

C
$E / 2$

D
zero

The correct answer is:
$E / 2$

Solution:
As the velocity at the highest point reduces to zero. The K.E. of the ball also becomes zero.

Topic: Physics, Work Power Energy, JEE Main
A ball whose kinetic energy is E, is projected at an angle of $45^{\circ}$ to the horizontal. The kinetic energy of the ball at the highest point of its flight will be

Options:
A
$\mathrm{E}$

B
$\mathrm{E} / \sqrt{2}$

C
$E / 2$

D
zero

The correct answer is:
$E / 2$

Solution:
As the velocity at the highest point reduces to zero. The K.E. of the ball also becomes zero.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2002
A ball whose kinetic energy is E, is thrown at an angle of 45Â° with the horizontal. Its kinetic energy at the highest point of its flight will be:

Options:
A E
B E 2
C E 2
D 0

The correct answer is:
E 2

Solution:
Kinetic energy is E

1 2 âm v 2 â=âEâââvâ=â 2E m

Velocity at highest point =vâcosâ45Â°

â´ââvâcosâ45Â°â=â 2E m â.â 1 2 â=â E m
[ âµ at the highest point there is only horizontal velocity]

â´â K.E. at highest point =â 1 2 âÃâmâ E m â=â E 2

Topic: Physics, Work Power Energy, JEE Main