datavorous/entrance-exam-dataset huggingface dataset:
A body is revolving with a constant speed along a circle. If its direction of motion is reversed but the speed remains the same. Then, which of the following statement is incorrect?

Options:
A the centripetal force will not suffer any change in magnitude
B the centripetal force will not suffer any change in direction
C the centripetal force will have its direction reversed
D None of these

The correct answer is:
the centripetal force will have its direction reversed

Solution:
Centripetal force will not suffer any change in magnitude because the speed is same in both cases
Also, centripetal force will not suffer any change in direction because it is directed toward the center in both cases

Topic: Physics, Rotational Motion, JEE Main
A body is taken to a height of $n R$ from the surface of the earth. The ratio of the acceleration due to gravity on the surface to that at the altitude is

Options:
A $(n+1)^{2}$
B $(n+1)^{-2}$
C $(n+1)^{-1}$
D $(n+1)$

The correct answer is:
$(n+1)^{2}$

Solution:
Aceleration due to gravity at a height $h$ above
the earth surface $g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}$
$$
\begin{array}{l}
\frac{g}{g^{\prime}}=\left(\frac{R+h}{R}\right)^{2} \\
\frac{g}{g^{\prime}}=\left(\frac{R+n R}{R}\right)^{2} \\
\frac{g}{g^{\prime}}=(1+n)^{2}
\end{array}
$$

Topic: Physics, Gravitation, JEE Main
A body is thrown from the surface of the earth with velocity u m s-1. The maximum height in meter above the surface of the earth up to which it will reach is (R= radius of the earth, g= acceleration due to gravity)

Options:
A u2R2gR-u2
B 2u2RgR-u2
C u2R22gR2-u2
D u2RgR-u2

The correct answer is:
u2R2gR-u2

Solution:
âGMm R + 1 2 m u 2 =0+ ( â GMm R+h )
GM R+h = GM R â u 2 2
GM ( R+h ) = 2GMâR u 2 2R
R+h GM = 2R 2GMâR u 2
h= 2GMR 2GMâR u 2 âR
= 2GMRâ2GMR+ R 2 u 2 2GMâR u 2
= R 2 u 2 2GMâR u 2 = R u 2 2gRâ u 2

Topic: Physics, Gravitation, JEE Main
A body is thrown from the surface of the earth with velocity 'V' $\frac{\mathrm{m}}{\mathrm{s}}$. The maximum height
above the earth's surface upto which it will reach is
$(\mathrm{R}=$ radius of earth, $\mathrm{g}=$ acceleration due to gravity)

Options:
A $\frac{\mathrm{VR}^{2}}{\boldsymbol{g R}-\mathbf{V}}$
B $\frac{\mathrm{V}^{2} \mathrm{R}}{2 \mathrm{gR}-\mathrm{V}^{2}}$
C $\frac{2 \sigma R}{V^{2}(R-1)}$
D $\frac{\mathrm{VR}}{2 \sigma R-V}$

The correct answer is:
$\frac{\mathrm{V}^{2} \mathrm{R}}{2 \mathrm{gR}-\mathrm{V}^{2}}$

Solution:
$\begin{aligned} \frac{1}{2} \mathrm{mu}^{2} &=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}-\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)=\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}} \\ &=\mathrm{GMm}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}\right] \\ \therefore \mathrm{u}^{2} &=2 \mathrm{GM}\left[\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}\right] \\ \mathrm{u}^{2} &=2 \mathrm{gR}^{2}\left[\frac{\mathrm{R}+\mathrm{h}-\mathrm{R}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\right] \\ \mathrm{u}^{2} &=2 \mathrm{gR}\left[\frac{\mathrm{h}}{\mathrm{R}+\mathrm{h}}\right] \\ \therefore \quad \frac{\mathrm{u}^{2}}{2 \mathrm{gR}} &=\left[\frac{\mathrm{h}}{\mathrm{R}+\mathrm{h}}\right] \quad \therefore \quad \frac{\mathrm{R}+\mathrm{h}}{\mathrm{R}}=\frac{2 \mathrm{gR}}{\mathrm{u}^{2}} \\ \frac{\mathrm{R}}{\mathrm{h}}+1 &=\frac{2 \mathrm{gR}}{\mathrm{u}^{2}} \\ \therefore \frac{\mathrm{R}}{\mathrm{h}} &=\frac{2 \mathrm{gR}}{\mathrm{u}^{2}}-1=\frac{2 \mathrm{gR}-\mathrm{u}^{2}}{\mathrm{u}^{2}} \therefore \mathrm{h}=\frac{\mathrm{Ru}^{2}}{2 \mathrm{gR}-\mathrm{u}^{2}} \end{aligned}$

Topic: Physics, Gravitation, JEE Main
A body is thrown vertically upwards from A, the top of the tower, reaches the ground in time $t_{1}$. If it is thrown vertically downwards from A with the same speed, it reaches the ground in time $\mathrm{t}_{2}$. If it is allowed to fall freely from $\mathrm{A},$ then the time it takes to reach the ground is given by

Options:
A $\mathrm{t}=\frac{\mathrm{t}_{1}+\mathrm{t}_{2}}{2}$
B $\mathrm{t}=\frac{\mathrm{t}_{1}-\mathrm{t}_{2}}{2}$
C $\quad \mathrm{t}=\sqrt{\mathrm{t}_{1} \mathrm{t}_{2}}$
D $\mathrm{t}=\sqrt{\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}}$

The correct answer is:
$\quad \mathrm{t}=\sqrt{\mathrm{t}_{1} \mathrm{t}_{2}}$

Solution:
Let the body is projected vertically upwards from A with a speed $u_{0}$ Using equation, $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

For case $(1)-\mathrm{h}=\mathrm{u}_{0} \mathrm{t}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2} \quad \ldots(1)$

For case (2) $-\mathrm{h}=\mathrm{u}_{0} \mathrm{t}_{2}-\frac{1}{2} \mathrm{gt}_{1}^{2}$

Subtracting eq (2) from (1), we get

$$

\begin{array}{l}

0=\mathrm{u}_{0}\left(\mathrm{t}_{2}+\mathrm{t}_{1}\right)+\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{2}^{2}-\mathrm{t}_{1}^{2}\right) \\

\Rightarrow \mathrm{u}_{0}=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)

\end{array}

$$

Putting the value of $\mathrm{u}_{0}$ in eq (2), we get

$$

-\mathrm{h}=-\left(\frac{1}{2}\right) \mathrm{g}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right) \mathrm{t}_{2}-\left(\frac{1}{2}\right) \mathrm{gt}_{2}^{2}

$$

$$

\Rightarrow \mathrm{h}=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1} \mathrm{t}_{2}\right)

$$

For case $3, \mathrm{u}_{0}=0, \mathrm{t}=?$

$$

-\mathrm{h}=0 \times \mathrm{t}-\left(\frac{1}{2}\right) \mathrm{gt}^{2}

$$

$\mathrm{h}=\left(\frac{1}{2}\right) \mathrm{gt}^{2}$

Comparing eq. (4) and $(5),$ we get

$$

\frac{1}{2} \mathrm{gt}^{2}=\frac{1}{2} \mathrm{gt}_{1} \mathrm{t}_{2} \quad \therefore \mathrm{t}=\sqrt{\mathrm{t}_{1} \mathrm{t}_{2}}

$$

Topic: Physics, Motion In One Dimension, JEE Main
A body is thrown vertically upwards. Which one of the following graphs correctly represents the velocityv vs time t?

Options:
A

B

C

D

The correct answer is:

Solution:
v=u-gt
y=c+mx
m=-g = slope
c=+u
Velocity becomes negative after some time and the negative slope is constant.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2017 (02 Apr)
A Body moves $6 m$ north. $8 m$ east and $10 m$ vertically upwards, what is its resultant displacement from initial position

Options:
A $10 \sqrt{2} m$
B $10 \mathrm{~m}$
C $\frac{10}{\sqrt{2}} m$
D $10 \times 2 m$

The correct answer is:
$10 \sqrt{2} m$

Solution:
$\begin{gathered}\bar{r}=x \hat{i}+y \hat{j}+z \hat{k} \quad \therefore r=\sqrt{x^2+y^2+z^2} \\ l=\sqrt{6^2+8^2+10^2}=10 \sqrt{2} \mathrm{~m}\end{gathered}$

Topic: Physics, Motion In One Dimension, JEE Main
A body moves in a circular path of the radius R with deceleration so that at any moment of time its tangential and normal acceleration are equal in magnitude. At the initial moment t=0 , the velocity of the body is v0 then the velocity of the body after it has moved S at any time will be -

Options:
A v=v0e-2SR
B v=v0e-SR
C v=v0eâSR
D v=v0eâ2SR

The correct answer is:
v=v0e-SR

Solution:
a T = a N
vdv ds = - v 2 R
(-ve because it is a retardation) and s is the distance. covered tangentially on circular path
on integrating we get
∫ v 0 v dv v = ∫ 0 s - ds R
log v v 0 v = 1 R - 1 s
log v v 0 = - s R
v = v 0 e - s R

Topic: Physics, Laws of Motion, JEE Main
A body moves on a frictionless plane starting from rest. If $S_n$ is distance moved between $t=n-1$ and $\mathrm{t}=\mathrm{n}$ and $\mathrm{S}_{\mathrm{n}-1}$ is distance moved between $\mathrm{t}=\mathrm{n}-2$ and $\mathrm{t}=\mathrm{n}-1$, then the ratio $\frac{\mathrm{S}_{\mathrm{n}-1}}{\mathrm{~S}_{\mathrm{n}}}$ is $\left(1-\frac{2}{x}\right)$ for $\mathrm{n}=10$. The value of $x$ is ______.

Options:

The correct answer is:
19

Solution:
$\begin{aligned} & S_n=\frac{1}{2} a(2 n-1)=\frac{19 a}{2} \\ & S_{n-1}=\frac{1}{2} a(2 n-3)=\frac{17 a}{2} \\ & \frac{S_{n-1}}{S_n}=\frac{17}{19}=1-\frac{2}{x} \Rightarrow x=19\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main
A body moves on a frictionless plane starting from rest. If $S_n$ is distance moved between $t=n-1$ and $\mathrm{t}=\mathrm{n}$ and $\mathrm{S}_{\mathrm{n}-1}$ is distance moved between $\mathrm{t}=\mathrm{n}-2$ and $\mathrm{t}=\mathrm{n}-1$, then the ratio $\frac{\mathrm{S}_{\mathrm{n}-1}}{\mathrm{~S}_{\mathrm{n}}}$ is $\left(1-\frac{2}{x}\right)$ for $\mathrm{n}=10$. The value of $x$ is ______.

Options:

The correct answer is:
19

Solution:
$\begin{aligned} & S_n=\frac{1}{2} a(2 n-1)=\frac{19 a}{2} \\ & S_{n-1}=\frac{1}{2} a(2 n-3)=\frac{17 a}{2} \\ & \frac{S_{n-1}}{S_n}=\frac{17}{19}=1-\frac{2}{x} \Rightarrow x=19\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (05 Apr Shift 1)
A body of capacity $4 \mu \mathrm{F}$ is charged to $80 \mathrm{~V}$ and another body of capacity $6 \mu \mathrm{F}$ is charged to $30 \mathrm{~V}$. When they are connected the energy lost by $4 \mu \mathrm{F}$ capacitor is

Options:
A $9.8 \mathrm{~mJ}$
B $4.6 \mathrm{~mJ}$
C $3.2 \mathrm{~mJ}$
D $2.5 \mathrm{~mJ}$

The correct answer is:
$9.8 \mathrm{~mJ}$

Solution:
$\mathrm{C}_2=4 \mu \mathrm{F}, V_1=80$ volt,
$C_2=6 \mu \mathrm{F}, V_2=30$ volt
Energy loss,
$\begin{aligned} \Delta U & =\frac{1}{2} \frac{C_1 C_2}{C_2+C_2}\left(V_1-V_2\right)^2 \\ & =\frac{1}{2} \times \frac{4 \times 10^{-6} \times 6 \times 10^{-4}}{4 \times 10^{-6}+6 \times 10^{-4}}(80-30)^2 \\ & =\frac{12 \times 10^{-6}}{10} \times 2500=1.2 \times 25 \times 10^{-4} \\ & =30 \times 10^{-4}=3 \times 10^{-3} \mathrm{~J}\end{aligned}$
Initial energy of $4 \mu \mathrm{F}$ capacitor.
$\begin{aligned} E_{\mathrm{n}} & =\frac{1}{2} C_1 V_1^2 \\ & =\frac{1}{2} \times 4 \times 10^{-4} \times(80)^2 \\ & =2 \times 10^{-4} \times 6400 \\ & =12.8 \times 10^{-3} \mathrm{~J}\end{aligned}$
Energy lost by $C_1=12.8 \times 10^{-1}-3 \times 10^{-1}$
$\begin{aligned} & =9.8 \times 10^{-3} \mathrm{~J} \\ & =9.8 \mathrm{~mJ}\end{aligned}$

Topic: Physics, Capacitance, JEE Main
A body of density $\rho$ is dropped from (at rest) height ' $h$ ' into a lake of density ' $\delta$ ' $(\delta>\rho)$. The maximum depth to which the body sinks before returning to float on the surface is [Neglect all dissipative forces]

Options:
A $\frac{(\delta-\rho)}{2 h \rho}$
B $\frac{2 h \rho}{(\delta-\rho)}$
C $\frac{h \rho}{2(\delta-\rho)}$
D $\frac{h \rho}{(\delta-\rho)}$

The correct answer is:
$\frac{h \rho}{(\delta-\rho)}$

Solution:
The velocity of the body when it reaches the surface of the lake is
$$
\begin{aligned}
& \frac{\mathrm{a}}{\mathrm{g}}=\frac{\delta-\rho}{\rho} \\
& \therefore \mathrm{a}=\left(\frac{\delta-\rho}{\rho}\right) \mathrm{g}
\end{aligned}
$$
If $a$ is the acceleration and retardation in the liquid then $\mathrm{v}^2=2 \mathrm{ad}$
by eq. (1) and (4)
$$
\begin{aligned}
& 2 \mathrm{ad}=2 \mathrm{gh} \\
& \therefore \mathrm{d}=\frac{\mathrm{g}}{\mathrm{a}} \mathrm{h} \quad \therefore \mathrm{d}=\frac{\rho}{\delta-\rho} \mathrm{h}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A body of mass $1 \mathrm{~g}$ and carrying a charge $10^{-8} \mathrm{C}$ passes from two points $P$ and $Q . P$ and $Q$ are at electric potentials. $600 \mathrm{~V}$ and $0 \mathrm{~V}$, respectively. The velocity of the body at $Q$ is $20 \mathrm{cms}^{-1}$. It velocity in $\mathrm{ms}^{-1}$ at $P$ is

Options:
A $\sqrt{0.028}$
B $\sqrt{0.056}$
C $\sqrt{0.56}$
D $\sqrt{5.6}$

The correct answer is:
$\sqrt{0.028}$

Solution:
$$
\begin{aligned}
m & =1 \mathrm{~g}=10^{-3} \mathrm{~kg}, \\
v & =20 \mathrm{~cm} / \mathrm{s}=20 \times 10^{-20} \mathrm{~m} / \mathrm{s} \\
q & =10^{-8} \mathrm{C} \\
V_p & =600 \mathrm{volt} \\
V_Q & =0 \text { volt }
\end{aligned}
$$
Work done in moving the charge from $P$ to $Q$
$$
\begin{aligned}
W & =q V_{P Q}=10^{-8}(600-0) \\
& =600 \times 10^{-8} \mathrm{~J}
\end{aligned}
$$
From work energy theorem
Work done $=$ change in $\mathrm{KE}$
$$
\begin{aligned}
600 \times 10^{-8} & =\frac{1}{2} m\left(20 \times 10^{-2}\right)^2-\frac{1}{2} m v^2 \\
600 \times 10^{-8} & =\frac{1}{2} \times 10^{-3} \times 4 \times 10^{-2}-\frac{1}{2} \times 10^{-3} \times v^2 \\
\frac{1}{2} \times 10^{-3} \times v^2 & =2 \times 10^{-5}-600 \times 10^{-8} \\
& =2 \times 10^{-5}-0.6 \times 10^{-5} \\
v^2 & =1.4 \times 10^{-2} \times 2=28 \times 10^{-2} \\
v & =\sqrt{28 \times 10^{-2}}=\sqrt{0.028} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main
A body of mass 1 kg begins to move under the action of a time dependent force F→=ti^+3t2j^ N, where i^ and j^ are the unit vectors along x and y axis. The power developed by above force, at the time t=2 s, will be _______W.

Options:

The correct answer is:
100

Solution:
Given, F→=ti^+3t2j^ N and m=1 kg

Now, we can write from Newton's second law, F→=ma→=mdv→dt

Then, dv→=ti^+3t2j^dt

Integrating the above,

∫0vdv→=∫0tti^+3t2j^dt⇒v→=t22i^+t3j^

Now, power developed by the force is P=F→·v→=ti^+3t2j^·t22i^+t3j^

=t32+3t5

At t=2 s, power is P=232+325=4+3×32=100 W

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (24 Jan Shift 2)
A body of mass 1 kg collides head on elastically with a stationary body of mass 3 kg. After collision, the smaller body reverses its direction of motion and moves with a speed of 2 m s-1. The initial speed of the smaller body before collision is ______ m s-1.

Options:

The correct answer is:
4

Solution:
Let u1 be the initial speed of the smaller body before collision.

Let v be the final speed of the bigger body after collision.

Applying conservation of linear momentum,

1×u1+0=-2+3v⇒u1=-2+3v ...1

We know, coefficient of restitution for elastic perfectly collision is 1.

⇒1=v+2u1⇒v+2=u1 ...2

Equating 1 and 2, we have

u1=4 m s-1.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2023 (25 Jan Shift 2)
A body of mass $1 \mathrm{~kg}$ falls freely from a height of $100 \mathrm{~m}$, on a platform of mass $3 \mathrm{~kg}$ which is mounted on a spring having spring constant $\mathrm{k}=1.25 \times 10^{6} \mathrm{~N} / \mathrm{m}$. The bodysticks to the platform and the spring's maximum compression is found to be $x$. Given that $g=10 \mathrm{~ms}^{-2},$ the value of $\mathrm{x}$ will be close to :

Options:
A $40 \mathrm{~cm}$
B $4 \mathrm{~cm}$
C $80 \mathrm{~cm}$
D $8 \mathrm{~cm}$

The correct answer is:
$4 \mathrm{~cm}$

Solution:
Velocity of $1 \mathrm{~kg}$ block just before it collides with $3 \mathrm{~kg}$

block $=\sqrt{2 \mathrm{gh}}=\sqrt{2000} \mathrm{~m} / \mathrm{s}$

Using principle of conservation of linear momentum just before and just after collision, we get

$1 \times \sqrt{2000}=4 \mathrm{v} \Rightarrow \mathrm{v}=\frac{\sqrt{2000}}{4} \mathrm{~m} / \mathrm{s}$

Initial compression of spring $1.25 \times 10^{6} \mathrm{x}_{0}=30 \Rightarrow \mathrm{x}_{0} \approx 0$

using work energy theorem, $\mathrm{W}_{\mathrm{g}}+\mathrm{W}_{\mathrm{sp}}=\Delta \mathrm{KE}$

$\Rightarrow 40 \times x+\frac{1}{2} \times 1.25 \times 10^{6}\left(0^{2}-x^{2}\right)$

$=0-\frac{1}{2} \times 4 \times v^{2}$

solving $x \approx 4 \mathrm{~cm}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A body of mass $1 \mathrm{~kg}$ falls freely from a height of $100 \mathrm{~m}$, on a platform of mass $3 \mathrm{~kg}$ which is mounted on a spring having spring constant $\mathrm{k}=1.25 \times 10^{6} \mathrm{~N} / \mathrm{m}$. The bodysticks to the platform and the spring's maximum compression is found to be $x$. Given that $g=10 \mathrm{~ms}^{-2},$ the value of $\mathrm{x}$ will be close to :

Options:
A $40 \mathrm{~cm}$
B $4 \mathrm{~cm}$
C $80 \mathrm{~cm}$
D $8 \mathrm{~cm}$

The correct answer is:
$4 \mathrm{~cm}$

Solution:
Velocity of $1 \mathrm{~kg}$ block just before it collides with $3 \mathrm{~kg}$

block $=\sqrt{2 \mathrm{gh}}=\sqrt{2000} \mathrm{~m} / \mathrm{s}$

Using principle of conservation of linear momentum just before and just after collision, we get

$1 \times \sqrt{2000}=4 \mathrm{v} \Rightarrow \mathrm{v}=\frac{\sqrt{2000}}{4} \mathrm{~m} / \mathrm{s}$

Initial compression of spring $1.25 \times 10^{6} \mathrm{x}_{0}=30 \Rightarrow \mathrm{x}_{0} \approx 0$

using work energy theorem, $\mathrm{W}_{\mathrm{g}}+\mathrm{W}_{\mathrm{sp}}=\Delta \mathrm{KE}$

$\Rightarrow 40 \times x+\frac{1}{2} \times 1.25 \times 10^{6}\left(0^{2}-x^{2}\right)$

$=0-\frac{1}{2} \times 4 \times v^{2}$

solving $x \approx 4 \mathrm{~cm}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2019 (11 Jan Shift 1)
A body of mass $1 \mathrm{~kg}$, initially at rest, explodes and breaks into three fragments of masses in the ratio 1:1:3. The two pieces of equal mass fly off perpendicular to each other with a speed of $30 \mathrm{~m} / \mathrm{sec}$ each. What is the velocity of the heavier fragment?

Options:
A $10 \sqrt{2} \mathrm{~m/s}$
B $5 \sqrt{3} \mathrm{~m/s}$
C $10 \sqrt{3} \mathrm{~m/s}$
D $5 \sqrt{2} \mathrm{~m/s}$

The correct answer is:
$10 \sqrt{2} \mathrm{~m/s}$

Solution:
Total mass of the body $M=1 \mathrm{~kg}$
So, mass of first fragment $\mathrm{m}_1=\frac{1}{5} \times 1=0.2 \mathrm{~kg}$
Mass of second fragment $m_2=\frac{1}{5} \times 1=0.2 \mathrm{~kg}$
Mass of third fragment $\mathrm{m}_3=\frac{3}{5} \times 1=0.6 \mathrm{~kg}$
Velocity of masses $m_1$ and $m_2, v=30 \mathrm{~m} / \mathrm{s}$
Momentum of first fragment
$\mathrm{p}_1=\mathrm{m}_1 \mathrm{v}=0.2 \times 30=6 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
Similarly, $p_2=m_2 v=0.2 \times 30=6 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
Third fragment will fly in a direction opposite to the direction of resultant of $p_1$ and $p_2$ so that total momentum of the system after explosion remains zero.
Momentum of the third fragment
$\begin{aligned} & p_3=\sqrt{p_1^2+p_2^2} \\ & \therefore p_3=\sqrt{6^2+6^2}=6 \sqrt{2} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\end{aligned}$
Thus speed of third fragment $v_3=\frac{p_3}{\mathrm{~m}_3}=\frac{6 \sqrt{2}}{0.6}=10 \sqrt{2} \mathrm{~m} / \mathrm{s}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A body of mass $1 \mathrm{~kg}$ is executing simple harmonic motion. Its displacement $y(\mathrm{~cm})$ at $t$ seconds is given by $y=6 \sin (100 t+\pi / 4)$. Its maximum kinetic energy is

Options:
A $6 \mathrm{~J}$
B $18 \mathrm{~J}$
C $24 \mathrm{~J}$
D $36 \mathrm{~J}$

The correct answer is:
$18 \mathrm{~J}$

Solution:
So $a=6 \mathrm{~cm} \omega=100 \mathrm{rad} / \mathrm{sec}$
$K_{\max }=\frac{1}{2} m \omega^2 a^2=\frac{1}{2} \times 1 \times(100)^2 \times\left(6 \times 10^{-2}\right)^2=18 \mathrm{~J}$

Topic: Physics, Oscillations, JEE Main
A body of mass 1 kg is executing simple harmonic motion. Its displacement ycm at t seconds is given by y=6sin100t+π4. Its maximum kinetic energy is

Options:
A 6 J
B 18 J
C 24 J
D 36 J

The correct answer is:
18 J

Solution:
Given, a=6 cm=6×10-2 m

ω=100 rad s-1,m=1 kg

Kmax=12mω2a2

=12×1×1002×6×10-22

=18 J

Topic: Physics, Oscillations, JEE Main
A body of mass $1 \mathrm{~kg}$ is moving with a velocity $10 \mathrm{~ms}^{-1}$ due to a constant force on a horizontal rough surface having coefficient of kinetic friction 0.4 . If the constant force is removed, the body comes to rest in a time (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

Options:
A $2.5 \mathrm{~s}$
B $4 \mathrm{~s}$
C $0.4 \mathrm{~s}$
D $0.25 \mathrm{~s}$

The correct answer is:
$2.5 \mathrm{~s}$

Solution:
Mass of body, $\mathrm{m}=1 \mathrm{~kg}$
velocity, $v=10 \mathrm{~m} / \mathrm{s}$
Coefficient of kinetic friction, $\mu=0.4$
from the equation of motion
$\begin{aligned}
& \mathrm{v}=\mathrm{u}+\mathrm{at} \\
& \mathrm{t}=\frac{\mathrm{v}}{\mathrm{a}} \quad \quad(\because \mathrm{F}=\mu \mathrm{mg}) \\
& =\frac{\mathrm{vm}}{\mathrm{F}}=\frac{\mathrm{vm}}{\mu \mathrm{mg}}=\frac{\mathrm{v}}{\mu \mathrm{g}}
\end{aligned}$
The body comes to rest in a time
$\mathrm{t}=\frac{\mathrm{v}}{\mu \mathrm{g}}=\frac{10}{0.4 \times 10}=2.5 \mathrm{~s}$

Topic: Physics, Laws of Motion, JEE Main
A body of mass $1 \mathrm{~kg}$ of thrown upwards with a velocity $20 \mathrm{~m} / \mathrm{s}$. It momenetarily comes to rest after attaining a height of $18 \mathrm{~m}$. How much energy is lost due to air friction ?
$$
\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)
$$

Options:
A $10 \mathrm{~J}$
B $20 \mathrm{~J}$
C $30 \mathrm{~J}$
D $40 \mathrm{~J}$

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

Solution:
Loss of energy $=\frac{1}{2} \mathrm{mv}^2-\mathrm{mgh}=20 \mathrm{~J}$

Topic: Physics, Work Power Energy, NEET, NEET 2009 (Mains)
A body of mass 1 kg rests on a horizontal floor with which it has a coefficient of static friction 13. It is desired to make the body move by applying the minimum possible force F N. The value of F will be _______.(Round off to the Nearest Integer) [Take g=10 m s-2 ]

Options:

The correct answer is:
5

Solution:
Fcosθ=μN

⇒Fsinθ+N=mg

⇒F=μmgcosθ+μsinθ

Fmin=μmg1+μ2=13×1023=5

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (17 Mar Shift 2)
A body of mass 1 kg starts moving from rest at t = 0, in a circular path of radius 8 m. Its kinetic energy varies as a function of times as KE =2t2 Joules, where t is in seconds. Then

Options:
A Tangential acceleration = 4 m s-2

B Power of all forces at t = 2 s is 8 W

C First-round is completed in 2 s.

D Tangential force at t = 2 s is 4 N.

The correct answer is:
Power of all forces at t = 2 s is 8 W

Solution:
12mv2=2t2
v2=4t2 ⇒ v=2t
at=dvdt ⇒ at=2 m s-2
Ft=mat ⇒ Ft=2 N
Power will be inverted only by tangential force
P=F→t+F→c.V→ since Fc⊥V & FtV
P=FtV ⇒ P=2×2t ⇒ P=4t ⇒ P=8 W

Topic: Physics, Laws of Motion, JEE Main
A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body?

Options:
A 1.8 kg
B 1.2 kg
C 1.0 kg
D 1.5 kg

The correct answer is:
1.2 kg

Solution:
From conservation of linear momentum
mU=mU4+MV . . . (i)
For elastic collision: e=1=V-U4u
V=5U4 . . . (ii)
From equation (i) & (ii),
mU=mU4+M5U4
m=m4+54M ⇒M=3m5=1.2 kg

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2019 (09 Apr Shift 1)
A body of mass 2 kg is driven by an engine delivering a constant power of 1 J s-1. the body starts from rest and moves in a straight line. After 9 s, the body has moved a distance (in m)….

Options:

The correct answer is:
18

Solution:
Given,

Power delivered by the engine P=1 J s-1

Mass of the body is m= 2 kg

Body starts from the rest, u=0

Using work energy Theorem,

W=∆K=Pt

Pt=12mv2-0

Substitute the values in the above equation, we get
(1)t=12×2×v2
⇒v=t=dsdt
⇒∫0tds=∫09tdt

⇒s=93232=23×27

⇒s=18 m

Topic: Physics, Work Power Energy, JEE Main
A body of mass 2 kg is driven by an engine delivering a constant power of 1 J s-1. the body starts from rest and moves in a straight line. After 9 s, the body has moved a distance (in m)….

Options:

The correct answer is:
18

Solution:
Given,

Power delivered by the engine P=1 J s-1

Mass of the body is m= 2 kg

Body starts from the rest, u=0

Using work energy Theorem,

W=∆K=Pt

Pt=12mv2-0

Substitute the values in the above equation, we get
(1)t=12×2×v2
⇒v=t=dsdt
⇒∫0tds=∫09tdt

⇒s=93232=23×27

⇒s=18 m

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2020 (05 Sep Shift 2)
A body of mass $2 \mathrm{~kg}$ is hung on a spring balance mounted vertically in a lift. If the lift descends with an acceleration equal to the acceleration due to gravity ' $g$ ', the reading on the spring balance will be

Options:
A $2 \mathrm{~kg}$
B $(4 \times g) \mathrm{kg}$
C $(2 \times g) \mathrm{kg}$
D Zero

The correct answer is:
Zero

Solution:
Reading on the spring balance $=m(g-a)$ and since $a=g \backslash$ Force $=0$

Topic: Physics, Laws of Motion, JEE Main
A body of mass 2 kg is initially at rest. It starts moving unidirectionally under the influence of a source of constant power P. Its displacement in 4 s is 13α2P m. The value of α will be ______.

Options:

The correct answer is:
4

Solution:
Energy provided will be used to increase kinetic energy of the body. Therefore,

12mV2=Pt

V=2Ptm

⇒dxdt=2Ptm⇒∫0xdx=2Pm∫04t12dt

⇒x=2Pm23t3204

⇒x=16P3=13×16P

Hence, required value of α=4.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (30 Jan Shift 2)