datavorous/entrance-exam-dataset huggingface dataset:
Two plates A and B have thermal conductivities 84 W m-1 K-1 and 126 W m-1 K-1 respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of A and B are kept at 100∘C and 0∘C respectively, then the temperature of the surface of contact in steady state is_____ C∘.

Options:

The correct answer is:
40

Solution:
Let the temperature of the contact surface be T,

So, HA=HB

KAA'(TA-T)L=KBA'(T-TB)L⇒84(100-T)=126(T-0)⇒2(100-T)=3T⇒T=40 °C

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Two plates identical in size, one of black and rough surface (B1) and the other smooth and polished (A2) are interconnected by a thin horizontal pipe with a mercury pellet at the centre. Two more plates A1 (identical to A2 ) and B2 (identical to B1) are heated to the same temperature and placed closed to the plates B1, and A2 as shown in the diagram. The mercury pellet

Options:
A Moves to the right
B Moves to the left
C Remains stationary
D Starts oscillating left and right

The correct answer is:
Remains stationary

Solution:
Smooth and polished plates are poor radiators of heat. Hence, heat coming out from A is small, even though B being a black and rough plate is a good absorber. Effectively the heat coming to the left of pellet P is small.
Black and rough plates are good radiators of heat. Hence, plate B2 radiates heat to a satisfactory level; however, plate A2, being smooth and polisher, is a bad absorber. Effectively, the heat coming to the right of P is also small.

Topic: Physics, Thermal Properties of Matter, NEET
Two plates of same area are placed in contact. Their thickness as well as their thermal conductivities are in the ratio $2: 3$. The outer surface of one plate is maintained at $10^{\circ} \mathrm{C}$ and that of the other at $0^{\circ} \mathrm{C}$. Temperature at the common surface is

Options:
A $0^{\circ} \mathrm{C}$
B $25^{\circ} \mathrm{C}$
C $5^{\circ} \mathrm{C}$
D $6.5^{\circ} \mathrm{C}$

The correct answer is:
$5^{\circ} \mathrm{C}$

Solution:
Given that, two plates are having same surface area $(A)$.
Let $K_1, K_2$ and $t_1, t_2$ be their conductivities and thickness respectively,
According to question,
$$
\frac{K_1}{K_2}=\frac{t_1}{t_2}=\frac{2}{3}...(i)
$$
Since, the rate of flow of heat through both plates be same, then
$$
\begin{aligned}
& \frac{\Delta Q}{\Delta t}=\frac{K_1 A \Delta T_1}{t_1}=\frac{K_2 A \Delta T_2}{t_2} \\
& \Rightarrow \frac{K_1}{K_2} \times \frac{t_2}{t_1} \times\left(T_1-T\right)=\left(T-T_2\right)
\end{aligned}
$$
[where $T$ be the temperature of common surface] Substituting the values, we get
$$
\begin{aligned}
\frac{2}{3} \times \frac{3}{2}(10-T) & =(T-0) \\
T & =5^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2021 (24 Aug Shift 1)
Two plots are shown below between concentration and time t. Which of the given orders are shown by the graph respectively?

Options:
A Zero order and first order
B First order and second order
C Zero order and second order
D First order and first order

The correct answer is:
Zero order and second order

Solution:
Linear plots are obtained in the graph of a-x vs t for zero order reaction and 1a-x vs t for second order reaction.

Topic: Chemistry, Chemical Kinetics, JEE Main
Two point charge \(\mathrm{Q}\) and \(-3 \mathrm{Q}\) are placed at some distance apart. If the electric field at the location of \(Q\) is \(\vec{E}\), the field at the location of \(-3 Q\) is:

Options:
A \(\vec{E}\)
B \(-\vec{E}\)
C \(+\frac{\vec{E}}{3}\)
D \(-\frac{\vec{E}}{3}\)

The correct answer is:
\(+\frac{\vec{E}}{3}\)

Solution:
Given Electric field at \(\mathrm{A}=\overrightarrow{\mathrm{E}}\)
\(\overrightarrow{\mathrm{E}}:\) Magnitude & direction
Magnitude : \(\mathrm{E}\)
direction : \((\mathrm{x}-\) axis)
Now, \(\mathrm{E}=\frac{\mathrm{K}(3 \mathrm{Q})}{\mathrm{r}^2}\{\) Electric field due to charge at \(\mathrm{B}\}\)
or \(\frac{\mathrm{KQ}}{\mathrm{r}^2}=\mathrm{E}^1\left\{\mathrm{E}^1\right.\) : Electric field magnitude at \(\mathrm{B}\) due to \(\left.\mathrm{A}\right\}\)
\(\therefore \mathrm{E}^1=\frac{\mathrm{E}}{3}\)
for direction at \(\mathrm{E}^1, \therefore \mathrm{Q}\) is a + ve charge particle & electric field will be along \(+\mathrm{x}-\) axis,
\(\therefore \overrightarrow{\mathrm{E}}=+\frac{\overrightarrow{\mathrm{E}}}{3}\) {or in direction of \(\overrightarrow{\mathrm{E}}\) itself}

Topic: Physics, Electrostatics, JEE Main
Two point charges $+8 q$ and $-2 q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the $x$ axis at which the net electric field due to these two point charges is zero is

Options:
A
$2 \mathrm{~L}$

B
$L / 4$

C
$8 \mathrm{~L}$

D
$4 \mathrm{~L}$

The correct answer is:
$2 \mathrm{~L}$

Solution:
$-\frac{\mathrm{k} 2 q}{(\mathrm{x}-\mathrm{L})^2}+\frac{\mathrm{k} 8 \mathrm{q}}{\mathrm{x}^2}=0$
$\Rightarrow \mathrm{x}=2 \mathrm{~L}$

Topic: Physics, Electrostatics, JEE Main
Two point charges $+8 q$ and $-2 q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the $x$ axis at which the net electric field due to these two point charges is zero is

Options:
A
$2 \mathrm{~L}$

B
$L / 4$

C
$8 \mathrm{~L}$

D
$4 \mathrm{~L}$

The correct answer is:
$2 \mathrm{~L}$

Solution:
$-\frac{\mathrm{k} 2 q}{(\mathrm{x}-\mathrm{L})^2}+\frac{\mathrm{k} 8 \mathrm{q}}{\mathrm{x}^2}=0$
$\Rightarrow \mathrm{x}=2 \mathrm{~L}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2005
Two point charges A and B of magnitude +8×10-6C and -8×10-6C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4×104 N C-1. The distance 'd' between the point charges A and B is

Options:
A 2 m
B 3 m
C 1 m
D 4 m

The correct answer is:
3 m

Solution:
Direction of electric field at mid-point due to both charge will be in same direction, which is towards right.

Therefore, E at mid-point =kqd22+kqd22

⇒6.4×104=2×9×109×8×10-6d2×4

⇒d=3 m

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Two point charges of \( 1 \mu \mathrm{C} \) and \( -1 \mu \mathrm{C} \) are separated by a distance of \( 100 Å \). A point \( P \) is at a distance of \( 10 \mathrm{~cm} \) from the mid-point and on the perpendicular bisector of the line joining the two charges. The electric field at \( P \) will be

Options:
A \( 9 \mathrm{~N} \mathrm{C}^{-1} \)
B \( 0.9 \mathrm{~V} \mathrm{~m}^{-1} \)
C \( 90 \mathrm{~V} \mathrm{~m}^{-1} \)
D \( 0.09 \mathrm{~N} \mathrm{C}^{-1} \)

The correct answer is:
\( 0.09 \mathrm{~N} \mathrm{C}^{-1} \)

Solution:
The point lies on the equatorial line of a short dipole.
∴ E=p4πε0r3=9×10910-6×10-810-13
=9×10-2=0.09 N C-1.

Topic: Physics, Electrostatics, JEE Main
Two point charges of $20 \mu \mathrm{C}$ and $80 \mu \mathrm{C}$ are $10 \mathrm{~cm}$ apart. Where will the electric field strength be zero on the line joining the charges from $20 \mu \mathrm{C}$ charge

Options:
A $0.1 \mathrm{~m}$
B $0.04 \mathrm{~m}$
C $0.033 \mathrm{~m}$
D $0.33 \mathrm{~m}$

The correct answer is:
$0.033 \mathrm{~m}$

Solution:
Let neutral point be obtained at a distance $x$ from $20 \mu \mathrm{C}$ charge. Hence at neutral point
$\frac{20}{(x)^2}=\frac{80}{(10-x)^2} \Rightarrow x=+0.033 \mathrm{~m}$

Topic: Physics, Electrostatics, JEE Main
Two point charges q110 μC and q2-25 μC are placed on the x -axis at x=1 m and x=4 m respectively. The electric field in V/m at a point y=3 m on y-axis is,
Take 14πϵ0=9×109 N m2C-2

Options:
A -81 i^+81 j^×102
B 81 i^-81 j^×102
C -63 i^+27 j^×102
D 63i^-27j^×102

The correct answer is:
63i^-27j^×102

Solution:
Electric field due to 10 μC
E→1=14π∈010×10-6r13 r→1
=14π∈010×10-6103-i^+3j^
Similarly, electric field due to -25 μC
E→2=14π∈025×10-6534i^-3j^
Net electric field,
E→=9×109×10-6-i^10+3j^10+45i^-35j^
=9×1037i^10-3j^10
=63i^-27j^×102 NC

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Two point charges $+\mathrm{q}_{1}$ and $+\mathrm{q}_{2}$ are placed a finite distance ' $\mathrm{d}$ ' apart. It is desired to put a third charge $\mathrm{q}_{3}$ in between these two charges so that $\mathrm{q}_{3}$ is in equilibrium. This is

Options:
A possible only if $q_{3}$ is negative.
B possible only if $\mathrm{q}_{3}$ is positive.
C possible irrespective of the sign of $q_{3}$.
D not possible at all.

The correct answer is:
possible irrespective of the sign of $q_{3}$.

Solution:
At null point $\vec{E}=0$
$\therefore$ any charge placed at that point will remain in equilibrium.

Topic: Physics, Electrostatics, WBJEE, WBJEE 2021
Two point charges +Q and -Q having masses M and 2M respectively, are projected from the same point with the same velocity v in the region of a uniform magnetic field B such that the velocity makes an angle Î¸ with the magnetic field. If we neglect the mutual interaction between the particles, the distance from the point of projection where they will meet again for the first time is

Options:
A 2πMvcosθQB
B 8πMvcosθQB
C πMvcosθQB
D 4ÏMvcosÎ¸QB

The correct answer is:
4πMvcosθQB

Solution:
Both the charged particle will follow the helical path and the particles will meet at the intersection of the two helix such that the pitch of one helix will be the integral multiple of the pitch of the other helix.
T1=2πMQB
T2=2π2MQB
T2=4πMQB
They will meet again when the first particle completes two revolutions and the second particle completes one revolution so distance will be given as
d=vcosθt=vcosθ.22πMQB=4πMvcosθQB

Topic: Physics, Magnetic Effects of Current, JEE Main
Two point charges +Q and -Q having masses M and 2M respectively, are projected from the same point with the same velocity v in the region of a uniform magnetic field B such that the velocity makes an angle θ with the magnetic field. If we neglect the mutual interaction between the particles, the distance from the point of projection where they will meet again for the first time is

Options:
A 2πMvcosθQB
B 8πMvcosθQB
C πMvcosθQB
D 4ÏMvcosÎ¸QB

The correct answer is:
4πMvcosθQB

Solution:
Here path will be helical
T1=2πMQB
T2=2π2MQB
T2=4πMQB
They will meet again when the first particle completes two revolutions and the second particle completes one revolution so distance will be given as
d=vcosθt=vcosθ.22πMQB=4πMvcosθQB

Topic: Physics, Magnetic Effects of Current, JEE Main
Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is:

Options:
A d
B d2
C d2
D d22

The correct answer is:
d22

Solution:
The distance between q and Q is r=x2+d22=x2+d24.

The force between q and Q will be F=kqQr2.

From the given figure we can see that only vertical components of the forces will add, and the horizontal components will get cancelled out. Therefore,

Fnet=2kqQr2cosθ=2kqQr2×xr=kqQxx2+d2432

For force to be maximum, dFnetdx=0.

⇒kqQx2+d2432-x×32x2+d2412×2xx2+d24322=0

⇒x2+d24=3x2⇒2x2=d24⇒x=d22

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Two point dipoles of dipole moment $\vec{p}_1$ and $\vec{p}_2$ are at a distance $x$ from each other and $\vec{p}_1 || \vec{p}_2$.
The force between the dipoles is :

Options:
A
$\frac{1}{4 \pi \varepsilon_0} \frac{4 p_1 p_2}{x^4}$

B
$\frac{1}{4 \pi \varepsilon_0} \frac{3 p_1 p_2}{x^3}$

C
$\frac{1}{4 \pi \varepsilon_0} \frac{6 p_1 p_2}{x^4}$

D
$\frac{1}{4 \pi \varepsilon_0} \frac{8 p_1 p_2}{x^4}$

The correct answer is:
$\frac{1}{4 \pi \varepsilon_0} \frac{3 p_1 p_2}{x^3}$

Solution:
Force of interaction
$\mathrm{F}=\frac{1}{4 \pi \epsilon_0} \cdot \frac{3 \mathrm{p}_1 \mathrm{p}_2}{\mathrm{r}^4}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2013 (09 Apr Online)
Two point masses connected by an ideal string are placed on a smooth horizontal surface as shown in the diagram. A sharp impulse of 10 kg m s-1 is given to the 5 kg mass at an angle of 60° to the line joining the masses. The velocity of the 10 kg mass just after the impulse will be

Options:
A 23 m s-1
B 13 m s-1
C 2 m s-1
D zero

The correct answer is:
13 m s-1

Solution:
Apply impulse equation, along string
Impulse = Îp
10 Ã 12=10 + 5v
v=13m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two point masses M are kept fixed on the x-axis at a distance a from the origin, another point mass m is moving in a circular path of radius R (in y-z plane) under the influence of gravitational force of attraction, then speed of m will be
(Assume no forces are acting on m other than the gravitational forces by two M)

Options:
A 2GMR2R2+a232
B GMR2R2+a232
C 3GMR2R2+a232
D 5GMR2R2+a232

The correct answer is:
2GMR2R2+a232

Solution:
FR=2Fsin⁡θ
mv2R=2GMmr2Rr
r=a2+R2
⇒ v=2GMR2R2+a232

Topic: Physics, Gravitation, JEE Main
Two point masses, m each carrying charges -q and +q are attached to the ends of a massless rigid non-conducting wire of length L. When this arrangement is placed in a uniform electric field, then it deflects through an angle Î¸. The minimum time needed by the rod to align itself along the field is

Options:
A 2πmLqE
B π2mL2qE
C π2mLqE
D 2π3mLqE

The correct answer is:
π2mL2qE

Solution:
When the wire is brought in a uniform field E, then the torque is given by
Ï=qELsinâ¡Î¸
=qELÎ¸ âµ Î¸ is very small
The moment of inertia of rod AB about O is
I=mL22+mL22=mL22
As Ï=IÎ±
So, Î±=ÏI=2qELÎ¸mL2
â Ï2Î¸=2qELÎ¸mL2 âµ Î±=Ï2Î¸
â Ï2=2qEmL
The time period of the wire is
T=2ÏÏ=2ÏmL22qE
The rod will become parallel to the field in time T4
So, t=T4=Ï2mL2qE

Topic: Physics, Magnetic Effects of Current, JEE Main
Two point masses of mass $m_1=f M$ and $m_2=(1-f) M(f < 1)$ are in outer space (far from gravitational influence of other objects) at a distance $R$ from each other. They move in circular orbits about their centre of mass with angular velocities $\omega_1$ for $m_1$ and $\omega_2$ for $m_2$. In that case

Options:
A
$(1-f) \omega_1=f \omega$

B
$\omega_1=\omega_2$ and independent of $f$

C
$f \omega_1=(1-f) \omega_2$

D
$\omega_1=\omega_2$ and depend on $f$

The correct answer is:
$\omega_1=\omega_2$ and independent of $f$

Solution:
Angular velocity is the angular displacement per unit time i.e., $\omega=\frac{\Delta \theta}{\Delta t}$ Here $w_1=w_2$ and independent of $f$.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2012 (19 May Online)
Two point sources $S_1$ and $S_2$ separated by a distance $10 \mu \mathrm{m}$ emit light waves of wavelength $4 \mu \mathrm{m}$ in phase. A circular wire of radius $40 \mu \mathrm{m}$ is placed around the sources as shown in figure, then ( $O$ is the centre of the circle and $O S_1=O S_2$ )

Options:
A points $A$ and $B$ are dark and points $C$ and $D$ are bright
B points $A$ and $B$ are bright and point $C$ and $D$ are dark
C points $A$ and $C$ are dark and points $B$ and $D$ are bright
D points $A$ and $C$ are bright and points $B$ and $D$ are dark

The correct answer is:
points $A$ and $C$ are dark and points $B$ and $D$ are bright

Solution:
Path difference at point $B$,
$$
\Delta x_B=S_1 B-S_2 B=0
$$
and at point $D$,
$$
\Delta x_D=S_1 D-S_2 D=0
$$
Phase difference,
$$
\phi=\frac{2 \pi}{\lambda} \times(0)=0
$$
So, intensity at points $B$ and $D$
$$
I=I_{\max } \cos ^2(\phi / 2)=I_{\max }
$$
So, points $B$ and $D$ are bright.
Path difference at point $A$,
$$
\Delta x_A=35 \mu \mathrm{m}
$$
$$
[40-5]
$$
But, figure shows $\Delta x_A=\Delta x_C$ is maximum for given circle. When path difference is maximum, intensity will be minimum.
So, points $A$ and $C$ are dark.

Topic: Physics, Wave Optics, JEE Main
Two point white dots are 1 mm apart on a black paper. They are viewed by an eye of pupil diameter 3 mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [Take wavelength of light = 500 nm]

Options:
A 6 m
B 3 m
C 1 m
D 5 m

The correct answer is:
5 m

Solution:
Resolution limit=1.22λd
Again resolution limit =sinθ=θ=yD
∴ yD=1·22λd

D=yd1.22λ
D=10-3×3×10-31.22×5×10-7=306.1≈5 m.

Topic: Physics, Wave Optics, JEE Main
Two point white dots are $1 \mathrm{~mm}$ apart on a black paper. They are viewed by eye of pupil diameter $3 \mathrm{~mm}$. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [ Take wavelength of light $=500 \mathrm{~nm}$ ]

Options:
A
$5 \mathrm{~m}$

B
$1 \mathrm{~m}$

C
$6 \mathrm{~m}$

D
$3 \mathrm{~m}$

The correct answer is:
$5 \mathrm{~m}$

Solution:
$\frac{1.22 \lambda}{(3 \mathrm{~mm})}=$ Re solution limit $=\frac{(1 \mathrm{~mm})}{\mathrm{R}}$
$\therefore \mathrm{R}=5 \mathrm{~m}$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2005
Two point-like charges of the same magnitude but opposite signs are fixed to either end of the base of an isosceles triangle as shown in the figure. The electric field intensity at the midpoint M has a magnitude EM. The electric field intensity at the point P (directly above the midpoint) has a magnitude EP. The ratio of these two fields, EM : EP=27. The value of cosâ¡Î± is

Options:
A 13
B 16
C 12
D 19

The correct answer is:
13

Solution:
EM=kq×2r2cos2α
EP=2kqcos⁡αr2
EMEP=1cos3⁡α=27
cos⁡α=1271/3⇒cos⁡α=1/3

Topic: Physics, Electrostatics, JEE Main
Two points $A$ and $B$ move from rest along a straight line with constant acceleration $f$ and $f$ ' respectively. If $A$ takes $m$ sec. more than $B$ and describes ' $n$ ' units more than $B$ in acquiring the same speed then

Options:
A
$\left(f-f^{\prime}\right) m^2=f f^{\prime} n$

B
$\left(f+f^{\prime}\right) m^2=f f^{\prime} n$

C
$\frac{1}{2}\left(f+f^{\prime}\right) m=f f^{\prime} n^2$

D
$\left(f^{\prime}-f\right) n=\frac{1}{2} f f^{\prime} m^2$

The correct answer is:
$\left(f^{\prime}-f\right) n=\frac{1}{2} f f^{\prime} m^2$

Solution:
$$
\begin{aligned}
& v^2=2 f(d+n)=2 f^{\prime} d \\
& v=f^{\prime}(t)=(m+t) f
\end{aligned}
$$
eliminate $d$ and $m$ we get
$$
\left(f^{\prime}-f\right) n=\frac{1}{2} f f^{\prime} m^2 \text {. }
$$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2005
. Two points are located at a distance of $10 \mathrm{~m}$ and $15 \mathrm{~m}$ from the source of oscillation. The period of oscillation is $0.05 \mathrm{sec}$ and the velocity of the wave is $300 \mathrm{~m} / \mathrm{sec}$. What is the phase difference between the oscillations of two points?

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

The correct answer is:
$\frac{2 \pi}{3}$

Solution:
Phase difference $\phi=\frac{2 \pi}{\lambda} \times$ path difference
$$
\begin{aligned}
& =\frac{2 \pi}{15} \times(15-10) \quad\{\lambda=v T=300 \times 0.05 \mathrm{~m}\} \\
& =\frac{2 \pi}{5}
\end{aligned}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2008 (Mains)
Two points on a travelling wave, having frequency 500 Hz and velocity 300 m s-1, are 60Â° out of phase. The minimum distance between the two points is

Options:
A 0.2
B 0.1
C 0.5
D 0.4

The correct answer is:
0.1

Solution:
v=nλ
∴ λ=vn=300500=35 m
Now, phase difference
=2πλ×path difference
∴ π3=2πλ×path difference
or π3=2π×53×path difference
path difference=3×60×π2π×5×180=0.1 m

Topic: Physics, Waves and Sound, JEE Main
Two points P and Q are lying on the curve y=log2x+3 in xy plane such that OPââi^=1 and OQââj^=3, then the value of OQâ-2OPâ is

(where, O is the origin)

Options:
A 6
B 7
C 8
D 10

The correct answer is:
10

Solution:
Let P x1,log2x1+3 and Q x2,log2x2+3
∵ OP→⋅i^=1⇒x1=1⇒P(1,2)
∵ OQ→⋅j^=3⇒log2x2+3=3⇒x2=5⇒Q5,3
OQ→-2OP→=5i^+3j^-2i^+2j^
=3i^-j^=10

Topic: Mathematics, Vector Algebra, JEE Main
Two points $P$ and $Q$ are maintained at the potentials of $10 \mathrm{~V}$ and $-4 \mathrm{~V}$ respectively. The work done in moving 100 electrons from $P$ to $Q$ is

Options:
A
$-19 \times 10^{-17} \mathrm{~J}$

B
$9.60 \times 10^{-17} \mathrm{~J}$

C
$-2.24 \times 10^{-16} \mathrm{~J}$

D
$2.24 \times 10^{-16} \mathrm{~J}$

The correct answer is:
$2.24 \times 10^{-16} \mathrm{~J}$

Solution:
$$
\begin{aligned}
& W=Q d V=Q\left(V_q-V_P\right)=-100 \times\left(1.6 \times 10^{-19}\right) \times(-4-10) \\
& =+100 \times 1.6 \times 10^{-19} \times 14=+2.2410^{-16} \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main
Two points $P$ and $Q$ are maintained at the potentials of $10 \mathrm{~V}$ and $-4 \mathrm{~V}$ respectively. The work done in moving 100 electrons from $P$ to $Q$ is

Options:
A
$-19 \times 10^{-17} \mathrm{~J}$

B
$9.60 \times 10^{-17} \mathrm{~J}$

C
$-2.24 \times 10^{-16} \mathrm{~J}$

D
$2.24 \times 10^{-16} \mathrm{~J}$

The correct answer is:
$2.24 \times 10^{-16} \mathrm{~J}$

Solution:
$$
\begin{aligned}
& W=Q d V=Q\left(V_q-V_P\right)=-100 \times\left(1.6 \times 10^{-19}\right) \times(-4-10) \\
& =+100 \times 1.6 \times 10^{-19} \times 14=+2.2410^{-16} \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2009
Two points P and Q in the argand plane represent the complex numbers z and 3z+2+i. If z=2, then Q moves on the circle, whose centre and radius are (here, i2=-1)

Options:
A -2+i,6
B 2-i,3
C 2+i,6
D 2+i,3

The correct answer is:
2+i,6

Solution:
z=2
x2+y2=4
3x+iy+2+i=h+ik
3x+2+i3y+1=h+ik
h=3x+2, k=3y+1
h-23=x, k-13=4
h-232+k-132=4
⇒h2+4-4h+k2+1-2k=36⇒h2+k2-4h-2k-31=0
Centre =2,1
Radius =4+1+31=6

Topic: Mathematics, Complex Number, JEE Main
Two polaroids A and B are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now, another polaroid C is placed between A and B bisecting angle between them. If intensity of unpolarised light is I0 then intensity of transmitted light after passing through polaroid B will be :

Options:
A I04
B I02
C I08
D Zero

The correct answer is:
I08

Solution:
As given here, the pass axis of polaroids A and B are inclined at angle of 90º with each other. Another polaroid Cis placed at angle of 45°with respect to A and B.

Therefore, intensity of light after falling on polaroid A, IA=I02

Intensity of light after falling on polaroid C, IC=IAcos245°=I02cos245°=I04

Intensity of light after falling on polaroid B, IB=ICcos245°=I04cos245°=I08

Topic: Physics, Wave Optics, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Two polaroids A and B are placed with their polaroid axes 30o to each other as shown in the figure. A plane polarised light passes through the polaroid A and after passing through it, intensity of light becomes I0. What is the intensity of finally transmitted light, after passing through the polaroid B ?

Options:
A 0.75I0
B 0.866I0
C 0.25I0
D 0.5I0

The correct answer is:
0.75I0

Solution:
The intensity I of the transmitted light varies directly as the square of the cosine of the angle between the transmission direction of polariser and the analyser.
I=I0cos2â¡Î¸
Intensity of finally transmitted light after passing through the polaroid B is given by I=I0cos2â¡Î¸ (Intensity)
I= I 0 â cos 2 30 o
I=0.75I0
where, I0 is the maximum intensity of transmitted light.

Topic: Physics, Wave Optics, JEE Main
Two polaroids are crossed. If now one of them is rotated through an angle of 30° and unpolarised light of intensity I0 is incident on the first polaroid, then the intensity of transmitted light will be

Options:
A I04
B 3I04
C 3I08
D I08

The correct answer is:
I08

Solution:
Intensity through first polaroid, I1=I02
∴ Intensity of transmitted light, I=I1cos2θ
=I02cos2(90o-30o)
=I02cos260o
=I08

Topic: Physics, Wave Optics, JEE Main
Two polaroids have their polarizing directions parallel so that the intensity of a transmitted light is maximum. The angle through which either polaroid must be turned if the intensity is to drop by one-half is

Options:
A
$135^{\circ}$

B
$90^{\circ}$

C
$120^{\circ}$

D
$180^{\circ}$

The correct answer is:
$135^{\circ}$

Solution:
For $I=\frac{I_0}{2}$ and $I=I_0 \cos ^2 \theta=\frac{I_0}{2}$ $\therefore \theta=45^{\circ}$
Therefore the angle through which either polaroids turned is $135^{\circ}\left(=180^{\circ}-45^{\circ}\right)$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2012 (07 May Online)