datavorous/entrance-exam-dataset huggingface dataset:
A point source of heat of power P is placed at the centre of a thin spherical shell of mean radius R. The material of the shell has thermal conductivity K. Calculate the thickness of the shell if the temperature difference between the outer and inner surfaces of the shell in steady-state is T.

Options:
A 4πR2KT2P
B 3πR4KTP
C 4πR2KTP
D 2πR2KT3P

The correct answer is:
4πR2KTP

Solution:
Consider a concentric spherical shell of radius r and thickness dr as shown in diagram. The radial rate of flow of heat through this shell in steady state will be,

H=dQdt=-KAdθdr
[Negative sign is used because, with an increase in r, θ decreases.]
Now as for spherical shell A=4πr2
SoH=-4πr2Kdθdror∫abdrr2=-4πKH∫θ1θ2dθ
which on integration and simplification gives
H=dQdt=K4πabθ1-θ2b-a...(i)
Now in steady state, as no heat is absorbed, the rate of loss of heat by conduction is equal to that of supply, i.e., H = P, and here
θ1-θ2=T and a≃b=R
So Equation (i) becomes,
P=4πR2KTb-a
i.e thickness of shell is b-a=4πR2KTP

Topic: Physics, Thermal Properties of Matter, JEE Main
A point source of light is placed at a distance h below the surface of a large deep lake. Neglecting the partial reflection of light during refraction, the percentage of light energy that escapes from the water surface is [Î¼water=43]

Options:
A 50%
B 25%
C 20%
D 17%

The correct answer is:
17%

Solution:
Solid angle ϕ =2π 1−cosθ
where θ= plane angle
escape energy = total energy 4 π × solid angle
escape energy total energy = 1 4 π × 2 π 1 - cosC
= 1 2 1 - cosC ....(1)
from law of refraction
μ w sin  i= μ a  sin  r
4 3 × sin C = 1 × sin 90 o
sin C=34⇒cos C=1-sin2C⇒cos C=74....2
from (1) and (2)
μ w sin i= μ a sin r
escaped energy total energy = 0.17
total % energy enscaped =17%

Topic: Physics, Ray Optics, JEE Main
A point source of light is used in an experiment of photo-electric effects. If the distance between the source and the photo-electric surface is doubled, which of the following may result?

Options:
A Stopping potential will be halved
B photo-electric current will decrease
C Maximum kinetic energy if photo-electrons will decrease
D Stopping potential will increase slightly

The correct answers are:
photo-electric current will decrease

Solution:
Hint:
Intensity $\mathrm{I} \propto \frac{1}{\text { d }}$
and photo-electric current $\propto$ intensity
$\therefore$ Current will decrease

Topic: Physics, Dual Nature of Matter, JEE Main
A point source of light of power P0 is placed at a distance of 4 m from the centre of a thin hemispherical shell as shown in the figure, The shell has a radius of 3 m and it behaves like a perfect black body. If the temperature of the hemisphere is related to the power of the source as T4=P0nÏÏ' where Ï is the Stefan's constant, then find the value of n10.

Options:

The correct answer is:
27

Solution:
Total power emitted by shell due to radiation
Pout=σ2πR2+πR2T4=3πR2σT4
Power gained by the shell from P0 :
Pin=P02πr2(1-cos⁡37°)4πr2=P010
Equating: P010=3πR2σT4
Gives T4=P0270πσ

n=27010=27

Topic: Physics, Thermal Properties of Matter, JEE Main
A point source of light, S is placed at a distance L in front of the center of plane mirror of width d which is hanging vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror, at a distance 2L as shown below. The distance over which the man can see the image of the light source in the mirror is:

Options:
A 3d
B 2d
C d
D d2

The correct answer is:
3d

Solution:
By similar triangles
ΔAEC~ΔIHA
y2L=d2L (by properties of similar triangles)
⇒y=d
∴Total distance CD=y+d2+d2+y=3d

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (12 Jan Shift 1)
A point source of light S, placed at a distance 60 cm infront of the centre of a plane mirror of width 50 cm, hangs vertically on a wall. A man walks infront of the mirror along a line parallel to the mirror at a distance 1.2 m from it (see in the figure). The distance between the extreme points where he can see the image of the light source in the mirror is__cm

Options:

The correct answer is:
150

Solution:
tanθ=2560=x180

x=75 cm

so distance between extreme point =2x=2×75=150 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (26 Feb Shift 2)
A point source S emitting light of wavelength 600 nm is placed at a very small height h above a flat reflecting surface AB (see figure). The intensity of the reflected light is 36% of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D from it.

If the intensity at point P corresponds to a maximum, then the minimum distance through which the reflecting surface AB should be shifted so that the intensity at P again becomes maximum is given by:-

Options:
A 300 nm
B 200 nm
C 400 nm
D None of these

The correct answer is:
300 nm

Solution:
Initially, path difference at P between two waves reaching from S and S' is 2h
Therefore, for maximum intensity at P:
2h=n-12Î» ....... (i)
Now, let the sources S is moved by a distance x, the path difference will be
2 h + 2 x or 2 h - 2 x So, for displaced by x ( away or towards the mirror) then maximum intensity at P,
2h+2x=n+1-12Î» ....... (ii)
2h-2x=n-1-12Î» .......(iii)
Solving Eqs. (i) and (ii) or Eqs. (i) and (iii), we get,
2x=n+12Î»-n-1â¡2Î»
=Î»
x=Î»2
x=Î»2=6002=300 nm
Note Here, we have taken the condition of maximum intensity at P as:
Path difference Îx=n-12Î»
Because the reflected beam suffers a phase difference of Ï.

Topic: Physics, Wave Optics, JEE Main
A point source S is placed at the bottom of a 12 mm high transparent block of diamond(refractive index = 2.4). The block is immersed in an optically rarer liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 18 mm on the top of the block. What is the refractive index of the liquid?

Options:

The correct answer is:
01.80

Solution:
As the light is passing from optically denser medium to rarer medium, let us assume θ is the critical angle.

From Pythagoras theorem,

SB=SO2+OB2=122+92=15 mm

∴ sinθ=OBSB=912=34

sinθ=μ2μ1=μ2.4

∴ 34=μ2.4

⇒μ=1.8

Topic: Physics, Ray Optics, JEE Main
A point z moves in the complex plane such that argz-2z+2=π4, then the minimum value of |z-92-2i|2 is equal to

Options:

The correct answer is:
98

Solution:
Given that

argz-2z+2=π4

Let z = x + iy
⇒argx+iy-2x+iy+2=π4

⇒argx+iy-2x+iy+2x+2-iyx+2-iy=π4

⇒argx2+y2-4+4iy(x+2)2+y2=π4

⇒x2+y2-4+4iy(x+2)2+y2=tanπ4

⇒x2+y2-44y=1

x2+(y-2)2=8 ……….(1)

Equation (1) represents circle, whose centre (0,2), radius 22.

Minimum distance d=|A C-r|
Given point A(92,2)

=|AC-r|=∣(92-22)=72

⇒ Square of distance =d2=(72)2=98

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (31 Aug Shift 1)
A polariser is used to

Options:
A Reduce intensity of light
B Produce polarised light
C Increase intensity of light
D Produce unpolarised light

The correct answer is:
Produce polarised light

Solution:
Polariser produced prolarised light

Topic: Physics, Wave Optics, JEE Main
A polarizer - analyser set is adjusted such that the intensity of light coming out of the analyser is just 36% of the original intensity. Assuming that the polarizer - analyser set does not absorb any light, the angle by which the analyser needs to be rotated further, to reduce the output intensity to zero, is sin-135=37°

Options:
A 53o
B 37o
C 90o
D 45o

The correct answer is:
37o

Solution:
I=I0cos2θ
0.36I0=I0cos2θ
cosθ=35

⇒θ=53°

angle rotated should be =90o-53o=37o

Topic: Physics, Wave Optics, JEE Main, JEE Main 2020 (07 Jan Shift 1)
A polarizer - analyser set is adjusted such that the intensity of light coming out of the analyser is just 36% of the original intensity. Assuming that the polarizer - analyser set does not absorb any light, the angle by which the analyser needs to be rotated further, to reduce the output intensity to zero, is sin-135=37°

Options:
A 53o
B 37o
C 90o
D 45o

The correct answer is:
37o

Solution:
I=I0cos2θ
0.36I0=I0cos2θ
cosθ=35

⇒θ=53°

angle rotated should be =90o-53o=37o

Topic: Physics, Wave Optics, JEE Main
A polaroid is placed at $45^{\circ}$ to an incoming light of intensity $I_0$. Now the intensity of light passing through polaroid after polarisation would be

Options:
A $I_0$
B $I_0 / 2$
C $I_0 / 4$
D Zero

The correct answer is:
$I_0 / 2$

Solution:
$I=I_0 \cos ^2 \theta=I_0 \cos ^2 45=\frac{I_0}{2}$

Topic: Physics, Wave Optics, JEE Main
A pole is situated at the centre of a regular hexagonal park. The angle of elevation of the top of the vertical pole when observed from each vertex of the hexagon is Ï3. If the area of the circle circumscribing the hexagon is 27 m2, then the height of the tower is

Options:
A 33ÏÂ m
B 3ÏÂ m
C 3ÏÂ m
D 9Ï m

The correct answer is:
9Ï m

Solution:
Let a be the side of the regular hexagon
Now, for the equilateral triangle FOA, we have
OF=OA=AF=a
Hence, area of circle, πa2=27⇒a=33π meters ... 1
In right-angled triangle AOT, right angled at O, we have
tanπ3=OTOA⇒OT=OA tanπ3⇒OT=33π×3
∴OT=9π meters
∴ Height of the tower =9π meters.

Topic: Mathematics, Heights and Distances, JEE Main
A pole is vertically submerged in swimming pool, such that it gives a length of shadow 2.15 m within water when sunlight is incident at an angle of 30° with the surface of water. If swimming pool is filled to a height of 1.5 m, then the height of the pole above the water surface in centimeters is (nw=43) ________.

Options:

The correct answer is:
50

Solution:
Let angle BAC is θ.

Applying Snell's law at E, we get 1×sin60°=43×sinθ

⇒sinθ=338⇒tanθ=sinθ1-sinθ2=3337

Now, BC=1.5×tanθ=1.28

Then, AB=2.15-BC=0.86 m

Required value, DF=AB tan30°=0.5015 m≈50 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (06 Apr Shift 1)
A pole stands vertically in the center of a square. When 45o is the elevation of the sun, the tip of its shadow just reaches the side of the square and is at a distance of 30 meters and 40 meters from the ends of that side. The height of the pole is

Options:
A 50 meters
B 25 meters
C 252 meters
D 502 meters

The correct answer is:
252 meters

Solution:
Let ABCD is a square and OP is the pole.
Now, ∠PQO=45o
Also, DQ=30 meters
QA=40 meters
Side of square =30+40=70 meters
Now in ∆POQ,

∠Q=45o and ∠O=90o
∴∠P=45o
∴OP=OQ=h (let)
Now, let a perpendicular is drawn from O to AD which is OM
In ∆OMQ,
QM=DM-QD=702-30=5 meters
OM=AB2=702=35 meters
∴OQ=QM2+OM2=52+352
=51+72=252 meters

Topic: Mathematics, Heights and Distances, JEE Main
A police car moving at 30 m s-1, chases a motorcyclist. The policeman sounds his horn at 180 Hz, while both of them move towards a stationary siren of frequency 160 Hz. If it is given that the motorcyclist does not observe any beats then, calculate the speed (in m s-1) of the motorcyclist round off to two decimal places? [speed of sound =330 m s-1]

Options:

The correct answer is:
35.02

Solution:
Speed of sound vS=330 m s-1
Case (i),
f1=f0Vs-u0Vs-us
=f0330-u330-30=180 (330-u)300
f1=180 330-u300 ... (i)
Case (ii), f2=160 330+u330 ... (ii)
∵ There are no beats
∴ f1=f2
180=330-u300=160330+u330
⇒ 9330-u30=8330-u33
⟹ 910330-u=811330+u
⟹32670-99u=26400+80u
∴ u=6270179=35.02 m s-1

Topic: Physics, Waves and Sound, JEE Main
A police jeep is chasing with, velocity of $45 \mathrm{~km} / \mathrm{h}$ a thief in another jeep moving with velocity $153 \mathrm{~km} / \mathrm{h}$. Police fires a bullet with muzzle velocity of $180 \mathrm{~m} / \mathrm{s}$. The velocity it will strike the car of the thief is

Options:
A 150 m/s
B 27 m/s
C 450 m/s
D 250 m/s

The correct answer is:
150 m/s

Solution:
Effective speed of the bullet
$=$ speed of bullet + speed of police jeep
$=180 \mathrm{~m} / \mathrm{s}+45 \mathrm{~km} / \mathrm{h}=(180+12.5) \mathrm{m} / \mathrm{s}=192.5 \mathrm{~m} / \mathrm{s}$
Speed of thief's jeep $=153 \mathrm{~km} / \mathrm{h}=42.5 \mathrm{~m} / \mathrm{s}$
Velocity of bullet w.r.t thief 's car $=192.5-42.5=150 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A police party is moving in a jeep at a constant speed $v$. They saw a thief at a distance $x$ on a motorcycle which is at rest. The moment the police saw the thief, the thief started at constant acceleration $a$. Which of the following relations is true if the police is able to catch the thief?

Options:
A $v^2 < a x$
B $v^2 < 2 a x$
C $v^2 \geq 2 a x$
D $v^2=a x$

The correct answer is:
$v^2 \geq 2 a x$

Solution:
Let the police party catch the thief after $t$ second.
$\therefore$ Distance travelled by police party in $t$ second $=v t$
and distance travelled by thief $=x+\frac{1}{2} \alpha t^2$
So, $\quad x+\frac{1}{2} \alpha t^2 \leq v t$
or $\quad \frac{\alpha t^2}{2}-v t+x=0$
or $\quad t=\frac{v \pm \sqrt{v^2-2 \alpha x}}{\alpha}$
For $t$ to be real, $v^2 \geq 2 \alpha x$

Topic: Physics, Motion In One Dimension, JEE Main
A polished metal plate with a rough black spot on it is heated to about 1400 K and quickly taken to a dark room. The spot will appear

Options:
A Darker than plate
B Brighter than plate
C Equally bright
D Equally dark

The correct answer is:
Brighter than plate

Solution:
On heating, black spot absorbs maximum radiation. Therefore, when plate is taken to a dark room the spot will emit more radiations than the rest of the plate. Hence, it will appear brighter than the plate.

Topic: Physics, Thermal Properties of Matter, JEE Main
A polyatomic gas has $f$ vibrational degrees of freedom, then the ratio of the specific heat at constant pressure to that at constant volume will be

Options:
A $\frac{4+f}{3+f}$
B $\frac{4-f}{3-f}$
C $\frac{3+f}{4+f}$
D $\frac{3-f}{4-f}$

The correct answer is:
$\frac{4+f}{3+f}$

Solution:
For a polyatomic atom value of specific heat at constant volume, $C_V=(3+f) R$ where, $f=$ vibration degree of freedom and $R=$ gas constant.
Now,
$$
\begin{aligned}
& C_p=C_V+R \\
& C_p=(3+f) R+R \\
& C_p=(4+f) R
\end{aligned}
$$
Now, ratio of $C_p$ and $C_V$
$$
\frac{C_p}{C_V}=\frac{4+f}{3+f}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A polyatomic gas with $n$ degrees of freedom has a mean energy per molecule given by

Options:
A $\frac{n k T}{N}$
B $\frac{n k T}{2 N}$
C $\frac{n k T}{2}$
D $\frac{3 k T}{2}$
( $N$ is Avogadro's number)

The correct answer is:
$\frac{n k T}{2}$

Solution:
Mean kinetic energy per molecule $E=\frac{f}{2} k T=\frac{n}{2} k T$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A polyatomic ideal gas has 24 vibrational modes. What is the value of γ?

Options:
A 1.03
B 1.30
C 1.37
D 10.3

The correct answer is:
1.03

Solution:
Since each vibrational mode has 2 degrees of freedom hence total vibrational degrees of freedom =48

f=3+3+48=54

γ=1+2f=2827=1.03

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (17 Mar Shift 1)
A polymer containing nitrogen is

Options:
A Bakelite
B Dacron
C Rubber
D Nylon-6,6

The correct answer is:
Nylon-6,6

Solution:
Nylon-6,6 is polymer of COOH -(CH2)4 -COOH
Adipic acid and H2N -(CH2)6 -NH2 (Hexamethylenediamine)
â´ Nylon-6,6 has nitrogen in it.

Topic: Chemistry, Polymers, JEE Main
A population that has not reached the carrying capacity is likely to:

Options:
A Decline in number
B Grow exponentially
C Remain stable in number
D Grow but not at an exponential rate

The correct answer is:
Grow exponentially

Solution:
Carrying capacity refers to the average population density or size of a species that can survive without depleting the life-sustaining resources. It is different for each species on account of their food, social and shelter requirements which is not uniform across. Survival of species is possible only until it reaches its carrying capacity of the environment. At this point, there will be a severe shortage of resources and thus will result in death.

Topic: Biology, Organisms and Populations, NEET
A porter lifts a heavy suitcase of mass 80 kg and at the destination lowers it down by a distance of 80 cm with a constant velocity. Calculate the work done by the porter in lowering the suitcase. (take g=9.8 ms-2 )

Options:
A -62720.0 J
B -627.2 J
C +627.2 J
D 784.0 J

The correct answer is:
-627.2 J

Solution:
WPorter +Wmg=ΔK.E.=0
WPorter =-Wmg=-mgh
=-80×9.8×.8=-627.2 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (22 Jul Shift 1)
A positive acute angle is divided into two parts whose tangents are $\frac{1}{2}$ and $\frac{1}{3}$. Then the angle is

Options:
A $\pi / 4$
B $\pi / 5$
C $\pi / 3$
D $\pi / 6$

The correct answer is:
$\pi / 4$

Solution:
Hints : Angle $\theta=\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \cdot \frac{1}{3}}\right)$
$=\tan ^{-1}\left(\frac{5 / 6}{5 / 6}\right)=\tan ^{-1}(1)=\pi / 4$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
A positive carbylamine test is given by

Options:
A $N, N$-dimethylaniline
B 2, 4-dimethylaniline
C N-methyl-o-methylaniline
D p-methylbenzylamine

The correct answer is:
2, 4-dimethylaniline

Solution:

Topic: Chemistry, Amines, JEE Main
A positive carbylamines test is given by

Options:
A N, N-dimethylaniline
B \( 2,4 \)-dimethylaniline
C N-methyl- \( o \)-methylaniline
D N-methylbenzylamine

The correct answer is:
\( 2,4 \)-dimethylaniline

Solution:
Only primary amines give positive carbylamine test.

Both aliphatic and aromatic primary amines on heating with chloroform and alcoholic KOH gives foul smelling isocyanides which are also known as carbylamines.

R-NH2+CHCl3+3KOHalc.→∆ R-N≡C+3KCl+3H2O

Topic: Chemistry, Amines, JEE Main
A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1×105 N C-1. If the charge on the particle is 40 μC and the initial velocity is 200 m s-1, how much distance it will travel before coming to the rest momentarily

Options:
A 0.5 m
B 1 m
C 5 m
D 10 m

The correct answer is:
0.5 m

Solution:
Force experienced by a particle with charge q in an electric field F=qE.

Now, the acceleration produced is given as a=Fm=qEm=40×10-6×1051×10-4

=4×104 m s-2 (As the particle is projected against the electric field, hence it is decelerated)

Using, v2=u2-2as

We have, 02=u2-2as

s=v22a=20022×4×104=0.5 m

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (29 Jun Shift 1)
A positive charge $\mathrm{Q}$ is uniformly distributed along a circular ring of radius $\mathrm{R}$. A small test charge $\mathrm{q}$ is placed at the centre of the ring figure. Then,

Options:
A
if $\mathrm{q}>0$ and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre

B
if $\mathrm{q} < 0$ and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring
C
if $q < 0$, it will perform $S H M$ for small displacement along the axis
D
$q$ at the centre of the ring is in an unstable equilibrium within the plane of the ring for $\mathrm{q}>0$

The correct answers are:
if $\mathrm{q}>0$ and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre
,
if $\mathrm{q} < 0$ and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring,
if $q < 0$, it will perform $S H M$ for small displacement along the axis

Solution:
The main concepts are used
(i) Interaction between charges.
(ii) Stable and unstable equilibrium.
The positive charge $\mathrm{Q}$ is uniformly distributed at the-outer surface of the ring. Thus, electric field inside the sphere is zero.
Then the positive charge of the ring will interact equally a charge placed at centre of ring will be in unstable equilibrium.
So the effect of electric field on charge q due to the positive charge $Q$ is zero.
Now, there are two forces are active between charges $(Q$ and q).
(i) attractive force. (ii) repulsive force.
There are two cases obtained :
Case I : When charge $q>0$ i.e., $q$ is a positive charge, there creates a repulsive force between charge $q$ and $Q$.
The repulsive forces of charge $Q$ from all around the charge $q$ will push it towards.ihe centre if it is displaced from the centre of the ring.
Case II : When charge $q < 0$ i.e., $q$ is a negative charge then there is an attractive force between charge $Q$ and $q$.
If $\mathrm{q}$ is displaced slightly small from the centre then the positive charges nearer to this charge will attract it towards itself and charge q will never return to the centre.

Topic: Physics, Electrostatics
A positive charge ' $\mathrm{q}$ ' of mass ' $\mathrm{m}$ ' is moving along the $+x$ axis. We wish to apply a uniform magnetic field B for time $\Delta t$ so that the charge reverses its direction crossing the $\mathrm{y}$ axis at a distance $\mathrm{d}$. Then:

Options:
A
$\mathrm{B}=\frac{\mathrm{mv}}{\mathrm{qd}}$ and $\Delta \mathrm{t}=\frac{\pi \mathrm{d}}{\mathrm{v}}$

B
$\mathrm{B}=\frac{\mathrm{mv}}{2 \mathrm{qd}}$ and $\Delta \mathrm{t}=\frac{\pi \mathrm{d}}{2 \mathrm{v}}$

C
$\mathrm{B}=\frac{2 \mathrm{mv}}{\mathrm{qd}}$ and $\Delta \mathrm{t}=\frac{\pi \mathrm{d}}{2 \mathrm{v}}$

D
$B=\frac{2 \mathrm{mv}}{\mathrm{qd}}$ and $\Delta \mathrm{t}=\frac{\pi \mathrm{d}}{\mathrm{v}}$

The correct answer is:
$\mathrm{B}=\frac{2 \mathrm{mv}}{\mathrm{qd}}$ and $\Delta \mathrm{t}=\frac{\pi \mathrm{d}}{2 \mathrm{v}}$

Solution:
The applied magnetic field provides the required centripetal force to the charge particle, so it can move in circular path of radius $\frac{\mathrm{d}}{2}$
$$
\therefore \quad B q v=\frac{m^2}{d / 2}
$$
or, $\quad B=\frac{2 \mathrm{mv}}{\mathrm{qd}}$
Time interval for which a uniform magnetic field is applied
$$
\Delta \mathrm{t}=\frac{\pi \cdot \frac{\mathrm{d}}{2}}{\mathrm{v}}
$$
(particle reverses its direction after time $\Delta \mathrm{t}$ by covering semi circle).
$$
\Delta \mathrm{t}=\frac{\pi \mathrm{d}}{2 \mathrm{v}}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2014 (12 Apr Online)
A positive point charge is placed at P in front of an earthed metal sheet S. Q and R are two points between P and S as shown in the figure. If the electric field strength at Q and R are respectively EQ, ER and potential at Q and R are respectively VQ, VR. Then

Options:
A EQ= ER
B EQ<Â ER
C VQ>VR
D VQ<Â VR

The correct answer is:
VQ>VR

Solution:
As we are are moving away from P toward sheet S spacing between electric lines of force is increasing. â´ER<EQ In direction of electric field potential decreases. â´VRâ<âVQ

Topic: Physics, Electrostatics, JEE Main
A positive point charge is released from rest at a distance r0 from a positive line charge with uniform charge density. The speed (v) of the point charge, as a function of instantaneous distance r from line charge, is proportional to

Options:
A v∝rr0
B v∝ln⁡rr0
C v∝ln⁡rr0
D v∝e+rr0

The correct answer is:
v∝ln⁡rr0

Solution:
12mv2=q∆V
12mv2=q×λ2πε0ln⁡rr0
v2=qλmπε0ln⁡rr0
v=qλmπε0ln⁡rr0

v∝ln⁡rr0

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (08 Apr Shift 2)
A positive point charge Q is brought near an isolated uncharged metal cube

Options:
A the cube becomes negatively charged.
B the cube becomes positively charged.
C the interior becomes positively charged and the surface becomes negatively charged.
D the cube remains neutral.

The correct answer is:
the cube remains neutral.

Solution:
Charge on an isolated cube remains conserved so it remains neutral.

Topic: Physics, Electrostatics, NEET