datavorous/entrance-exam-dataset huggingface dataset:
A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is 1.878×10–4. The mass of the particle is close to :

Options:
A 4.8×10-27 kg
B 9.1×10-31 kg
C 1.2×10-28 kg
D 9.7×10-28 kg

The correct answer is:
9.7×10-28 kg

Solution:
λ=hmv
⇒ m'm=vλv'λ'=15×11.878×10-4×9.1×10-31
m'=9.7×10-28kg

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2020 (02 Sep Shift 2)
A particle is moving along a circular path with a constant speed of $10 \mathrm{~ms}^{-1}$. What is the magnitude of the change in velocity of the particle, when it moves through an angle of $60^{\circ}$ around the centre of the circle?

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

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

Solution:
Change in velocity,
$$
\begin{array}{l}
|\Delta \overline{\mathrm{v}}|=\sqrt{\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}+2 \mathrm{v}_{1} \mathrm{v}_{2} \cos (\pi-\theta)} \\
=2 \mathrm{v} \sin \frac{\theta}{2} \quad\left(\because\left|\overrightarrow{\mathrm{v}}_{1}\right|=\left|\overrightarrow{\mathrm{v}}_{2}\right|\right)=\mathrm{v} \\
=(2 \times 10) \times \sin \left(30^{\circ}\right)=2 \times 10 \times \frac{1}{2} \\
=10 \mathrm{~m} / \mathrm{s}
\end{array}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2019 (11 Jan Shift 1)
A particle is moving along a line $y=x+a \mid$ with a constant velocity $v$. Find the angular momentum of the particle about the origin.

Options:
A $m v a$
B $m v a \sqrt{2}$
C $\frac{m v a}{\sqrt{2}}$
D $\frac{m v y}{x \sqrt{2}}$

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

Solution:
$\angle P O Q=45^{\circ}$ So, $O Q=a \cos 45^{\circ}=a / \sqrt{2}$ $L=m v(O Q)=m v a / \sqrt{2}$

Topic: Physics, Rotational Motion, JEE Main
A particle is moving eastwards with a velocity of $5 \mathrm{~m} / \mathrm{s}$ in 10 seconds the velocity changes to $5 \mathrm{~m} / \mathrm{s}$ northwards. The average acceleration in this time is

Options:
A
$\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^2$ towards north-east

B
$\frac{1}{2} \mathrm{~m} / \mathrm{s}^2$ towards north.

C
zero

D
$\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^2$ towards north-west

The correct answer is:
$\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^2$ towards north-west

Solution:
$\overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{V}}_{\mathrm{f}}-\overrightarrow{\mathrm{V}}_{\mathrm{i}}}{\mathrm{t}}$
$=\frac{5 \hat{\mathrm{j}}-5 \hat{\mathrm{i}}}{10}=\frac{1}{2}(\hat{\mathrm{j}}-\hat{\mathrm{i}})$
$\therefore \mathrm{a}=\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}$ towards north west

Topic: Physics, Mathematics in Physics, JEE Main
A particle is moving eastwards with a velocity of $5 \mathrm{~m} / \mathrm{s}$ in 10 seconds the velocity changes to $5 \mathrm{~m} / \mathrm{s}$ northwards. The average acceleration in this time is

Options:
A
$\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^2$ towards north-east

B
$\frac{1}{2} \mathrm{~m} / \mathrm{s}^2$ towards north.

C
zero

D
$\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^2$ towards north-west

The correct answer is:
$\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^2$ towards north-west

Solution:
$\overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{V}}_{\mathrm{f}}-\overrightarrow{\mathrm{V}}_{\mathrm{i}}}{\mathrm{t}}$
$=\frac{5 \hat{\mathrm{j}}-5 \hat{\mathrm{i}}}{10}=\frac{1}{2}(\hat{\mathrm{j}}-\hat{\mathrm{i}})$
$\therefore \mathrm{a}=\frac{1}{\sqrt{2}} \mathrm{~ms}^{-2}$ towards north west

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2005
A particle is moving in a circle of radius R in such a way that at any instant the tangential retardation of the particle and the normal acceleration of the particle are equal. If its speed at t=0 is v0, the time taken to complete the first revolution is

Options:
A Rv0
B Rv0(1-e-2π)
C Rv0e-2π
D 2πRv0

The correct answer is:
Rv0(1-e-2π)

Solution:
When a particle moves in a circular motion, it is acted upon by centripetal force directed towards the centre. Hence, centripetal acceleration is
aN=dvdt=v2R
or ∫0tdtR=∫v0vdvv2
or t=-R1vv0v
v=v0RR-v0t
Also drdt=v0RR-v0t
∫02π Rdr=v0R ∫0TdtR-v0t
⟹ T=Rv01-e-2π

Topic: Physics, Laws of Motion, JEE Main
A particle is moving in a circle of radius r under the action of a force F=αr2 which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy=0 for r=0):

Options:
A 56αr3
B αr3
C 12αr3
D 43αr3

The correct answer is:
56αr3

Solution:
Force F=Î±r2 will act as centripetal force
âÎ±r2=mv2râmv2=Î±r3
â´ Kinetic energy K.E. =12mv2
âK.E.=Î±r32
Also dU=-Fâ.drâ
âU=-â«Fâ.drâ
=-â«Î±r2-dr
=Î±â«0rr2dr
U-0=Î±r33
âU=Î±r33
So total mechanical energy
E=U+K
=Î±r33+Î±r32
E=5Î±r36

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2015 (11 Apr Online)
A particle is moving in a circular path of radius a under the action of an attractive potential U=-k2r2. Its total energy is:

Options:
A -32ka2
B –k4a2
C k2a2
D Zero

The correct answer is:
Zero

Solution:
U=-K2r2⇒F=-dUdr=2K2r3=Kr3⇒mv2r=Kr3⇒v2=Kmr2
⇒12mv2=K2r2⇒E=K2r2-K2r2=0.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2018 (08 Apr)
A particle is moving in a circular path of radius a, with a constant velocity $\mathrm{v}$ as shown in the figure. The centre of circle is marked by ' $\mathrm{C}$ '. The angular momentum from the origin $\mathrm{O}$ can be written as:

Options:
A
va $(1+\cos 2 \theta)$

B
va $(1+\cos \theta)$

C
va $\cos 2 \theta$

D
va

The correct answer is:
va $(1+\cos 2 \theta)$

Solution:
No Solution Available

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2014 (12 Apr Online)
A particle is moving in a straight line such that its velocity is increasing at 5 m s-1 per meter. The acceleration of the particle is _____ m s-2 at a point where its velocity is 20 m s-1.

Options:

The correct answer is:
100

Solution:
According to the question, velocity is increasing at 5 m s-1. Therefore, dvds=5 s-1.

Now acceleration a=dvdt=vdvds

⇒a=vdvds=20×5=100 m s-2

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (25 Jul Shift 2)
A particle is moving in a straight line. The variation of position x as a function of time t is given as x=t3-6t2+20t+15 m. The velocity of the body when its acceleration becomes zero is:

Options:
A 4 m s-1
B 8 m s-1
C 10 m s-1
D 6 m s-1

The correct answer is:
8 m s-1

Solution:
Given the instantaneous position is

x=t3-6t2+20t+15 ...1

The velocity of the particle is given by

v=dxdt=3t2-12t+20 ...2

And, the acceleration of the particle is given by

a=dvdt=6t-12 ...3

When a=0, it implies

6t-12=0⇒t=2 s

At t=2 s, the velocity can be calculated as follows:

v=3(2)2-12(2)+20=8 m s-1

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (29 Jan Shift 2)
A particle is moving in a straight line with initial velocity and uniform acceleration a. If the sum of the distance travelled in $t^{\text {th }}$ and $(t+1)^{\text {th }}$ seconds is $100 \mathrm{~cm}$, then its velocity after $t$ seconds, in $\mathrm{cm} / \mathrm{s}, \mathrm{is}$

Options:
A 80
B 50
C 20
D 30

The correct answer is:
50

Solution:
No Solution available

Topic: Physics, Motion In One Dimension, JEE Main
A particle is moving in a uniform circular motion on a horizontal surface. Particle position and velocity at time t=0 are shown in the figure in the coordinate system. Which of the indicated variable on the vertical axis is incorrectly matched by the graph shown alongside for particle's motion-

Options:
A y component of force keeping particle moving in a circle

B x component of velocity

C Angular velocity of the particle

D x coordinate of the particle

The correct answer is:
x component of velocity

Solution:
where, Fx=FcosÎ¸, and Fy=FsinÎ¸
at time t, in Uniform circular motion only centripetal force exists constant in magnitude
So, Y - component of force is â Fy â â=âFsinÎ¸ââ j ^
Fy=âFsinÏtââ j ^ (as, Î¸=Ït )
So given graph is correct for Fy vs t
In Uniform circular motion. V= Ï r = constant
â V x â =âVsinÏâtâ j ^
So given graph for V x ât is incorrect.
In Uniform circular motion., Ï= constant
â´âÏât graph is correct
âµâx=rcosÎ¸ââ=ârcosÏât
ââxât graph is correct

Topic: Physics, Laws of Motion, JEE Main
A particle is moving in one dimension (along x axis) under the action of a variable force. It's initial position was 16 m right of origin. The variation of its position x with time t is given as x=–3t3+18t2+16t, where x is in m and t is in s. The velocity of the particle when its acceleration becomes zero is _________ m s-1.

Options:

The correct answer is:
52

Solution:
Given:

x=-3t3+18t2+16t

As we know, v=dxdt=-9t2+36t+16

Now, a=dvdt=-18t+36

For a=0, time till be t=2 s

Therefore, required velocity v=-9(2)2+36×2+16

v=52 m s-1

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (01 Feb Shift 1)
A particle is moving in the xy-plane along a curve C passing through the point 3,3. The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with

Options:
A length of latus rectum 3
B length of latus rectum 6
C focus 43,0
D focus 0,33

The correct answer is:
length of latus rectum 3

Solution:
Let Point Px,y

So equation of line passing through Px,y will be

Y-y=mX-x

Point on x-axis is given by Qx-ym,0

Mid-point of PQ lies on y axis so its x co-ordinate will be zero,

x-ym+x=0

m=y2·x⇒2dyy=dxx

2lnn=lnx+lnk

y2=kx

It passes through 3,3⇒k=3

curve c⇒y2=3x

Length of L.R. =3

Focus=34,0

Topic: Mathematics, Parabola, JEE Main, JEE Main 2022 (24 Jun Shift 2)
A particle is moving in xy-plane with a constant speed v0 such that its y displacement is given by y=Î±e-2vx3v0, where vx is component of velocity along the x-axis. If at some instant x component of it velocity is positive and the slope of the tangent on its path is -13, then the displacement of the particle in y-direction at the instant is

Options:
A Î± e - 1
B α e - 2
C Zero
D α 2 e

The correct answer is:
α e - 1

Solution:
Here, particle is moving in two - dimensional (x â y) plane, with constant speed so slope of
graph d y d x = v y v x and v 0 = v x 2 + v y 2 .
As slope, dy dx = - 1 3 = v y v x
So, v x = - v y 3
On squaring and adding
v x 2 + v y 2 = v 0 2
3 v y 2 + v y 2 = v 0 2
v y = Â± v 0 2
v x = â 3 v 0 2
So, y = Î± e - 2 3 v 0 Ã 3 v 0 2
y = Î± e - 1

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle is moving on a circular path of radius $r$ with uniform speed $V$. The magnitude of change in velocity when the particle moves from $\mathrm{P}$ to $\mathrm{Q}$ is :$\left(\angle \mathrm{P~POQ}=40^{\circ}\right)$

Options:
A $2 v \cos 40^{\circ}$
B $2 v \sin 40^{\circ}$
C $2 v \sin 20^{\circ}$
D $2 v \cos 20^{\circ}$

The correct answer is:
$2 v \sin 20^{\circ}$

Solution:
Let the velocity at point $P$
$\overrightarrow{v_1}=v \hat{1}$
So the velocity at point $\mathrm{Q}$ will be
$\overrightarrow{v_2}=v \cos 40^{\circ} \hat{i}-v \sin 40^{\circ} \hat{j}$
Change in velocity
$\overrightarrow{\Delta v}=\overrightarrow{v_2}-\overrightarrow{v_1}=\left(v \cos 40^{\circ} \hat{i}-v \sin 40^{\circ} \hat{j}\right)-v \hat{i}$
Magnitude of change of velocity
$|\Delta \mathrm{v}|=\sqrt{\left(\mathrm{v} \cos 40^{\circ}-\mathrm{v}\right)^2+\left(-\mathrm{v} \sin 40^{\circ}\right)^2}$
On solving we get
$|\Delta \mathrm{v}|=2 \mathrm{v} \sin 20^{\circ}$

Topic: Physics, Mathematics in Physics, JEE Main
A particle is moving on a circular path with a constant speed $v$. Its change of velocity as it moves from $A$ to $B$ in the figure is

Options:
A $2 v \sin \frac{\theta}{2}$
B $v \sin \theta$
C $\frac{v \sin 2 \theta}{2}$
D $2 v \sin \theta$

The correct answer is:
$2 v \sin \frac{\theta}{2}$

Solution:
The given situation is shown below
From above figure we can observe that velocity vector (tangential to circle) can be resolved into horizontal $\left(v \cos \frac{\theta}{2}\right)$ and vertical $\left(v \sin \frac{\theta}{2}\right)$ components.
As, particle is executing uniform circular motion, magnitudes of components remains same but direction of vertical component is reversed.
So, Initial velocity
$$
v_1=v \cos \frac{\theta}{2} \hat{i}+v \sin \frac{\theta}{2} \hat{j} \text { and }
$$
final velocity, $v_2=v \cos \frac{\theta}{2} \hat{i}-v \sin \frac{\theta}{2} \hat{j}$
Change in velocity
$$
\begin{aligned}
& =\text { final velocity }- \text { Initial velocity } \\
& =v_2-v_1 \\
& =\left(v \cos \frac{\theta}{2} \hat{i}-v \sin \frac{\theta}{2} \hat{j}\right)-\left(v \cos \frac{\theta}{2} \hat{i}+v \sin \frac{\theta}{2} \hat{j}\right) \\
& =-2 v \sin \frac{\theta}{2} \hat{j}
\end{aligned}
$$
Magnitude of change of velocity will be
$$
\begin{aligned}
|\Delta \mathbf{v}| & =\left|v_2-v_1\right| \\
& =2 v \sin \frac{\theta}{2}
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, AP EAMCET, AP EAMCET 2022 (07 Jul Shift 2)
A particle is moving three time as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is \( 1.813 \times 10^{-4} \). The mass of the particle is (Mass of electron \( =9.1 \times 10^{-31} \mathrm{~kg} \) )

Options:
A \( 1.67 \times 10^{-27} \mathrm{~kg} \)
B \( 1.67 \times 10^{-31} \mathrm{~kg} \)
C \( 1.67 \times 10^{-30} \mathrm{~kg} \)
D \( 1.67 \times 10^{-32} \mathrm{~kg} \)

The correct answer is:
\( 1.67 \times 10^{-27} \mathrm{~kg} \)

Solution:
de Broglie wavelength of a moving particle, having mass m and velocity v is given by
λ = h p = h mv⁡ or m = h λ v
For an electron
λ e = h m e v e = or m e = h λ e v e

for particle

λ=hmv=orm=hλv
Given : v v e = 3 , λ λ e⁡ = 1 · 8 1 3 × 1 0 - 4
Mass of the particle,mme=vevλeλm=mevevλeλ
Substituting the values, we get
m= 9 · 1 × 1 0 - 3 1 kg × 1 3 × 1 1 · 8 1 3 × 1 0 - 4
m = 1.67 × 10–27 kg

Topic: Physics, Dual Nature of Matter, JEE Main
A particle is moving unidirectional on a horizontal plane under the action of a constant power supplying energy source. The displacement (s) – time (t) graph that describes the motion of the particle is (graphs are drawn schematically and are not to scale):

Options:
A
B
C
D

The correct answer is:

Solution:
Here Power is constant,
Hence, P=Fv=mav=av=constant.

av=kvdvdx×v=k∫0vv2dv=k∫0xdx∴v=3k1/3×x1/3⇒v=Ax1/3 where A=3k1/3=constant ....1

Also, v=dxdt∴dxdt=Ax1/3⇒∫0ddxx1/3=A∫0tdt

Simplifying further we get,

⇒d=At3/2 ...2

From equations 1 & 2 its clear that the graph won't be straight line so first & third options cant be correct.
Now, from equation 1 its clear that when x is increased then the slope dsdt should also increase which is possible in second option. Hence option 2 is correct.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2020 (03 Sep Shift 2)
A particle is moving uniformly along a straight line as shown in the figure. During the motion of
the particle from A to B, the angular momentum of the particle about 'O'

Options:
A remains constant
B increases
C first increases then decreases
D decreases

The correct answer is:
remains constant

Solution:
Angular momentum L=mvr
where r is the perpendicular distance between velocity and reference point
Here since the line AB is straight, there would only 1 perpendicular distance from reference point i.e OA
So,
Angular momentum at A =mvr
Angular momentum at P =mvr
Angular momentum at B =mvr
Since, perpendicular distance between the line AB and O will always be R
So all the points lying on the line AB will have same angular momentum equals to mvr.

Topic: Physics, Rotational Motion, KCET, KCET 2019
A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then:

Options:
A T∝Rn/2
B T∝R3/2 for any n
C T∝Rn2+1
D T∝Rn+12

The correct answer is:
T∝Rn+12

Solution:
F=KRn⇒mV2r=KRn⇒V2∝KRn-1
⇒V∝1Rn-12⇒T=2πRV∝RV
⇒T∝R1Rn-12⇒T∝Rn+12

Topic: Physics, Gravitation, JEE Main, JEE Main 2018 (08 Apr)
A particle is moving with a uniform speed $v$ in a circular path of radius $r$ with the centre at $O$. When the particle moves from a point $P$ to $Q$ on the circle such that $\angle P O Q=\theta$, then the magnitude of the change in velocity is

Options:
A $2 \mathrm{v} \sin (2 \theta)$
B zero
C $2 v \sin \left(\frac{\theta}{2}\right)$
D $2 v \cos \left(\frac{\theta}{2}\right)$

The correct answer is:
$2 v \sin \left(\frac{\theta}{2}\right)$

Solution:
$\therefore$ Change in magnitude of velocity
$\begin{aligned}
|\Delta v| &=\sqrt{v^{2}+v^{2}-2 v^{2} \cos \theta} \\
&=2 v \sin \frac{\theta}{2}
\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main
A particle is moving with a velocity v→=Kyi^+xj^, where K is a constant.
The general equation for its path is:

Options:
A y2=x+constant
B xy=constant
C y=x2+constant
D y2=x2+constant

The correct answer is:
y2=x2+constant

Solution:
v→=Kyi^+xj^
vx=y=dxdt
vy=x=dydt
dxdy=yx
∫x dx=∫y dy
x22=y22+C
or y2=x2+constant

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A particle is moving with constant speed in a circular path. When the particle turns by an angle 90°, the ratio of instantaneous velocity to its average velocity is π:x2. The value of x will be

Options:
A 2
B 5
C 1
D 7

The correct answer is:
2

Solution:
In circular motion with constant speed, magnitude of instantaneous velocity, vins=ωr.

Time taken to turn by 90° will be =π2ω=π2ω.

Displacement will be equal to the length of AB=r2+r2=r2

Average velocity

vavg=displacementtime=r2×2ωπ=22πωr

Hence, required ratio vins vavg =ωrπ22ωr=π22

Therefore, x=2.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (06 Apr Shift 1)
A particle is moving with speed v=bx along positive x- axis. Calculate the speed of the particle at time t=τ (assume that the particle is at origin at t=0 )

Options:
A b2τ
B b2τ2
C b2τ2
D b2τ4

The correct answer is:
b2τ2

Solution:
v=bx … (1)
⇒dxdt=bx
dxx=bdt
∫0xdxx=∫0tbdt
2x=bt … (2)
From equation (1) and (2)
v=bbt2
∴at t=τ
v=b2τ2

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A particle is moving with uniform speed along the circumference of a circle of radius R under the action of a central fictitious force F which is inversely proportional to R3. Its time period of revolution will be given by :

Options:
A T∝R43
B T∝R52
C T∝R32
D T∝R2

The correct answer is:
T∝R2

Solution:
F∝1R3

KR3=mω2R

ω2=Km×1R4

2πT2=Km×1R4

T2∝R4

T∝R2

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (26 Feb Shift 1)
A particle is moving with velocity $\overrightarrow{\mathrm{v}}=\mathrm{K}(\mathrm{y} \hat{\mathrm{i}}+\mathrm{x} \hat{\mathrm{j}})$, where $\mathrm{K}$ is a constant. The general equation for its path is

Options:
A
$y=x^2+$ constant

B
$y^2=x+$ constant

C
$x y=$ constant

D
$y^2=x^2+$ constant

The correct answer is:
$y^2=x^2+$ constant

Solution:
$\vec{v}=K y \hat{i}+K x \hat{j}$
$\frac{d x}{d t}=K y, \quad \frac{d y}{d t}=K x$
$\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{K x}{K y}$
$y d y=x d x$
$y^2=x^2+c$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2010
A particle is oscillating according to the equation $X=7 \cos 0.5 \pi t$, where $t$ is in second. The point moves from the position of equilibrium to maximum displacement in time

Options:
A $4.0 \mathrm{sec}$
B $2.0 \mathrm{sec}$
C $1.0 \mathrm{sec}$
D $0.5 \mathrm{sec}$

The correct answer is:
$1.0 \mathrm{sec}$

Solution:
A particle is oscillating according to the equation $X=7 \cos 0.5 \pi t$ ... (i)
We know that,
$X=A \cos \omega t$
$T=2 \pi \omega \backslash$ Where $T$ is the total time taken
On comparing eq (I) we get $\omega=0.5 \pi$
Or, $T=2 \pi / 0.5 \pi$
Or, $T=4 \mathrm{sec}$
To move from the position of equilibrium to maximum displacement time needed $=T / 4=$ $1 \sec ^2$

Topic: Physics, Oscillations, JEE Main
A particle is oscillating on the x-axis with an amplitude 2 cm about the point x0=10 cm with a frequency. A concave mirror of focal length 5 cm is placed at the origin (see figure).

Identify the correct statements?
(i) The image executes periodic motion.
(ii) The image executes non-periodic motion.
(iii) The turning points of the image are asymmetric with respect to the image of the point at X=10 cm.
(iv) The distance between the turning points of the oscillation of the image is 10021 cm.

Options:
A (ii, iii)
B (i, iii, iv)
C (i, iv)
D (ii, iv)

The correct answer is:
(i, iii, iv)

Solution:
Recall the proper formula of mirror formula, 1f=1v+1u, given f=-10 cm and u=-5 cm then the formula of the concave mirror, -110=1v-15⇒v=-10 cm. As, image copies the time period of the object A is right as well. It will be a periodic motion. For one extreme,

⇒-18+1v=1-5⇒v=-403 cm, right arrow.

For other extreme,

⇒-112+1v=1-5⇒v=-760 cm. These points are asymmetrical about x0=10 cm. Now amplitude of oscillation of image, ⇒403-607=10021 cm.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2018 (15 Apr)
A particle is oscillating on the $\mathrm{X}$-axis with an amplitude $2 \mathrm{~cm}$ about the point $x_0=10 \mathrm{~cm}$ with a frequency $\omega$. A concave mirror of focal length 5 $\mathrm{cm}$ is placed at the origin (see figure)
Identify the correct statements:
(A) The image executes periodic motion
(B) The image executes non-periodic motion
(C) The turning points of the image are asymmetric w.r.t the image of the point at $x$ $=10 \mathrm{~cm}$
(D) The distance between the turning points of the oscillation of the image is $\frac{100}{21}$

Options:
A
(B), (D)

B
(B), (C)

C
(A), (C), (D)

D
(A), (D)

The correct answer is:
(A), (C), (D)

Solution:
When object is at $8 \mathrm{~cm}$
Image $V_1=\frac{\mathrm{f} \times \mathrm{u}}{\mathrm{u}-\mathrm{f}}=\frac{5 \times 8}{8-5}=-\frac{40}{3} \mathrm{~cm}$
When object is at $12 \mathrm{~cm}$
Image $\mathrm{V}_2=\frac{\mathrm{f} \times \mathrm{u}}{\mathrm{u}-\mathrm{f}}=\frac{5 \times 12}{12-5}=-\frac{60}{7} \mathrm{~cm}$
Separation $=\left|\mathrm{V}_1-\mathrm{V}_2\right|=\frac{40}{3}-\frac{60}{7}=\frac{100}{21} \mathrm{~cm}$
So $\mathrm{A}, \mathrm{C}$ and $\mathrm{D}$ are correct statements.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A particle is placed at rest inside a hollow hemisphere of radius $R$. The coefficient of friction between the particle and the hemisphere is $\mu=\frac{1}{\sqrt{3}}$. The maximum height upto which the particle can remain stationary is

Options:
A $\frac{R}{2}$
B $\left(1-\frac{\sqrt{3}}{2}\right) R$
C $\frac{\sqrt{3}}{2} R$
D $\frac{3R}{8}$

The correct answer is:
$\left(1-\frac{\sqrt{3}}{2}\right) R$

Solution:
Using $h_{\max }=\left(1-\frac{1}{\sqrt{\mu^2+1}}\right) R$
$\begin{aligned} & =\left(1-\frac{1}{\sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+1}}\right) \times R \quad\left(\because \mu=\frac{1}{\sqrt{3}}\right) \\ & =\left(1-\frac{1}{\sqrt{\frac{1}{3}+1}}\right) \times R=\left(1-\frac{1}{\sqrt{\frac{1+3}{3}}}\right) \times R \\ & \text { or, } h_{\max }=\left(1-\frac{\sqrt{3}}{2}\right) R\end{aligned}$

Topic: Physics, Laws of Motion, AP EAMCET, AP EAMCET 2016
A particle is placed at the origin and a force F=kx is acting on it (where k is a positive constant). IF U(0)=0, the graph of U(x) versus x will be (where U is the potential energy function)

Options:
A
B
C
D

The correct answer is:

Solution:
F = - dU dx ⇒ ∫ dU = - ∫ Fdx
U 2 - U 1 = - ∫ kxdx ⇒ U = - 1 2 kx 2

Topic: Physics, Work Power Energy, JEE Main
A particle is placed at the point A of a frictionless track ABC as shown in figure. It is gently pushed towards right. The speed of the particle when it reaches the point B is: (Take g=10 m s-2.

Options:
A 20 m s-1
B 10 m s-1
C 210 m s-1
D 10 m s-1

The correct answer is:
10 m s-1

Solution:
Applying conservation of mechanical energy between point A & B, we get

KEA+UA=KEB+UB

⇒0+mg(1)=12mv2+mg×0.5

⇒v=g=10 m s-1

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2024 (30 Jan Shift 1)