datavorous/entrance-exam-dataset huggingface dataset:
A particle moves in x-y plane. The position vector of the particle at any time t is râ=2ti^+2t2j^ m. The rate of change of Î¸ at time t=2 s (where Î¸ is the angle which its velocity vector makes with positive x -axis ) is

Options:
A 217 rad s-1
B 114 rad s-1
C 47 rad s-1
D 65 rad s-1

The correct answer is:
217 rad s-1

Solution:
x=2t ⇒vx=dxdt=2
y=2t2 ⇒vy=dydt=4t
∴ tan⁡θ=vyvx= 4t2=2t
Differentiating with respect to time we get,
sec2⁡θ dθdt=2
or 1+ tan2⁡θ dθdt=2
or 1+4t2 dθdt=2
or dθdt=21+4t2
dθdtat t=2 s is dθdt=21+422=217 rad s-1

Topic: Physics, Laws of Motion, JEE Main
A particle moves in $x-y$ plane under the influence of a force $\vec{F}$ such that its linear momentum is $\overrightarrow{\mathrm{p}}(\mathrm{t})=\hat{i} \cos (\mathrm{kt})-\hat{j} \sin (\mathrm{kt})$. If $\mathrm{k}$ is constant, the angle between $\overrightarrow{\mathrm{F}}$ and $\overrightarrow{\mathrm{p}}$ will be :

Options:
A $\frac{\pi}{4}$
B $\frac{\pi}{6}$
C $\frac{\pi}{2}$
D $\frac{\pi}{3}$

The correct answer is:
$\frac{\pi}{2}$

Solution:
$\begin{aligned} & \overrightarrow{\mathrm{P}}=\cos (\mathrm{kt}) \hat{\mathrm{i}}-\sin (\mathrm{kt}) \hat{\mathrm{j}} ;|\overrightarrow{\mathrm{P}}|=1 \\ & \because \overrightarrow{\mathrm{P}}=\mathrm{m} \overrightarrow{\mathrm{v}} \\ & \therefore \hat{\mathrm{P}}=\hat{\mathrm{v}} \\ & \Rightarrow \hat{\mathrm{v}}=\cos (\mathrm{kt}) \hat{\mathrm{i}}-\sin (\mathrm{kt}) \hat{\mathrm{j}} \\ & \hat{\mathrm{a}}=\frac{-\mathrm{k} \sin (\mathrm{kt}) \hat{\mathrm{i}}-\mathrm{k} \cos (\mathrm{kt}) \hat{\mathrm{j}}}{\mathrm{k}} \\ & \Rightarrow \hat{\mathrm{a}}=-\sin k t \hat{\mathrm{i}}-\cos \mathrm{kt} \hat{\mathrm{j}} \\ & \because \hat{\mathrm{F}}=\hat{\mathrm{a}}=-\sin \mathrm{kt} \hat{\mathrm{i}}-\cos \mathrm{kt} \hat{\mathrm{j}} \\ & \cos \theta=\frac{\hat{\mathrm{F}} \cdot \hat{\mathrm{P}}}{|\hat{\mathrm{F}}||\hat{\mathrm{P}}|}=-\frac{\sin \mathrm{kt} \cos \mathrm{t}+\sin \mathrm{kt} \cos \mathrm{t}}{1 \times 1}=0 \\ & \Rightarrow \theta=\frac{\pi}{2}\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2024 (05 Apr Shift 2)
A particle moves in $X Y$-plane with $x$ and $y$ varying with time $t$ as $x(t)=5 t, y(t)=5 t\left(27-t^2\right)$. At what time in seconds, the direction of velocity and acceleration will be perpendicular to each other?

Options:
A $5 \sqrt{\frac{27}{2}}$
B 5
C $5 \sqrt{12}$
D 3

The correct answer is:
3

Solution:
Velocity in $x$-direction is
$v_x=\frac{d}{d t}(x)=\frac{d}{d t}(5 t)=5 \mathrm{~ms}^{-1}$
Velocity in $y$-direction is
$v_y=\frac{d}{d t}(y)=\frac{d}{d t}\left(5 \times 27 t-5 t^3\right)=5 \times 27-5 \times 3 t^2$
As, $v_x=$ constant and $v_y=$ time dependent.
$\therefore$ This is a case similar to projectile motion and velocity and acceleration are perpendicular, when $v_y=0$.
$\Rightarrow \quad 5 \times 27-5 \times 3 t^2=0 \Rightarrow t^2=9 \Rightarrow t=3 \mathrm{~s}$

Topic: Physics, Mathematics in Physics, JEE Main
A particle moves such that its position vector r→t=cos⁡ωti^+sin⁡ωtj^ where ω is a constant and t is time. Then which of the following statements is true for the velocity v→t and acceleration a→t of the particle:

Options:
A v→ is perpendicular to r→ and a→ is directed away from the origin
B v→ and a→ both are perpendicular to r→
C v→ and a→ both are parallel to r→
D v→ is perpendicular to r→ and a→ is directed towards the origin

The correct answer is:
v→ is perpendicular to r→ and a→ is directed towards the origin

Solution:
r→=cos⁡ωti^+sin⁡ωtj^
v→=dr→dt=ω-sin⁡ωti^+cos⁡ωtj^
a→=dv→dt=-ω2cos⁡ωti^+sin⁡ωtj^
a→=-ω2r→∴a→ is antiparallel to r→
v→.r→=ω-sin⁡ωtcos⁡ωt+cos⁡ωtsin⁡ωt=0
So v→⊥r→

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2020 (08 Jan Shift 2)
A particle moves towards east from a point $A$ to a point $B$ at the rate of $4 \mathrm{~km} / \mathrm{h}$ and then towards north from $B$ to $C$ at the rate of $5 \mathrm{~km} / \mathrm{h}$. If $A B=12 \mathrm{~km}$ and $B C=5 \mathrm{~km}$, then its average speed for its journey from $A$ to $C$ and resultant average velocity direct from $A$ to $C$ are respectively

Options:
A
$\frac{17}{4} \mathrm{~km} / \mathrm{h}$ and $\frac{13}{4} \mathrm{~km} / \mathrm{h}$

B
$\frac{13}{4} \mathrm{~km} / \mathrm{h}$ and $\frac{17}{4} \mathrm{~km} / \mathrm{h}$

C
$\frac{17}{9} \mathrm{~km} / \mathrm{h}$ and $\frac{13}{9} \mathrm{~km} / \mathrm{h}$

D
$\frac{13}{9} \mathrm{~km} / \mathrm{h}$ and $\frac{17}{9} \mathrm{~km} / \mathrm{h}$

The correct answer is:
$\frac{17}{4} \mathrm{~km} / \mathrm{h}$ and $\frac{13}{4} \mathrm{~km} / \mathrm{h}$

Solution:
Time $T_1$ from $A$ to $B=\frac{12}{4}=3 \mathrm{hrs}$
$T_2$ from $B$ to $C=\frac{5}{5}=1 \mathrm{hrs}$.
Total time $=4 \mathrm{hrs}$.
Average speed $=\frac{17}{4} \mathrm{~km} / \mathrm{hr}$.
Resultant average velocity $=\frac{13}{4} \mathrm{~km} / \mathrm{hr}$.

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2004
A particle moves with a velocity $6 \hat{i}-4 \hat{j}+3 \hat{k} m / s$ under the influence of a constant force $\hat{F}=20 \hat{i}+15 \hat{j}-5 \hat{k} N$. The instantaneous power applied to the particle is

Options:
A $35 \mathrm{~J} / \mathrm{s}$
B $45 \mathrm{~J} / \mathrm{s}$
C $25 \mathrm{~J} / \mathrm{s}$
D $1.95 \mathrm{~J} / \mathrm{s}$

The correct answer is:
$45 \mathrm{~J} / \mathrm{s}$

Solution:
$P=\vec{F} \cdot \vec{v}=20 \times 6+15 \times(-4)+(-5) \times 3$
$=120-60-15=120-75=45 \mathrm{~J} / \mathrm{s}$

Topic: Physics, Mathematics in Physics, JEE Main
A particle moves with a velocity of 4i^â2j^+2k^ m s-1 under the influence of a constant force Fâ=10i^+3j^-2k^ N. The instantaneous power applied to the particle is

Options:

The correct answer is:
30

Solution:
P=F→⋅v→
=4i^-2j^+2k^·10i^+3j^-2k^ =40-6-4
∴P=30 W

Topic: Physics, Work Power Energy, JEE Main
A particle moves with constant acceleration along a straight line. If v1, v2 and v3 are the average velocities in the three successive intervals t1, t2 and t3 of time, then the correct relation is

Options:
A v1-v2v2-v3=t1-t2t2+t3
B v1-v2v2-v3=t1-t2t1- t3
C v1-v2v2-v3=t1-t2t2- t3
D v1-v2v2-v3=t1+t2t2+t3

The correct answer is:
v1-v2v2-v3=t1+t2t2+t3

Solution:
Let \(u\) be the initial velocity.
\(\therefore v_{1}=u+a t_{1}, v_{2}=u+d\left(t_{3}+t_{2}\right) \mathrm{and}\)
\(v_{3}=u+a\left(t_{1}+t_{3}+t_{3}\right)\)
Now, \(v_{1}=\frac{u w_{1}}{2}, \frac{u\left(u+a t_{1}\right)}{2}=u+\frac{1}{2}\) at,
\(v_{1}=\frac{v_{1}+c_{2}}{2}+u+a t_{1}+\frac{1}{2} a t_{2}\)
\(v_{3}=\frac{v_{2}+v_{2}}{2}=u+a t_{1}+a t_{2}+\frac{1}{2} a t_{3}\)
So, \(\left(y_{1}-v_{2}\right)=-\frac{1}{2} a\left(t_{3}+t_{2}\right)\)
and \(\left(v_{2}-v_{3}\right)=-\frac{1}{2}\) a \(\left(t_{2}+t_{3}\right)\)
\(\therefore \frac{\left(v_{1}-v_{2}\right)}{\left(v_{2}-v_{3}\right)}=\frac{\left(t_{1}+t_{2}\right)}{\left(t_{2}+t_{3}\right)}\)

Topic: Physics, Motion In One Dimension, JEE Main
A particle moves with constant angular velocity in a circle. During the motion its

Options:
A Energy is conserved
B Momentum is conserved
C Energy and momentum both are conserved
D None of the above is conserved

The correct answer is:
Energy is conserved

Solution:
In uniform circular motion (constant angular velocity) kinetic energy remains constant but due to change in velocity of particle its momentum varies. So energy is conserved.

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle moves with simple harmonic motion in a straight line. In first τ s , after starting from rest it travels a distance a, and in next τ s it travels 2a, in same direction, then :

Options:
A Amplitude of motion is 3a
B Time period of oscillations is 8 τ
C Amplitude of motion is 4a
D Time period of oscillations is 6 τ

The correct answer is:
Time period of oscillations is 6 τ

Solution:
x = - A cos ω t
- A + a = - A cos ω τ ...(i)
- A + 3 a = - A cos ω 2 τ ...(ii)
∴ - 1 + a A = - cos ω τ ...(iii)
and - 1 + 3 a A = - cos 2 ω τ ...(iv)
Adding (iii) and (iv)
2 = 3 cos ω τ - 2 cos 2 ω τ + 1
∴ 2 cos 2 ω τ - 3 cos ω τ + 1 = 0
∴ cos ω τ = 1 2
∴ ω τ = π 3
∴ T = 6 τ

Topic: Physics, Oscillations, JEE Main, JEE Main 2014 (06 Apr)
A particle moving along X-axis has acceleration f, at time t given byf=f01-tT, where f0 and T are constants. The particle at t=0 and the instant when f=0, the particle's velocity vx is

Options:
A f0T
B 12f0T2
C f0T2
D 12f0T

The correct answer is:
12f0T

Solution:
Acceleration

f=dvdt=f01+tT

or dv=f01-tT·dt ......... Eq(i)

Integrating Eq. (i) on both sides, we get

v=f0t-f0T·t22+C

After applying boundary conditions

v=0 at t=0

We get, C=0

⇒ v=f0t-f0T·t22 .......... Eq(ii)

As f=f01-tT

When f0=0,t=T

Substituting t=T in Eq. (ii), then velocity

vx=f0T-f0T·T22=12f0T

Topic: Physics, Motion In One Dimension, JEE Main
A particle moving in a circle of radius R with uniform speed takes time T to complete one revolution. If this particle is projected with the same speed at an angle θ to the horizontal, the maximum height attained by it is equal to 4R. The angle of projection θ is then given by :

Options:
A sin−12gT2π2R12
B sin−1π2R2gT212
C cos−12gT2π2R12
D cos−1πR2gT212

The correct answer is:
sin−12gT2π2R12

Solution:
Time period of circular motion is given by,

2πRv=T⇒v=2πRT

Now, maximum height H=v2sin2θ2g

⇒4R=4π2R2T2sin2θ2g

⇒sinθ=2gT2π2R

⇒θ=sin−12gT2π2R12

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2024 (01 Feb Shift 1)
A particle moving in a straight line covers half the distance with speed $6 \mathrm{~m} / \mathrm{s}$. The other half is covered in two equal time intervals with speeds $9 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$ respectively. The average speed of the particle during the motion is :

Options:
A $10 \mathrm{~m} / \mathrm{s}$
B $8 \mathrm{~m} / \mathrm{s}$
C $9.2 \mathrm{~m} / \mathrm{s}$
D $8.8 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$8 \mathrm{~m} / \mathrm{s}$

Solution:
$\begin{aligned} & \mathrm{BD} \Rightarrow \mathrm{S}=9 \mathrm{t}+15 \mathrm{t}=24 \mathrm{t} \\ & \mathrm{AB} \Rightarrow \mathrm{S}=6 \mathrm{t}_1=24 \mathrm{t} \Rightarrow \mathrm{t}_1=4 \mathrm{t} \\ & \begin{aligned} < \text { speed }> & =\frac{\text { dist. }}{\text { time }}=\frac{48 \mathrm{t}}{2 \mathrm{t}+\mathrm{t}_1} \\ & =\frac{48 \mathrm{t}}{2 \mathrm{t}+4 \mathrm{t}} \Rightarrow \frac{48 \mathrm{t}}{6 \mathrm{t}} \Rightarrow 8 \mathrm{~m} / \mathrm{s}\end{aligned}\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (09 Apr Shift 1)
A particle moving in the xy-plane experiences a velocity dependent force F→=kυyi^+υxj^, where υx and υy are the x and y components of its velocity υ→. If a→ is the acceleration of the particle, then which of the following statements is true for the particle ?

Options:
A quantity υ→×a→ is constant in time
B F→ arises due to a magnetic field
C kinetic energy of particle is constant in time
D quantity υ→.a→ is constant in time

The correct answer is:
quantity υ→×a→ is constant in time

Solution:
F→=kv→
⇒ F→‖v→
⇒ a→‖v→
⇒ v→×a→=0 always

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2020 (06 Sep Shift 2)
A particle moving with kinetic energy E has de Broglie wavelength λ . If energy ΔE is added to its energy, the wavelength become λ2 . Value of ΔE, is:

Options:
A E
B 4E
C 3E
D 2E

The correct answer is:
3E

Solution:
λ=h2KEm⇒λ∝1KE
λλ2=KEfKEi
4KEi=KEf
⇒ΔE=4KEi-KEi=3KE=3E

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2020 (09 Jan Shift 1)
A particle moving with uniform speed in a circular path maintains:

Options:
A Constant acceleration
B Constant velocity but varying acceleration
C Varying velocity and varying acceleration
D Constant velocity

The correct answer is:
Varying velocity and varying acceleration

Solution:
A particle moving with uniform speed in a circular path maintains varying velocity and varying acceleration. It is because direction of both velocity as well as acceleration will change continuously.

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2024
A particle moving with velocity v→=i^+3j^ produces an electric field E→=2k^ at some point. It will produce a magnetic field at that point, which is equal to

Options:
A 6i^-2j^c2
B 6i^+2j^c2
C zero
D data insufficient

The correct answer is:
6i^-2j^c2

Solution:
B→=μ04πqv→×r→r3⇒E→=14πε0qr→r3
B→=μ0ε0v→×E→=v→×E→c2
B→=6i^-2j^c2

Topic: Physics, Magnetic Effects of Current, JEE Main
A particle moving with velocity V→ is acted by three forces shown by the vector triangle PQR. The velocity of the particle will

Options:
A increase
B decrease
C remain constant
D change according to the smallest force QR→

The correct answer is:
remain constant

Solution:
We know that sum of vector in a loop arranged in tip-to tail manner is zero.

So, here the sum of forces arranged in tip-to-tail manner is zero.

Fnet→=0
Since, the sum of forces is zero.acceleration will be zero and Velocity will remain constant.

Topic: Physics, Mathematics in Physics, JEE Main
A particle moving with velocity V→ is acted by three forces shown by the vector triangle PQR. The velocity of the particle will

Options:
A increase
B decrease
C remain constant
D change according to the smallest force QR→

The correct answer is:
remain constant

Solution:
We know that sum of vector in a loop arranged in tip-to tail manner is zero.

So, here the sum of forces arranged in tip-to-tail manner is zero.

Fnet→=0
Since, the sum of forces is zero.acceleration will be zero and Velocity will remain constant.

Topic: Physics, Mathematics in Physics, NEET, NEET 2019 (All India)
A particle of charge $16 \times 10^{-16} \mathrm{C}$ moving with velocity $10 \mathrm{~ms}^{-1}$ along $x$-axis enters a region where magnetic field of induction
$\vec{B}$ is along the $y$-axis and an electric field of magnitude $10^4 \mathrm{Vm}^{-1}$ is along the negative $z$-axis.
If the charged particle continues moving along $x$-axis, the magnitude of $\vec{B}$ is :

Options:
A
$16 \times 10^3 \mathrm{~Wb} \mathrm{~m}^{-2}$

B
$2 \times 10^3 \mathrm{~Wb} \mathrm{~m}^{-2}$

C
$1 \times 10^3 \mathrm{~Wb} \mathrm{~m}^{-2}$

D
$4 \times 10^3 \mathrm{~Wb} \mathrm{~m}^{-2}$

The correct answer is:
$1 \times 10^3 \mathrm{~Wb} \mathrm{~m}^{-2}$

Solution:
Since particle is moving undeflected
So, $q_E=q v B$
$$
\Rightarrow B=\frac{E}{V}=\frac{10^4}{10}=10^3 \mathrm{wb} / \mathrm{m}^2
$$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2013 (23 Apr Online)
A particle of charge $-16 \times 10^{-18}$ coulomb moving with velocity $10 \mathrm{~ms}^{-1}$ along the $\mathrm{x}$-axis enters a region where a magnetic field of induction $\mathrm{B}$ is along the $\mathrm{y}$-axis, and an electric field of magnitude $10^4 \mathrm{~V} / \mathrm{m}$ is along the negative z-axis. If the charged particle continues moving along the $\mathrm{x}$-axis, the magnitude of $\mathrm{B}$ is

Options:
A
$10^3 \mathrm{~Wb} / \mathrm{m}^2$

B
$10^5 \mathrm{~Wb} / \mathrm{m}^2$

C
$10^{16} \mathrm{~Wb} / \mathrm{m}^2$

D
$10^{-3} \mathrm{~Wb} / \mathrm{m}^2$

The correct answer is:
$10^3 \mathrm{~Wb} / \mathrm{m}^2$

Solution:
Since there is no deviation in the path of the charged particle, so net force due to presence of electric and magnetic field must be zero $\Rightarrow \mathrm{vB}=\mathrm{qE} \Rightarrow \mathrm{B}=\frac{\mathrm{E}}{\mathrm{V}}=\frac{10^4}{10}=10^3 \mathrm{~Wb} / \mathrm{m}^2$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2003
A particle of charge equal to that of an electron, \( -e \), and mass \( 208 \) times the mass of electron (called \( \mu-\operatorname{meson} \) ) moves in a circular orbit around a nucleus of charge \( +3 \mathrm{e} \). (Take the mass of the nucleus to be infinite). Assuming that Bohr model of the atom is applicable to this system: Derive an expression for the radius of the \( \mathrm{n}^{\text {th }} \) Bohr orbit

Options:
A \( \frac{\varepsilon_{0} \mathrm{n}^{2} \mathrm{~h}^{2}}{208 \pi \mathrm{m}_{\mathrm{e}} \mathrm{e}^{2}} \)
B \( \frac{\varepsilon_{0} \mathrm{n}^{2} \mathrm{~h}^{2}}{3 \pi \mathrm{m}_{\mathrm{e}} \mathrm{e}^{2}} \)
C \( \frac{\varepsilon_{0} \mathrm{n}^{2} \mathrm{~h}^{2}}{624 \pi \mathrm{m}_{\mathrm{e}} \mathrm{e}^{2}} \)
D \( \frac{\varepsilon_{0} \mathrm{n}^{2} \mathrm{~h}^{2}}{64 \pi \mathrm{m}_{\mathrm{e}} \mathrm{e}^{2}} \)

The correct answer is:
\( \frac{\varepsilon_{0} \mathrm{n}^{2} \mathrm{~h}^{2}}{208 \pi \mathrm{m}_{\mathrm{e}} \mathrm{e}^{2}} \)

Solution:
Electrostatic force will provide required centripetal force for circular motion in bohr orbit.
rn=n2h2Ïµ0 Ï me2 Z
Put Z=3;m=208 me

Topic: Physics, Atomic Physics, JEE Main
A particle of charge per unit mass Î± is released from the origin with velocity v â = v 0 i ^ in the magnetic field
B â = - B 0 k ^ for x â¤ 3 2 v 0 B 0 Î±
and B â = 0 for x > 3 2 v 0 B 0 Î±
The x-coordinate of the particle at time t > Ï 3 B 0 Î± would be

Options:
A 3 2 v 0 B 0 α + 3 2 v 0 t - π B 0 α
B 3 2 v 0 B 0 α + v 0 t - π 3 B 0 α
C 3 2 v 0 B 0 α + v 0 2 t - π 3 B 0 α
D 3 2 v 0 B 0 α + v 0 t 2

The correct answer is:
3 2 v 0 B 0 α + v 0 2 t - π 3 B 0 α

Solution:
r = mv 0 B 0 q = v 0 B 0 α , x r = 3 2 = sin θ ⇒ θ = 6 0 â
t OA = T 6 = π 3 B 0 α

Therefore x-coordinate of particle at any time t > π 3 B 0 α will be
x = 3 2 v 0 B 0 α + v 0 t - π 3 B 0 α cos 6 0 â
= 3 2 v 0 B 0 α + v 0 2 t - π 3 B 0 α

Topic: Physics, Magnetic Effects of Current, JEE Main
A particle of charge q and mass m is moving with a velocity -vi^(v≠0) towards a large screen placed in the Y-Z plane at distance d. If there is magnetic field B→=B0k^, the minimum value of v for which the particle will not hit the screen is :

Options:
A qdB03m
B 2qdB0m
C qdB0m
D qdB02m

The correct answer is:
qdB0m

Solution:
The radius of circular path in given magnetic field is

r=d=mvqB0

so v=qB0dm

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (06 Sep Shift 1)
A particle of charge q and mass m is subjected to an electric field E=E01–ax2 in the x-direction, where a and E0 are constants. Initially the particle was at rest at x=0. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :

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

The correct answer is:
3a

Solution:
Wex=ΔK K1-K1=0

∫0xqEdx=0

q ∫0xE01-Qx2dx=0

qE0∫0x1-ax2dx=0

x-ax33=0

1-ax23=0

ax23=1

x2=3a

x=3a

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2020 (04 Sep Shift 2)
A particle of charge -q and mass m moves in a circle of radius r around an infinitely long line charge of linear density +λ. Then time period will be given as:
(Consider k as Coulomb's constant)

Options:
A T2=4π2 m2kλqr3
B T=2πrm2kλq
C T=12πrm2kλq
D T=12π2kλqm

The correct answer is:
T=2πrm2kλq

Solution:
Attractive electrostatic force due to line charge on the charged particle will provide the required centripetal force. Therefore,

q2kλr=mω2r

⇒ω2=2kλqmr2

⇒2πT2=2kλqmr2

⇒T=2πrm2kλq

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (30 Jan Shift 2)
A particle of mass $1 \times 10^{-26} \mathrm{~kg}$ and charge $1.6 \times 10^{-19} \mathrm{C}$ travelling with a velocity $1.28 \times 10^6 \mathrm{~ms}^{-1}$ along the positive $X$-axis enters a region in which a uniform electric field $\mathbf{E}$ and a uniform magnetic field of induction $\mathbf{B}$ are present. If $\mathbf{E}=-102.4 \times 10^3 \hat{\mathbf{k}} \mathrm{NC}^{-1}$ and $\mathbf{B}=8 \times 10^{-2} \hat{\mathbf{j}} \mathrm{Wbm}^{-2}$ the direction of motion of the particles is

Options:
A along the positive $X$-axis
B along the negative $X$-axis
C at $45^{\circ}$ to the positive $X$-axis
D at $135^{\circ}$ to the positive $X$-axis

The correct answer is:
along the positive $X$-axis

Solution:
$\begin{aligned}
m & =1 \times 10^{-26} \mathrm{~kg}, q=1.6 \times 10^{-19} \mathrm{C} \\
v & =1.28 \times 10^6 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Electric field, $\mathbf{E}=-102.4 \times 10^3 \hat{\mathbf{k}} \mathrm{N} / \mathrm{C}$
Magnetic field, $\mathbf{B}=8 \times 10^{-2} \hat{\mathbf{j}} \mathrm{Wbm}^{-2}$
$\begin{aligned}
\frac{|\mathbf{E}|}{|\mathbf{B}|} & =\frac{102.4 \times 10^3}{8 \times 10^{-2}} \\
& =\frac{10.24 \times 10^6}{8} \\
& =1.28 \times 10^6
\end{aligned}$
Hence,
$|\mathbf{v}|=\frac{|\mathbf{E}|}{|\mathbf{B}|}$
So, particle will remain undeflected, hence direction of motion of particle is along the positive $X$-axis.

Topic: Physics, Magnetic Effects of Current, AP EAMCET, AP EAMCET 2005
A particle of mass $1 \times 10^{-26} \mathrm{~kg}$ and charge $1.6 \times 10^{-19} \mathrm{C}$ travelling with a velocity $1.28 \times 10^6 \mathrm{~ms}^{-1}$ along the positive $X$-axis enters a region in which a uniform electric field $\mathbf{E}$ and a uniform magnetic field of induction $\mathbf{B}$ are present. If $\mathbf{E}=-102.4 \times 10^3 \hat{\mathbf{k}} \mathrm{NC}^{-1}$ and $\mathbf{B}=8 \times 10^{-2} \hat{\mathbf{j}} \mathrm{Wbm}^{-2}$ the direction of motion of the particles is

Options:
A along the positive $X$-axis
B along the negative $X$-axis
C at $45^{\circ}$ to the positive $X$-axis
D at $135^{\circ}$ to the positive $X$-axis

The correct answer is:
along the positive $X$-axis

Solution:
$\begin{aligned}
m & =1 \times 10^{-26} \mathrm{~kg}, q=1.6 \times 10^{-19} \mathrm{C} \\
v & =1.28 \times 10^6 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Electric field, $\mathbf{E}=-102.4 \times 10^3 \hat{\mathbf{k}} \mathrm{N} / \mathrm{C}$
Magnetic field, $\mathbf{B}=8 \times 10^{-2} \hat{\mathbf{j}} \mathrm{Wbm}^{-2}$
$\begin{aligned}
\frac{|\mathbf{E}|}{|\mathbf{B}|} & =\frac{102.4 \times 10^3}{8 \times 10^{-2}} \\
& =\frac{10.24 \times 10^6}{8} \\
& =1.28 \times 10^6
\end{aligned}$
Hence,
$|\mathbf{v}|=\frac{|\mathbf{E}|}{|\mathbf{B}|}$
So, particle will remain undeflected, hence direction of motion of particle is along the positive $X$-axis.

Topic: Physics, Magnetic Effects of Current, TS EAMCET, TS EAMCET 2005
A particle of mass 1 kg and charge 13 Î¼C is projected towards a non conducting fixed spherical shell having the same charge uniformly distributed on its surface. The minimum initial velocity V0 of projection of particle required if the particle just grazes the shell is

Options:
A 23 m s-1
B 2 23 m s-1
C 23 m s-1
D 1 m s-1

The correct answer is:
2 23 m s-1

Solution:
From conservation of angular momentum,
mv0r2=mvr ⇒v=v02
From conservation of energy,
12mv02=12mv2+14 π ϵ0q2r
v0=83 m s-1

Topic: Physics, Electrostatics, JEE Main
A particle of mass 1 kg is hanging from a spring of force constant 100 N m-1. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is Tn. The value of n is ________.

Options:

The correct answer is:
8

Solution:
KE=PE
12kA2-x2=12kx2
x2=A22
x=±A2
y=Asinωt

A2=Asinωt
sinπ4=sinωt
ωt=π4;t=T8. Thus n=8

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (31 Aug Shift 1)
A particle of mass $1 \mathrm{~kg}$ is projected with an initial velocity $10 \mathrm{~ms}^{-1}$ at an angle of projection $45^{\circ}$ with the horizontal. The average torque acting on the projectile, between the time at which it is projected and the time at which it strikes the ground, about the point of projection in newton-metre is

Options:
A $25$
B $50$
C $75$
D $100$

The correct answer is:
$50$

Solution:
$\begin{aligned} \tau=\frac{d L}{d t} & =m\left[u^2 \cos ^2 \theta\right] \\ & =(1)(10)^2 \cos ^2 45^{\circ}=50 \mathrm{Nm} .\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A particle of mass 1 mg and charge q is lying at the mid-point of two stationary particles kept at a distance 2 m when each is carrying same charge q. If the free charged particle is displaced from its equilibrium position through distance x x<<1 m. The particle executes SHM. Its angular frequency of oscillation will be _______ ×105 rad s-1 (if q2=10C2)

Options:

The correct answer is:
6000

Solution:
Net force on free charged particle,

F=kq2d+x2-kq2d-x2

F=-kq24dxd2-x22

a=-4kq2dmxd4

a=-4kq2md3x

So, angular frequency

ω=4 kq2md3

ω=4×9×109×101×10-6×13

ω=6×108 rad sec-1

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (25 Jul Shift 1)