datavorous/entrance-exam-dataset huggingface dataset:
A body of mass $2 \mathrm{~kg}$ is kept by pressing to a vertical wall by a force of $100 \mathrm{~N}$. The coefficient of friction between wall and body is 0.3 . Then the frictional force is equal to

Options:
A $6 \mathrm{~N}$
B $20 \mathrm{~N}$
C $600 \mathrm{~N}$
D $700 \mathrm{~N}$

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

Solution:
For the given condition, Static friction
$\begin{aligned}
& =\text { Applied force = Weight of body } \\
& =2 \times 10=20 \mathrm{~N}
\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main
A body of mass $2 \mathrm{~kg}$ is projected vertically upwards with a velocity of $2 \mathrm{~m~} \mathrm{sec}^{-1}$. The K.E. of the body just before striking the ground is

Options:
A $2 \mathrm{~J}$
B $1 \mathrm{~J}$
C $4 \mathrm{~J}$
D $8 \mathrm{~J}$

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

Solution:
If particle is projected vertically upward with velocity of $2 \mathrm{~m} / \mathrm{s}$ then it returns with the same velocity.
So its kinetic energy $=\frac{1}{2} m v^2=\frac{1}{2} \times 2 \times(2)^2=4 \mathrm{~J}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A body of mass $2 \mathrm{~kg}$ is thrown up vertically with K.E. of 490 joules. If the acceleration due to gravity is $9.8 \mathrm{m} / \mathrm{s}^2$, then the height at which the K.E. of the body becomes half its original value is given by

Options:
A 50 m
B 12.5 m
C 25 m
D 10 m

The correct answer is:
12.5 m

Solution:
Let $h$ is that height at which the kinetic energy of the body becomes half its original value i.e. half of its kinetic energy will convert into potential energy
$\therefore m g h=\frac{490}{2} \Rightarrow 2 \times 9.8 \times h=\frac{490}{2} \Rightarrow h=12.5 m$.

Topic: Physics, Work Power Energy, JEE Main
A body of mass $2 \mathrm{~kg}$ is thrown up vertically with K.E. of 490 joules. If the acceleration due to gravity is $9.8 \mathrm{~m} / \mathrm{s}^2$, then the height at which the K.E. of the body becomes half its original value is given by

Options:
A $50 \mathrm{~m}$
B $12.5 \mathrm{~m}$
C $25 \mathrm{~m}$
D $10 \mathrm{~m}$

The correct answer is:
$12.5 \mathrm{~m}$

Solution:
Let $h$ is that height at which the kinetic energy of the body becomes half its original value i.e. half of its kinetic energy will convert into potential energy
$\therefore m g h=\frac{490}{2} \Rightarrow 2 \times 9.8 \times h=\frac{490}{2} \Rightarrow h=12.5 m \text {. }$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A body of mass 2 kg moves under a force of 2i^+3j^+5k^N It starts from rest and was at the origin initially. After 4 s, its new coordinates are 8,b,20. The value of b is ______.

(Round off to the Nearest Integer)

Options:

The correct answer is:
12

Solution:
a→=F→m=2i^+3j^+5k^2

=i^+1.5j^+2.5k^

τ→=u→t+12a→t2

=0+12i^+1.5j^+2.5k^16

=8i^+12j^+20k^

b=12

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (16 Mar Shift 2)
A body of mass 2 kg moving with a speed of 4 m s-1 makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial speed. The speed of the two body centre of mass is $x/10$. Find the value of x.

Options:

The correct answer is:
25

Solution:
pi=pf

2×4=2×1+m2×v2

m2v2=6 …(1)

By coefficient of restitution,
1=v2-14⇒v2=5 m s-1

by (i)

m2×5=6

m2=1.2 kg

vcm=m1v1+m2v2 m1+m2

vcm=2×1+1.2×52+1.2=83.2=2510

x=25

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2021 (25 Jul Shift 1)
A body of mass $2 \mathrm{~kg}$ moving with a velocity of $3 \mathrm{~m} / \mathrm{sec}$ collides head on with a body of mass $1 \mathrm{~kg}$ moving in opposite direction with a velocity of $4 \mathrm{~m} / \mathrm{sec}$. After collision, two bodies stick together and move with a common velocity which in $\mathrm{m} / \mathrm{sec}$ is equal to

Options:
A $1 / 4$
B $1 / 3$
C $2 / 3$
D $3 / 4$

The correct answer is:
$2 / 3$

Solution:
$\begin{gathered}m_1 v_1-m_2 v_2=\left(m_1+m_2\right) v \\ \Rightarrow 2 \times 3-1 \times 4=(2+1) v \Rightarrow v=\frac{2}{3} m / s\end{gathered}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A body of mass $2 \mathrm{~kg}$ slides down with an acceleration of $3 \mathrm{~m} / \mathrm{s}^2$ on a rough inclined plane having a slope of $30^{\circ}$. The external force required to take the same body up the plane with the same acceleration will be: $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$

Options:
A
$4 \mathrm{~N}$

B
$14 \mathrm{~N}$

C
$6 \mathrm{~N}$

D
$20 \mathrm{~N}$

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

Solution:
Equation of motion when the mass slides down
$$
\begin{aligned}
&M g \sin \theta-f=M a \\
&\Rightarrow 10-f=6 \\
&\quad\left(M=2 \mathrm{~kg}, a=3 \mathrm{~m} / \mathrm{s}^2, \theta=30^{\circ} \text { given }\right) \\
&\therefore f=4 \mathrm{~N}
\end{aligned}
$$
{ Equation of motion when the block is }
pushed up
Let the external force required to take the block up the plane with same acceleration be $F$
$$
\begin{aligned}
&F-M g \sin \theta-f=M a \\
&\Rightarrow F-10-4=6 \\
&F=20 \mathrm{~N}
\end{aligned}
$$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A body of mass 2m moving with velocity v makes a head-on elastic collision with another body of mass m which is initially at rest. Loss of kinetic energy of the colliding body (mass 2m) is

Options:
A 19 of its initial kinetic energy
B 16 of its initial kinetic energy
C 12 of its initial kinetic energy
D 89 of its initial kinetic energy

The correct answer is:
89 of its initial kinetic energy

Solution:
Initial K.E of ball of mass 2m= K1
=12Ã2mÃv2
=mv2
Collision is elastic so both K.E and momentum are conserved. Let velocities of balls are v1 and v2 after collision.
So, KE is conserved
122mv2=122mv12+12mv22
â v2=v12+12v22
2( v 2 − v 1 2 )= v 2 2 ....( i )
And, momentum is conserved
2mv+m0=2mv1+mv2
â2v=2v1+v2 ... (ii)
Now,
v2=2v-v1
From (i)
2( v 2 − v 1 2 ) 2( v− v 1 ) = v 2 2 v 2
v+ v 1 = v 2
v− v 1 = v 2 2
2 v 1 = v 2 2
v 1 = v 2 4 = v 3
So, final K.E of ball of mass 2m ,
k2=122mv12=12Ã2mÃv29=19k1
Hence, loss of K.E. of Ist ball
=K1-19K1=89K1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A body of mass 2M splits into four masses {m,M-m,m,M-m}, which are rearranged to form a square as shown in the figure. The ratio of Mm for which, the gravitational potential energy of the system becomes maximum is x : 1. The value of x is ________.

Options:

The correct answer is:
2

Solution:
U=Gm2a+G(M-m)2a+2Gm(M-m)a+2Gm(M-m)2a

For maxima or minima of potential energy dUdm=0

dUdm=Ga(2 m-2(M-m)+(M-2 m)2+2(M-2 m)

(4m-4m-22m)+(-2M+2M+2M)=0; 2M=22m; Mm=2

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (27 Aug Shift 1)
A body of mass \(2 \mathrm{~kg}\) moves in a horizontal circular path of radius \(5 \mathrm{~m}\). At an instant, its speed is \(2 \sqrt{5} \mathrm{~m} / \mathrm{s}\) and is increasing at the rate of \(3 \mathrm{~m} / \mathrm{s}^2\). The magnitude of force acting on the body at the instant is,

Options:
A \(6 \mathrm{~N}\)
B \(8 \mathrm{~N}\)
C \(14 \mathrm{~N}\)
D \(10 \mathrm{~N}\)

The correct answer is:
\(10 \mathrm{~N}\)

Solution:
Hint: \(F=\) ma
\(\begin{aligned}
& =m \sqrt{a_c^2+a_T^2} \\
& =m \sqrt{\left(\frac{20}{5}\right)^2+9} \\
& =2 \sqrt{16+9}=2 \times 5=10 \mathrm{~N}
\end{aligned}\)

Topic: Physics, Laws of Motion, JEE Main
A body of mass 3 kg collides elastically with another body at rest and then continues to move in the original direction with one-half of its original speed. What is the mass of the target body (in kg)?

Options:

The correct answer is:
1

Solution:
Conservation of linear momentum :
3u=(3u/2)+mv ⇒mv=(3u/2)…(1)

Newton's experimental law :
u=–e(u/2–v)

Putting e=1 for elastic collision we obtain,
v=3u2…..(2)
Using (1) and (2), we obtain m=1 kg.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A body of mass $3 \mathrm{~kg}$ is under a constant force which causes a displacement $s$ in metres in it, given by the relation $s=\frac{1}{3} t^2$, where $t$ is in seconds. Work done by the force in 2 seconds is:

Options:
A $\frac{19}{5} \mathrm{~J}$
B $\frac{5}{19} \mathrm{~J}$
C $\frac{3}{8} \mathrm{~J}$
D $\frac{8}{3} \mathrm{~J}$

The correct answer is:
$\frac{8}{3} \mathrm{~J}$

Solution:
It is given that:
$s=\frac{t^2}{3}$
Differentiating w.r.t. $t$ we have
$U=\frac{d s}{d t}=\frac{2 t}{3}$
Again differentiating w.r.t. $t$ we have
$a=\frac{d^2 s}{d t^2}=\frac{2}{3}$
Now $\mathrm{W}=\int \mathrm{F} d s=\int \mathrm{mads}$
$\begin{aligned}
& =\int m \times \frac{2}{3} \times \frac{2 t}{3} d t \\
& =\frac{4}{9} m \int t d t \\
W & =\frac{4}{9} m \times\left[\frac{t^2}{2}\right]_0^2 \\
& =\frac{4}{9} \times 3 \times 2=\frac{8}{3} \mathrm{~J}
\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
A body of mass 4 kg experiences two forces F→1=5i^+8j^+7k^ and F→2=3i^-4j^-3k.^ The acceleration acting on the body is:

Options:
A -2i^-j^-k^
B 4i^+2j^+2k^
C 2i^+j^+k^
D 2i^+3j^+3k^

The correct answer is:
2i^+j^+k^

Solution:
Net force F→=F→1+F→2=8i^+4j^+4k^

Therefore, required acceleration a→=F→m=2i^+j^+k^

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (01 Feb Shift 2)
A body of mass $4 \mathrm{~kg}$ moving with velocity $12 \mathrm{~m} / \mathrm{s}$ collides with another body of mass $6 \mathrm{~kg}$ at rest. If two bodies stick together after collision, then the loss of kinetic energy of system is

Options:
A Zero
B $288 \mathrm{~J}$
C $172.8 \mathrm{~J}$
D $144 \mathrm{~J}$

The correct answer is:
$172.8 \mathrm{~J}$

Solution:
Loss in K.E. =
$\begin{aligned}
& \frac{m_1 m_2}{2\left(m_1+m_2\right)}\left(u_1-u_2\right)^2 \\
&=\frac{4 \times 6}{2 \times 10} \times(12-0)^2=172.8
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A body of mass 4m is lying in x - y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass m move perpendicular to each other with equal speeds υ. The total kinetic energy generated due to explosion is:

Options:
A mυ2
B 32mυ2
C 2 mυ2
D 4 mυ2

The correct answer is:
32mυ2

Solution:
The total momentum of the system will be conserved before and after explosion.

Then, pi=pf
O=mvi^+mvj^+2mv→
v→=-v2i^-v2j^
v→=v2
Total kinetic energy generated due to explosion is KE=v2mv2+v2mv2+v22mv22=mv2+mv22=32mv2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A body of mass (5±0.5) kg is moving with a velocity of (20±0.4) m s-1. Its kinetic energy will be

Options:
A (1000±0.14) J
B (500±0.14) J
C (500±140) J
D (1000±140) J

The correct answer is:
(1000±140) J

Solution:
The kinetic energy is given by,

K=12mv2

Differentiating the above equation,

dK=12(v2dm+2mvdv)

Dividing by K,

⇒dKK=dmm+2dvv

⇒dK1000=0.55+2×0.420

⇒dK=100+40=140

⇒KE=(1000±140)J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (13 Apr Shift 1)
A body of mass $5 \mathrm{~kg}$ explodes at rest into three fragments with masses in the ratio $1: 1: 3$. The fragments with equal masses fly in mutually perpendicular directions with speeds of $21 \mathrm{~m} / \mathrm{s}$. The velocity of the heaviest fragment will be

Options:
A $11.5 \mathrm{~m} / \mathrm{s}$
B $14.0 \mathrm{~m} / \mathrm{s}$
C $7.0 \mathrm{~m} / \mathrm{s}$
D $9.89 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$9.89 \mathrm{~m} / \mathrm{s}$

Solution:
$\begin{aligned}
& P_x=m \times v_x=1 \times 21=21 \mathrm{~kg} \mathrm{~m} / \mathrm{s} \\
& P_y=m \times v_y=1 \times 21=21 \mathrm{kgm} / \mathrm{s} \\
& \therefore \text { Resultant }=\sqrt{P_x^2+P_y^2}=21 \sqrt{2} \mathrm{~kg} \mathrm{~m} / \mathrm{s}
\end{aligned}$
The momentum of heavier fragment should be numerically equal to resultant of $\vec{P}_x$ and $\vec{P}_y$.
$3 \times v=\sqrt{P_x^2+P_y^2}=21 \sqrt{2} \therefore v=7 \sqrt{2}=9.89 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A body of mass 5 kg is moving with a momentum of 10 kg m s-1. Now a force of 2 N acts on the body in the direction of its motion for 5 s. The increase in the Kinetic energy of the body is _____ J.

Options:

The correct answer is:
30

Solution:
The formula to calculate the change on momentum of the object is given by

pf-pi=∫Fdt ...1

Substitute the values of the known parameters into equation (1) and solve to calculate the final momentum of the object.

pf-10 kg m s-1=2 N×5 s⇒pf=10+10 kg m s-1= 20 kg m s-1

The formula to calculate increase in kinetic energy ∆K of the particle is given by

∆K=12mpf2-pi2 ...2

Substitute the values of the known parameters into equation (2) to calculate the required increase in kinetic energy.

∆K=12×5 kg×202-102 kg2 m2 s-2= 30 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (08 Apr Shift 2)
A body of mass $0.05 \mathrm{~kg}$ is observed to fall with an acceleration of $9.5 \mathrm{~ms}^{-2}$. The opposing force of air on the body is $\left(g=9.8 \mathrm{~ms}^{-2}\right)$

Options:
A $0.015 \mathrm{~N}$
B $0.15 \mathrm{~N}$
C $0.030 \mathrm{~N}$
D zero

The correct answer is:
$0.015 \mathrm{~N}$

Solution:
From, Newton's laws of motion,
or
$\begin{aligned}
\mathrm{mg}-\mathrm{F}_{\mathrm{air}} &=\mathrm{ma} \\
\mathrm{F}_{\mathrm{air}} &=\mathrm{m}(\mathrm{g}-\mathrm{a}) \\
&=0.05(9.8-9.5) \\
&=0.015 \mathrm{~N}
\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main
A body of mass $5 \mathrm{~kg}$ is suspended by a spring balance on an inclined plane as shown in figure. The spring balance measure

Options:
A $50 \mathrm{~N}$
B $25 \mathrm{~N}$
C $500 \mathrm{~N}$
D $10 \mathrm{~N}$

The correct answer is:
$25 \mathrm{~N}$

Solution:
Since downward force along the inclined plane
$=m g \sin \theta=5 \times 10 \times \sin 30^{\circ}=25 \mathrm{~N}$

Topic: Physics, Laws of Motion, JEE Main
A body of mass $5 \mathrm{~kg}$ is thrown vertically up with a kinetic energy of $490 \mathrm{~J}$. The height at which the kinetic energy of the body becomes half of the original value is
(acceleration due to gravity $=9.8 \mathrm{~ms}^{-2}$ )

Options:
A $5 \mathrm{~m}$
B $2.5 \mathrm{~m}$
C $10 \mathrm{~m}$
D $12.5 \mathrm{~m}$

The correct answer is:
$5 \mathrm{~m}$

Solution:
Given, $\quad m=5 \mathrm{~kg}$
and
$\mathrm{KE}=490 \mathrm{~J}$
By the law of conservation of energy
$$
\begin{aligned}
\frac{1}{2} m u^{2} &=\frac{1}{2} m v^{2}+m g h \\
490 &=245+5 \times 9.8 \times h \\
h &=\frac{490-245}{5 \times 9.8} \\
h &=\frac{245}{49} \\
&=5 \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Work Power Energy, JEE Main
A body of mass $5 \mathrm{~kg}$ makes an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal to $\frac{1}{10}$ th of its original velocity. Then the mass of the second body is

Options:
A 4.09 kg
B 0.5 kg
C 5 kg
D 5.09 kg

The correct answer is:
4.09 kg

Solution:
Mass of the first body $m_1=5 \mathrm{~kg}$, for elastic collision $e=1$.
Suppose initially body $m_1$ moves with velocity $v$ after collision velocity becomes $\left(\frac{u}{10}\right)$.
Let after collision velocity of $M$ block becomes $\left(v_2\right)$.
By conservation of momentum
$\begin{aligned}
m_1 u_1+m_2 u_2 & =m_1 v_1+m_2 v_2 \\
\text { or } \quad 5 u+M \times 0 & =5 \times \frac{u}{10}+M v_2
\end{aligned}$
Since,
$\begin{aligned}
v_1-v_2 & =-e\left(u_1-u_2\right) \\
\frac{u}{10}-v_2 & =-1(u)
\end{aligned}$
or $\quad \frac{u}{10}+u=v_2$
Substituting value of $v_2$ in Eq. (i) from Eq. (ii), we get
$5 u=\frac{u}{2}+M\left(\frac{11 u}{10}\right)$
or $5-\frac{1}{2}=M\left(\frac{11}{10}\right)$
or $M=\frac{9 \times 10}{2 \times 11}$
or $\begin{aligned}
M & =\frac{45}{11} \\
& =4.09 \mathrm{~kg}
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A body of mass 5 kg moving with a uniform speed 32 m s-1 in X-Y plane along the line y=x+4. The angular momentum of the particle about the origin will be _______kg m2 s-1.

Options:

The correct answer is:
60

Solution:
Equation of line given: y-x-4=0.

Let d1 be the perpendicular distance of given line from origin.

d1=-412+12=22

So, |L→|=mvd1=5×32×22 kg m2 s-1

=60 kg m2 s-1

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2024 (29 Jan Shift 2)
A body of mass 5 kg starts from the origin with an initial velocity u=30iË^+40jË^ m s-1. If a constant force -6iË^-5jË^ N acts on the body, the time (in seconds) in which the y component of the velocity becomes zero is

Options:

The correct answer is:
40

Solution:
From Newton's second law of motion,

Fy=â³Pyâ³t
â³t=â³PyFy
=0-5Ã40-5=40 s

Topic: Physics, Laws of Motion, JEE Main
A body of mass $5 \mathrm{~kg}$ under the action of constant force $\vec{F}=F_x \hat{i}+F_y \hat{j}$ has velocity at $\mathrm{t}=0 \mathrm{~s}$ as $\overrightarrow{\mathrm{v}}=(6 \hat{\mathrm{i}}-2 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s})$ and at $\mathrm{t}=10 \mathrm{~s}$ as $\overrightarrow{\mathrm{v}}=+6 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}$. The force $\overrightarrow{\mathrm{F}}$ is:

Options:
A
$(-3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}) \mathrm{N}$

B
$\left(-\frac{3}{5} \hat{i}+\frac{4}{5} \hat{j}\right) N$

C
$(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}) \mathrm{N}$

D
$\left(\frac{3}{5} \hat{\mathrm{i}}-\frac{4}{5} \hat{\mathrm{j}}\right) \mathrm{N}$

The correct answer is:
$(-3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}) \mathrm{N}$

Solution:
From question,
Mass of body, $m=5 \mathrm{~kg}$
Velocity at $t=0$,
$$
u=(6 \hat{i}-2 \hat{j}) \mathrm{m} / \mathrm{s}
$$
Velocity at $t=10 \mathrm{~s}$,
$$
v=+6 \hat{j} \mathrm{~m} / \mathrm{s}
$$
Force, $F=$ ?
Acceleration, $a=\frac{v-u}{t}$
$$
=\frac{6 \hat{j}-(6 \hat{i}-2 \hat{j})}{10}=\frac{-3 \hat{i}+4 \hat{j}}{5} \mathrm{~m} / \mathrm{s}^2
$$
Force, $F=m a$
$$
=5 \times \frac{(-3 \hat{i}+4 \hat{j})}{5}=(-3 \hat{i}+4 \hat{j}) N
$$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2014 (11 Apr Online)
A body of mass \( 5 \mathrm{~kg} \) is suspended by a spring balance on an inclined plane as shown in the figure. Measure the force applied on spring balance,

Options:
A \( 50 \mathrm{~N} \)
B \( 25 \mathrm{~N} \)
C \( 500 \mathrm{~N} \)
D \( 10 \mathrm{~N} \)

The correct answer is:
\( 25 \mathrm{~N} \)

Solution:
Acceleration of the body down the rough inclined plane=gsinθ.

∴ Force applied on the spring balance

=mgsinθ=5×10×sin30°

=5×10×12=25 N.

Topic: Physics, Laws of Motion, JEE Main
A body of mass 8 kg and another of mass 2 kg are moving with equal kinetic energy. The ratio of their respective momenta will be

Options:
A 1:1
B 2:1
C 1:4
D 4:1

The correct answer is:
2:1

Solution:
Given: The mass of the objects m1=8 kg and m2=2 kg.

Let the momentum of the 8 kg and 2 kg masses is p1 and p2, respectively, then according to question, their kinetic energy are equal, i.e.,

K1=K2

p122m1=p222m2

p1p2=m1m2=41=21

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2022 (26 Jul Shift 2)
A body of mass $10 \mathrm{~kg}$ is acted upon by a force given by equation $F=\left(3 t^2-30\right)$ newtons. The initial velocity of the body is $10 \mathrm{~m} / \mathrm{s}$. The velocity of the body after $5 \mathrm{~s}$ is

Options:
A $4.5 \mathrm{~m} / \mathrm{s}$
B $6 \mathrm{~m} / \mathrm{s}$
C $7.5 \mathrm{~m} / \mathrm{s}$
D $5 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$7.5 \mathrm{~m} / \mathrm{s}$

Solution:
Body is acted upon by a force given by equation
$$
\begin{aligned}
& F=\left(3 t^2-30\right) \mathrm{N} \\
& \therefore m(v-u)=\int F d t \\
& =\frac{3 t^3}{3}-30 t=t^3-30 t \\
& =5^3-30(5)=125-150=-25 \\
& \therefore \quad m[v-10]=-25 \\
& \Rightarrow \text { Final velocity } \quad v=7.5 \mathrm{~m} / \mathrm{s} \\
&
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, JEE Main
A body of mass $10 \mathrm{~kg}$ is moving with a constant velocity of $10 \mathrm{~m} / \mathrm{s}$. When a constant force acts for 4 seconds on it, it moves with a velocity $2 \mathrm{~m} / \mathrm{sec}$ in the opposite direction. The acceleration produced in it is

Options:
A $3 \mathrm{~m} / \mathrm{sec}^2$
B $-3 \mathrm{~m} / \mathrm{sec}^2$
C $0.3 \mathrm{~m} / \mathrm{sec}^2$
D $-0.3 \mathrm{~m} / \mathrm{sec}^2$

The correct answer is:
$-3 \mathrm{~m} / \mathrm{sec}^2$

Solution:
$v=u+a t \Rightarrow-2=10+a \times 4 \Rightarrow a=-3 \mathrm{~m} / \mathrm{sec}^2$

Topic: Physics, Motion In One Dimension, JEE Main
A body of mass 10 kg is moving with an initial speed of 20 m s-1. The body stops after 5 s due to friction between body and the floor. The value of the coefficient of friction is: (Take acceleration due to gravity g=10 m s-2)

Options:
A 0.2
B 0.3
C 0.5
D 0.4

The correct answer is:
0.4

Solution:
Friction force wil act opposite to the motion of the body.

F=μN=μmg

⇒a=-μg-ve sign for direction relative to the motion of body

Now, applying equation of motion, we get

v=u+at

⇒0=20+-μ×10×5

⇒50μ=20

⇒μ=0.4

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2023 (31 Jan Shift 2)
A body of mass $10 \mathrm{mg}$ is moving with a velocity of $100 \mathrm{~ms}^{-1}$. The wavelength of de-Broglie wave associated with it would be ( $h=6.63 \times 10^{-34} \mathrm{Js}$ )

Options:
A $6.63 \times 10^{-35} \mathrm{~m}$
B $6.63 \times 10^{-34} \mathrm{~m}$
C $6.63 \times 10^{-31} \mathrm{~m}$
D $6.63 \times 10^{-37} \mathrm{~m}$

The correct answer is:
$6.63 \times 10^{-31} \mathrm{~m}$

Solution:
$m=10 \mathrm{mg}=10 \times 10^{-6} \mathrm{~kg}$
$$
\begin{aligned}
v &=100 \mathrm{~ms}^{-1} \\
\lambda &=\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{10 \times 10^{-6} \times 100} \\
&=6.63 \times 10^{-31} \mathrm{~m}
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main
A body of mass $50 \mathrm{~kg}$ is lifted to a height of $20 \mathrm{~m}$ from the ground in the two different ways as shown in the figures. The ratio of work done against the gravity in both the respective cases, will be :

Options:
A $1: 2$
B $\sqrt{3}: 2$
C $2: 1$
D $1: 1$

The correct answer is:
$1: 1$

Solution:
Work done by gravity is independent of path. It depends only on vertical displacement so work done in both cases will be same.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2024 (05 Apr Shift 1)
A body of mass \( 100 \mathrm{~g} \) moves at the speed of \( 36 \mathrm{~km} \mathrm{~h}^{-1} \). The de Broglie wavelength related to it is of the order
\( \mathrm{m}\left(h=6.626 \times 10^{-34} \mathrm{Js}\right) \)

Options:
A \( 10^{-14} \)
B \( 10^{-24} \)
C \( 10^{-34} \)
D \( 10^{-44} \)

The correct answer is:
\( 10^{-34} \)

Solution:
It is given that a body of mass m=100 g moves at the speed of v=36 km h-1.

The de Broglie wavelength related to it,

λ==hp=hmv, here h is Planck's constant and p is momentum.

Substituting given values,

λ=6.626×10-340.1×36×10003600≃10-34

Topic: Physics, Dual Nature of Matter, JEE Main
A body of mass 200 g is tied to a spring of spring constant 12.5 N m-1, while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad s-1, then the ratio of extension in the spring to its natural length will be :

Options:
A 1:2
B 1:1
C 2:3
D 2:5

The correct answer is:
2:3

Solution:
Let the extension in spring be x and the natural length of spring be l.

Here the centrifugal force will stretch the spring.

So, centrifugal force=spring force

kx=ml+xω2⇒12.5x=15l+x25⇒1.5x=l⇒xl=11.5=23

⇒x:l=2:3

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2023 (24 Jan Shift 2)
A body of mass 200 gram is tied to a spring of spring constant $12.5 \mathrm{~N} / \mathrm{m}$, while other end of spring is fixed at point ' $\mathrm{O}$ '. If the body moves about ' $\mathrm{O}$ ' in a circular path on a smooth horizontal surface with constant angular speed $5 \mathrm{rad} / \mathrm{s}$ then the ratio of extension in the spring to its natural length will be

Options:
A $1: 2$
B $1:1$
C $2:3$
D $2:5$

The correct answer is:
$2:3$

Solution:
Let the normal length be $\mathrm{L}$ and the extension be $\mathrm{x}$.
$\therefore \quad$ Restoring Force $=$ Centripetal Force
$\begin{aligned}
& \mathrm{kx}=\mathrm{m}(\mathrm{L}+\mathrm{x}) \omega^2 \\
& 12.5 \mathrm{x}=0.2(\mathrm{~L}+\mathrm{x}) 25 \quad \ldots(\because \omega=5 \mathrm{rad} / \mathrm{s})
\end{aligned}$
$\begin{aligned} & 12.5 x=5(L+x) \\ & 7.5 x=5 L \\ \therefore \quad & \frac{x}{L}=\frac{5}{7.5}=\frac{2}{3}=2: 3\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main