datavorous/entrance-exam-dataset huggingface dataset:
A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of $3 \times 10^6 \mathrm{~ms}^{-1}$. The velocity of the particle is
(Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ )

Options:
A $2.7 \times 10^{-18} \mathrm{~ms}^{-1}$
B $9 \times 10^{-2} \mathrm{~ms}^{-1}$
C $3 \times 10^{-31} \mathrm{~ms}^{-1}$
D $2.7 \times 10^{-21} \mathrm{~ms}^{-1}$

The correct answer is:
$2.7 \times 10^{-18} \mathrm{~ms}^{-1}$

Solution:
Wavelength of a particle is given by
$\lambda=\frac{h}{p}$
where $h$ is Planck's constant.
and wavelength of an electron is given by
$\begin{aligned}
\lambda_e & =\frac{h}{p_e} \\
\lambda & =\lambda_e \\
p & =p_e \\
m v & =m_e y_e \\
y & =\frac{m_e v_e}{m}
\end{aligned}$
So, Putting the undergiven data, we get
$\begin{aligned}
& m_e=9.1 \times 10^{-31} \mathrm{~kg}, v_e=3 \times 10^6 \mathrm{~m} / \mathrm{s} \\
& m=1 \mathrm{mg}= 1 \times 10^{-6} \mathrm{~kg} \\
& y=\frac{9.1 \times 10^{-31} \times 3 \times 10^6}{1 \times 10^{-6}} \\
&=2.7 \times 10^{-18} \mathrm{~ms}^{-1}
\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main
A particle of mass $1 \mathrm{mg}$ has the same wavelength as an electron moving with a velocity of $3 \times 10^6 \mathrm{~ms}^{-1}$. The velocity of the particle is (mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ )

Options:
A $2.7 \times 10^{-21} \mathrm{~ms}^{-1}$
B $2.7 \times 10^{-18} \mathrm{~ms}^{-1}$
C $9 \times 10^{-2} \mathrm{~ms}^{-1}$
D $3 \times 10^{-31} \mathrm{~ms}^{-1}$

The correct answer is:
$2.7 \times 10^{-18} \mathrm{~ms}^{-1}$

Solution:
Same momentum of both particles
$$
\begin{aligned}
& 1 \times 10^{-3} \times v=9 \times 10^{-31} \times 3 \times 10^6 \\
& v=2.7 \times 10^{-18} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2008 (Mains)
A particle of mass $2 \times 10^{-27} \mathrm{~kg}$ has de-Broglie wavelength of $3.3 \times 10^{-10} \mathrm{~m}$. The kinetic energy of this particle is
(Plank's Constant $\left.\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$

Options:
A $5 \times 10^{-20} \mathrm{~J}$
B $8 \times 10^{-20} \mathrm{~J}$
C $1 \times 10^{-21} \mathrm{~J}$
D $6 \times 10^{-22} \mathrm{~J}$

The correct answer is:
$1 \times 10^{-21} \mathrm{~J}$

Solution:
As $\lambda=\frac{\mathrm{h}}{\mathrm{P}} \Rightarrow \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{~K} . \mathrm{E})}} \Rightarrow \mathrm{K} \cdot \mathrm{E}=\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda^2}$
So, K.E $=\frac{\left(6.6 \times 10^{-34}\right)^2}{2 \times 2 \times 10^{-27} \times\left(3.3 \times 10^{-10}\right)^2}$
$=1 \times 10^{-21} \mathrm{~J}$

Topic: Physics, Dual Nature of Matter, JEE Main
A particle of mass 2/3 kg with velocity v=-15 m/s at t=-2 s is acted upon by a force F=k-βt2. Here, k=8 N and β=2 N/s2. The motion is one-dimensional. Then, the speed at which the particle acceleration is zero again, is

Options:
A 1 m/s
B 16 m/s
C 17 m/s
D 32 m/s

The correct answer is:
17 m/s

Solution:
Force on the object is F=k-βt2

Acceleration of the particle,

a=Fm=k-βt2m

Acceleration is zero when k=βt2 or t2=kβ or t2=82⇒t=2 s

Now, a=dvdt=k-βt2m⇒dv=k-βt2m·dt

Integrating between given limits, we have

⇒∫v=-15vt=2 sdv=∫t=-2 st=2 sk-βt2mdt

=32∫t=-2t=28-2t2dt

vt=2 s--15=328t-2t33-22

⇒vat t=2s=-15+32×82--2-238--23

⇒vat t=2s=-15+3232-323

=-15+32643

=-15+32=17 ms-1

Topic: Physics, Motion In One Dimension, JEE Main
A particle of mass 2Ã10-5 kg moves horizontally between the plates of a parallel plate capacitor which produce an electric field of 200 N C-1 in the vertically upward direction. A magnetic induction of 2.0 T is applied at right angles to the electric field in a direction normal to both E â and v â . If g is 9.8 m s-2 and the charge on the particle is 10-6 C, then the velocity of the charged particle so that it continues to move horizontally is

Options:
A 2 m s-1
B 20 m s-1
C 0.2 m s-1
D 100 m s-1

The correct answer is:
2 m s-1

Solution:
Net force on the particle should be zero.

qE = 1 0 - 6 × 2 0 0 = 2 × 1 0 - 4 N
mg = 2 × 1 0 - 5 × 9.8 = 1.96 × 1 0 - 4 N
Since qE>mg, so magnetic force qvB should act downwards to balance the forces.
qE = mg + qvB ⇒ 2 × 1 0 - 4 = 1.96 × 1 0 - 4 + 1 0 - 6 v × 2
⇒ v = 2 m/s

Topic: Physics, Magnetic Effects of Current, JEE Main
A particle of mass $2 \mathrm{~kg}$ is moving such that at time $t$, its position, in meter, is given by $\vec{r}(t)=5 \hat{i}-2 t^2 \hat{j}$. The angular momentum of the particle at $t=2 s$ about the origin in $\mathrm{kg} \mathrm{m}^{-2} \mathrm{~s}^{-1}$ is :

Options:
A
$-80 \hat{k}$

B
$(10 \hat{i}-16 \hat{j})$

C
$-40 \hat{k}$

D
$40 \hat{k}$

The correct answer is:
$-80 \hat{k}$

Solution:
Angular momentum $\mathrm{L}=\mathrm{m}(\mathrm{v} \times \mathrm{r})$
$$
\begin{aligned}
& =2 \mathrm{~kg}\left(\frac{\mathrm{dr}}{\mathrm{dt}} \times \mathrm{r}\right)=2 \mathrm{~kg}\left(4 \mathrm{t} \times \times 5 \mathrm{i}-2 \mathrm{t}^2 \hat{\mathrm{j}}\right) \\
& =2 \mathrm{~kg}(-20 \mathrm{t} \hat{\mathrm{k}})=2 \mathrm{~kg} \times-20 \times 2 \mathrm{~m}^{-2} \mathrm{~s}^{-1} \hat{\mathrm{k}} \\
& =-80 \hat{\mathrm{k}}
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2013 (23 Apr Online)
A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is 12 rad s-1 , the magnitude of its angular momentum about a point on the ground right under the center of the circle is:

Options:
A 14.4 kg m2 s-1
B 11.52 kg m2 s-1
C 20.16 kg m2 s-1
D 8.64 kg m2 s-1

The correct answer is:
14.4 kg m2 s-1

Solution:
p→=mv j^=2×v j^=2×0.6×12=7.2×2=14.4j^ v=rω
r→=(0.8 k^+0.6 i^)⇒L→=r→×p→
=0.6 i^+0.8 k^×14.4 j^
=0.6×14.4 k^-0.8×14.4 i^
=14.40.6 k^-0.8 i^
⇒|L→|=14.4 0.62+0.82
=14.40.36+0.64
=14.4×1
=14.4 kg m2 s-1

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2015 (11 Apr Online)
A particle of mass $2 \mathrm{~kg}$ is on a smooth horizontal table and moves in a circular path of radius $0.6 \mathrm{~m}$. The height of the table from the ground is 0.8 $\mathrm{m}$. If the angular speed of the particle is $12 \mathrm{rad} \mathrm{s}^{-1}$ the magnitude of its angular momentum about a point on the ground right under the centre of the circle is

Options:
A $14.4 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$
B $8.64 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$
C $20.16 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$
D $11.52 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$

The correct answer is:
$14.4 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$

Solution:
Angular momentum, $L_0=m v r \sin 90^{\circ}$
$=2 \times 0.6 \times 12 \times 1 \times 1 \mid\left[\right.$ As $\left.V=r \omega, \operatorname{Sin} 90^{\circ}=1\right]$
So, $L_0=14.4 \mathrm{kgm}^2 / \mathrm{s}$

Topic: Physics, Rotational Motion, JEE Main
A particle of mass 2 kg moving with a speed of 6 m/s collides elastically with another particle of mass 4 kg traveling in same direction with a speed of 2 m/s . The maximum possible deflection of the 2 kg particle is

Options:
A 37 0
B 450
C 530
D 600

The correct answer is:
530

Solution:
Let situations after collision for balls are
According to question 2v1+4v3=20
2v2=4v4
12Ã2v12+v22+12Ã4v32+v42=12Ã2Ã62+12Ã4Ã22=44
i.e. v12+v22+2v32+v222=44
i.e. v12+32v22+10-v122 2=44
i.e. 2v12+3v22+100-20v1+v12=88
i.e. 3v12+3v22-20v1+12=0
i.e. 12v12-20v1+3v22v12+3=0
Quadratic in 1v1 which is real
â´Dâ¥0 â´-202-4Ã121+v22v12Ã3â¥0
â´ 1+v22v12â¤400144
v22v12â¤256144
-1612â¤v2v1â¤1612âk=tanâ¡Î¸â¤43âÎ¸â¤53o

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A particle of mass 2m is projected at an angle of 45Â° with horizontal with a velocity of 202 m s-1. After 1 s explosion takes place and the particle is broken into two equal pieces. As a result of the explosion, one part comes to rest. The maximum height attained by the other part is (Take g=10 m s-2)

Options:
A 35 m

B 40 m
C 15 m
D 20 m

The correct answer is:
35 m

Solution:
Applying conservation of linear momentum at the time of the collision, or at t=1 s,
m v → + m 0 = 2 m 2 0 i ^ + 1 0 j ^
∴ v → = 4 0 i ^ + 2 0 j ^
At 1 sec, masses will be at height :
h 1 = u y t + 1 2 v y t 2 = 2 0 1 + 1 2 - 1 0 1 2 = 15 m
After explosion other mass will further rise to a height :
h 2 = u y 2 2 g = 2 0 2 2 × 1 0 = 2 0 m : u y = 20 m/s just after the collision.
∴ Total height h=h1+h2= 35 m

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A particle of mass $0.3 \mathrm{~kg}$ is subjected to a force $F=-k x$ with $k=15 \mathrm{~N} / \mathrm{m}$. What will be its initial acceleration if it is released from a point $20 \mathrm{~cm}$ away from the origin?

Options:
A
$3 \mathrm{~m} / \mathrm{s}^2$

B
$15 \mathrm{~m} / \mathrm{s}^2$

C
$5 \mathrm{~m} / \mathrm{s}^2$

D
$10 \mathrm{~m} / \mathrm{s}^2$

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

Solution:
$a=\frac{k x}{m}=10 \mathrm{~m} / \mathrm{s}^2$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2005
A particle of mass 4M at rest disintegrates into two particles of mass M and 3M, respectively, having non zero velocities. The ratio of de-Broglie wavelength of particle of mass M to that of mass 3M will be:

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

The correct answer is:
1:1

Solution:
λ=hp

Both the particles will move with momentum same in magnitude and opposite in direction.

So de Broglie wavelength of both will be same, i.e., ratio 1:1.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (25 Jul Shift 1)
A particle of mass $4 m$ explodes into three pieces of masses $m, m$ and $2 \mathrm{~m}$. The equal masses move along $X$-axis and $Y$-axis with velocities $4 \mathrm{~ms}^{-1}$ and $6 \mathrm{~ms}^{-1}$ respectively. The magnitude of the velocity of the heavier mass is

Options:
A $\sqrt{17} \mathrm{~ms}^{-1}$
B $2 \sqrt{13} \mathrm{~ms}^{-1}$
C $\sqrt{13} \mathrm{~ms}^{-1}$
D $\frac{\sqrt{13}}{2} \mathrm{~ms}^{-1}$

The correct answer is:
$\sqrt{17} \mathrm{~ms}^{-1}$

Solution:
Let third mass particle $(2 \mathrm{~m}$ ) moves making angle $\theta$ with $X$-axis.
The horizontal component of velocity of $2 \mathrm{~m}$ mass particle $=u \cos \theta$ and vertical component $=u \sin \theta$
From conservation of linear momentum in $X$-direction
$\begin{array}{rlrl}\text { or } & m_1 u_1+m_2 u_2 =m_1 v_1+m_2 v_2 \\ \text { or } & 0 =m \times 4+2 m(u \cos \theta) \\ \text { or } & -4 =2 u \cos \theta\end{array}$
Again, applying law of conservation of linear momentum in $Y$-direction.
$\begin{aligned}
0 & =m \times 6+2 m(u \sin \theta) \\
\Rightarrow \quad-\frac{6}{2} & =u \sin \theta
\end{aligned}$
or $\quad-3=u \sin \theta \quad \ldots$ (ii)
Squaring Eqs. (i) and(ii) and adding, we get
$\begin{aligned}
\text { (4) } +(9) & =u^2 \cos ^2 \theta+u^2 \sin ^2 \theta \\
& =u^2\left(\cos ^2 \theta+\sin ^2 \theta\right)
\end{aligned}$
$\begin{aligned} \text { or } 13=u^2 \\ \text { or } u=\sqrt{13} \mathrm{~ms}^{-1}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A particle of mass $5 \mathrm{~m}$ at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular direction with each speed v. The energy released during the process is

Options:
A $\frac{3}{5} m v^2$
B $\frac{5}{3} m v^2$
C $\frac{3}{2} m v^2$
D $\frac{4}{3} m v^2$

The correct answer is:
$\frac{4}{3} m v^2$

Solution:
The particle of mass $5 \mathrm{~m}$ breaks in three fragments of mass $m, m$ and $3 \mathrm{~m}$ respectively. Two fragments of mass meach, move in perpendicular direction with velocity $v$ and the left fragment will move in a direction with velocity $v^{\prime}$ such that the total momentum of the system must remain conserved.
By law of conservation of momentum,
$$
\begin{aligned}
5 \mathrm{~m} \times 0 & =\mathrm{mv} \hat{\mathrm{i}}+\mathrm{mv} \hat{\mathrm{j}}+3 \mathrm{mv}^{\prime} \\
\Rightarrow \quad \mathrm{v}^{\prime} & =-\frac{\mathrm{v}}{3} \hat{\mathrm{i}}-\frac{\mathrm{v}}{3} \hat{\mathrm{j}} \\
\therefore \quad\left|\mathrm{v}^{\prime}\right| & =\sqrt{\left(-\frac{\mathrm{v}}{3}\right)^2+\left(-\frac{\mathrm{v}}{3}\right)^2}=\frac{\mathrm{v} \sqrt{2}}{3}
\end{aligned}
$$
$\therefore$ Energy released
$$
\begin{aligned}
\mathrm{E} & =\frac{1}{2} m v^2+\frac{1}{2} m v^2+\frac{1}{2} \times 3 \mathrm{~m}\left(\frac{\mathrm{v} \sqrt{2}}{3}\right)^2 \\
& =m v^2+\frac{m v^2}{3}=\frac{4}{3} m v^2
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A particle of mass $5 \mathrm{~m}$ at rest suddenly breaks on its own into three fragments. Two fragments of mass m each move along mutually perpendicular direction with each speed v. The energy released during the process is

Options:
A $\frac{3}{5} m v^2$
B $\frac{5}{3} m v^2$
C $\frac{3}{2} m v^2$
D $\frac{4}{3} m v^2$

The correct answer is:
$\frac{4}{3} m v^2$

Solution:
The particle of mass $5 \mathrm{~m}$ breaks in three fragments of mass $m, m$ and $3 \mathrm{~m}$ respectively. Two fragments of mass meach, move in perpendicular direction with velocity $v$ and the left fragment will move in a direction with velocity $v^{\prime}$ such that the total momentum of the system must remain conserved.
By law of conservation of momentum,
$$
\begin{aligned}
5 \mathrm{~m} \times 0 & =\mathrm{mv} \hat{\mathrm{i}}+\mathrm{mv} \hat{\mathrm{j}}+3 \mathrm{mv}^{\prime} \\
\Rightarrow \quad \mathrm{v}^{\prime} & =-\frac{\mathrm{v}}{3} \hat{\mathrm{i}}-\frac{\mathrm{v}}{3} \hat{\mathrm{j}} \\
\therefore \quad\left|\mathrm{v}^{\prime}\right| & =\sqrt{\left(-\frac{\mathrm{v}}{3}\right)^2+\left(-\frac{\mathrm{v}}{3}\right)^2}=\frac{\mathrm{v} \sqrt{2}}{3}
\end{aligned}
$$
$\therefore$ Energy released
$$
\begin{aligned}
\mathrm{E} & =\frac{1}{2} m v^2+\frac{1}{2} m v^2+\frac{1}{2} \times 3 \mathrm{~m}\left(\frac{\mathrm{v} \sqrt{2}}{3}\right)^2 \\
& =m v^2+\frac{m v^2}{3}=\frac{4}{3} m v^2
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2019 (Odisha)
A particle of mass 9.1×10-31 kg travels in a medium with a speed of 106 m s-1 and a photon of radiation of linear momentum 10-27 kg m s-1 travels in a vacuum. The wavelength of the photon is ____ times the wavelength of the particle.

Options:

The correct answer is:
910

Solution:
For photon λ1=hP=6.6×10-3410-27
For particle λ2=hmv=6.6×10-349.1×10-31×106
∴ λ1λ2=910

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (27 Jul Shift 1)
A particle, of mass 10-3 kg and charge 1.0 C, is initially at rest. At time t=0, the particle comes under the influence of an electric field Eât=E0sinâ¡Ït i^ where E0=1.0 N C-1 and Ï=103 rad s-1. Consider the effect of only the electrical force on the particle. Then the maximum speed, in ms-1, attained by the particle at subsequent times is ____________.

Options:

The correct answer is:
2

Solution:
m=10-3 kg q=1 C t=0
Force on particle will be
F=qE=qE0sinâ¡Ït
At vmax, F=0 âqE0sinâ¡Ï t o =0
â w t o =Ï
F=qE0sinâ¡Ït
dvdt=qE0msinâ¡Ït
â«0vdv=â«0 ÏâÏ qE0msinâ¡Ïtdt
v-0=qE0mÏ-cosâ¡Ït0 ÏâÏ
v-0=qE0mÏ-cosâ¡Ï--cosâ¡0
v=1Ã1mÏÃ2=2 m/s
=2.00

Topic: Physics, Electrostatics, JEE Main
A particle of mass $10 \mathrm{~g}$ is kept on the surface of a uniform sphere of mass $100 \mathrm{~kg}$ and radius $10 \mathrm{~cm}$. Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^2 / \mathrm{kg}^2$ )

Options:
A
$13.34 \times 10^{-10} \mathrm{~J}$

B
$3.33 \times 10^{-10} \mathrm{~J}$

C
$6.67 \times 10^{-9} \mathrm{~J}$

D
$6.67 \times 10^{-10} \mathrm{~J}$

The correct answer is:
$6.67 \times 10^{-10} \mathrm{~J}$

Solution:
$\mathrm{w}=\mathrm{GMm} / \mathrm{R}=6.67 \times 10^{-10} \mathrm{~J}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2005
A particle of mass $10 \mathrm{~g}$ moves along a circle of radius $6.4 \mathrm{~cm}$ with a constant tangential acceleration. If the kinetic energy of the particle becomes $8 \times 10^{-4} \mathrm{~J}$ by the end of the second revolution after the beginning of the motion, the magnitude of the tangential acceleration is

Options:
A 0.6 m/s
B 0.4 m/s
C 0.1 m/s
D 0.3 m/s

The correct answer is:
0.1 m/s

Solution:
Consider the following diagram:
The tangential acceleration is given by,
$\begin{aligned}
& a_t=\frac{\mathrm{d} v}{\mathrm{~d} t}=\text { Constant } \\
& \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=a_t \\
& \Rightarrow \frac{\mathrm{d} s}{\mathrm{~d} t} \cdot \frac{\mathrm{d} v}{\mathrm{~d} s}=a t \\
& \Rightarrow v=\frac{\mathrm{d} v}{\mathrm{~d} t}=a_t \\
& \Rightarrow \int_0^v \frac{v^2}{2} \mathrm{~d} v=\int_0^{(4 \pi r)} a_t \mathrm{~d} s
\end{aligned}$
We know,
Now, kinetic energy is given by,
$K E=\frac{m v^2}{2}$
Now, using eq ${ }^{\mathrm{n}}(1) \&(2)$
$\begin{aligned}
& \left(\frac{2 K E}{m}\right)=(8 \pi r) a_t \\
& \Rightarrow a_t=\left(\frac{K E}{(4 \pi r) m}\right)
\end{aligned}$
Given, $K E=8 \times 10^{-4} \mathrm{~J}, r=6.4 \times 10^{-2} \mathrm{~m}$ and $m=1 \times 10^{-2} \mathrm{~kg}$.
$\Rightarrow a_t=\frac{\left(8 \times 10^{-4}\right) \mathrm{J}}{4 \pi\left(6.4 \times 10^{-2} \mathrm{~m}\right)\left(10^{-2} \mathrm{~kg}\right)}=0.1 \mathrm{~m} / \mathrm{s}^2$

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle of mass $10 \mathrm{~g}$ moves along a circle of radius $6.4 \mathrm{~cm}$ with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to $8 \times 10^{-4} \mathrm{~J}$ boy the end of the second revolution after the beginning of the motion?

Options:
A $0.2 \mathrm{~m} / \mathrm{s}^2$
B $0.1 \mathrm{~m} / \mathrm{s}^2$
C $0.15 \mathrm{~m} / \mathrm{s}^2$
D $0.18 \mathrm{~m} / \mathrm{s}^2$

The correct answer is:
$0.1 \mathrm{~m} / \mathrm{s}^2$

Solution:
kinetic energy $=8 \times 10^{-4} \mathrm{~J}$
$\begin{aligned}& \text { or, } \frac{1}{2} \mathrm{mv} 2=8 \times 10^{-4} \\& \text { or, } \frac{1}{2} \times 10 \times 10^{-3} \mathrm{v}^2=8 \times 10^{-4} \\& \text { or, } \mathrm{v}^2=16 \times 10^{-2}=\gt\mathrm{v}=0.4 \mathrm{~m} / \mathrm{s}\end{aligned}$
initial velocity of particle, $\mathbf{u}=\mathrm{om} / \mathrm{s}$
we have to find Tangential acceleration at the end of 2 nd revolution.
total distance covered, $s=2(2 \pi r)=4 \pi r$
$\begin{aligned}& \text { so, } v^2=2 \mathrm{as} \\& \mathrm{a}=\frac{\mathrm{v}^2}{2 \mathrm{~s}}=\frac{(0.4)^2}{2(4 \pi r)}
\end{aligned}$
$\begin{aligned} & =\frac{16 \times 10^{-2}}{\left(8 \times 3.14 \times 6.4 \times 10^{-} 2\right)} \\ & =0.0995 \mathrm{~m} / \mathrm{s}^2 \approx 0.1 \mathrm{~m} / \mathrm{s}^2\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
A particle of mass 10 g moves along a circle of the radius 1Ï cm with a constant tangential acceleration. What is the magnitude of this acceleration (in m s-1) if the kinetic energy of the particle becomes equal to 8Ã10-4 J by the end of the second revolution after the beginning of the motion? (Particle starts from rest)

Options:

The correct answer is:
2

Solution:
Tangential acceleration at=rα= constant =K
α=Kr
At the end of the second revolution, angular velocity is ω then
ω2-ω02=2αθ
ω2-02=2Kr4π
ω2=8πKr
K.E. of the particle is =K.E.=12mv2
K.E.=12mr2ω2
K.E.=12mr28πKr
8×10-4=12×10-4×8×1π×π×K
⇒ K=2 m s-2

Topic: Physics, Laws of Motion, JEE Main
A particle of mass 20 g is released with an initial velocity 5 m s-1 along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be: (Take g=10 m s-2)

Options:
A 3 kg m2 s-1
B 2 kg m2 s-1
C 6 kg m2 s-1
D 8 kg m2 s-1

The correct answer is:
6 kg m2 s-1

Solution:
Applying conservation of energy,
mgh=12mvB2-12mvA2
⇒vB=2gh+vA2
⇒vB=2×10×10+25
⇒vB=15 m s-1
Angular momentum about O,

LO=mvBh+a
⇒LO=20×10-3×15×20 kg m2 s-1
⇒LO=6 kg m2 s-1

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2019 (12 Jan Shift 2)
A particle of mass $20 \mathrm{~g}$ is released with an initial velocity $5 \mathrm{~ms}^{-1}$ along the curve from the point $A$, as shown in the figure. The point $A$ is at height $h$ from point $B$. The particle slides along the frictionless surface. When the particle reaches point $B$, its angular momentum about $\mathrm{O}$ will be: (Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )

Options:
A $8 \mathrm{~kg}-\mathrm{m}^2 \mathrm{~s}^{-1}$
B $3 \mathrm{~kg}-\mathrm{m}^2 \mathrm{~s}^{-1}$
C $6 \mathrm{~kg}-\mathrm{m}^2 \mathrm{~s}^{-1}$
D $2 \mathrm{~kg}-\mathrm{m}^2 \mathrm{~s}^{-1}$

The correct answer is:
$6 \mathrm{~kg}-\mathrm{m}^2 \mathrm{~s}^{-1}$

Solution:
$\begin{aligned}& m g h=\frac{1}{2} m v_B^2-\frac{1}{2} m v_A^2 \\& \Rightarrow 2 g h=v_B^2-v_A^2 \\& \Rightarrow 2 \times 10 \times 10=v_B^2-5^2 \\& \Rightarrow v_B=15 m s^{-1}\end{aligned}$
Angular momentum about O,
$L_O=m v_B r$
Here, $r=20 \mathrm{~m}$
$\begin{aligned}& \Rightarrow L_O=20 \times 10^{-3} \times 15 \times 20 \\& \Rightarrow \mathrm{L}_{\mathrm{O}}=6 \mathrm{~kg}-\mathrm{m}^2 \mathrm{~s}^{-1} \\&\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A particle of mass $0.50 \mathrm{~kg}$ executes simple harmonic motion under force $F=-50\left(\mathrm{Nm}^{-1}\right) x$. The time period of oscillation is $\frac{x}{35} \mathrm{~s}$. The value of $x$ is _______
(Given $\pi=\frac{22}{7}$ )

Options:

The correct answer is:
22

Solution:
$\begin{aligned} & \mathrm{m}=0.5 \mathrm{~kg} \\ & \mathrm{~F}=-50(\mathrm{x}) \\ & \mathrm{ma}=(-50 \mathrm{x}) \\ & 0.5 \mathrm{a}=-50 \mathrm{x} \\ & \mathrm{a}=(-100 \mathrm{x}) \\ & \mathrm{W}^2=100 \Rightarrow(\mathrm{w}=10) \\ & \mathrm{T}=\frac{2 \pi}{10}=\left(\frac{\pi}{5}\right)=\frac{22}{7 \times 15}=\left(\frac{22}{35}\right) \\ & \frac{\pi}{35}=\frac{22}{35} \Rightarrow \mathrm{x}=22\end{aligned}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2024 (09 Apr Shift 2)
A particle of mass 100 g is projected at time t=0 with a speed 20 m s–1 at an angle 45° to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time t=2 s is found to be K kg m2 s-1. The value of K is ______.

(Take g=10 m s-2)

Options:

The correct answer is:
800

Solution:
As torque due to gravitational force will be variable, therefore we can use

ΔL=∫0tτdt.

Now, torque of mg at time t =vxtmg and horizontal velocity of the projectile will remain constantvx=vcos45°=20×12=102.

L0=∫02mgvxtdt

=mgvxt22=(0.1)(10)(102)222

=202

=800 kg m2 s-1

Therefore, K=800.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (29 Jan Shift 2)
A particle of mass $100 \mathrm{~g}$ is thrown vertically upwards with a speed of $5 \mathrm{~m} / \mathrm{s}$. the work done by the force of gravity during the time the particle goes up is

Options:
A
$0.5 \mathrm{~J}$

B
$-0.5 \mathrm{~J}$

C
$-1.25 \mathrm{~J}$

D
$1.25 \mathrm{~J}$

The correct answer is:
$-1.25 \mathrm{~J}$

Solution:
$-m g h=-m g\left(\frac{v^2}{2 g}\right)=-1.25 \mathrm{~J}$

Topic: Physics, Work Power Energy, JEE Main
A particle of mass $100 \mathrm{~g}$ is thrown vertically upwards with a speed of $5 \mathrm{~m} / \mathrm{s}$. the work done by the force of gravity during the time the particle goes up is

Options:
A
$0.5 \mathrm{~J}$

B
$-0.5 \mathrm{~J}$

C
$-1.25 \mathrm{~J}$

D
$1.25 \mathrm{~J}$

The correct answer is:
$-1.25 \mathrm{~J}$

Solution:
$-m g h=-m g\left(\frac{v^2}{2 g}\right)=-1.25 \mathrm{~J}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2006
A particle of mass $200 \mathrm{gm}$ executes S.H.M. The restoring force is provided by a spring of force constant $80 \mathrm{~N} / \mathrm{m}$. The time period of oscillations is

Options:
A $0.31 \mathrm{sec}$
B $0.15 \mathrm{sec}$
C $0.05 \mathrm{sec}$
D $0.02 \mathrm{sec}$

The correct answer is:
$0.31 \mathrm{sec}$

Solution:
$T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{0.2}{80}}=0.31 \mathrm{sec}$

Topic: Physics, Oscillations, JEE Main