datavorous/entrance-exam-dataset huggingface dataset:
A mass falls from a height ' $h$ ' and its time of fall ' $t$ ' is recorded in terms of time period $T$ of a simple pendulum. On the surface of earth it is found that $t=2 \mathrm{~T}$. The entire set $u$ is taken on the surface of another planet whose mass is half of earth and radius the same. Same experiment is repeated and corresponding times noted as $\mathrm{t}^{\prime}$ and $\mathrm{T}^{\prime}$.

Options:
A $t^{\prime}=\sqrt{2} T^{\prime}$
B $t^{\prime}>2 T^{\prime}$
C $t^{\prime} < 2 T^{\prime}$
D $t^{\prime}=2 T^{\prime}$

The correct answer is:
$t^{\prime}=2 T^{\prime}$

Solution:
The distance covered by the mass falling from height ' $h$ ' during its time of fall ' $t$ ' is given by
$$
\begin{array}{r}
\mathrm{s}=\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^2 \\
\text { As, } \mathrm{u}=0 \Rightarrow \mathrm{h}=\frac{1}{2} \mathrm{gt}^2 \Rightarrow \mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}
\end{array}
$$
The time period of simple pendulum is
$$
\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{l}}{\mathrm{g}}}
$$
where, lis the length of the pendulum. From Eq. (i) and (ii), since ' $h$ ' and ' $l$ ' are constant so, we can conclude that,
$$
\mathrm{t} \propto \frac{\mathrm{l}}{\sqrt{\mathrm{g}}} \text { and } \mathrm{T} \propto \frac{\mathrm{l}}{\sqrt{\mathrm{g}}} \quad \therefore \frac{\mathrm{t}}{\mathrm{T}}=\mathrm{l}
$$
Thus, the ratio of time of fall and time period of pendulum is independent of value of gravity $(\mathrm{g})$ or any other parameter like mass and radius of the planet. Thus, the relation between $\mathrm{t}^{\prime}$ and $\mathrm{T}^{\prime}$ on another planet irrespective of its mass or radius will remains same as it was on earth i.e.
$$
\mathrm{t}^{\prime}=2 \mathrm{~T}^{\prime}
$$

Topic: Physics, Oscillations, NEET, NEET 2019 (Odisha)
A mass is whirled in a circular path with constant angular velocity and its linear velocity is v. If the string is now halved keeping the angular momentum same, the linear velocity is

Options:
A 2v
B v2
C v
D v2

The correct answer is:
2v

Solution:
Angular velocity = ω
Linear velocity = v
Length of string = Radius = r

If string is halved, r'=r2
Angular momentum, L=r×p
where, r = radius vector
and p = linear momentum
=r×mv∵θ=90o
=mvr
As L remain constant,
mvr = constant
⇒ If r'→r2'v'=2v such that
mv'r'=m2v×r2
Hence, linear velocity will be 2 v.

Topic: Physics, Rotational Motion, MHT CET, MHT CET 2019 (Shift 1)
A mass m1=1 kg connected to a horizontal spring performs S.H.M. with amplitude A. While mass m1 is passing through mean position another mass m2=3 kg is placed on it so that both the masses move together with amplitude A1. The ratio of A1A is pq12, where p and q are the smallest integers. Then what is the value of p+q?

Options:

The correct answer is:
5

Solution:
P.E of the oscillating mass is given by
E= 1 2 mÏ x 2 = 1 2 k x 2 where k=m Ï 2
Let k=m Ï 2 be the velocity at mean position. When mass M 2 is attached v 2 = m 1 v 1 m 1 + m 2
If A 1 be the new amplitude 1 2 k A 1 2 = 1 2 ( m 1 + m 2 ) v 2 2
= 1 2 âk A 2 Ã m 1 m 1 + m 2
â A 1 A = m 1 m 1 + m 2 =14

Topic: Physics, Oscillations, JEE Main
A mass m1 connected to a horizontal spring performs S.H.M. with amplitude A. While mass m1 is passing through mean position another mass m2 is placed on it so that both the masses move together with amplitude A1. The ratio of A1A is (m2<m1)

Options:
A m1m1+m212
B m1+m2m112
C m2m1+m212
D m1+m2m212

The correct answer is:
m1m1+m212

Solution:
P.E of the oscillating mass is given by
E= 1 2 mω x 2 = 1 2 k x 2 where k=m ω 2
Let k=m ω 2 be the velocity at mean position. When mass M 2 is attached v 2 = m 1 v 1 m 1 + m 2
If A 1 be the new amplitude 1 2 k A 1 2 = 1 2 ( m 1 + m 2 ) v 2 2
1 2 ( m 1 + m 2 ) m 1 2 v 1 2 ( m 1 + m 2 ) 2
= 1 2   m 1 v 1 2 × m 1 m 1 + m 2
= 1 2  k A 2 × m 1 m 1 + m 2
⇒ A 1 A = m 1 m 1 + m 2

Topic: Physics, Oscillations, JEE Main
A mass $m=1.0 \mathrm{~kg}$ is put on a flat pan attached to a vertical spring fixed on the ground. The mass of the spring and the pan is negligible. When pressed slightly and released, the mass executes simple harmonic motion. The spring constant is $500 \mathrm{~N} / \mathrm{m}$. What is the amplitude A of the motion, so that the mass $m$ tends to get detached from the pan ? (Take $g=10 \mathrm{~m} / \mathrm{s}^2$ ).
The spring is stiff enough so that it does not get distorted during the motion.

Options:
A
$\mathrm{A}>2.0 \mathrm{~cm}$

B
$\mathrm{A}=2.0 \mathrm{~cm}$

C
$\mathrm{A} < 2.0 \mathrm{~cm}$
D
$\mathrm{A}=1.5 \mathrm{~cm}$

The correct answer is:
$\mathrm{A} < 2.0 \mathrm{~cm}$

Solution:
As $\mathrm{F}=-\mathrm{kx}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2013 (22 Apr Online)
A mass $M$, attached to a horizontal spring, executes S.H.M. with amplitude $A_1$. When the mass $M$ passes through its mean position then a smaller mass $\mathrm{m}$ is placed over it and both of them move together with amplitude $A_2$. The ratio of $\left(\frac{A_1}{A_2}\right)$ is :

Options:
A
$\frac{M+m}{M}$

B
$\left(\frac{M}{M+m}\right)^{1 / 2}$

C
$\left(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{M}}\right)^{1 / 2}$

D
$\frac{M}{M+m}$

The correct answer is:
$\left(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{M}}\right)^{1 / 2}$

Solution:
Energy of simple harmonic oscillator is constant.
$$
\begin{aligned}
& \Rightarrow \frac{1}{2} M \omega^2 A_1^2=\frac{1}{2}(m+M) \omega^2 A_2^2 \\
& \frac{A_1^2}{A_2^2}=\frac{M+m}{M} \\
& \therefore \frac{A_1}{A_2}=\sqrt{\frac{M+m}{M}}
\end{aligned}
$$

Topic: Physics, Oscillations, JEE Main, JEE Main 2011
A mass M attached to a spring oscillates with a period of 2 s . If the mass is increased by 2 kg the period increases by one sec. Find the initial mass M (In kg ) assuming that Hookeâs Law is obeyed.

Options:

The correct answer is:
1.6

Solution:
T = 2 Ï M K or T â M , T 2 T 1 = M 2 M 1
â´ 3 2 = M + 2 M
Solving this equation we get, M = 1.6 kg .

Topic: Physics, Oscillations, JEE Main
A mass m attached to free end of a spring executes SHM with a period of 1 s. If the mass is increased by 3 kg the period of oscillation increases by one second, the value of mass m is ________ kg.

Options:

The correct answer is:
1

Solution:
Time period of oscillation is given by T=2πmk, where k is the force constant of the spring.

∵ Spring is same, therefore, spring constant will be same.

Putting the different values,

In the first case T=1=2πmk

In the second case T'=2=2πm+3k

Dividing the equations, we get

TT'=12=mm+3

⇒mm+3=14⇒4m=m+3⇒m=1

Thus, m=1 kg

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (24 Jan Shift 2)
A mass M hangs on a massless rod of length l which rotates at a constant angular frequency. The mass M moves with steady speed in a circular path of constant radius. Assume that the system is in steady circular motion with constant angular velocity ω. The angular momentum of M about point A is LA which lies in the positive z direction and the angular momentum of M about B is LB. The correct statement for this system is:

Options:
A LA and LB are both constant in magnitude and direction
B LB is constant in direction with varying magnitude
C LB is constant, both in magnitude and direction
D LA is constant, both in magnitude and direction

The correct answer is:
LA is constant, both in magnitude and direction

Solution:
We know, L→=m(r→×v→)

Now with respect to A, we always get direction
of L→ along +ve z-axis and also constant magnitude as mvr. But with respect to B, we get constant magnitude but continuously changing direction.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (17 Mar Shift 1)
A mass $\mathrm{m}$ hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass $\mathrm{m}$ and radius $\mathrm{R}$. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass $m$, if the string does not slip on the pulley, is

Options:
A
$\mathrm{g}$

B
$\frac{2}{3} \mathrm{~g}$

C
$\frac{\mathrm{g}}{3}$

D
$\frac{3}{2} g$

The correct answer is:
$\frac{2}{3} \mathrm{~g}$

Solution:
$$
\begin{aligned}
& M g-T=M a \quad \quad \ldots \ldots (1)\\
& T \times R=I \alpha=\frac{1}{2} M^2 \alpha \\
& T=\frac{1}{2} M a \quad(a=\alpha R) \quad \quad \ldots \ldots (2)
\end{aligned}
$$
From (1) and (2) $a=\frac{2 g}{3}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2011
A mass $m$ is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:

Options:
A inclined at a $60^{\circ}$ angle from vertical
B the mass is at the highest point
C the wire is horizontal
D the mass is at the lowest point.

The correct answer is:
the mass is at the lowest point.

Solution:
We know that,
$T-m g=\frac{m u^2}{l}$
$\Longrightarrow T=m g+\frac{m u^2}{l}$
The tension is maximum at the lowest position of mass, so the chance of breaking is maximum.
Hence option 4 is the correct answer

Topic: Physics, Laws of Motion, JEE Main
A mass m is attached to two springs as shown in figure. The spring constants of two springs are K1 and K2. For the frictionless surface, the time period of oscillation of mass m is

Options:
A 2πm K1+K2
B 12πK1-K2m
C 2πm K1-K2
D 12πK1+K2m

The correct answer is:
2πm K1+K2

Solution:
For the given combination, if the block is displaced towards right, elongation in one spring will be equal to the compression in other and the direction of force due to both spring will be the same and restoring in nature. Hence, both springs are effectively in parallel combination.

Therefore,

⇒Keff=K1+K2 and

⇒T=2πmKeff=2πmK1+K2

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2023 (06 Apr Shift 1)
A mass \(m\) is in rest on an inclined plane of mass \(M\) which is further resting on a smooth horizontal plane. Now, if the mass \(m\) starts moving under gravity, the position of centre of mass of system will

Options:
A remain unchanged
B change along the horizontal direction
C move up in vertical direction
D move down in the vertical direction and changes along the horizontal

The correct answer is:
move up in vertical direction

Solution:
The given situation is shown in the following figure.
Clearly when mass \(m\) starts moving under gravity, hence net force on the system in horizontal direction is zero. Therefore, centre of mass of the system will not move in horizontal direction. Now, for vertical direction there is a net force that is the mass of the wedge and the block and hence the centre of mass changes in this direction.

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 1)
A mass m is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular moment with respect to the origin

Options:
A Is zero
B Remains constant
C Goes on increasing
D Goes on decreasing

The correct answer is:
Remains constant

Solution:
Angular momentum of the particle about origin O will be given by
L â = r â Ã p â = mr â Ã V â
or L â = L = mrV sin Î¸
= mVr sin Î¸ = mVh
Now, m,Vand hall are constant

Therefore, angular momentum of particle about origin will remain constant. The direction of r â Ã v â also remains the same (negative z direction)

Topic: Physics, Rotational Motion, NEET
A mass m is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

Options:
A 2 g 3
B g 2
C 5 g 6
D g

The correct answer is:
g 2

Solution:
mg - T = ma ----- 1
TR = Iα
TR =mR 2 α
â∴T=mRα ----2
For no slippping
a = Rα ---3
putting (3) in (2)
T = ma ---4
Adding 1 and 4
mg = 2ma
∴ a = g 2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2014 (06 Apr)
A mass $m$ is suspended by means of two coiled spring which have the same length in unstretched condition as in figure. Their force constant are $k_1$ and $k_2$ respectively. When set into vertical vibrations, the period will be

Options:
A $2 \pi \sqrt{\left(\frac{m}{k_1 k_2}\right)}$
B $2 \pi \sqrt{m\left(\frac{k_1}{k_2}\right)}$
C $2 \pi \sqrt{\left(\frac{m}{k_1-k_2}\right)}$
D $2 \pi \sqrt{\left(\frac{m}{k_1+k_2}\right)}$

The correct answer is:
$2 \pi \sqrt{\left(\frac{m}{k_1+k_2}\right)}$

Solution:
Given spring system has parallel combination, so $k_{\text {eq }}=k_1+k_2$ and time period $T=2 \pi \sqrt{\frac{m}{\left(k_1+k_2\right)}}$

Topic: Physics, Oscillations, JEE Main
A mass m is suspended from a spring of negligible mass and the system oscillates with a frequency f1. The frequency of oscillations if a mass 9m is suspended from the same spring is f2. The value of f1f2 is _____ .

Options:

The correct answer is:
3

Solution:
Frequency of oscillation of a spring-block system is given by, f=12πkm

Therefore, for both cases, we can write

f1=12πkm and

f2=12πk9m

Required ratio, f1f2=91=31

Topic: Physics, Oscillations, JEE Main, JEE Main 2024 (01 Feb Shift 2)
A mass $\mathrm{M}$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by $\mathrm{m}$, the time period becomes $\frac{5 \mathrm{~T}}{3}$. Then the ratio of $\frac{\mathrm{m}}{\mathrm{M}}$ is

Options:
A
$\frac{3}{5}$

B
$\frac{25}{9}$

C
$\frac{16}{9}$

D
$\frac{5}{3}$

The correct answer is:
$\frac{16}{9}$

Solution:
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{K}}}$
$\frac{5 \mathrm{~T}}{3}=2 \pi \sqrt{\frac{\mathrm{M}+\mathrm{m}}{\mathrm{K}}}$
Dividing equation (ii) by equation (i), $\frac{5}{3}=\sqrt{\frac{\mathrm{M}+\mathrm{m}}{\mathrm{M}}}$. Squaring both the sides
$\frac{25}{9}=\frac{M+m}{M}=1+\frac{m}{M} \Rightarrow \frac{m}{M}=\frac{25}{9}-1=\frac{16}{9}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2003
A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period become 5T3. Then the ratio of mM is

Options:
A 3/5
B 25/9
C 16/9
D 5/3

The correct answer is:
16/9

Solution:
T=2πMk when mass is increased by m then ...(i)
T=2πM+mk
⇒ 5T3=2πM+mk ...(ii)
Dividing Eq. (i) by Eq. (ii), we get
35=MM+m
925=MM+m
⇒ 9M+9m=25M
⇒ 16M=9m
mM=169

Topic: Physics, Oscillations, JEE Main
A mass 'm' is tied to one end of a spring and whirled in a horizontal circle with
constant angular velocity. The elongation in the spring is $1 \mathrm{~cm} .$ If the angular speed
is doubled, the elongation in the spring is $6 \mathrm{~cm}$. The original length of the spring is

Options:
A 1. 3 Cm
B 9 $\mathrm{Cm}$
C 6 Cn
D $12 \mathrm{Cm}$

The correct answer is:
9 $\mathrm{Cm}$

Solution:
Let $\ell$ be the original length of the spring.
Let the initial angular velocity be $\omega$ and the corresponding elongation $\mathbf{e}_{1}=1 \mathrm{~cm}$.
When the angular velocity is doubled the elongation $e_{2}=6 \mathrm{~cm}$.
If $k$ is the spring constant then we have
$$
\begin{array}{l}
\mathrm{m}\left(\ell+e_{1}\right) \omega^{2}=\mathrm{ke}_{1} \\
\text { and } \mathrm{m}\left(\ell+e_{2}\right)(2 \omega)^{2}=\mathrm{ke}_{2} \\
\text { or } \quad \mathrm{m}\left(\ell+e_{2}\right) \cdot 4 \omega^{2}=\mathrm{ke}_{2}
\end{array}
$$
Dividing Eq(1) by Eq(2), we get
$$
\frac{\ell+e_{1}}{4\left(\ell+e_{2}\right)}=\frac{e_{1}}{e_{2}}
$$
solving we get $\ell=9 \mathrm{~cm}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A mass ' $m$ ' moves with a velocity $v$ and collides inelastically with another identical mass. After collision the $1^{\text {st }}$ mass moves with velocity $v / \sqrt{3}$ in a direction perpendicular to the initial direction of motion. Find the speed of the $2^{\text {nd }}$ mass after collision

Options:
A
v

B
$\sqrt{3} v$

C
$2 v / \sqrt{3}$

D
$v / \sqrt{3}$

The correct answer is:
$2 v / \sqrt{3}$

Solution:
$m v=m v_1 \cos \theta$
$0=\frac{m v}{\sqrt{3}}-m v_1 \sin \theta$
$\therefore v_1=\frac{2}{\sqrt{3}} v$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2005
A mass ' $m$ ' moves with a velocity ' $v$ ' and collides inelastically with another identical mass. After collision the Ist mass moves with velocity $\frac{v}{\sqrt{3}}$ in a direction perpendicular to the initial direction of motion. Find the speed of the $2^{\text {nd }}$ mass after collision

Options:
A $\frac{2}{\sqrt{3}} v$
B $\frac{v}{\sqrt{3}}$
C $v$
D $\sqrt{3} v$

The correct answer is:
$\frac{2}{\sqrt{3}} v$

Solution:
Let mass $A$ moves with velocity $v$ and collides inelasticalfy with mass B, which is at rest.
According to problem mass $A$ moves in a perpendicular direction and let the mass $B$ moves at angle $\theta$ with the horizontal with velocity $v$.
Initial horizontal momentum of system (before collision) $=\mathrm{mv}$
Final horizontal momentum of system (after collision) $\quad=m V \cos \theta$
...(ii)
From the conservation of horizontal linear momentum $\quad m v=m V \cos \theta \Rightarrow V=V \cos \theta$ ....(iii)
Initial vertical momentum of system (before collision) is zero.
Final vertical momentum of system $\frac{m v}{\sqrt{3}}-m V \sin \theta$
From the conservation of vertical linear momentum $\frac{m v}{\sqrt{3}}-m V \sin \theta=0 \Rightarrow \frac{v}{\sqrt{3}}=V \sin \theta$ ....(iv)
By solving (iii) and (iv)
$\begin{aligned}
& v^2+\frac{v^2}{3}=V^2\left(\sin ^2 \theta+\cos ^2 \theta\right) \\
& \Rightarrow \frac{4 v^2}{3}=V^2 \Rightarrow V=\frac{2}{\sqrt{3}} v .
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A mass $m$ moving horizontally (along the $x$-axis) with velocity $v$ collides and sticks to mass of $3 m$ moving vertically upward (along the $y$-axis) with velocity $2 v$. The final velocity of the combination is

Options:
A $\frac{1}{4} v \hat{\mathbf{i}}+\frac{3}{2} y \hat{\mathbf{j}}$
B $\frac{1}{3} v \hat{\mathbf{i}}+\frac{2}{3} v \hat{\mathbf{j}}$
C $\frac{2}{3} v \hat{\mathbf{i}}+\frac{1}{3} \hat{\mathbf{j}}$
D $\frac{3}{2} v \hat{\mathbf{i}}+\frac{1}{4} v \hat{\mathbf{j}}$

The correct answer is:
$\frac{1}{4} v \hat{\mathbf{i}}+\frac{3}{2} y \hat{\mathbf{j}}$

Solution:
From the law of conservation of linear momentum

$\begin{aligned}
m v+(3 m)(2 v) & =(4 m) v^{\prime} \\
m v \hat{i}+6 m v \hat{j} & =4 m v^{\prime} \\
v^{\prime} & =\frac{1}{4} v \hat{i}+\frac{6}{4} v \hat{j} \\
y^{\prime} & =\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2011 (Mains)
A mass m on the surface of the Earth is shifted to a target equal to the radius of the Earth. If R is the radius and M is the mass of the Earth, then work done in this process is

Options:
A mgR2
B mgR
C 2 mgR
D mgR4

The correct answer is:
mgR2

Solution:
Work done =âPE=-GMmR+R--GMmR
=GMm2R=mgR2

Topic: Physics, Work Power Energy, JEE Main
A mass \( \mathrm{m} \) is suspended from a massless pulley, which it self is suspended by a massless string and a spring as shown in figure. The force constant of spring is \( \mathrm{K} \). What will be the time period of oscillation of \( \mathrm{m} \) :

Options:
A \( 2 \pi \sqrt{\frac{m}{2 K}} \)
B \( 2 \pi \sqrt{\frac{m}{K}} \)
C \( \pi \sqrt{\frac{m}{K}} \)
D \( \pi \sqrt{\frac{m}{2 K}} \)

The correct answer is:
\( \pi \sqrt{\frac{m}{K}} \)

Solution:
F=4 K x=ma

T=2πm4K=πmK.

Topic: Physics, Oscillations, JEE Main
A mass of 2 kg is whirled in a horizontal circle by means of a string at an initial speed of 5 revolutions per minute. Keeping the radius constant, if we want the tension in the string to be doubled, then the new speed should nearly be

Options:
A 2.25 rpm

B 7 rpm

C 10 rpm
D 14 rpm

The correct answer is:
7 rpm

Solution:
F = m Ï 2 r
= mr 2 n Ï 2
thus F Ã n 2 n = resolution per second.
thus 2 F F = n â² n 2
or n â² = 2 n
= 5 2 rpm
= 7 rpm

Topic: Physics, Laws of Motion, JEE Main
A mass of $0.5 \mathrm{~kg}$ moving with a speed of $1.5 \mathrm{~m} / \mathrm{s}$ on a horizontal smooth surface, collides with a nearly weightless spring of force constant $k=50 \mathrm{~N} / \mathrm{m}$. The maximum compression of the spring would be:

Options:
A $0.15 \mathrm{~m}$
B $0.12 \mathrm{~m}$
C $1.5 \mathrm{~m}$
D $0.5 \mathrm{~m}$

The correct answer is:
$0.15 \mathrm{~m}$

Solution:
According to the question
$$
\begin{aligned}
\frac{1}{2} m v^2 & =\frac{1}{2} k x^2 \\
\Rightarrow \quad x & =v \sqrt{\frac{m}{k}}=1.5 \sqrt{\frac{0.5}{50}} \\
& =0.15 \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Work Power Energy, JEE Main
A mass of $0.5 \mathrm{~kg}$ moving with a speed of $1.5 \mathrm{~m} /$ $s$ on a horizontal smooth surface, collides with a nearly weightless spring of force constant $\mathrm{k}=$ $50 \mathrm{~N} / \mathrm{m}$. The maximum compression of the spring would be

Options:
A $0.5 \mathrm{~m}$
B $0.15 \mathrm{~m}$
C $0.12 \mathrm{~m}$
D $1.5 \mathrm{~m}$

The correct answer is:
$0.15 \mathrm{~m}$

Solution:
Apply law of conservation of energy,
$$
\begin{array}{l}
\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{kx}^{2} \\
\Rightarrow \mathrm{mv}^{2}=\mathrm{kx}^{2} \text { or } 0.5 \times(1.5)^{2}=50 \times \mathrm{x}^{2} \\
\therefore \mathrm{x}=0.15 \mathrm{~m}
\end{array}
$$

Topic: Physics, Work Power Energy, JEE Main
A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of 45° with the vertical. Then F equals: (Take g=10 m s-2 and the rope to be massless)

Options:
A 100N
B 90N
C 70N
D 75N

The correct answer is:
100N

Solution:
Let the tension in the string is T.

Applying the condition of equilibrium in the vertical direction,

Tcos45°=100

⇒ T2=100 ...1

In the horizontal direction,

Tsin45°=F

⇒ T2=F

Put the value of T from equation 1,

⇒ F=100 N

So, the horizontal force applied on the rope will be 100 N.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2020 (07 Jan Shift 2)
A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is g=10 m s-2

Options:
A 100 N
B 200Â N
C 140Â N
D 70Â N

The correct answer is:
100 N

Solution:
Tcos⁡45°=mg
Tsin⁡45°=F
tan 45°=Fmg
F=mg=100 N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2019 (09 Jan Shift 2)
A mass of 10 kg is suspended vertically by a rope of length 5 m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is α=tan-1x×10-1. The value of x is _____ .

(Given, g=10 m s-2)

Options:

The correct answer is:
3

Solution:
⇒Tsinθ=100 and Tcosθ=30

Therefore, tanθ=103

Then, tanα=3 10⇒α=tan-13×10-1

Hence, value of x=3.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (27 Jun Shift 2)
A mass of $20 \mathrm{~kg}$ moving with a speed of $10 \mathrm{~m} / \mathrm{s}$ collides with another stationary mass of $5 \mathrm{~kg}$. As a result of the collision, the two masses stick together. The kinetic energy of the composite mass will be

Options:
A 600 Joule
B 800 Joule
C 1000 Joule
D 1200 Joule

The correct answer is:
800 Joule

Solution:
$\begin{aligned} & \qquad m_1 v_1+m_2 v_2=\left(m_1+m_2\right) v_{\text {sys }} \\ & 20 \times 10+5 \times 0=(20+5) v_{\text {sys. }} \Rightarrow v_{\text {sys }}=8 \mathrm{~m} / \mathrm{s} \\ & \text { K.E. of composite mass }=\frac{1}{2}(20+5) \times(8)^2=800 \mathrm{~J}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main