datavorous/entrance-exam-dataset huggingface dataset:
A point is in motion along a hyperbola $y=\frac{10}{x}$ so that its abscissa $x$ increases uniformly at a rate of 1 unit per second. Then, the rate of change of its ordinate when the point passes through (5,2)

Options:
A increases at the rate of $\frac{1}{2}$ unit per second
B decreases at the rate of $\frac{1}{2}$ unit per second
C decreases at the rate of $\frac{2}{5}$ unit per second
D increases at the rate of $\frac{2}{5}$ unit per second

The correct answer is:
decreases at the rate of $\frac{2}{5}$ unit per second

Solution:
We have,
$$
\begin{array}{r}
\frac{d x}{d t}=\operatorname{l and} y=\frac{10}{x} \\
\text { Now, } y=\frac{10}{x}
\end{array}
$$
On differentiating both the sides w.r.t. $t$, we get
$$
\begin{aligned}
\Rightarrow \quad \frac{d y}{d t} &=\frac{-10}{x^{2}} \times \frac{d x}{d t} \\
&=\frac{-10}{(5)^{2}} \times 1
\end{aligned}
$$
$$
=\frac{-10}{25}=\frac{-2}{5} \quad\left[x=5, \frac{d x}{d t}=1\right]
$$
$\therefore y$ decreases at the rate of $\frac{2}{5}$ unit per second.

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2019
A point moves, so that the sum of squares of its distance from the points (1,2) and (-2,1) is always $6 .$ Then, its locus is

Options:
A the straight line $y-\frac{3}{2}=-3\left(x+\frac{1}{2}\right)$
B a circle with centre $\left(-\frac{1}{2}, \frac{3}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$
C a parabola with focus (1,2) and directrix passing through (-2,1)
D an ellipse with foci (1,2) and (-2,1)

The correct answer is:
a circle with centre $\left(-\frac{1}{2}, \frac{3}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$

Solution:
Let $P$ be any point, whose coordinate is $(h, k)$
Given.

P moves, so that the sum of squares of its distances from the points $A(1,2)$ and $B(-2,1)$ is $6 .$
1.e., $(P A)^{2}+(P B)^{2}=6$
$\Rightarrow \quad(h-1)^{2}+(k-2)^{2}+(h+2)^{2}+(k-1)^{2}=6$
$\Rightarrow \quad h^{2}+1-2 h+k^{2}+4-4 k+h^{2}+4+4 h$
$$
+k^{2}+1-2 k=6
$$
$\Rightarrow \quad 2 h^{2}+2 k^{2}+2 h-6 k+4=0$
$\Rightarrow \quad h^{2}+k^{2}+h-3 k+2=0$
$\therefore$ Required locus is
$$
x^{2}+y^{2}+x-3 y+2=0
$$
Which represent a circle. Whose centre is $\left(\frac{-1}{2}, \frac{3}{2}\right)$ and radius $=\sqrt{\frac{1}{4}+\frac{9}{4}-2}=\sqrt{\frac{5}{2}-2}=\frac{1}{\sqrt{2}}$

Topic: Mathematics, Circle, WBJEE, WBJEE 2013
A point object is held above a thin equiconvex lens at its focus. The focal length is $0.1 \mathrm{m}$ and the lens rests on a horizontal thin plane mirror. The final image will be formed at

Options:
A infinite distance above the lens
B $0.1 \mathrm{m}$ above the center of the lens
C infinite distance below the lens
D $0.1 \mathrm{m}$ below the center of the lens

The correct answer is:
$0.1 \mathrm{m}$ above the center of the lens

Solution:
As the lens is placed on a horizontal thin plane mirror.

So due to role of plane mirror the image will be formed at the same point $O$ above $0.1 \mathrm{m}$ distance above the center of the lens.

Topic: Physics, Ray Optics, WBJEE, WBJEE 2017
A point object is placed on the axis of a thin convex lens of focal length $0.05 \mathrm{m}$ at a distance of $0.2 \mathrm{m}$ from the lens and its image is formed on the axis. If the object is now made to oscillate along the axis with a small amplitude of $A \mathrm{cm},$ then what is the amplitude of oscillation of the image? $\left[\right.$ you may assume, $\frac{1}{1+x} \approx 1-x,$ where $\left.x< < 1\right]$

Options:
A $\frac{4 A}{9} \times 10^{-2} m$
B $\frac{5 A}{9} \times 10^{-2} m$
C $\frac{A}{3} \times 10^{-2} m$
D $\frac{A}{9} \times 10^{-2} \mathrm{m}$

The correct answer is:
$\frac{A}{9} \times 10^{-2} \mathrm{m}$

Solution:
According to the question, we can draw the following diagram

Given, $u=-0.2 \mathrm{m}$ and $f=0.05 \mathrm{m}$
As we know that,
$$
\begin{array}{l}
\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\
\frac{1}{v}=\frac{1}{f}+\frac{1}{u} \\
\frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.2} \\
\frac{1}{v}=\frac{100}{5}-\frac{10}{2} \\
\frac{1}{v}=20-5 \Rightarrow v=\frac{1}{15} \mathrm{m}
\end{array}
$$
Now, differentiating eq. (i), we get
$$
\begin{array}{l}
-\frac{d v}{v^{2}}=-\frac{d u}{u^{2}} \quad \therefore \quad d v=d u \frac{v^{2}}{u^{2}} \\
A_{\max }=A \times\left(\frac{1}{15}\right)^{2} \times\left(\frac{1}{(-0.2)}\right)^{2} \\
A_{\max }=A \times \frac{1}{225} \times 25 \\
A_{\max }=\frac{A}{9}
\end{array}
$$
Here $A$ is in $\mathrm{cm} .$
Hence, $A_{\max }=\frac{A}{9} \times 10^{-2} \mathrm{m}$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2019
A point $P$ lies on a line through $Q(1,-2,3)$ and is parallel to the line $\frac{x}{1}=\frac{y}{4}=\frac{z}{5},$ If $P$ lies on the plane $2 x+3 y-4 z+22=0,$ then segment PQ equals

Options:
A $\sqrt{42}$ units
B $\sqrt{32}$ units
C 4 units
D 5 units

The correct answer is:
$\sqrt{42}$ units

Solution:
Equation of line through $Q(1,-2,3)$ and
parallel to the line $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ is $\frac{x-1}{1}=\frac{y+2}{4}=\frac{z-3}{5}=\lambda \quad$ (say)
since, point $P$ lies on above line.
$\therefore$
$$
P(\lambda+1,4 \lambda-2,5 \lambda+3)
$$
since, $P$ lies on the given plane. $\therefore \quad 2(\lambda+1)+3(4 \lambda-2)-4(5 \lambda+3)+22=0$
$\Rightarrow \quad 2 \lambda+2+12 \lambda-6-20 \lambda-12+22=0$
$$
\Rightarrow \quad-6 \lambda+6=0
$$
$\Rightarrow$
$\lambda=1$
$\therefore$
$P(2,2,8)$
$\therefore \quad P Q=\sqrt{(2-1)^{2}+(2+2)^{2}+(3-8)^{2}}$
$\Rightarrow \quad P Q=\sqrt{1+16+25}=\sqrt{42}$

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2018
A point $P$ lies on the circle $x^{2}+y^{2}=169$. If $Q=(5,12)$ and $R=(-12,5),$ then the $\angle Q P R$ is

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

The correct answer is:
$\frac{\pi}{4}$

Solution:
Given equation of circle is
$$
x^{2}+y^{2}=169
$$
Its centre $=(0,0)$ and radius $=13$

Now, slope of $O R=\frac{-5}{12}=m_{1}\left(\because\right.$ slope $\left.=\frac{y_{2}-y}{x_{2}-x}\right)$
and slope of $O Q=\frac{12}{5}=m_{2}$
$\because m_{1} \cdot m_{2}=-1 \Rightarrow \angle R O Q=\frac{\pi}{2}$
We know that, angle made by the chord of circle at circumference is equal to the half of the angle made by the same chord at the centre of circle.
$\therefore \quad \angle Q P R=\frac{1}{2} \cdot \angle R O Q=\frac{1}{2} \times \frac{\pi}{2}=\frac{\pi}{4}$

Topic: Mathematics, Circle, WBJEE, WBJEE 2013
A point source is placed at coordinates (0,1) in $x y$ -plane. A ray of light from the source is reflected on a plane mirror placed along the
$X$ -axis and perpendicular to the $x y$ -plane. The reflected ray passes through the point $(3,3) .$ What is the path length of the ray from (0,1) to (3,3)$?$

Options:
A 5
B $\sqrt{13}$
C $2 \sqrt{3}$
D $1+2 \sqrt{3}$

The correct answer is:
5

Solution:
Ray diagram for the question.

$I$ is image of source $S$ by the plane mirror placed perpendicularly along $X$ -axis.
$S M=I M$
$\therefore P M+M S=P M+M I=P I$
$\therefore \quad P I=\sqrt{(3-0)^{2}+(3+1)^{2}}$
$=\sqrt{9+16}=\sqrt{25}=5$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2018
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, WBJEE, WBJEE 2020
A poker hand consists of 5 cards drawn at random from a well-shuffled pack of 52 cards. Then, the probability that a poker hand consists of a pair and a triple of equal face values (for example, 2 sevens and 3 kings or 2 aces and 3 queens, etc. ) is

Options:
A $\frac{6}{4165}$
B $\frac{23}{4165}$
C $\frac{1797}{4165}$
D $\frac{1}{4165}$

The correct answer is:
$\frac{6}{4165}$

Solution:
Required probability
\(\frac{{ }^{13} c_1 \times{ }^4 c_2 \times{ }^{12} c_1 \times{ }^4 c_3}{{ }^{52} c_5}\)
\(\frac{13 \times 6 \times 12 \times 4}{\frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1}}\)
$=\frac{13 \times 6 \times 12 \times 4}{52 \times 51 \times 10 \times 49 \times 2}$
$=\frac{6}{4165}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2014
A polygon has 44 diagonals. The number of its sides is

Options:
A 10
B 11
C 12
D 13

The correct answer is:
11

Solution:
Hints: ${ }^n \mathrm{C}_2-n=44$
$$
\begin{aligned}
& \frac{n(n-1)}{2}-n=44 \\
& n\left[\frac{n-1}{2}-1\right]=44 \\
& n(n-3)=88 \\
& n(n-3)=11 \times 8 \\
& n=11
\end{aligned}
$$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2009
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, WBJEE, WBJEE 2009
A positive charge $Q$ is situated at the centre of a cube. The electric flux through any face of the cube is (in SI units)

Options:
A $\frac{Q}{6 \varepsilon_{0}}$
B $4 \pi Q$
C $\frac{Q}{4 \pi \varepsilon_{0}}$
D $\frac{Q}{6 \pi \varepsilon_{0}}$

The correct answer is:
$\frac{Q}{6 \varepsilon_{0}}$

Solution:
The positive charge $Q$ is situated at the centre of a cube

According to Gauss theorem, the flux from any closed surface
$$
\phi_{E}=\frac{Q}{\varepsilon_{0}}
$$
Cube has 6 faces, so electric flux from any face $\phi_{z}=\frac{Q}{6 \varepsilon_{0}}$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2017
A problem in mathematics is given to 4 students whose chances of solving individually are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and $\frac{1}{5},$ The probability that the problem will be solved at least by one student is

Options:
A $\frac{2}{3}$
B $\frac{3}{5}$
C $\frac{4}{5}$
D $\frac{3}{4}$

The correct answer is:
$\frac{4}{5}$

Solution:
$$
\begin{aligned}
&\text { (c) Let } P(A)=\frac{1}{2}, P(B)=\frac{1}{3}\\
&\begin{aligned}
P(C) &=\frac{1}{4}, P(D)=\frac{1}{5} \\
\text { Now, } P(A&\cup B \cup C \cup D) \\
&=1-P(\bar{A} \cap \bar{B} \cap \bar{C} \cap \bar{D}) \\
&=1-P(\bar{A}) P(\bar{B}) P(\bar{C}) P(\bar{D}) \\
&=1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right) \\
&=1-\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)\left(\frac{3}{4}\right)\left(\frac{4}{5}\right)=1-\frac{1}{5}=\frac{4}{5}
\end{aligned}
\end{aligned}
$$

Topic: Mathematics, Probability, WBJEE, WBJEE 2019
A progressive wave moving along $x$ -axis is represented by $y=A \sin \left[\frac{2 \pi}{\lambda}(v t-x)\right]$ The
wavelength $(\lambda)$ at which the maximum particle velocity is 3 times the wave velocity is

Options:
A $\frac{\mathrm{A}}{A}$
B $2 \mathrm{A}(3 \pi)$
C $\left(\frac{3}{4}\right) \pi A$
D $(2 / 3) \pi A$

The correct answer is:
$(2 / 3) \pi A$

Solution:
$y=A \sin \left[\frac{2 \pi}{\lambda}(v t-x)\right]$
$v=\frac{2 \pi}{\lambda} V A$
$$
\begin{aligned}
\left(v_{p}\right)_{\max } &=3 v \\
\lambda &=\frac{2 \pi}{3} A
\end{aligned}
$$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2012
A projectile thrown with an initial velocity of $10 \mathrm{ms}^{-1}$ at an angle $\alpha$ with the horizontal. has a range of $5 \mathrm{m}$. Taking $g=10 \mathrm{ms}^{-2}$ and neglecting air resistance, what will he the estimated value of $\alpha$?

Options:
A $15^{\circ}$
B $30^{\circ}$
C $45^{\circ}$
D $75^{\circ}$

The correct answers are:
$15^{\circ}$, $75^{\circ}$

Solution:
Given, initial velocity of particle,
$$
u=10 \mathrm{ms}^{-1}
$$
Range, $R=5 \mathrm{m}$ Gravitational acceleration,
$$
g=10 \mathrm{ms}^{-2}
$$
As we known that,
Range of projectile, $R=\frac{u^{2} \sin 2 \theta}{g}$
or
$$
\begin{aligned}
5 &=\frac{(10)^{2} \sin 2 \alpha}{10} \\
\sin 2 \alpha &=\frac{5}{10} \\
\sin 2 \alpha &=\frac{1}{2} \\
\alpha &=15^{\circ} \\
\left(90^{\circ}-15^{\circ}\right.&\left.=75^{\circ}\right)
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2019
A proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is 2000 times heavier than an electron, what will be the relation between the de Broglie wavelength of the proton $\left(\lambda_{p}\right)$ and that of electron $\left(\lambda_{c}\right) ?$

Options:
A $\lambda_{p}=2000 \lambda_{e}$
B $\lambda_{p}=\frac{\lambda_{e}}{2000}$
C $\lambda_{p}=20 \sqrt{5} \lambda_{0}$
D $\lambda_{p}=\frac{\lambda_{0}}{20 \sqrt{5}}$

The correct answer is:
$\lambda_{p}=\frac{\lambda_{0}}{20 \sqrt{5}}$

Solution:
As we know that de-Broglie wavelength is given as $\lambda=\frac{h}{p}=\frac{h}{m v}$
where, $h=$ Planck constant
$$
\begin{array}{l}
p=\text { momentum of particle } \\
v=\text { velocity of particle }
\end{array}
$$
and $m=$ mass of the particle.
Eq.(i) can be written as,
$$
\lambda=\frac{h}{2 m(\mathrm{KE})}=\frac{h}{\sqrt{2 m q v}} \quad[\because \mathrm{KE}=q v]
$$
where, $\mathrm{KE}=$ Kinetic energy of particle
Hence, $\lambda \propto \frac{1}{\sqrt{m}}$
$$
\text { Now, } \frac{\lambda_{\text {proton }}}{\lambda_{\text {electron }}}=\sqrt{\frac{m_{\text {electron }}}{m_{\text {proton }}}}
$$
Given, mass of proton, $m_{\text {proton }}=2000 \mathrm{m}_{\text {electron }}$
$\frac{\lambda_{\text {proton }}}{\lambda_{\text {electron }}}=\sqrt{\frac{1}{2000}}$
$\frac{\lambda_{\text {proton }}}{\lambda_{\text {electron }}}=\frac{1}{20 \sqrt{5}} \Rightarrow \lambda_{p}=\frac{\lambda_{e}}{20 \sqrt{5}}$

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2019
A proton is moving with a uniform velocity of $10^{6} \mathrm{ms}^{-1}$ along the $Y$ -axis, under the joint action of a magnetic field along $Z$ -axis and an electric field of magnitude $2 \times 10^{4} \mathrm{Vm}^{-1}$ along the negative $X$ -axis. If the electric field is switched off, the proton starts moving in a circle. The radius of the circle is nearly given: $\frac{e}{m}$ ratio for proton $=10^{4} \mathrm{Ckg}^{-1}$

Options:
A $0.5 \mathrm{m}$
B $0.2 \mathrm{m}$
C $0.1 \mathrm{m}$
D $0.05 \mathrm{m}$

The correct answer is:
$0.5 \mathrm{m}$

Solution:
Velocity of proton $=10^{6} \mathrm{m} / \mathrm{s}$ along $y$ -direction

$$
\text { Electric field }=2 \times 10^{4} \mathrm{V} / \mathrm{m}
$$
$\frac{e}{=}$ for proton $=10^{-8} \mathrm{C} / \mathrm{kg}$
it
The magnetic field,
$$
\begin{aligned}
B &=\frac{E}{v} \quad\left(\because q, v B=q_{\rho} E\right) \\
&=\frac{2 \times 10^{4}}{10^{6}}=2 \times 10^{-2} \mathrm{T}
\end{aligned}
$$
The radius of circular path
$$
\begin{aligned}
\gamma &=\frac{m v}{q_{p} \cdot B}=\frac{10^{-6} \times 10^{6}}{2 \times 10^{-2}} \\
&=\frac{1}{2}=0 \cdot 5 \mathrm{m}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2017
A proton of mass $m$ and charge $q$ is moving in a plane with kinetic energy $E$. If there exists a uniform magnetic field $B$, perpendicular to the plane of the motion, the proton will move in a circular path of radius

Options:
A $\frac{2 E m}{q B}$
B $\frac{\sqrt{2 E m}}{q B}$
C $\frac{\sqrt{E m}}{2 q B}$
D $\sqrt{\frac{2 E q}{m B}}$

The correct answer is:
$\frac{\sqrt{2 E m}}{q B}$

Solution:
Given, Kinetic energy $=E$
Mass $=m$
Magnetic field $=B$
Charge $=q$ We know that
$$
F=q v B \sin \theta
$$
(motion of a charged particle in a uniform magnetic field) If
$$
\theta=90^{\circ}
$$
Then
$$
F=q v B
$$
We know that also (centripetal force)
$$
F=\frac{m v^{2}}{r}
$$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
q v B &=\frac{m v^{2}}{r}, r=\frac{m v}{q B}\left[\because E=\frac{1}{2} m v^{2}, v=\sqrt{\frac{2 E}{m}}\right] \\
\therefore \quad r &=\frac{m \sqrt{\frac{2 E}{m}}}{q B} \Rightarrow r=\frac{\sqrt{2 E m}}{q B}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2014
A proton of mass $m$ moving with a speed $v$ ( $< < $ c. velocity of light in vacuum) completes circular orbit in time $T$ in a uniform Whetic ficld. If the speed of the proton is increased to $\sqrt{2}$ v, what will be time needed to plete the circular orbit?

Options:
A $\sqrt{T}$
B $T$
C $\frac{T}{\sqrt{2}}$
D $\frac{T}{2}$

The correct answer is:
$T$

Solution:
$\because$ When a charged particle moves at a circular path in uniform magnetic field, it time period is independent from its speed.
Time period of proton will not change with change in its speed.

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2018
A radioactive atom \(\mathrm{X}_{\mathrm{Y}}\) emits two \(\alpha\) particles and one \(\beta\) particle successively. The number of neutrons in the nucleus of the product will be

Options:
A \(\mathrm{X}-4-\mathrm{Y}\)
B \(\mathrm{X}-\mathrm{Y}-5\)
C \(\mathrm{X}-\mathrm{Y}-3\)
D \(\mathrm{X}-\mathrm{Y}-6\)

The correct answer is:
\(\mathrm{X}-\mathrm{Y}-5\)

Solution:
Hints:
\(\begin{aligned}
\text { Number of neutrons } & =\text { Mass no. }- \text { Atomic no. } \\
& =\mathrm{X}-8-\mathrm{Y}+3 \\
& =\mathrm{X}-\mathrm{Y}-5
\end{aligned}\)

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2011
A radioactive nucleus of mass number A, initially at rest, emits an \(\alpha\)-particle with a speed \(v\). The recoil speed of the daughter nucleus will be

Options:
A \(\frac{2 v}{A-4}\)
B \(\frac{2 \mathrm{v}}{\mathrm{A}+4}\)
C \(\frac{4 v}{A-4}\)
D \(\frac{4 v}{\mathrm{~A}+4}\)

The correct answer is:
\(\frac{4 v}{A-4}\)

Solution:
Hints : From conservation of momentum \(4 . \mathrm{V}=(\mathrm{A}-4) \mathrm{V}_1 ; \mathrm{V}=\frac{4 \mathrm{~V}}{\mathrm{~A}-4}\)

Topic: Physics, Nuclear Physics, WBJEE, WBJEE 2011
A ray of light along $x+\sqrt{3}y=\sqrt{3}$ gets reflected upon reaching $x$-axis, the equation of the reflected ray is

Options:
A $y=x+\sqrt{3}$
B $\sqrt{3} \mathrm{y}=\mathrm{x}-\sqrt{3}$
C $y=\sqrt{3} x-\sqrt{3}$
D $\sqrt{3} y=x-1$

The correct answer is:
$\sqrt{3} \mathrm{y}=\mathrm{x}-\sqrt{3}$

Solution:
$\mathrm{A} \equiv(\sqrt{3}, 0)$
Let, B $(0,1)$
$\therefore \mathrm{B}^{\prime} \equiv(0,-1)$
$\therefore$ Equation of $\mathrm{AB}^{\prime}: \frac{\mathrm{y}-0}{0+1}=\frac{\mathrm{x}-\sqrt{3}}{\sqrt{3}-0} \Rightarrow \sqrt{3} \mathrm{y}=\mathrm{x}-\sqrt{3}$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2021
A ray of light is incident at an angle $i$ on a glass slab of refractive index $\mu$. The angla between reflected and refracted light is $90^{\circ}$ Then, the relationship between $i$ and $\mu$ is

Options:
A $i=\tan ^{-1}\left(\frac{1}{\mu}\right)$
B $\tan i=\mu$
C $\sin i=\mu$
D $\cos i=\mu$

The correct answer is:
$\tan i=\mu$

Solution:
As situation can be diagrammatically as below

From law of reflection
$$
i=\theta
$$
Now, $\quad \theta+r+90^{\circ}=180^{\circ}$
$$
\begin{aligned}
\Rightarrow \quad i+r+90^{\circ} &=180^{\circ} \\
r &=90^{\circ}-i
\end{aligned}
$$
Also, from Snell's law $\frac{\sin i}{\sin r}=\mu \Rightarrow \frac{\sin i}{\sin \left(90^{\circ}-i\right)}=\frac{\sin i}{\cos i}=\mu$
$\Rightarrow \tan i=\mu$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2015
A ray of light is incident on a right angled isosceles prism parallel to its base as shown in the figure. Refractive index of the material of the prism is $\sqrt{2}$. Then. which of the following statement(s) is/are true?

Options:
A The refiection at $P$ is total internal.
B The reflection at $Q$ is total internal.
C The ray emerging at $R$ is parallel to the ray incident at $S .$
D Total deviation of the ray is $150^{\circ}$

The correct answers are:
The refiection at $P$ is total internal., The ray emerging at $R$ is parallel to the ray incident at $S .$

Solution:
Critical angle $\theta_{C}=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}$
Snell's law at points, $\left(\because n_{1} \sin \theta_{1}=n_{2} \sin \theta_{2}\right)$
$1 \sin 45^{\circ}=\sqrt{2} \sin \theta$
$\therefore \quad \theta=30^{\circ}$
At $P \because$ incidence angle at $P=75^{\circ}>\theta_{C}$
$\therefore$ TIR at point $P$
At $Q \because$ incidence angle at $Q=15^{\circ} < \theta_{C}$
$\therefore$ Partial reflection and refraction at $Q .$ As $S P \| Q R$, emergent ray at $R$ is parallel to incid,
ray at $S$.
$\therefore$ Net deviation $=180^{\circ}$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2018
A ray of light is reflected by a plane mirror. $\hat{e}_{0}, \hat{e}$ and $\hat{n}$ be the unit vectors along the incident ray, reflected ray and the normal to the reflecting surface respectively. Which of the following gives an expression for ê?

Options:
A $\hat{e}_{0}+2\left(\hat{e}_{0} \cdot \hat{n}\right) \hat{n}$
B $\hat{\boldsymbol{e}}_{0}-2\left(\hat{\boldsymbol{e}}_{0} \cdot \hat{\mathbf{n}}\right) \hat{\mathbf{n}}$
C $\hat{\mathbf{e}}_{0}-\left(\hat{\boldsymbol{\theta}}_{0} \cdot \hat{\mathbf{n}}\right) \hat{\mathbf{n}}$
D \(\mathrm{d}) \hat{\mathrm{e}}_{0}+\left(\hat{e}_{0} \cdot \hat{n}\right) \hat{\mathrm{n}\)

The correct answer is:
$\hat{\boldsymbol{e}}_{0}-2\left(\hat{\boldsymbol{e}}_{0} \cdot \hat{\mathbf{n}}\right) \hat{\mathbf{n}}$

Solution:
$\text { (b) According to the question, }$
We know that incident ray, reflected ray and normal lie in the same plane. and angle of incidence $=$ angle of reflection
Therefore $\hat{\mathrm{n}}$ will be along the angle bisector of e and $-\hat{e}_{0}$
i.e. $\hat{\mathbf{n}}=\frac{\hat{e}+\left(-\hat{\mathbf{e}}_{0}\right)}{\left|\hat{e}-\hat{\mathbf{e}}_{0}\right|}$
[$\because$ Bisector will along a vector dividing in same ratio as the ratio of sides forming that angle ]
But $\hat{\mathbf{n}}$ is a unit vector where $\left|\mathbf{e}-\hat{\mathbf{e}}_{0}\right|=O C$
$\begin{aligned}
&=20 P=2|\hat{e}| \\
\cos \theta &=2 \cos \theta
\end{aligned}$
Substituting this value in Eq. (i), we get
$\hat{\mathbf{n}}=\frac{\hat{\mathbf{e}}-\hat{\mathbf{e}}_{0}}{2 \cos \theta}$
$\begin{array}{l}
\hat{e}=\hat{e}_{0}+(2 \cos \theta) \hat{n} \\
\left.\hat{e}=\hat{e}_{0}-2 \hat{n} \cdot \hat{e}_{0}\right) \hat{n}\left[\because \hat{n} \cdot \hat{e}_{0}=-\cos \theta\right]
\end{array}$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2019
A ray of light strikes a glass plate at an angle of $60^{\circ}$. If the reflected and refracted rays are perpendicular to each other, the refractive index of glass is

Options:
A $\frac{\sqrt{3}}{2}$
B $\frac{3}{2}$
C $\frac{1}{2}$
D $\sqrt{3}$

The correct answer is:
$\sqrt{3}$

Solution:
Here, angle of incidence is $60^{\circ}$

$$
\mu=\frac{\sin i}{\sin r}
$$
$\sin i=\mu \sin r$
here,
$$
\begin{aligned}
r &=90-i \\
\sin i &=\mu \sin \left(90^{\circ}-i\right)=\mu \cos i \\
\mu &=\frac{\sin i}{\cos i}=\tan i=\tan 60^{\circ} \\
\mu &=\sqrt{3}
\end{aligned}
$$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2016
A ray of monochromatic light is incident on the plane surface of separation between two media \(X\) and \(Y\) with angle of incidence 'i' in medium \(X\) and angle of refraction ' \(r\) ' in medium \(Y\). The given graph shows the relation between sin \(i\) and \(\sin r\). If \(V_x\) and \(V_{\mathrm{Y}}\) are the velocities of the ray in media \(X\) and \(Y\) respectively, then which of the following is true?

Options:
A \(V_X=\frac{1}{\sqrt{3}} V_Y\)
B \(\mathrm{V}_{\mathrm{X}}=\sqrt{3} \mathrm{~V}_Y\)
C Total internal reflection can happen when the light is incident in medium \(X\).
D \(v_X=\sqrt{3} v_Y\), where \(v_X\) and \(v_Y\) are frequencies of the light in medium \(X\) and \(Y\) respectively.

The correct answer is:
\(\mathrm{V}_{\mathrm{X}}=\sqrt{3} \mathrm{~V}_Y\)

Solution:
Hint :
\(\begin{aligned} & \mu_1 \sin i=\mu_2 \sin r \\ & \frac{C}{V_X} \sin i=\frac{C}{V_Y} \sin r \\ & \frac{\sin i}{\sin r}=\frac{V_X}{V_Y}=\sqrt{3} \\ & V_Y=\frac{V_X}{\sqrt{3}} \\ & V_X=\sqrt{3} V_Y\end{aligned}\)
\(y=m x\)
\(\sin r=\left(\tan 30^{\circ}\right) \sin \mathrm{i}\)
\(\frac{\sin i}{\sin r}=\frac{1}{\tan 30^{\circ}}=\frac{\sqrt{3}}{1}\)

Topic: Physics, Ray Optics, WBJEE, WBJEE 2023