datavorous/entrance-exam-dataset huggingface dataset:
A particle of mass m collides head-on with another stationary particle of mass M such that the second particle starts moving and the first particle comes to rest after the collision. Which of the following conditions is valid if the coefficient of restitution is e?

Options:
A e=0
B e=mM≤1
C e=mM≥1
D e=Mm≤1

The correct answer is:
e=mM≤1

Solution:
The collision is not perfectly elastic or inelastic since when one is at rest the other is in motion with different velocities
mu=Mv
⇒ mM=vu
But e=vu≤1
e=mM≤1

Topic: Physics, Center of Mass Momentum and Collision, NEET
A particle of mass m describes a circle of radius r. The centripetal acceleration of the particle is 4r2. The momentum of the particle is

Options:
A 4mr
B 2mr
C Â 4mr
D 2mr

The correct answer is:
2mr

Solution:
Given, centripetal acceleration = 4r2

∴v2r=4r2

or v=2r
Hence, p=2mr

Topic: Physics, Center of Mass Momentum and Collision, NEET
A particle of mass m, initial speed u and angle of projection Î¸, is projected as shown in the figure. Average torque on the projectile between initial and final positions P and Q about the point of projection is

Options:
A mu2sin2Î¸2
B mu2cosθ
C mu2sinθ
D mu2cosθ2

The correct answer is:
mu2sin2θ2

Solution:
τ→αv×Δt=ΔL→ ...(1)
Here, Δt = time of flight = 2usinθg
Δ→L=L→f-L→i about the point of projection.
=musinθrange-0
=musinθu2sin2θg
∴ τ→av=ΔLΔt=mu3sinθsin2θgg2usinθ
=mu2sin2θ2

Topic: Physics, Motion In Two Dimensions, NEET
A particle of mass m is acted upon by a force F=t2-kx . Initially, the particle is at rest at the origin. Then

Options:
A its displacement will be in simple harmonic
B its velocity will be in simple harmonic
C its acceleration will be in simple harmonic
D particle will move with constant velocity

The correct answer is:
its acceleration will be in simple harmonic

Solution:
a=t2m-kmx
dadt=122t-km.dxdt
d2adt2=2m-kma
d2adt2+kma-2m=0
This is second order differential equation for acceleration, hence a will be in simple harmonic.

Topic: Physics, Oscillations, NEET
A particle of mass m is driven by a machine that delivers a constant power k watts. If the particle starts from rest the force on the particle at time t is:

Options:
A mk2 t−12
B mk t-12
C 2mk t-12
D 12mk t-12

The correct answer is:
mk2 t−12

Solution:
\(\begin{aligned}
& F v=\text { constant }=\mathrm{k} \\
& \mathrm{So}, \\
& \frac{\mathrm{mdv}}{\mathrm{dt}} \mathrm{v}=\mathrm{k} \\
\Rightarrow & \int \mathrm{vdv}=\int \frac{\mathrm{k}}{\mathrm{m}} \mathrm{dt} \\
\Rightarrow & \frac{\mathrm{v}^2}{2}=\frac{\mathrm{k}}{\mathrm{m}} \mathrm{t} \\
\Rightarrow & \mathrm{v}=\sqrt{\frac{2 \mathrm{k}}{\mathrm{m}} \mathrm{t}}
\end{aligned}\)
Now,
\(\begin{aligned}
& F=\frac{\mathrm{mdv}}{\mathrm{dt}} \\
& =\mathrm{m} \sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}} \frac{1}{2} \mathrm{t}^{-\frac{1}{2}} \\
& =\sqrt{\frac{\mathrm{mk}}{2}} \mathrm{t}^{-\frac{1}{2}}
\end{aligned}\)

Topic: Physics, Work Power Energy, NEET, NEET 2015 (Phase 1)
A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is Ux=kx3 , where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

Options:
A Proportional to 1a
B Independent to a
C Proportional to a
D Proportional to a32

The correct answer is:
Proportional to 1a

Solution:
U=kx3 ⇒ F=-dUdx=-3kx2
Also, for SHM x=asinωt and d2xdt2+ω2x=0
⇒ Acceleration a=d2xdt2=-ω2x ⇒F=ma
=md2xdt2=-mω2x
From equation (i) and (ii) we get ω=3kxm
⇒T=2πω=2πm3kx=2πm3kasinωt ⇒T∝1a .

Topic: Physics, Oscillations, NEET
A particle of mass m is executing oscillations about the origin on the x-axis with amplitude A. Its potential energy is given as Ux=Î± x4, where Î± is a positive constant. The x-coordinate of mass where potential energy is one-third the kinetic energy of the particle is

Options:
A ±A2
B ±A2
C Â±A3
D Â±A3

The correct answer is:
±A2

Solution:
The energy of oscillation is E=αA4
∴ the kinetic energy of mass at x = x is
K=E-U=αA4-x4
Given, K = 3U
∴ αA4-x4=3 α x4
Or x=±A2

Topic: Physics, Oscillations, NEET
A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length l. The system is rotated about the other end of the spring with an angular velocity ω, in gravity-free space. Then increase in the length of the spring will be:

Options:
A m ω2lk
B m ω2lk-m ω2
C m ω2lk+m ω2
D None of these

The correct answer is:
m ω2lk-m ω2

Solution:
Let x be increase in length of the spring. The particle would move in a circular path of radius (l+x). Centripetal force = force due to the spring
ml+xω2=kx
∴ x=mω2lk-mω2

Topic: Physics, Laws of Motion, NEET
A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length l. The system is rotated about the other end of the spring with an angular velocity ω, in gravity free space. Then increase in length of the spring will be:

Options:
A m ω2lk
B m ω2lk-m ω2
C m ω2lk+m ω2
D None of these

The correct answer is:
m ω2lk-m ω2

Solution:
Let x be increase in length of the spring. The particle would move in a circular path of radius (l+x). Centripetal force = force due to the spring
ml+xω2=kx
∴ x=mω2lk-mω2

Topic: Physics, Laws of Motion, NEET
A particle of mass ' $m$ ' is kept at rest at a height $3 R$ from the surface of earth, where ' $R$ ' is radius of earth and ' $M$ ' is mass of earth. The minimum speed with which is should be projected, so that it does not return back, is $(g$ is acceleration due to gravity on the surface of earth)

Options:
A $\left(\frac{G M}{2 R}\right)^{1 / 2}$
B $\left(\frac{g R}{4}\right)^{1 / 2}$
C $\left(\frac{2 g}{R}\right)^{1 / 2}$
D $\left(\frac{G M}{R}\right)^{\frac{1}{2}}$

The correct answer is:
$\left(\frac{G M}{2 R}\right)^{1 / 2}$

Solution:
Escape velocity $V_e=\sqrt{\frac{2 \mathrm{GM}}{(\mathrm{R}+h)}}$ from the question $h=3 R$
$$
\begin{aligned}
& \therefore V_e^{\prime}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}+3 \mathrm{R}}} \\
& =\sqrt{\frac{2 G M}{4 R}}=\sqrt{\frac{G M}{2 R}}=\sqrt{\frac{g R}{2}} \\
& \text { where } g=\frac{G M}{R^2}
\end{aligned}
$$

Topic: Physics, Gravitation, NEET, NEET 2013 (Karnataka)
A particle of mass m is located in a field such that its potential energy is given by Ux=U01-cos⁡ax where U0 and a are positive constants. The period of small oscillations is

Options:
A 2πU0ma2
B 2πmU0a2
C 2πa0mU0
D 2πmU0a2

The correct answer is:
2πmU0a2

Solution:
Given potential energy is
Ux=U01-cos⁡ax
Therefore force on the particle is given by
F=-dUdx=-U0asin(ax)
Using small angle approximation, sin(ax)≈ax
Therefore F=-U0a2x
Time period of SHM is,
T=2πmk where k is force constant.
Therefore time period for given force is,
T=2πmU0a2

Topic: Physics, Oscillations, NEET
A particle of mass m is moving along the x-axis under the potential V(x)= k x 2 2 + Î» x â² where k and x â² are positive constants of appropriate dimensions. The particle is slightly displaced from its equilibrium position. The particle oscillates with the angular frequency Ï given by

Options:
A 3km
B 3mk
C km
D 3mk

The correct answer is:
km

Solution:
Given, V(x)= k x 2 2 + λ x ′
F=-∂V(x)∂x= -kx
a=-kmx
ω2=km
ω=km

Topic: Physics, Oscillations, NEET
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac=k2 rt2 , where k is a constant. The power delivered to the particle by the force acting on it is

Options:
A 2π mk2r2
B mk2 r2t
C mk4 r2t53
D Zero

The correct answer is:
mk2 r2t

Solution:
ac=k2 rt2
or v2r= k2 rt2
or v=krt
Therefore, tangential acceleration, at =dvdt= kr
or Tangential force,
Ft=mat=mkr
Only tangential force does work.
Power=Ft v=mkrkrt
or Power=mk2 r2 t

Topic: Physics, Work Power Energy, NEET
A particle of mass m is moving with speed 2v and collides with a mass 2m moving with speed v in the same direction. After the collision, the first mass is stopped completely while the second one splits into two particles each of mass m, which move at an angle 45o with respect to the original direction. The speed of each of the moving particle will be

Options:
A 2v
B v2
C 22v
D v22

The correct answer is:
22v

Solution:
As 2m splits into equal parts, v1=v2 and θ=450
From conservation of linear momentum,
m2v+2mv=2mv1cos⁡45o
⇒v1=v2=22v

Topic: Physics, Center of Mass Momentum and Collision, NEET
A particle of mass m is projected at an angle α to the horizontal with an initial velocity u. The work done by gravity during the time it reaches its highest point is -

Options:
A mu 2 sin 2 α
B mu2cos2Î±2
C mu2sin2Î±2
D âmu2sin2Î±2

The correct answer is:
–mu2sin2α2

Solution:
W g =âÎU
Work done by gravity = decrease in potential energy
W=ââmgH=âââmg u 2 sin 2 Î± 2g
= âmu2sin2Î±2

Topic: Physics, Work Power Energy, NEET
A particle of mass m is projected from the ground with initial linear momentum p (magnitude) to have maximum possible range. Its minimum kinetic energy will be

Options:
A p&ApplyFunction; 2 2 m&ApplyFunction;
B p⁡ 2 4 m⁡
C p&ApplyFunction; 2 m&ApplyFunction;
D p⁡ 2 3 m⁡

The correct answer is:
p⁡ 2 4 m⁡

Solution:
At the time of projection θ = 4 5 o and momentum p⁡ = m⁡ u⁡
⇒ u = p m
Minimum KE will be when projectile is at the highest point.
Velocity will be is u cos θ
∴ K⁡ min = 1 2 m⁡ u cos θ 2
= m⁡ 2 p⁡ 2 m⁡ 2 1 2 = p⁡ 2 4m⁡

Topic: Physics, Motion In Two Dimensions, NEET
A particle of mass m is projected upwards with velocity v=ve2, where ve is the escape velocity then at the maximum height the potential energy of the particle is : (R is radius of earth and M is mass of earth)

Options:
A -GMm2R
B -GMm4R
C -3GMm4R
D -2GMm3R

The correct answer is:
-3GMm4R

Solution:
When only conservative forces are acting, mechanical energy is conserved and at maximum height speed is zero.
-GMmR+12mve22=U+0
-GMmR+12mGM2R=U
U=-GMmR+GMm4R=-3GMm4R

Topic: Physics, Gravitation, NEET
A particle of mass m is projected with a velocity v=kVe (k<1) from the surface of the earth (Ve= escape velocity). The maximum height above the surface reached by the particle is___

Options:
A Rk1-k2
B Rk1+k2
C Rk21-k2
D R2k1+k

The correct answer is:
Rk21-k2

Solution:
v=kVe(k<1)

12mv2-GMmR=0-GMmR+h

12mk2×2GMR-GMmR=-GMmR+h ∵Ve=2GMR

GMmk2R=-GMm1R+h-1R

k2R=hR(R+h)⇒1+Rh=1k2

⇒Rh=1k2-1

⇒h=RK21-k2

Topic: Physics, Gravitation, NEET, NEET 2021
A particle of mass $m$ is projected with velocity $y$ making an angle of $45^{\circ}$ with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be

Options:
A $2 m v$
B $m v / \sqrt{2}$
C $m v \sqrt{2}$
D zero

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

Solution:
Key Idea: Required momentum is the difference of final and initial momentum. The situation is shown in figure.

Change in momentum,
$\begin{aligned}
\Delta \overrightarrow{\mathbf{p}} & =\overrightarrow{\mathbf{p}}_f-\overrightarrow{\mathbf{p}}_i \\
& =m\left(\overrightarrow{\mathbf{v}}_f-\overrightarrow{\mathbf{v}}_i\right) \\
& =m\left[\left(v \cos 45^{\circ} \hat{\mathbf{i}}-v \sin 45^{\circ} \hat{\mathbf{j}}\right)\right. \\
& \left. -\left(v \cos 45^{\circ} \hat{\mathbf{i}}+v \sin 45^{\circ} \hat{\mathbf{j}}\right)\right] \\
& =m\left[\left(\frac{v}{\sqrt{2}} \hat{\mathbf{i}}-\frac{v}{\sqrt{2}} \hat{\mathbf{j}}\right)-\left(\frac{v}{\sqrt{2}} \hat{\mathbf{i}}+\frac{v}{\sqrt{2}} \hat{\mathbf{j}}\right)\right] \\
& =-\sqrt{2} m v \hat{\mathbf{j}} \\
\therefore & {[\Delta \overrightarrow{\mathbf{p}}]=\sqrt{2} m v }
\end{aligned}$
Alternative :

The horizontal momentum does not change. The change in vertical momentum is $m v \sin \theta-(-m v \sin \theta)=2 m v \frac{1}{\sqrt{2}}=\sqrt{2} m v$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2008 (Screening)
A particle of mass $m$ is released from rest and follows a parabolic path as shown. Assuming that the displacement of the mass from the origin is small, which graph correctly depicts the position of the particle as a function of time?

Options:
A

B

C

D

The correct answer is:

Solution:
Potential $V(x)$ versus $x$ is parabolic, SHM starts from extreme position and $x$ versus $t$ should be cosin curve so option (4) is correct.

Topic: Physics, Oscillations, NEET, NEET 2011 (Screening)
A particle of mass $M$ is situated at the centre of a spherical shell of same mass and radius a. The gravitational potential at a point situated at a/ 2 distance from the centre, will be

Options:
A $-\frac{3 G M}{a}$
B $-\frac{2 \mathrm{GM}}{\mathrm{a}}$
C $-\frac{\mathrm{GM}}{\mathrm{a}}$
D $-\frac{4 \mathrm{GM}}{\mathrm{a}}$

The correct answer is:
$-\frac{3 G M}{a}$

Solution:
Gravitational potential at point a/2 distance from centre
$$
\begin{aligned}
& =-\frac{\mathrm{GM}}{\mathrm{a}}-\frac{\mathrm{GM}}{\mathrm{a} / 2} \\
& =-\frac{3 \mathrm{GM}}{\mathrm{a}}
\end{aligned}
$$

Topic: Physics, Gravitation, NEET, NEET 2010 (Screening)
A particle of mass m moves along line PC with velocity Î½ as shown. What is the angular momentum of the particle about P?

Options:
A mvL
B mvl
C mvr
D zero

The correct answer is:
zero

Solution:
The particle moves with linear velocity v along line PC. The line of motion is through P. Hence angular momentum is zero.

Topic: Physics, Rotational Motion, NEET
A particle of mass m moves along the quarter section of the circular path whose centre is at the origin. The radius of the circular path is a. A force Fâ=yi^-xj^ N acts on the particle where x, y denote the coordinates of the position of the particle. Calculate the work done by this force, in taking the particle from point A (a,0) to point B (0, a), along the circular path.

Options:
A -2 a2 J
B -π a24 J
C -a2 J
D -π a22 J

The correct answer is:
-π a22 J

Solution:
The given tangential force opposite to the motion
F¯t=x2+y2=a
w=∫Ftdscos180o -a∫ds=-a.πa2=-πa22 J

Topic: Physics, Work Power Energy, NEET
A particle of mass $m$ moves in the $X Y$ plane with a velocity $v$ along the straight line $\mathrm{AB}$. If the angular momentum of the particle with respect to origin $\mathrm{O}$ is $L_A$ when it is at A and $L_B$ when it is at B, then:

Options:
A $L_A=L_B$
B the relationship between $L_A$ and $L_B$ depends upon the slope of the line $\mathrm{AB}$
C $L_A < L_B$
D $L_A>L_{B^*}$

The correct answer is:
$L_A=L_B$

Solution:
Moment of Momentum is angular momentum. OP is the same whether the mass is at $\mathrm{A}$ or $\mathrm{B}$

Topic: Physics, Rotational Motion, NEET, NEET 2007
A particle of mass m moves in the XY plane with a velocity v along the straight line AB. If the angular momentum of the particle with respect to the origin O is LA when it is at A and LB when it is at B, then

Options:
A LA>LB
B LA=LB
C The relationship between LA and LB depends upon the slope of the line AB
D LA<LB

The correct answer is:
LA=LB

Solution:
From the definition of angular momentum,
L=r×p=rmvsin⁡ϕ(-k^)
Therefore, the magnitude of L is
L=mvrsin⁡ϕ=mvd
where d=rsin⁡ϕ is the distance of closest approach of the particle to the origin. As d is same for both the particles, hence LA=LB.

Topic: Physics, Rotational Motion, NEET
A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:

Options:
A 62%
B 44%
C 50%
D 56%

The correct answer is:
56%

Solution:
The initial momentum of system is Piâ=m2Vi^+2mV j^
According to question as
On perfectly inelastic collision the particles stick to each other.
Pfâ=3m Vfâ
By conservation of linear momentum
Pfâ=Piâ â3m Vfâ=m2V i^+2mVj^
âVfâ=2V3i^+j^â | V f | =223V
â´ loss in KE. of system =Kinitial-Kfinal
=12m2V2+122mV2-123m 2 2 Vâ¡ 3 2
=2mV2+mV2-43mV2=3mV2-4mV23=53mV2
% Loss in KE =100ÃâKKi=53mv23mV2=59Ã100=56%

Topic: Physics, Center of Mass Momentum and Collision, NEET
A particle of mass m moving with velocity 1 m s-1 collides perfectly elastically with another stationary particle of mass 2 m. If the incident particle is deflected by 90Â°. The heavy mass will make an angle Î¸ with the initial direction of m equal to

Options:
A 60o
B 45o
C 15o
D 30o

The correct answer is:
30o

Solution:
Conservation of momentum in initial direction of motion of m (let x- direction)
m×1i^=2mvi^
vx=12i^ m s-1
0=mvj^+2mvyj^
vy=-v2
KEBefore=KEAfter
12m12+0=12mv2+122mvx2+vy2
1=v2+2122+-v22
1=v2+12+v22
12=3v22
v=13 m s-1
Let v be the speed of m after it is deflected by 90o .
Conservation of momentum in perpendicular direction of motion of m (let y- direction)
As the collision is elastic therefore,
tan⁡θ=vyvx
tan⁡θ=v212=v
tan⁡θ=13
θ=30o

Topic: Physics, Center of Mass Momentum and Collision, NEET
A particle of mass m oscillates along the horizontal diameter AB inside a smooth spherical shell of the radius R . At any instant the kinetic energy of the particle is K. Then the force applied by particle on the shell at this instant is

Options:
A KR
B 2KR
C 3KR
D K2R

The correct answer is:
3KR

Solution:
Let the velocity of the particle at the point P is v.
From the conservation of mechanical energy
12mv2=K=mgh
Let N be the normal reaction between the particle and the shell at this instant. Then
N- mg sin⁡θ=mv2Rmv2R=2KR
or N=mghR+2KR=KR+2KR mgh= K
∴ N=3KR= force on shell

Topic: Physics, Laws of Motion, NEET
A particle of mass $m$ oscillates along $x$-axis according to equation $x=a$ sin $\omega t$. The nature of the graph between momentum and displacement of the particle is:

Options:
A Circle
B Hyperbola
C Ellipse
D Straight line passing through origin

The correct answer is:
Ellipse

Solution:
According to the question
$$
x=\operatorname{arsin} \omega t \rightarrow \frac{x}{a}=\sin \omega t
$$
Now velocity $\mathrm{V}=\frac{d x}{d t}=a \omega \cos \omega t$ $\Rightarrow \quad \frac{\mathrm{V}}{a \omega}=\cos \omega t$
From eq. (i) and (ii)
$$
\begin{gathered}
\frac{x^2}{a^2}+\frac{v^2}{a^2 \omega^2}=\sin ^2 \omega t+\cos ^2 \omega t \\
\frac{x^2}{a^2}+\frac{v^2}{a^2 \omega^2}=1
\end{gathered}
$$
$\Rightarrow$ This is the equation of ellipse. So the graph between moment and displacement of particle is ellipse.

Topic: Physics, Oscillations, NEET, NEET 2013 (Karnataka)
A particle of mass $m$ oscillates with simple harmonic motion between points $x_1$ and $x_2$, the equilibrium position being $\mathrm{O}$. Its potential energy is plotted. It will be as given below in the graph:

Options:
A

B

C

D

The correct answer is:

Solution:
$\mathrm{PE}$ in $\mathrm{SHM}=\frac{1}{2} K x^2 \rightarrow$ equation of parabola

Topic: Physics, Oscillations, NEET, NEET 2003
A particle of mass m rotates in a circle of the radius a with a uniform angular speed Ï0 . It is viewed from a frame rotating about the z-axis with a uniform angular speed Ï. The centrifugal force on the particle is

Options:
A mω2a
B mω02a
C mω+ω022a
D mωω0

The correct answer is:
mω2a

Solution:
Centrifugal force does not depend upon angular velocity of particle. It depends upon angular velocity of frame of reference.
∴ Centrifugal force is F=mω2a

Topic: Physics, Laws of Motion, NEET
A particle originally at rest at the highest point of a smooth vertical circle is slightly displaced. It will leave the circle at a vertical distance h below the highest point, such that

Options:
A h = R
B h = R/3
C h = R/2
D h = 2R

The correct answer is:
h = R/3

Solution:
mg cos θ - N = mv 2 R

to leave the contact N = 0
mg cos θ = mv 2 R ⇒ g cos θ = v 2 R … 1
cos θ = R - h R … 2
from energy conservation ⇒ 1 2 mv 2 = mgh … 3
from 1, 2, 3 1 2 gR × R - h R = gh ⇒ R - h = 2 h ⇒ h = R / 3

Topic: Physics, Laws of Motion, NEET
A particle P is moving in a circle of the radius r with a uniform speed v. C is the centre of the circle and AB is the diameter. The angular velocity of P about A and C is in the ratio

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

The correct answer is:
1:2

Solution:
From properties of Δ
If ∠CAP=θ
then ∠BCP=2 θ
ω A = dθ dt
ω C =2 dθ dt
ω A : ω C =1:2

Topic: Physics, Laws of Motion, NEET
A particle P is sliding down a frictionless hemispherical bowl. It passes point A at t=0. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at t=0 along the horizontal string AB, with the speed v. Friction between the bead and the string may be neglected. Let tp and tQ be the, respective, times taken by P and Q to reach the point B. Then

Options:
A tP<tQ
B tP=tQ
C tP>tQ
D none of these

The correct answer is:
tP<tQ

Solution:
For particle P, motion between AC will be an accelerated one while between CB, a retarded one. But in any case, the horizontal component of its velocity will be greater than or equal to v. On the other hand, in the case of particle Q, it is always equal to v. Horizontal displacement for both the particles is equal. Therefore, tP<tQ.

Topic: Physics, Laws of Motion, NEET
A particle performing uniform circular motion has angular momentum L. If its angular frequency is doubled and its kinetic energy halved, then the new angular momentum will be

Options:
A 2 L
B 4 L
C L / 2
D L / 4

The correct answer is:
L / 4

Solution:
Angular momentum L = I Ï
Rotational kinetic energy K = 1 2 I Ï 2
â´ L K = I Ï Ã 2 I Ï 2 = 2 Ï â L = 2 K Ï
or L 1 L 2 = K 1 K 2 Ã Ï 2 Ï 1 = 2 Ã 2 = 4
â´ L 2 = L 1 4 = L 4 .

Topic: Physics, Laws of Motion, NEET
A particle performing uniform circular motion has

Options:
A Radial velocity and radial acceleration
B A radial velocity and transverse acceleration
C Transverse velocity and radial acceleration
D Transverse velocity and transverse acceleration

The correct answer is:
Transverse velocity and radial acceleration

Solution:
A particle performing a uniform circular motion has a transverse velocity and radial acceleration

Topic: Physics, Laws of Motion, NEET
A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point

Options:
A $B$
B $C$
C $D$
D $A$

The correct answer is:
$C$

Solution:
Particle has maximum instantaneous velocity at that point at which its slope is maximum. Therefore, $y_{\max }=\frac{d x}{d t}=$ maximum slope From figure it is obvious that at point $C$, slope is maximum, hence at this point velocity is maximum.

Topic: Physics, Motion In One Dimension, NEET, NEET 2008 (Screening)
A particle starting from rest, moves in a circle of radius ' $r$ '. It attains a velocity of $\mathrm{v}_0 \mathrm{~m} / \mathrm{s}$ in the $\mathrm{n}^{\text {th }}$ round. Its angular acceleration will be

Options:
A $\frac{v_0}{n} \mathrm{rad} / \mathrm{s}^2$
B $\frac{v_0^2}{2 \pi n r^2} \mathrm{rad} / \mathrm{s}^2$
C $\frac{v_0^2}{4 \pi n r^2} \mathrm{rad} / \mathrm{s}^2$
D $\frac{v_0^2}{4 \pi n r} \mathrm{rad} / \mathrm{s}^2$

The correct answer is:
$\frac{v_0^2}{4 \pi n r^2} \mathrm{rad} / \mathrm{s}^2$

Solution:
From third equation of motion for circular motion
$$
\omega^2-\omega_0^2=2 \alpha \theta
$$
where, $\omega=$ final angular velocity of particle
$\omega_0=$ initial angular velocity
$\alpha=$ angular acceleration and
$\theta=$ angular displacement
Here, $\omega=\frac{\mathrm{v}_0}{\mathrm{r}} \mathrm{rad} / \mathrm{s}$ (where, $\mathrm{r}$ radius of the circle)
$\omega_0=0$ (initially particle is at rest)
$\theta=2 \pi n$ (for $n$ rounds)
Substituting these values in Eq. (i), we get
$$
\left(\frac{\mathrm{v}_0}{\mathrm{r}}\right)^2-0=2 \alpha(2 \pi \mathrm{n}) \Rightarrow \alpha=\frac{\mathrm{v}_0^2}{4 \pi \mathrm{nr}^2} \mathrm{rad} / \mathrm{s}^2
$$

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2019 (Odisha)
A particle starting from the origin $(0,0)$ moves in a straight line in the $(x, y)$ plane. Its coordinates at a later time are $(\sqrt{3}, 3)$. The path of the particle makes with the $x$-axis an angle of:

Options:
A $45^{\circ}$
B $60^{\circ}$
C $0^{\circ}$
D $30^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
Particle starts moving from \((0, 0)\) and goes till \((\sqrt{3}, 3)\).
So total displacement along x axis is \(\sqrt{3}\) unites and along y axis is 3 units.
We know that:
\(\begin{aligned}
& \Rightarrow \tan (\theta)=\frac{y}{x} \\
& \Rightarrow \tan (\theta)=\frac{3}{\sqrt{3}} \\
& \Rightarrow \tan (\theta)=\sqrt{3} \\
& \Rightarrow \theta=\frac{\pi}{3}=60^{\circ}
\end{aligned}\)

Topic: Physics, Mathematics in Physics, NEET, NEET 2007