datavorous/entrance-exam-dataset huggingface dataset:
A body of mass 500 g moves along x-axis such that it's velocity varies with displacement x according to the relation v=10x m s-1 the force acting on the body is:

Options:
A 125 N
B 25 N
C 166 N
D 5 N

The correct answer is:
25 N

Solution:
It is given that

v=10x m s-1

Force is defined by F=mdvdt ...(i)

Multiplying and dividing equation (i) by dx

F=mdvdtdxdx⇒F=mdxdtdvdx=mvdvdx ...(ii)

Finding the values of the terms in equation (ii)
dvdx=102x=5x
vdvdx=a=10x×5x=50 m s-2

Hence, the force is
F=12×50=25 N.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2023 (11 Apr Shift 2)
A body of mass 1000 kg is moving horizontally with a velocity 6 m s-1. If 200 kg extra mass is added, the final velocity (in m s-1) is:

Options:
A 6
B 2
C 3
D 5

The correct answer is:
5

Solution:
From the conservation of linear momentum, it can be written that

Mu=M+mv ...1

where, M is the initial mass, u is the initial velocity, m is the extra mass added and v is the final velocity of the composition.

From equation (1), it follows that

1000×6=1200×v⇒v=5 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2024 (27 Jan Shift 1)
A body of mass is taken from earth surface to the height h equal to twice the radius of earth (Re), the increase in potential energy will be : (g = acceleration due to gravity on the surface of earth)

Options:
A 3mgRe
B 13mgRe
C 23mgRe
D 12mgRe

The correct answer is:
23mgRe

Solution:
The potential energy of a particle at a distance r (>Re) from the centre of Earth is,

U=-GMemr

Initially, the particle is at a height Re from the surface. Therefore, r=Re.

∴Ui=-GMemRe

When height is increased to 2Re from surface, r=2Re+Re=3Re

∴Uf=-GMem3Re

Thus, the change in potential energy is

∆U=Uf-Ui=-GMem3Re+GMemRe=23GMemRe=23GMeRe2mRe=23mgRe as g=GMeRe2

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (25 Jan Shift 2)
A body of mass m=1 kg is moving in a medium and experiences a frictional force F=-kv, where v is the speed of the body. The initial speed is v0=10 m s-1 and after 10 s, its energy becomes half of the initial energy. Then, the value of k is

Options:
A 10 ln2
B ln2
C 120ln2
D 10 ln 2

The correct answer is:
120ln2

Solution:
âµ According to the question,
12mvf2=12Ã12mvi2
vi, vf=initial and final speeds of the body
or vf2=vi22 or vf=102
Given, F=-kv
or ma= -kvâmdvdt=-kv
ââ«101021vdv= -â«010kdt âµ m=1kg
â ln v10102=-k10
â ln 102- ln 10=-k10
â k=110 ln10102=110 ln2=lnâ¡220

Topic: Physics, Work Power Energy, JEE Main
A body of mass m1 moving with an unknown velocity of v1 i^, undergoes a collinear collision with a body of mass m2 moving with a velocity v2 i^. After the collision, m1 and m2 move with velocities of v3 i^ and v4 i^, respectively. If m2=0.5 m1 and v3=0.5 v1, then v1 is:

Options:
A v4-v24
B v4-v22
C v4+v2
D v4-v2

The correct answer is:
v4-v2

Solution:
From the conservation of linear momentum,
m1v1+m2v2=m1v3+m2v4
m1v1+0.5 m1 v2=m10.5 v1+0.5 m1 v4
0.5 v1+0.5 v2=0.5 v4
v4=v1+v2
v1=v4-v2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2019 (08 Apr Shift 2)
A body of mass m=10−2 kg is moving in a medium and experiences a frictional force F=−k v 2 . Its initial speed is v0=10 m s−1. After 10 s its kinetic energy is 18mv02, then value of k will be:-

Options:
A 10−1 kg m-1 s-1
B 10−3 kg m-1
C 10−3 kg s-1
D 10−4 kg m-1

The correct answer is:
10−4 kg m-1

Solution:
After 10 sec, 12mv2=18mv02
⇒v=v02=5 m s-1
Acceleration, a=− k v 2 m
⇒ dv dt =− k v 2 m
⇒ ∫105dvv2=−km ∫010dt
-15+110=-k10-2×10
k=10−4 kg m-1

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2017 (02 Apr)
A body of mass m=10 kg is attached to one end of a wire of length 0.3 m. What is the maximum angular speed (in rad s-1) with which it can be rotated about its other end in a space station without breaking the wire?
[Breaking stress of wire σ=4.8×107 N m-2 and area of cross-section of the wire=10-2 cm2]

Options:

The correct answer is:
4

Solution:
TA=σ………………….1
T=mω2l………………….2
Solving we get,
ω=4 rad s-1

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2020 (09 Jan Shift 1)
A body of mass m=10 kg is attached to one end of a wire of length 0.3 m. What is the maximum angular speed (in rad s-1) with which it can be rotated about its other end in a space station without breaking the wire?
[Breaking stress of wire=4.8Ã107 N m-2 and area of cross-section of the wire=10-2 cm2]

Options:

The correct answer is:
4

Solution:
TA=σ………………….1
T=mω2l………………….2
Solving we get,
ω=4 rad s-1

Topic: Physics, Mechanical Properties of Solids, JEE Main
A body of mass $m=3.513 \mathrm{~kg}$ is moving along the $x$-axis with a speed of $5.00 \mathrm{~ms}^{-1}$. The magnitude of its momentum is recorded as

Options:
A
$17.6 \mathrm{~kg} \mathrm{~ms}^{-1}$

B
$17.565 \mathrm{~kg} \mathrm{~ms}^{-1}$

C
$17.56 \mathrm{~kg} \mathrm{~ms}^{-1}$

D
$17.57 \mathrm{~kg} \mathrm{~ms}^{-1}$

The correct answer is:
$17.6 \mathrm{~kg} \mathrm{~ms}^{-1}$

Solution:
$$
P=m v=3.513 \times 5.00 \approx 17.6
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2008
A body of mass m accelerates uniformly from rest to v1 in time t1. The instantaneous power delivered to the body as a function of the t is

Options:
A mâ¡ ν 1 tâ¡ tâ¡ 1
B mâ¡ ν 1 2 tâ¡ tâ¡ 1 2
C mâ¡ ν 1 tâ¡ 2 tâ¡ 1
D mâ¡ ν 1 2 tâ¡ tâ¡ 1

The correct answer is:
mâ¡ ν 1 2 tâ¡ tâ¡ 1 2

Solution:
Acceleration a = ν 1 tâ¡ 1
∴ Velocity ( ν ) = 0 + a tâ¡ = ν 1 tâ¡ 1 tâ¡
Power Pâ¡ = Force × velocity = mâ¡ a ν
or Pâ¡ = mâ¡ ( ν 1 tâ¡ 1 ) × ( ν 1 tâ¡ tâ¡ 1 ) = mâ¡ ν 1 2 tâ¡ tâ¡ 1 2

Topic: Physics, Work Power Energy, JEE Main
A body of mass $m$, accelerates uniformly from rest to $v_1$ in time $t_1$. The instantaneous power delivered to the body as a function of time $t$ is

Options:
A
$\frac{\mathrm{mv}_1 \mathrm{t}}{\mathrm{t}_1}$

B
$\frac{\mathrm{mv}_1^2 \mathrm{t}}{\mathrm{t}_1^2}$

C
$\frac{m v_1 t^2}{t_1}$

D
$\frac{m v_1^2 t}{t_1}$

The correct answer is:
$\frac{\mathrm{mv}_1^2 \mathrm{t}}{\mathrm{t}_1^2}$

Solution:
Power $P=\vec{F} \cdot \vec{v}=\operatorname{mav}=m\left(\frac{v_1}{t_1}\right)\left(\frac{v_1}{t_1} t\right)=\frac{m_1^2 t}{t_1^2}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2004
A body of mass M and charge q is connected to a spring of spring constant k. It is oscillating along x-direction about its equilibrium position in the horizontal plane, taken to be at x=0, with an amplitude A. An electric field E is applied along the x-direction. Which of the following statements is correct?

Options:
A The total energy of the system is 12mω2A2+12q2E2k
B The new equilibrium position is at a distance 2qEk from x= 0
C The new equilibrium position is at a distance qE2k from x= 0
D The total energy of the system is 12mω2A2-12q2E2k .

The correct answer is:
The total energy of the system is 12mω2A2+12q2E2k

Solution:
The equilibrium position will shift to a point where the resultant force is equal to zero. kxeq=qE ⇒xeq=qEk. Energy =12mω2A2+qEk2=12mω2A2+12q2E2k.

Topic: Physics, Oscillations, JEE Main
A body of mass M and charge q is connected to a spring of spring constant k. It is oscillating along x-direction about its equilibrium position in the horizontal plane, taken to be at x=0, with an amplitude A. An electric field E is applied along the x-direction. Which of the following statements is correct?

Options:
A The total energy of the system is 12mω2A2+12q2E2k
B The new equilibrium position is at a distance 2qEk from x= 0
C The new equilibrium position is at a distance qE2k from x= 0
D The total energy of the system is 12mω2A2-12q2E2k .

The correct answer is:
The total energy of the system is 12mω2A2+12q2E2k

Solution:
The equilibrium position will shift to a point where the resultant force is equal to zero. kxeq=qE ⇒xeq=qEk. Energy =12mω2A2+qEk2=12mω2A2+12q2E2k.

Topic: Physics, Oscillations, JEE Main, JEE Main 2018 (15 Apr)
A body of mass $M$ and charge $q$ is connected to a spring of spring constant $k$. It is oscillating along $\mathrm{x}$-direction about its equilibrium position, taken to be at $x=0$, with an amplitude $A$. An electric field $E$ is applied along the $\mathrm{x}$-direction. Which of the following statements is correct?

Options:
A
The total energy of the system is
$$
\frac{1}{2} m \omega^2 A^2+\frac{1}{2} \frac{q^2 E^2}{k}
$$

B
The new equilibrium position is at a distance: $\frac{2 q E}{k}$ from $x=0$

C
The new equilibrium position is at a distance: $\frac{q E}{2 k}$ from $\mathrm{x}=0$

D
The total energy of the system is $\frac{1}{2} m \omega^2 A^2-\frac{1}{2} \frac{q^2 E^2}{k}$

The correct answer is:
The total energy of the system is
$$
\frac{1}{2} m \omega^2 A^2+\frac{1}{2} \frac{q^2 E^2}{k}
$$

Solution:
Equilibrium position will shift to point where resultant force $=0$
$$
\mathrm{kx}_{\text {eq }}=\mathrm{qE} \Rightarrow \mathrm{x}_{\text {eq }}=\frac{\mathrm{qE}}{\mathrm{k}}
$$
Total energy $=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2+\frac{1}{2} \mathrm{kx}_{\text {eq }}^2$
Total energy $=\frac{1}{2} m \omega^2 A^2+\frac{1}{2} \frac{q^2 E^2}{k}$

Topic: Physics, Electrostatics, JEE Main
A body of mass $M$ and charge $q$ is connected to a spring of spring constant $k$. It is oscillating along $\mathrm{x}$-direction about its equilibrium position, taken to be at $x=0$, with an amplitude $A$. An electric field $E$ is applied along the $\mathrm{x}$-direction. Which of the following statements is correct?

Options:
A
The total energy of the system is
$$
\frac{1}{2} m \omega^2 A^2+\frac{1}{2} \frac{q^2 E^2}{k}
$$

B
The new equilibrium position is at a distance: $\frac{2 q E}{k}$ from $x=0$

C
The new equilibrium position is at a distance: $\frac{q E}{2 k}$ from $\mathrm{x}=0$

D
The total energy of the system is $\frac{1}{2} m \omega^2 A^2-\frac{1}{2} \frac{q^2 E^2}{k}$

The correct answer is:
The total energy of the system is
$$
\frac{1}{2} m \omega^2 A^2+\frac{1}{2} \frac{q^2 E^2}{k}
$$

Solution:
Equilibrium position will shift to point where resultant force $=0$
$$
\mathrm{kx}_{\text {eq }}=\mathrm{qE} \Rightarrow \mathrm{x}_{\text {eq }}=\frac{\mathrm{qE}}{\mathrm{k}}
$$
Total energy $=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2+\frac{1}{2} \mathrm{kx}_{\text {eq }}^2$
Total energy $=\frac{1}{2} m \omega^2 A^2+\frac{1}{2} \frac{q^2 E^2}{k}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A body of mass M at rest explodes into three pieces, in the ratio of masses 1:1:2. Two smaller pieces fly off perpendicular to each other with velocities of 30 m s-1 and 40 m s-1 respectively. The velocity of the third piece will be

Options:
A 35 m s-1
B 50 m s-1
C 25 m s-1
D 15 m s-1

The correct answer is:
25 m s-1

Solution:
Let the masses of the three parts be m', m' and 2m'.

Momentum of first part p1=m'×30=30m' and momentum of second part p2=m'×40=40m'.

Third part will fly in a direction opposite to the direction of resultant of p1 and p2, so that total momentum of the system after explosion remains zero.

Momentum of the third part is p3=p12+p22=30m'2+40m'2=50m'

Velocity of third part, v=50m'2m'=25 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2022 (29 Jun Shift 1)
A body of mass m dropped from a height h reaches the ground with a speed of 0.8gh. The value of work done by the air-friction is:

Options:
A -0.68mgh
B mgh
C 0.64mgh
D 1.64mgh

The correct answer is:
-0.68mgh

Solution:
K1+U1=K2+U2+Wf
Wf: work done against friction
0+mgh=12mv2+0+Wf
mgh=12m0.64gh+Wf
Wf=0.68mgh
Work done by friction Wf'=-Wf=-0.68mgh

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (01 Sep Shift 2)
A body of mass $m$ is accelerated uniformly from rest to a speed $v$ in a time $T$. The instantaneous power delivered to the body as a function time is given by

Options:
A
$\frac{\mathrm{mv}^2}{\mathrm{~T}^2} \cdot \mathrm{t}$

B
$\frac{m v^2}{T^2} \cdot t^2$

C
$\frac{1}{2} \frac{\mathrm{mv}^2}{\mathrm{~T}^2} \cdot \mathrm{t}$

D
$\frac{1}{2} \frac{\mathrm{mv}^2}{\mathrm{~T}^2} \cdot \mathrm{t}^2$

The correct answer is:
$\frac{\mathrm{mv}^2}{\mathrm{~T}^2} \cdot \mathrm{t}$

Solution:
$\mathrm{P}=(\mathrm{ma}) \cdot \mathrm{v}$
$=m a^2 t$
$=m \frac{V^2}{T^2} t$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2005
A body of mass, m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3 s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5 s. The value of m in kg is

Options:
A 34
B 43
C 169
D 916

The correct answer is:
916

Solution:
T1=3=2πmK.

Then, T2=5=2πm+1K.

By dividing both the equations, we get,

35=mm+1

925=mm+1

9m+9=25 m

16m=9

m=916.

Topic: Physics, Oscillations, JEE Main
A body of mass $m$ is kept on a rough horizontal surface (coefficient of friction $=\mu$ ). Horizontal force is applied on the body, but it does not move. The resultant of normal reaction and the frictional force acting on the object is given $\mathrm{F}$, where $\mathrm{F}$ is

Options:
A $|\mathrm{F}|=m g+\mu m g$
B $|\mathrm{F}|=\mu m g$
C $|\mathrm{F}| \leq m g \sqrt{1+\mu^2}$
D $|\mathrm{F}|=m g$

The correct answer is:
$|\mathrm{F}| \leq m g \sqrt{1+\mu^2}$

Solution:
$$
\text { The situation can be drawn as }
$$
The frictional force, $\mathrm{f}=\mu \mathrm{N}=\mu \mathrm{mg} \quad[\because \mathrm{N}=\mathrm{mg}]$ From Free body diagram (FBD), the resultant force is
$$
\begin{aligned}
|F| & =\sqrt{\mathrm{N}^2+\mathrm{f}^2} \\
& =\sqrt{(\mathrm{mg})^2+(\mu \mathrm{mg})^2}=\mathrm{mg} \sqrt{1+\mu^2}
\end{aligned}
$$
This is the minimum force required to move the object. But as the body is not moving
$$
\therefore \quad|\mathrm{F}| \leq \mathrm{mg} \sqrt{1-\mu^2}
$$

Topic: Physics, Laws of Motion, JEE Main
A body of mass m is launched up on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of friction between the body and plane is x5 if the time of ascent is half of the time of descent. The value of x is

Options:

The correct answer is:
3

Solution:
ta=12td

2 saa=122 sad …i

aa=gsinθ+μgcosθ

=g2+32μg

ad=gsinθ-μgcosθ

=g2-32μg

Using the above values of aa and ad and putting in
equation i, we will get, μ=35

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (20 Jul Shift 2)
A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius R2 . And the other mass, in a circular orbit of radius 3R2. The difference between the final and the initial total energies is

Options:
A +Gm6R
B -Gm2R
C -Gm6R
D Gm2R

The correct answer is:
-Gm6R

Solution:
Ei=-GMm2R,
Ef=-GMm/22R2-GMm/223R2=-GMm2R-GMm6R=-4GMm6R=-2GMm3R
Ef-Ei= GMm2R-23+12= -GMm6R.

Topic: Physics, Gravitation, JEE Main, JEE Main 2018 (15 Apr)
A body of mass $m$ is moving in a circular orbit of radius $R$ about a planet of mass $M$. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius $\frac{R}{2}$, and the other mass, in a circular orbit of radius $\frac{3 R}{2}$. The difference between the final and initial total energies is:

Options:
A
$-\frac{G M m}{2 R}$

B
$+\frac{G M m}{6 R}$

C
$-\frac{G M m}{6 R}$

D
$\frac{G M m}{2 R}$

The correct answer is:
$-\frac{G M m}{6 R}$

Solution:
Initial gravitational potential energy,
$$
E_i=-\frac{G M m}{2 R}
$$
Final gravitational potential energy,
$$
\begin{aligned}
E_f &=-\frac{G M m / 2}{2\left(\frac{R}{2}\right)}-\frac{G M m / 2}{2\left(\frac{3 R}{2}\right)} \\
&=-\frac{G M m}{2 R}-\frac{G M m}{6 R} \\
&=-\frac{4 G M m}{6 R}=-\frac{2 G M m}{3 R}
\end{aligned}
$$
$\therefore$ Difference between initial and final energy,
$$
E_f-E_i=\frac{G M m}{R}\left(-\frac{2}{3}+\frac{1}{2}\right)=-\frac{G M m}{6 R}
$$

Topic: Physics, Gravitation, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A body of mass ' $\mathrm{m}$ ' is moving with speed ' $\mathrm{V}$ ' along a circular path of radius ' $r$ '. Now the speed is reduced to $\frac{\mathrm{V}}{2}$ and radius is increased to ' $3 r$ '. For this change, initial centripetal force needs to be

Options:
A increased by $\frac{7}{12}$ times.
B increased by $\frac{10}{12}$ times.
C increased by $\frac{11}{12}$ times.
D increased by $\frac{1}{12}$ times.

The correct answer is:
increased by $\frac{11}{12}$ times.

Solution:
$\begin{aligned} & \text { Centripetal force, } \mathrm{F}=\frac{\mathrm{mv}^2}{\mathrm{r}} \\ & \mathrm{F}_1=\frac{\mathrm{mv}_1^2}{\mathrm{r}_1} \text { and } \mathrm{F}_2=\frac{\mathrm{mv}_2^2}{\mathrm{r}_2} \\ & \frac{\mathrm{F}_2}{\mathrm{~F}_1}=\frac{\mathrm{v}_2^2}{\mathrm{v}_1^2} \cdot \frac{\mathrm{r}_1}{\mathrm{r}_2} \\ & \mathrm{v}_2=\frac{\mathrm{v}_1}{2} \text { and } \mathrm{r}_2=3 \mathrm{r}_1 \\ & \frac{\mathrm{F}_2}{\mathrm{~F}_1}=\left(\frac{1}{2}\right)^2 \cdot \frac{1}{3}=\frac{1}{12} \\ & \mathrm{v}_2=\frac{\mathrm{v}_1}{2} \text { and } \mathrm{r}_2=3 \mathrm{r}_1 \\ & \frac{\mathrm{F}_2}{\mathrm{~F}_1}=\left(\frac{1}{2}\right)^2 \cdot \frac{1}{3}=\frac{1}{12} \\ & \mathrm{~F}_2=\frac{\mathrm{F}_1}{12}\end{aligned}$
Change in force $=F_2-F_1=\frac{F_1}{12}-F_1=-\frac{11}{12} F_1$ Hence force is decreased by $\frac{11}{12}$ times.

Topic: Physics, Motion In Two Dimensions, JEE Main
A body of mass m is projected with a speed u making an angle of 45o with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as 2mu3Xg. The value of X is_______.

Options:

The correct answer is:
8

Solution:
Consider projection as shown in above diagram.

At the topmost point, velocity of the body will be ucos45°.

Therefore, we can write the required angular momentum as

L=muH×H=mucosθu2sin2θ2g

=mu3×142g⇒X=8

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2024 (31 Jan Shift 2)
A body of mass m is projected with velocity λve in vertically upward direction from the surface of the earth into space. It is given that ve is escape velocity and λ<1. If air resistance is considered to the negligible, then the maximum height from the centre of earth, to which the body can go, will be (R : radius of earth)

Options:
A R1+λ2
B R1-λ2
C R1-λ
D λ2R1-λ2

The correct answer is:
R1-λ2

Solution:
Escape velocity is given by, ve=2GMeRe

If the body goes to the height h from the surface of the earth, using mechanical energy conservation,

⇒12mλ2GMeRe2-GMemRe=-GMemRe+h

⇒λ2GMemRe-GMemRe=-GMemRe+h

⇒Re+h=Re1-λ2

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (27 Jul Shift 2)
A body of mass $m$ is thrown vertically upward with speed $\sqrt{3} v_e$, where $v_e$ is the escape velocity of a body from earth surface. The final velocity of the body is

Options:
A 0
B $2 v_e$
C $\sqrt{3} v_e$
D $\sqrt{2} \mathrm{v}_{\mathrm{e}}$

The correct answer is:
$\sqrt{2} \mathrm{v}_{\mathrm{e}}$

Solution:
$-\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{~m}\left(\sqrt{3} \mathrm{v}_{\mathrm{e}}\right)^2=0+\frac{1}{2} \mathrm{mv}^2, \quad \mathrm{v}=\sqrt{2} \mathrm{v}_{\mathrm{e}}$

Topic: Physics, Gravitation, JEE Main
A body of mass ' $m$ ' is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is $1 \mathrm{~cm}$. If the angular velocity is doubled, the elongation in the spring is $5 \mathrm{~cm}$. The original length of the spring is :

Options:
A
$15 \mathrm{~cm}$

B
$12 \mathrm{~cm}$

C
$16 \mathrm{~cm}$

D
$10 \mathrm{~cm}$

The correct answer is:
$15 \mathrm{~cm}$

Solution:
No Solution Available

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2013 (23 Apr Online)
A body of mass $m$ moving along a straight line covers half the distance with a speed of $2 \mathrm{~ms}^{-1}$. The remaining half of the distance is covered in two equal time intervals with a speed of $3 \mathrm{~ms}^{-1}$ and $5 \mathrm{~ms}^{-1}$ respectively. The average speed of the particle for the entire journey is

Options:
A $\frac{3}{8} \mathrm{~ms}^{-1}$
B $\frac{8}{3} \mathrm{~ms}^{-1}$
C $\frac{4}{3} \mathrm{~ms}^{-1}$
D $\frac{16}{3} \mathrm{~ms}^{-1}$

The correct answer is:
$\frac{8}{3} \mathrm{~ms}^{-1}$

Solution:
Let the total distance travelled by the body is $2 S$. If $\mathrm{t}_{1}$ is the time taken by the body to travel first half of the distance, then
$\mathrm{t}_{1}=\frac{\mathrm{S}}{2}$
Let $\mathrm{t}_{2}$ be the time taken by the body for each time interval for the remaining half journey.
$\therefore \quad \mathrm{S}=3 \mathrm{t}_{2}+5 \mathrm{t}_{2}=8 \mathrm{t}_{2}$
So, average speed $=\frac{\text { Total distance travelled }}{\text { Total time taken }}$
$=\frac{2 S}{t_{1}+2 t_{2}}$
$=\frac{2 S}{\frac{S}{2}+\frac{S}{4}}$
$=\frac{8}{3} \mathrm{~ms}^{-1}$

Topic: Physics, Motion In One Dimension, KCET, KCET 2009
A body of mass m moving with velocity v collides head-on elastically with another body of mass 2m which is initially at rest. The ratio of K.E. of mass m before and after the collision will be

Options:

The correct answer is:
9

Solution:
K.E. of the colliding body before the collision =12mv2
After collision its velocity becomes
v'=m1-m2m1+m2v=m3mv=v3
∴K.E. after collision 12mv'2=12mv29
The ratio of kinetic energy =K.EbeforeK.Eafter=12mv212mv29=9:1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A body of mass M moving with velocity v m s-1 suddenly breaks into two pieces. One part having mass M4 remains stationary. The velocity of the other part will be

Options:
A v
B 2v
C 3v4
D 4v3

The correct answer is:
4v3

Solution:
Mv=M4v1+34Mv2
Mv=34Mv2 (∴v1=0)
v2=4v3

Topic: Physics, Center of Mass Momentum and Collision, NEET
A body of mass $m$ rises to height $h=R / 5$ from the earth's surface, where $R$ is earth's radius. If $g$ is acceleration due to gravity at earth's surface, the increase in potential energy is

Options:
A $m g h$
B $\frac{4}{5} m g h$
C $\frac{5}{6} m g h$
D $\frac{6}{7} m g h$

The correct answer is:
$\frac{5}{6} m g h$

Solution:
$\Delta U=\frac{m g h}{1+h / R}$
Substituting $R=5 h$ we get
$\Delta U=\frac{m g h}{1+1 / 5}=\frac{5}{6} m g h$

Topic: Physics, Gravitation, JEE Main
A body of mass m starts moving from rest along x-axis so that its velocity varies as v=as where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t second after the start of the motion is

Options:
A 8 ma4t2
B 14Â ma4t2Â
C 4Â ma4t2
D 18ma4t2

The correct answer is:
18ma4t2

Solution:
v=as=dsdt
⇒dss=a.dt
⇒2s=at
⇒s=a2t24, v=dsdt=a2t2
⇒Wall=∆K=Kf-Ki
⇒Wall=12mv22-v12
⇒v1=0, v2=a2t2
⇒Wall=12ma2t22-0
⇒Wall=18ma4t2
⇒Wall=∆K

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2018 (16 Apr Online)
A body of mass $M$ thrown horizontally with velocity $v$ from the top of the tower of height $H$ touches the ground at a distance of $100 \mathrm{~m}$ from the foot of the tower. A body of mass $2 \mathrm{M}$ thrown at a velocity $\frac{v}{2}$ from the top of the tower of height $4 \mathrm{H}$ will touch the ground at a distance of _____$\mathrm{m}$.

Options:

The correct answer is:
100

Solution:
$\begin{aligned} & 100=v \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}} ; \quad \mathrm{x}=\frac{v}{2} \sqrt{\frac{2(4 \mathrm{H})}{\mathrm{g}}}=v \sqrt{\frac{2 \mathrm{H}}{\mathrm{g}}} \\ & \Rightarrow \mathrm{x}=100\end{aligned}$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2024 (08 Apr Shift 2)