datavorous/entrance-exam-dataset huggingface dataset:
A thin convex lens is placed just above an empty vessel of depth $80 \mathrm{~cm}$. The image of a coin kept at the bottom of the vessel is thus formed $20 \mathrm{~cm}$ above the lens. If now water is poured in the vessel up to a height of $64 \mathrm{~cm}$, what will be the approximate new position of the image. Assume that refractive index of water is $4 / 3$.

Options:
A $21.33 \mathrm{~cm}$ above the lens
B $6.67 \mathrm{~cm}$ below the lens
C $33.67 \mathrm{~cm}$ above the lens
D $24 \mathrm{~cm}$ above the lens

The correct answer is:
$21.33 \mathrm{~cm}$ above the lens

Solution:
Hint:
$\mathrm{u}=-80 \mathrm{~cm} \quad \mathrm{v}=+20 \mathrm{~cm}$
$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{20}+\frac{1}{80}=\frac{5}{80}$
$f=16 \mathrm{~cm}$
$\mathrm{u}^{\prime}=16+\frac{64 \times 3}{4}=64 \mathrm{~cm}$
$\frac{1}{16}=\frac{1}{v}+\frac{1}{64} \quad \therefore \frac{1}{v}=\frac{1}{16}-\frac{1}{64}=\frac{3}{64}$
$v=21.33$ above the lens

Topic: Physics, Ray Optics, WBJEE, WBJEE 2020
A thin glass rod is bent in a semicircle of radius \(\mathrm{R}\). A change is non-uniformly distributed along the rod with a linear charge density \(\lambda=\lambda_0 \sin \left(\lambda_0\right.\) is a positive constant). The electric field at the centre \(P\) of the semicircle is,

Options:
A \(-\frac{\lambda_0}{8 \pi \varepsilon_0 R} \hat{j}\)
B \(\frac{\lambda_0}{8 \pi \varepsilon_0 R} \hat{j}\)
C \(\frac{\lambda_0}{8 \pi \varepsilon_0 R} \hat{i}\)
D None of these

The correct answer is:
None of these

Solution:
Hint :
\(\begin{aligned}
& \lambda=\lambda_0 \sin \theta \\
& \mathrm{dE}_{\mathrm{y}}=\frac{\mathrm{k} \lambda}{\mathrm{R}} \sin \theta \mathrm{d} \theta \\
& =\frac{\mathrm{k} \lambda_0}{\mathrm{R}} \int_{-\pi / 2}^{+\pi / 2} \sin ^2 \theta \mathrm{d} \theta=\frac{\mathrm{k} \lambda_0}{2 \mathrm{R}} \int_{-\pi / 2}^{+\pi / 2}(1-\cos 2 \theta) \mathrm{d} \theta \\
& \therefore \overrightarrow{\mathrm{E}}=-\frac{\lambda_0}{8 \varepsilon_0 \mathrm{R}} \hat{\mathrm{j}}
\end{aligned}\)

Topic: Physics, Electrostatics, WBJEE, WBJEE 2023
A thin plastic sheet of refractive index 1.6 is used to cover one of the slits of a clouble slit arrangement. The central point on the scireen is now occupied by what would have been the 7th bright fringo before the plastic was used. If the wavelength of light is $600 \mathrm{nm}$, what is the thickness (in $\mu \mathrm{m}$ ) of the plastic sheet?

Options:
A 7
B 4
C 8
D 6

The correct answer is:
7

Solution:
The change in path length $=(\mu-1) t$ This path length change the position of bright fringe upto seventh bright fringe. The thickness of plastic substance inserted in the path of rays will must be 7 (on assuming fringe width to be units).

Topic: Physics, Wave Optics, WBJEE, WBJEE 2015
A train approaching a railway platform with a speed of $20 \mathrm{ms}^{-1}$ starts blowing the whistle. Speed of sound in air is $340 \mathrm{ms}^{-1}$. If the frequency of the emitted sound from the whistle is $640 \mathrm{Hz}$, the frequency of sound to a person standing on the platform will appear to be

Options:
A $600 \mathrm{Hz}$
B $640 \mathrm{Hz}$
C $680 \mathrm{Hz}$
D $720 \mathrm{Hz}$

The correct answer is:
$680 \mathrm{Hz}$

Solution:
Given $\quad v=340 \mathrm{ms}^{-1}, \quad u_{5}=20 \mathrm{ms}^{-1} \quad$ and
$v_{0}=640 \mathrm{Hz}$
From Doppler's law,
$$
\begin{array}{l}
v=\left(\frac{340}{340-20}\right) 640 \\
v=680 \mathrm{Hz}
\end{array}
$$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2012
A train is moving along the tracks at a constant speed \(\mathrm{u}\). A girl on the train throws a ball of mass \(m\) straight ahead along the direction of motion of the train with speed \(v\) with respect to herself. Then

Options:
A Kinetic energy of the ball as measured by the girl on the train is \(\mathrm{mv}^2 / 2\)
B Work done by the girl in throwing the ball is \(\mathrm{mv}^2 / 2\)
C Work done by the train is mvu
D The gain in kinetic energy of the ball as measured by a person standing by the rail track is \(\mathrm{mv}^2 / 2\)

The correct answers are:
Kinetic energy of the ball as measured by the girl on the train is \(\mathrm{mv}^2 / 2\), Work done by the girl in throwing the ball is \(\mathrm{mv}^2 / 2\), Work done by the train is mvu

Solution:
Hint : w.r.t. the girl \(E_k=\frac{1}{2} m v^2\)
\(\therefore \mathrm{W}=\Delta \mathrm{E}_{\mathrm{k}}=1 / 2 \mathrm{mv}^2\)
Work by the train \(=\left\{\frac{1}{2}(v+u)^2-\frac{1}{2} m u^2\right\}-\frac{1}{2} m v^2\)
\(\begin{aligned}
& =\frac{1}{2} m\left(v^2+u^2+2 v u\right)-\frac{1}{2} m\left(v^2+u^2\right) \\
& =m v u
\end{aligned}\)
Gain in \(E_k=\frac{1}{2} m(v+u)^2-\frac{1}{2} m u^2=\frac{1}{2} m v^2+m v u\)
measured from rail track

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2023
$A$ train is moving with a uniform speed $c^{2}$ $33 \mathrm{m} / \mathrm{s}$ and an observer is approaching the train with the same speed. If the train blows a whistle of frequency $1000 \mathrm{Hz}$ and the velocity of sound is $333 \mathrm{m} / \mathrm{s},$ then the apparent frequency of the sound that the observer hears is

Options:
A $1220 \mathrm{Hz}$
B $1099 \mathrm{Hz}$
C $1110 \mathrm{Hz}$
D $1200 \mathrm{Hi}$

The correct answer is:
$1220 \mathrm{Hz}$

Solution:
We have,
Apparent frequency,
$\mathrm{v} =\left(\frac{v+u_{0}}{v-u_{s}}\right)\mathrm{v}_{0}=\left(\frac{333+33}{333-33}\right) \times 1000 \\=\frac{366}{300} \times 1000=1220 \mathrm{Hz}$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2015
A train moves from rest with acceleration and in time $t_{1}$ covers a distance $x$. It then decelerates to rest at constant retardation $\beta$ for distance $y$ in time $t_{2}$. Then,

Options:
A $\frac{x}{y}=\frac{\beta}{\alpha}$
B $\frac{\beta}{\alpha}=\frac{t_{1}}{t_{2}}$
C $x=y$
D $\frac{x}{y}=\frac{\beta t_{1}}{\alpha t_{2}}$

The correct answers are:
$\frac{x}{y}=\frac{\beta}{\alpha}$, $\frac{\beta}{\alpha}=\frac{t_{1}}{t_{2}}$

Solution:
Shape of the graph $\mathrm{OP}$ shows the acceleration in train
$\therefore \tan \theta=$ acceleration.
$$
\boldsymbol{\alpha}=\frac{V_{0}}{t_{1}}
$$
Shape of the graph pt shows the deceleration in train.
$\therefore$ Similarly. $\quad \beta=\frac{v_{0}}{t_{2}}$
$\therefore \quad \frac{\boldsymbol{\beta}}{\boldsymbol{\alpha}}=\frac{\boldsymbol{v}_{0} / \boldsymbol{t}_{\boldsymbol{2}}}{\mathrm{v}_{0} / t_{1}}=\frac{t_{1}}{t_{2}}$
Displacement = area of graph obtained between v-t
$$
\begin{array}{l}
x=\frac{1}{2} t_{1}, v_{0} \\
y=\frac{1}{2} t_{2}, v_{0}
\end{array}
$$
$\therefore \frac{x}{y}=\frac{t_{1}}{t_{2}}
\Rightarrow \quad \frac{x}{y}=\frac{t_{1}}{t_{2}}=\frac{\beta}{a}$

Topic: Physics, Motion In One Dimension, WBJEE, WBJEE 2016
A travelling acoustic wave frequency $500 \mathrm{Hz}$ is moving along the positive $x$ -direction with a velocity of $300 \mathrm{ms}^{-1}$. The phase difference between two points $x_{1}$ and $x_{2}$ is $60^{\circ} .$ Then the minimum separation between the two points is

Options:
A $1 \mathrm{mm}$
B $1 \mathrm{cm}$
C $10 \mathrm{cm}$
D $10 \mathrm{mm}$

The correct answer is:
$10 \mathrm{cm}$

Solution:
By using the relation, $v=v \lambda$
We have, $\quad \lambda=\frac{V}{v}=\frac{300}{500}=\frac{3}{5} \mathrm{m}$
The phase difference, $\phi=\frac{2 \pi}{\lambda}(\Delta x)$ $\Rightarrow \quad \frac{\pi}{3}=\frac{2 \pi \times 5}{3}(\Delta x) \Rightarrow \Delta x=\frac{1}{10} m=10 \mathrm{cm}$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2013
A uniform capillary tube of length $l$ and inner radius $r$ with its upper end sealed is submerged vertically into water. The outside pressure is $p_{0}$ and surface tension of water is $\gamma$. When a length $x$ of the capillary is submerged into water, it is found that water levels inside and outside the capillary coincide. The value of $x$ is

Options:
A $\frac{l}{\left(1+\frac{p_{0} r}{4 \gamma}\right)}$
B $l\left(1-\frac{p_{0} r}{4 \gamma}\right)$
C $l\left(1-\frac{p_{0} r}{2 \gamma}\right)$
D $\frac{l}{\left(1+\frac{p_{0} r}{2 \gamma}\right)}$

The correct answer is:
$\frac{l}{\left(1+\frac{p_{0} r}{2 \gamma}\right)}$

Solution:
When the sealed capillary tube is submerged vertically into water, the pressure inside the tube changes
For the inside capillary,
$$
p_{1} V_{1}=p_{2} v_{2}
$$
$\therefore \quad P_{0}(l A)=p^{\prime}(l-x) A$
where $p^{\prime}$ is pressure in capillary after being submerged
$\therefore$
$$
p^{\prime}=\frac{p_{0}l}{l-x}
$$
According to question, since level of water inside capillary coincides with outside.
$$
\begin{array}{l}
\therefore \quad p^{\prime}-p_{0}=\frac{2 \gamma}{r} \\
\therefore \quad \frac{p_{d}}{l-x}-p_{0}=\frac{2 \gamma}{r}=\frac{l}{\left(1+\frac{p_{0} r}{2 \gamma}\right)}
\end{array}
$$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2017
A uniform current is flowing along the length dinfinite, straight. thin. hollow cylinder dradius $R$ The magnetic field $B$ produced at perpendicular distance $d$ from the axis of the cylinder is plotied in a graph. Which of the following figures looks like the plot?

Options:
A
B
C
D

The correct answer is:

Solution:
$\because$ Magnetic field due to a hollow cylinder of radius $r$

$$
\begin{aligned}
B_{\text {inside }} &=0(\text { for } d < r) \\
B_{\text {outside }} &=\frac{\mu_{0}}{2 \pi} \cdot \frac{i}{d}(\text { for } d \geq r)
\end{aligned}
$$
So, graph between $B$ and $d$.

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2018
A uniform long tube is bent into a circle of radius $\mathrm{R}$ and it lies in vertical plane. Two liquids of same volume but densities $\rho$ and $\delta$ fill half the tube. The angle $\theta$ is

Options:
A $\tan ^{-1}\left(\frac{\rho-\delta}{\rho+\delta}\right)$
B $\tan ^{-1} \frac{\rho}{\delta}$
C $\tan ^{-1} \frac{\delta}{\rho}$
D $\tan ^{-1}\left(\frac{\rho+\delta}{\rho-\delta}\right)$

The correct answer is:
$\tan ^{-1}\left(\frac{\rho-\delta}{\rho+\delta}\right)$

Solution:
$$
\begin{aligned}
& \text { Hints: } \delta g R(\cos \theta+\sin \theta)=\rho g R(\cos \theta-\sin \theta) \\
& \delta \cos \theta+\delta \sin \theta=\rho \cos \theta-\rho \sin \theta \\
& \sin \theta(\delta+\rho)=\cos \theta(\rho-\delta) \\
& \tan \theta=\frac{\rho-\delta}{\rho+\delta}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2010
A uniform magnetic field \(B\) exists in a region. An electron of charge \(q\) and mass \(m\) moving with velocity \(v\) enters the region in a direction perpendicular to the magnetic field. Considering Bohr angular momentum quantization, which of the following statement(s) is/are true?

Options:
A The radius of \(\mathrm{n}^{\text {th }}\) orbit \(\mathrm{r}_{\mathrm{n}} \propto \sqrt{\mathrm{n}}\)
B The maximum velocity of the electron is \(\frac{\sqrt{\mathrm{qB} \hbar}}{\mathrm{m}}\)
C Energy of the \(n^{\text {th }}\) level \(E_n \propto n\)
D Transition frequency \(\omega\) between two successive levels is independent of \(n\)

The correct answers are:
The radius of \(\mathrm{n}^{\text {th }}\) orbit \(\mathrm{r}_{\mathrm{n}} \propto \sqrt{\mathrm{n}}\), The maximum velocity of the electron is \(\frac{\sqrt{\mathrm{qB} \hbar}}{\mathrm{m}}\), Energy of the \(n^{\text {th }}\) level \(E_n \propto n\), Transition frequency \(\omega\) between two successive levels is independent of \(n\)

Solution:
Hint: \(r=\frac{m v}{q B}\)
\(m v r=\frac{n h}{2 \pi}\)
\(m v=\frac{n h}{2 \pi r}\)
\(r=\frac{n h}{2 \pi r q B}\)
\(\begin{aligned} & r^2=\frac{n h}{2 \pi q B} \Rightarrow r=\sqrt{\frac{n h}{2 \pi q B}} \\ & r \propto \sqrt{n} \\ & v=\frac{q B r}{m}=\frac{q B}{m} \sqrt{\frac{n h}{2 \pi q B}} \\ & \underset{m i n}{v}=\sqrt{\frac{q^2 B^2}{m^2} \times \frac{n h}{2 \pi q B}}=\sqrt{\frac{n \mathrm{qBh}}{2 \pi m^2}}=\frac{1}{m} \sqrt{q B} \hbar\end{aligned}\)
\(\begin{aligned} E & =\frac{1}{2} m v^2 \\ & =\frac{1}{2} m\left[\frac{n q B h}{2 \pi m^2}\right]\end{aligned}\)
\(E \propto n\)
\(\begin{aligned} & \mathrm{E}_2-\mathrm{E}_1=\left(\mathrm{n}_2-\mathrm{n}_1\right) \frac{\mathrm{qBh}}{4 \pi \mathrm{m}} \\ & \mathrm{hf}=\left(\mathrm{n}_2-\mathrm{n}_1\right) \frac{\mathrm{qBh}}{4 \pi \mathrm{m}} \\ & \mathrm{n}_2-\mathrm{n}_1=1 \text { for successive levels }\end{aligned}\)

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2023
A uniform rod is suspended horizontally from its mid-point. A piece of metal whose weight is $w$ is suspended at a distance $l$ from the mid-point. Another weight $W_{1}$ is suspended on the other side at a distance $l_{1}$ from the mid-point to bring the rod to a horizontal position. Wben $w$ is completely immersed in water, $w_{1}$ needs to be kept at a distance $l_{2}$ from the mid-point to get the rod back into horizontal position. The specific gravity of the metal piece is

Options:
A $\frac{w}{w_{1}}$
B $\frac{w l_{1}}{w l-w_{1} l_{2}}$
C $\frac{L_{1}}{l_{1}-l_{2}}$
D $\frac{l_{1}}{l_{2}}$

The correct answer is:
$\frac{L_{1}}{l_{1}-l_{2}}$

Solution:
$$
W-F_{B}=W\left(1-\frac{1}{\rho}\right)
$$
where $\rho=$ specific gravity $W\left(1-\frac{1}{\rho}\right)=W_{1} b$
$\left(1-\frac{1}{\rho}\right)=\frac{W_{1} L_{2}}{W}$
$\left(1-\frac{1}{\rho}\right)=\frac{W_{1} b_{2}}{W_{1} L_{1}}$
$1-\frac{1}{\rho}=\frac{L_{2}}{L_{1}} \Rightarrow \frac{1}{\rho}=1-\frac{b}{L_{1}}$
$\frac{1}{\rho}=\frac{l_{1}-L_{2}}{l_{1}}$ or $\rho=\frac{l_{1}}{l_{1}-L_{2}}$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2014
A uniform rod of length $L$ pivoted at one end $P$ is freely rotated in a horizontal plane with an angular velocity $\omega$ about a vertical axis passing through $\mathrm{P}$. If the temperature of the system is increased by $\Delta \mathrm{T}$, angular velocity becomes $\frac{\omega}{2}$. If coefficient of linear expansion of the rod is $\alpha(\alpha< < 1)$, then $\Delta \mathrm{T}$ will be

Options:
A $\frac{1}{\alpha}$
B $\frac{1}{2 \alpha}$
C $\frac{1}{4 \alpha}$
D $\alpha$

The correct answer is:
$\frac{1}{2 \alpha}$

Solution:
$\mathrm{I} \omega=\mathrm{I}_{1} \frac{\omega}{2}$
$\quad \mathrm{I}_{1}=2 \mathrm{I}$
$\quad \frac{\mathrm{ml}_{1}^{2}}{3}=2 \frac{\mathrm{ml}_{0}^{2}}{3}$
$\left\{\mathrm{I}_{0}^{2}(1+\alpha \Delta \mathrm{T})\right\}^{2}=2 \mathrm{I}_{0}^{2}, \therefore 1+2 \alpha \Delta \mathrm{T}=2 \therefore \Delta \mathrm{T}=\frac{1}{2 \alpha}$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2021
A uniform rope of length \(4 \mathrm{~m}\) and mass \(0.4 \mathrm{~kg}\) is held on a frictionless table in such a way that \(0.6 \mathrm{~m}\) of the rope is hanging over the edge. The work done to pull the hanging part of the rope on the to the table is, (Assume \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) )

Options:
A \(0.36 \mathrm{~J}\)
B \(0.24 \mathrm{~J}\)
C \(0.12 \mathrm{~J}\)
D \(0.18 \mathrm{~J}\)

The correct answer is:
\(0.18 \mathrm{~J}\)

Solution:
\(\begin{aligned}
& \text {Hint : } W=\frac{\mathrm{mgL}}{2}=\frac{0.4}{4} \times 0.6 \times 10 \times \frac{0.6}{2} \\
& =0.1 \times 0.6 \times 10 \times 0.3 \\
& =0.18 \mathrm{~J}
\end{aligned}\)

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2023
A uniform solid spherical ball is rolling down a smooth inclined plane from a height $h$. The velocity attained by the ball when it reaches the bottom of the inclined plane is v. If the ball is now thrown vertically upwards with the same velocity $v$, the maximum height to which the ball will rise is

Options:
A $\frac{5 h}{8}$
B $\frac{3 h}{5}$
C $\frac{5 h}{7}$
D $\frac{7 h}{9}$

The correct answer is:
$\frac{5 h}{7}$

Solution:
We know that total kinetic energy of a body rolling without slipping
$K_{\text {total }}=K_{\text {rot }}+K_{\text {trans }}$ For solid spherical ball, $I=\frac{2}{5} m R^{2}$ (along to diameter) and $v=R \omega$ where $R$ is radius of spherical ball
So.
$$
K_{\text {total }}=\frac{1}{2}\left(\frac{2}{5} m R^{2}\right) \omega^{2}+\frac{1}{2} m R^{2} \omega^{2}
$$
$$
\begin{array}{l}
=\frac{7}{10} m R^{2} \omega^{2} \\
K=\frac{7}{10} m v^{2}
\end{array}
$$
Potential energy = Kinetic energy
$$
\begin{aligned}
m g h &=\frac{7}{10} m v^{2} \\
v^{2} &=\frac{10}{7} g h
\end{aligned}
$$
For vertical projection,
$$
\begin{array}{c}
v^{2}=u^{2}+2 g h^{\prime} \\
\frac{10}{7} g h=0+2 g h^{\prime} \Rightarrow h^{\prime}=\frac{5}{7} h
\end{array}
$$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2014
A uniform string of length $L$ and mass $M$ is fixed at both ends while it is subject to a
tension $T$. It can vibrate at frequencies $(v)$ given by the formula (where $n=1,2,3, \ldots)$

Options:
A $v=\frac{n}{2} \sqrt{\frac{T}{M L}}$
B $v=\frac{n}{2 L} \sqrt{\frac{T}{M}}$
C $v=\frac{1}{2 n} \sqrt{\frac{T}{M L}}$
D $v=\frac{n}{2} \sqrt{\frac{T L}{M}}$

The correct answer is:
$v=\frac{n}{2} \sqrt{\frac{T}{M L}}$

Solution:
The formula of frequency of 3 vibrations produced in tensed wire is
$$
v=\frac{n}{2} \sqrt{\frac{T}{m}}
$$
$m=$ Mass per unit length of the wire $=\frac{M}{L}$
$M=$ Mass of wire
$L=$ Length of wire $n=$ Number of loops produced in wire $v=\frac{n}{2 L} \sqrt{\frac{T}{M / L}}=\frac{n}{2} \sqrt{\frac{T}{M L}}$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2017
A uniform thin rod of length $L$, mass $m$ is lying on a smooth horizontal table. A horizontal impulse $P$ is suddenly applied perpendicular to the rod at one end. The total energy of the rod after the impulse is

Options:
A $\frac{\mathrm{P}^{2}}{\mathrm{M}}$
B $\frac{7 \mathrm{P}^{2}}{8 \mathrm{M}}$
C $\frac{13 \mathrm{P}^{2}}{2 \mathrm{M}}$
D $\frac{2 P^{2}}{M}$

The correct answer is:
$\frac{2 P^{2}}{M}$

Solution:
$\mathrm{P}=(\mathrm{mv}-0)$
angular impulse $\mathrm{J}=\mathrm{P} \times \frac{\mathrm{L}}{2}=\mathrm{L}_{\mathrm{f}}-0$
$\mathrm{LE}_{\mathrm{f}}=\frac{\mathrm{PL}}{2}$
$=\frac{\mathrm{K}_{\mathrm{r}}}{2 \mathrm{~m}}+\mathrm{K}_{\mathrm{t}}=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}}+\frac{\mathrm{L}_{\mathrm{f}}^{2}}{2 \mathrm{I}}$
$=\frac{(\mathrm{PL} / 2)^{2}}{2 \times \frac{\mathrm{mL}^{2}}{12}}$
$=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}}+\frac{\mathrm{P}^{2} \mathrm{~L}^{2}}{4 \times \frac{\mathrm{mL}^{2}}{6}}=\frac{\mathrm{P}}{2 \mathrm{~m}}+\frac{3 \mathrm{P}^{2}}{2 \mathrm{~m}}=\frac{2 \mathrm{P}^{2}}{\mathrm{~m}}$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2021
A unit negative charge with mass $M$ resides at the mid-point of the straight line of length $2 a$ adjoining two fixed charges of magnitude $+Q$ each. If it is given a very small displacement $x(x< < a)$ in a direction perpendicular to the straight line, it will

Options:
A come back to its original position and stay there
B execute oscillations with frequency
$$
\frac{1}{2 \pi} \sqrt{\frac{Q}{4 \pi \varepsilon_{0} M a^{3}}}
$$
C None of the above
D execute oscillations with frequency
$$
\frac{1}{2 \pi} \sqrt{\frac{Q}{4 \pi \varepsilon_{0} M a^{2}}}
$$

The correct answer is:
None of the above

Solution:
From figure the net force
$\begin{aligned} F_{\text {bet }} &=-F \cos \theta+(-F \cos \theta) \\ &=-2 F \cos \theta \\ &=-2 \times \frac{k Q \times 1}{\left(x^{2}+a^{2}\right)} \times \frac{x}{\sqrt{x^{2}+a^{2}}} \\ &=-\frac{2 k Q}{\left(x^{2}+a^{2}\right)^{3 / 2}} \cdot x \\ F_{\text {net }} &=-\left(\frac{2 k Q}{a^{3}}\right) \cdot x \end{aligned}$
Frequency of oscillation $=\frac{1}{2 \pi} \sqrt{\frac{2 k Q}{M a^{3}}}$
$=\frac{1}{2 \pi} \sqrt{\frac{2 \times \frac{1}{4 \pi \varepsilon_{0}} \cdot Q}{M a^{3}}}$
$=\frac{1}{2 \pi} \sqrt{\frac{Q}{2 \pi e_{0} M a^{3}}}$
$\Rightarrow\left(^{*}\right.$ None of the option matches)

Topic: Physics, Oscillations, WBJEE, WBJEE 2017
A van der Waals' gas may behave ideally when

Options:
A the volume is very low
B the temperature is very high
C the pressure is very low
D the temperature, pressure and volume all are very high

The correct answer is:
the pressure is very low

Solution:
At very low pressure, compressibility factor, $Z$ is approximatedly one.
$\therefore$ van der Waals' gas may behave ideally

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2013
A variable circle passes through the fixed point $A(p, q)$ and touches $X$ -axis. The locus of the other end of the diameter through $A$ is

Options:
A $(x-p)^{2}=4 q y$
B $(x-q)^{2}=4 p y$
C $(y-p)^{2}=4 q x$
D $(y-q)^{2}=4 p x$

The correct answer is:
$(x-p)^{2}=4 q y$

Solution:
In a circle, $A B$ is a diameter where the coordinate of $A$ is $(p, q)$ and let the coordinate of $B\left(x_{1}, y_{1}\right)$
Equation of circle in diameter form is
$$
(x-p)\left(x-x_{1}\right)+(y-q)\left(y-y_{1}\right)=0
$$
$\Rightarrow x^{2}-\left(p+x_{1}\right) x+p x_{1}+y^{2}-\left(y_{1}+q\right) y+q y_{1}=0$
$\Rightarrow x^{2}-\left(p+x_{1}\right) x+y^{2}-\left(y_{1}+q\right) y+p x_{1}+q y_{1}=0$
Since, this circle touches $X$ -axis
$\therefore$
$$
y=0
$$
$\Rightarrow \quad x^{2}-\left(p+x_{1}\right) x+p x_{1}+q y_{1}=0$
Also, the discriminant of above equation will be equal to zero because circle touches $X$ -axis.
$\therefore \quad\left(p+x_{1}\right)^{2}=4\left(p x_{1}+q y_{1}\right)$
$\Rightarrow p^{2}+x_{1}^{2}+2 p x_{1}=4 p x_{1}+4 q y_{1}$
$\Rightarrow x_{1}^{2}-2 p x_{1}+p^{2}=4 q y_{1}$
$\Rightarrow \quad\left(x_{1}-p\right)^{2}=4 q y_{1}$
Therefore, the locus of point $B$ is
$$
(x-p)^{2}=4 q y
$$

Topic: Mathematics, Circle, WBJEE, WBJEE 2019
A variable line passes through a fixed point $\left(x_{1}, y_{1}\right)$ and meets the axes at $A$ and $B$. If the rectangle $O A P B$ be completed, the locus of $P$ is, $(O$ being the origin of the system of axes).

Options:
A $\left(y-y_{1}\right)^{2}=4\left(x-x_{1}\right)$
B $\frac{x_{1}}{x}+\frac{y_{1}}{y}=1$
C $x^{2}+y^{2}=x_{1}^{2}+y_{1}^{2}$
D $\frac{x^{2}}{2 x_{1}^{2}}+\frac{y^{2}}{y_{1}^{2}}=1$

The correct answer is:
$\frac{x_{1}}{x}+\frac{y_{1}}{y}=1$

Solution:
Let the equation of line be $\frac{x}{a}+\frac{y}{b}=1$ since, the line passing through a fixed point $\left(x_{1}, y_{1}\right)$.

$\therefore \quad \frac{x_{1}}{a}+\frac{y_{1}}{b}=1$
since, $O A P B$ is a rectangle, therefore the coordinate of $P$ will be $(a, b)$
Hence, locus of $P$ is
$\frac{x_{1}}{x}+\frac{y_{1}}{y}=1$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2019
A variable line passes through the fixed point $(\alpha, \beta) .$ The locus of the foot of the perpendicular from the origin on the line is

Options:
A $x^{2}+y^{2}-\alpha x-\beta y=0$
B $x^{2}-y^{2}+2 a x+2 \beta y=0$
C $\alpha x+\beta y \pm \sqrt{\left(\alpha^{2}+\beta^{2}\right)}=0$
D $\frac{x^{2}}{\alpha^{2}}+\frac{y^{2}}{\beta^{2}}=1$

The correct answer is:
$x^{2}+y^{2}-\alpha x-\beta y=0$

Solution:
Let $(\alpha, \beta)$ be the given point, let $Q(x, y)$ be the foot of the perpendicular, and let $O$ be the origin. The line can have any direction
$$
\angle P Q O=90^{\circ}
$$
Point $Q$ lies on the circle having diameter $O P$.
The locus of point $Q$.
$$
(x-0)(x-\alpha)+(y-0)(y-\beta)=0
$$
$\Rightarrow \quad x^{2}-x \alpha+y^{2}-y \beta=0$
$$
\Rightarrow \quad x^{2}+y^{2}-\alpha x-\beta y=0
$$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2019
A vehicle registration number consists of 2 letters of English alphabet followed by 4 digits, where the first digit is not zero. Then, the total number of vehicles with distinct registration numbers is

Options:
A $26^{2} \times 10^{4}$
B ${ }^{26} \mathrm{P}_{2} \times{ }^{10} \mathrm{P}_{4}$
C ${ }^{26} \mathrm{P}_{2} \times 9 \times{ }^{10} \mathrm{P}_{3}$
D $26^{2} \times 9 \times 10^{3}$

The correct answer is:
$26^{2} \times 9 \times 10^{3}$

Solution:
The total number of arrangements of 2 letters of English alphabet
$$
=26 \times 26
$$
The total number of arrangements of 4 digits number in which first digit is not zero
$$
=9 \times 10 \times 10 \times 10
$$
$\therefore$ The total number of vehicles with distinct registration number
$$
\begin{array}{l}
=26 \times 26 \times 9 \times 10 \times 10 \times 10 \\
=26^{2} \times 9 \times 10^{3}
\end{array}
$$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2012