datavorous/entrance-exam-dataset huggingface dataset:
A rectangle \(A B C D\) has its side parallel to the line \(y=2 x\) and vertices \(A, B, D\) are on lines \(y=1, x=1\) and \(x=-1\) respectively. The coordinate of \(C\) can be

Options:
A \((3,8)\)
B \((-3,8)\)
C \((-3,-1)\)
D None of these

The correct answers are:
None of these

Solution:
Hint: Case I
line \(y=2 x\ldots(1)\)
\(A D\) parallel to \(y=2 x\)
\(\Rightarrow \mathrm{M}_{\mathrm{AD}}=2 ~\&~ \mathrm{M}_{\mathrm{AB}}=-\frac{1}{2}\)
for \(a=3\)
\(A(3,1), B(1,2), C(-3,-6), D(-1,-7)\)
\(\begin{aligned} & \text { for } a=-3 \\ & C(3,3), A(-3,1), B(1,-1), D(-1,5)\end{aligned}\)
Case II AD perpendicular to line (1)
\(\Rightarrow M_{A D}=\frac{1}{2}, M_{A B}=2 \Rightarrow b=3-2 a \Rightarrow d=\frac{a+1}{2}\)
for \(a=3, C(-3,-2)\)
for \(a=-3, C(3,7)\)
\({ }^*\) In both cases if abscissa of \(C: 3\) or -3
then co-ordinates of \(C\) can be \((3,3)\) or \((3,7)\) or \((-3,-6)\) or \((-3,-2)\)

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2023
A rectangular coil carrying-current is placed in a non-uniform magnetic field. On that coil, the total

Options:
A force is non-zero
B force is zero
C torque is zero
D torque is non-zero

The correct answers are:
force is non-zero, torque is non-zero

Solution:
Current-carrying rectangular coil is placed in a non-uniform magnetic field. So, the force on each arm of the coil will different giving the resultant force as non-zero.
Thus, option (a) is correct.
Also, the torque acting on current-carrying coil in non-uniform magnetic field will be non-zero.
Thus, option (d) is also correct.
following reason.

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2016
A relation $\rho$ on the set of real number $R$ is defined as $\{x \rho y: x y>0\} .$ Then, which of the following is/are true?

Options:
A $\rho$ is reflexve and symmetric
B $\rho$ is symmetric but not reflexive
C $\rho$ is symmetric and transitive
D $\rho$ is an equivalence relation

The correct answers are:
$\rho$ is symmetric but not reflexive, $\rho$ is symmetric and transitive

Solution:
We have, $x p y: x y > 0$
(i) Reflexive Suppose $x \rho x \in R$
$\Rightarrow x^{2} > 0$
which is not true when $x=0$. Hence, relation is not reflexive.
(ii) Symmetric $x p y \in R$ $\Rightarrow \quad x y>0$
$\Rightarrow \quad y x>0$
$\Rightarrow \quad y \rho x \in R$
If $(x, y) \in R,$ then, relation is symmetric.
(iii) Transitive $\mathrm{xp} \mathrm{y} \in \mathrm{R}$
$x y>0$
and $\quad y \rho z \in R$ $y z>0$
$\begin{array}{ll}
\Rightarrow x y^{2} z > 0 \\
\Rightarrow x z > 0 \\
\Rightarrow (x, z) \in R \\
\text {If }(x, y) \in R, \text { then }(y, z) \in R \\
\Rightarrow (x, z) \in R
\end{array}$
Hence, relation is transitive.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2015
A rifleman is firing at a distant target and has only $10 \%$ chance of hitting it. The least number of rounds he must fire to have more than $50 \%$ chance of hitting it at least once, is

Options:
A 5
B 7
C 9
D 11

The correct answer is:
7

Solution:
Hint:
$\mathrm{P}$ (hitting a target) $=\frac{1}{10}$
$\therefore \mathrm{P}($ not hitting a target $)=\frac{9}{10}$
$\therefore$ Let number of trials $=n$
So, $\mathrm{P}($ hitting at least once $)=1-\mathrm{P}$ (missing all) $=1-\left(\frac{9}{10}\right)^{n} \geq \frac{1}{2}$
$\Rightarrow(0.9)^{n} \leq 0.5$
$(0.9)^{6}=0.531441,(0.9)^{7}=0.4782969 \Rightarrow \mathrm{n}=7$

Topic: Mathematics, Probability, WBJEE, WBJEE 2020
A sample of hydrogen atom in its ground state is radiated with photons of $10.2 \mathrm{eV}$ energies. The radiation from the sample is absorbed by excited ionized $\mathrm{He}^{+}$. Then which of the following statement/s is / are true ?

Options:
A $\mathrm{He}^{+}$electron moves from $n=2$ to $n=4$
B In the $\mathrm{He}^{+}$emission spectra, there will be 6 lines
C Smallest wavelength of $\mathrm{He}^{+}$spectrum is obtained when transition taken place from $n=4$ to $n=3$
D $\mathrm{He}^{+}$electron moves from $n=2$ to $n=3$

The correct answers are:
$\mathrm{He}^{+}$electron moves from $n=2$ to $n=4$, Smallest wavelength of $\mathrm{He}^{+}$spectrum is obtained when transition taken place from $n=4$ to $n=3$

Solution:
$\begin{gathered}
E=13.6 \times 2^2\left(\frac{1}{n^2}-\frac{1}{m^2}\right) \\
n=2 \text { to } m=4 \\
E=13.6 \times 2^2 \times\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=10.2 \mathrm{eV}
\end{gathered}$
So $\mathrm{He}^{+}$electron can move from $n=2$ to $n=4$
Number of spectral lines $=\frac{n(n-1)}{2}=\frac{4}{2}(4-1)=6$
Smallest wavelength correspond to $n=4$ to $n=3$ transition

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2022
A sample of $\mathrm{MgCO}_3$ is dissolved in dil. $\mathrm{HCl}$ and the solution is neutralized with ammonia and buffered with $\mathrm{NH} \mathrm{Cl}_4 /$ $\mathrm{NH}_4 \mathrm{OH}$. Disodium hydrogen phosphate reagent is added to the resulting solution. A white precipitate is formed. What is the formula of the precipitate?

Options:
A $\mathrm{Mg}_3\left(\mathrm{PO}_4\right)_2$
B $\mathrm{Mg}\left(\mathrm{NH}_4\right) \mathrm{PO}_4$
C $\mathrm{MgHPO}_4$
D $\mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7$

The correct answer is:
$\mathrm{Mg}\left(\mathrm{NH}_4\right) \mathrm{PO}_4$

Solution:

Topic: Chemistry, s Block Elements, WBJEE, WBJEE 2022
A satellite has kinetic energy $K$, potential energy $V$ and total energy $E .$ Which of the following statements is true?

Options:
A $K=-V / 2$
B $K=V / 2$
C $E=K / 2$
D $E=-K / 2$

The correct answer is:
$K=-V / 2$

Solution:
As we know that for the earth and satellite system
Kinetic energy, $\quad K=\frac{G M m}{2 a}$
Potential energy, $V=-\frac{G M m}{a}$
and total energy, $E=-\frac{G M m}{2 a}$
where, $a=$ radius of the orbit of the satellite and $m=$ mass of the satellite
On the basis of above three expressions for the energies, we have
$K=\frac{G M m}{2 a}=-\left(-\frac{G M m}{2}\right)=-\frac{V}{2}$

Topic: Physics, Gravitation, WBJEE, WBJEE 2015
A saturated solution of $\mathrm{BaSO}_{4}$ at $25^{\circ} \mathrm{C}$ is $4 \times 10^{-5} \mathrm{M}$. The solubility of $\mathrm{BaSO}_{4}$ in $0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}$ at this temperature will be

Options:
A $1.6 \times 10^{-9} \mathrm{M}$
B $1.6 \times 10^{-8} \mathrm{M}$
C $4 \times 10^{-6} \mathrm{M}$
D $4 \times 10^{-4} \mathrm{M}$

The correct answer is:
$4 \times 10^{-4} \mathrm{M}$

Solution:
$\begin{array}{r}
\mathrm{BaSO}_{4} \rightleftharpoons \mathrm{Ba}^{2+}+\mathrm{SO}_{4}^{2-} \\
\mathrm{s} \quad \mathrm{s}+0.1 \\
\mathrm{Na}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Na}^{+}+\mathrm{SO}_{4}^{2-} \\
0.2 \quad 0.1
\end{array}$
$\begin{array}{l}
\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{+2}\right]\left[\mathrm{SO}_{4}^{-2}\right] \\
\therefore 4 \times 10^{-5}=\mathrm{s} \times(\mathrm{s}+0.1)
\end{array}$
or $4 \times 10^{-5} \approx \mathrm{s} \times 0.1 \quad \therefore \mathrm{s}=4 \times 10^{-4}$ as $\mathrm{s}<< < 0.1 \quad \therefore \mathrm{s}+0.1 \approx 0.1$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2021
A scientist proposes a new temperature scale in which the ice point is $25 \mathrm{X}(\mathrm{X}$ is the new unit of temperature) and the steam point is $305 \mathrm{X}$. The specific heat capacity of water in this new scale is $\left(\operatorname{in} J \mathrm{kg}^{-1} \mathrm{X}^{-1}\right)$

Options:
A $4.2 \times 10^{3}$
B $3.0 \times 10^{3}$
C $1.2 \times 10^{3}$
D $1.5 \times 10^{3}$

The correct answer is:
$1.5 \times 10^{3}$

Solution:
Given, $305 \times-25 X=100^{\circ} \mathrm{C}$
(: $\mathrm{X}$ is the new unit of temperature
$$
(305-25) \mathrm{X}=100^{\circ} \mathrm{C} \Rightarrow 280 \mathrm{X}=100^{\circ} \mathrm{C}
$$
$\therefore \quad 1^{\circ} \mathrm{C}=2.8 \mathrm{x}$
The specinc heat capacity of water
$\begin{aligned}=4200 \frac{j 0 u l e}{k g \cdot^{\circ} C} &=4200 \times \frac{j \text { oule }}{k g \times 2.8 \times} \\ &=1500 J / \mathrm{kg}-\mathrm{X} \\ &=1.5 \times 10^{3} \mathrm{J} \mathrm{kg}^{-1} \mathrm{X}^{-1} \end{aligned}$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2014
A shell of mass $5 M$, acted upon by no external force and initially at rest, bursts into three fragments of masses $M, 2 M$ and $2 M$ respectively. The first two fragments move in opposite directions with velocities of magnitudes 2v and $v$ respectively. The third fragment will

Options:
A move with a velocily $v$ in a direction perpendicular To the other two
B move with a velocity $2 v$ in the direction of velocity of the first fragment
C be at rest
D move with velocity $v$ in the direction of velocity of the second fragment

The correct answer is:
be at rest

Solution:
Let the velocity of third fragment is $v'$. Then by the conservation of linear momentum
$$
\begin{aligned}
& 5(0)=M \times 2 \mathrm{v}-2 M \times \mathrm{v}+2 \mathrm{M} \times \mathrm{v}^{\prime} \\
\Rightarrow \quad & \mathrm{v}^{\prime}=0
\end{aligned}
$$
i.e., the third fragment will be at rest.

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2013
A shown in the figure, a liquid is at same levels in two arms of a U-tube of uniform cross-section when at rest. If the \(U\)-tube moves with an acceleration ' \(f\) ' towards right, the difference between liwuid height between two arms of the \(\mathrm{U}\)-tube will be, (acceleration due to gravity \(=\mathrm{g}\) )

Options:
A \(\frac{f}{g} a\)
B \(\frac{g}{f} a\)
C \(a\)
D 0

The correct answer is:
\(\frac{f}{g} a\)

Solution:
Hint :
\(\begin{aligned} & \tan \theta=\frac{\text { acceleration }}{\mathrm{g}} \\ & \frac{\mathrm{h}}{\mathrm{a}}=\frac{\text { acceleration }}{\mathrm{g}} \\ & \frac{\mathrm{h}}{\mathrm{a}}=\frac{\mathrm{f}}{\mathrm{g}} \\ & \mathrm{h}=\frac{\mathrm{fa}}{\mathrm{g}}\end{aligned}\)

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2023
A simple pendulum, consisting of a small ball of mass $m$ attached to a massless string hanging vertically from the ceiling, is oscillating with an amplitude such that $T_{\max }=2 T_{\min }$ where $T_{\max }$ and $T_{\min }$ are the maximum and minimum tension in the string respectively. The value of maximum tension $\mathrm{T}_{\max }$ in the string is

Options:
A $\frac{3 m g}{2}$
B $\mathrm{mg}$
C $\frac{3 m g}{4}$
D $3 \mathrm{mg}$

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

Solution:
$\mathrm{T}_{\min }=m g \cos \theta$
$\mathrm{T}_{\max }=\mathrm{mg}+\frac{\mathrm{mv}^{2}}{\mathrm{I}}=\mathrm{mg}+\frac{\mathrm{mv}^{2}}{\mathrm{I}}$
$\frac{1}{2} \mathrm{mv}^{2}=\mathrm{mgl}(1-\cos \theta)$
$\mathrm{T}_{\max }=2 \mathrm{~T}_{\min }$
$\mathrm{mg}+\frac{\mathrm{mv}^{2}}{\mathrm{I}}=2 \mathrm{mg} \cos \theta$
$\therefore \cos \theta=\frac{3}{4}$
$\mathrm{v}^{2}=2 \mathrm{gl}\left(1-\frac{3}{4}\right)=\frac{\mathrm{gl}}{2}$
$\mathrm{~T}_{\max }=\mathrm{mg}+\frac{\mathrm{m}}{\mathrm{I}} \times \frac{\mathrm{g} \mathrm{l}}{2}=\frac{3 \mathrm{mg}}{2}$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2021
A simple pendulum of length $\ell$ is displaced so that its taught string is horizontal and then released. A uniform bar pivoted at one end is simultaneously released from its horizontal position. If their motions are synchronous, what is the length of the bar?

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

The correct answers are:
$\frac{3 \ell}{2}$

Solution:
Hint:
$v=\sqrt{2 g h}$
$=\sqrt{2 g \mid \sin \theta}$

$\omega_{1}=\frac{v}{l}=\sqrt{\frac{2 g \sin \theta}{1}}$
$m g h=\frac{1}{2} l \omega^{2}$

$\Rightarrow \mathrm{mg} \cdot \frac{\mathrm{l}^{\prime} \sin \theta}{2}=\frac{1}{2} \times \frac{\mathrm{ml}^{2}}{3} \times \omega_{2}^{2}$
$\omega_{2}=\sqrt{\frac{3 \mathrm{~g} \sin \theta}{\mathrm{l}}}=\omega_{1} \Rightarrow \frac{2}{\mathrm{l}}=\frac{3}{\mathrm{l}}$
$\Rightarrow \mathrm{I}^{\prime}=\frac{3}{2}I$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2020
A simple pendulum of length $L$ swings in a vertical plane. The tension of the string when it makes an angle $\theta$ with the vertical and the bob of mass $m$ moves with a speed $v$ is $\mathrm{fg}$ is the gravitational acceleration)

Options:
A $m v^{2} / L$
B $m g \cos \theta+m v^{2} / L$
C $m g \cos \theta-m v^{2} / L$
D $m g \cos \theta$

The correct answer is:
$m g \cos \theta+m v^{2} / L$

Solution:
The situation is given below

For motion along a vertical circular track, the required centripetal force is along the radius and towards the centre of the circle is given by $T-m g \cos \theta=\frac{m v^{2}}{L}$
$\Rightarrow \quad T=\frac{m v^{2}}{L}+m g \cos \theta$

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2015
A small bar magnet of dipole moment $M$ is moving with speed $v$ along $x$ direction towards a small closed circular conducting loop of radius ' $a$ ' with its centre $O$ at $x=0$ (see figure). Assume $x>>a$ and the coil has a resistance $R$. Then which of the following statemenmt(s) is/are true?

Options:
A Magnetic field at the centre $\mathrm{O}$ of the circular coil due to the bar magnet is $\frac{\mathrm{M}}{\mathrm{x}^{3}}$
B Induced EMF is proportional to $\frac{1}{x^{4}}$
C The magnetic moment $\mu$ due to induced current in the coil is proportional to $\mathrm{a}^{4}$
D The heat produced is proportional to $\frac{1}{x^{6}}$

The correct answers are:
Induced EMF is proportional to $\frac{1}{x^{4}}$, The magnetic moment $\mu$ due to induced current in the coil is proportional to $\mathrm{a}^{4}$

Solution:
Magnetic field at an axial point is $2\left(\frac{\mu_{0}}{4 \pi}\right) \frac{M}{x^{3}}$
$$
\phi=2\left(\frac{\mu_{0}}{4 \pi}\right) \frac{M}{x^{3}} \cdot \pi a^{2}
$$
$\mathrm{EMF}=\frac{-\mathrm{d} \phi}{\mathrm{dt}}=-2\left(\frac{\mu_{0}}{4 \pi}\right) \mathrm{M} \pi \mathrm{a}^{2}\left(\frac{-3}{\mathrm{x}^{4}}\right) \frac{\mathrm{dx}}{\mathrm{dt}}=6\left(\frac{\mu_{0}}{4 \pi}\right) \frac{\mathrm{M} \pi \mathrm{a}^{2}}{\mathrm{x}^{4}} \mathrm{~V} \quad \mathrm{EMF} \propto \frac{1}{\mathrm{x}^{4}}$
Current in coil $=\frac{E M F}{R} \quad i=6\left(\frac{\mu_{0}}{4 \pi}\right) \frac{M \pi a^{2}}{x^{4}} \frac{V}{R}$
Magnetic moment $\mu=\mathrm{i} \pi \mathrm{a}^{2}, \mu \propto \mathrm{a}^{4}$
Rate at which heat is produced $=\mathrm{i}^{2} \mathrm{R} . \propto \frac{1}{\mathrm{x}^{8}}$

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2021
A small mass $m$, attached to one end of a spring with a negligible mass and an unstretched length $L$, executes virtual oscillation with angular frequency $\omega_{0}$. When the mass is rotated with an angular speed $\omega$ by holding the other end of the spring at a fixed point, the mass moves uniformly in a circular path in a horizontal plane. Then the increase in length of the spring during the rotation is

Options:
A \(\frac{\omega^{2} L}{\omega_{0}^{2}-\omega^{2}}\)
B \(\frac{\omega_{0}^{2} L}{\omega^{2}-\omega_{0}^{2}}\)
C \(\frac{\omega^{2} L}{\omega_{0}^{2}}\)
D \(\frac{\omega_{0}^{2} L}{\omega^{2}}\)

The correct answer is:
\(\frac{\omega^{2} L}{\omega_{0}^{2}-\omega^{2}}\)

Solution:
The given situation can be shown as

From figure, $K\left(x \sin \theta=m \omega^{2}(L+x) \sin \theta\right.$
$\Rightarrow \quad K x=\operatorname{m\omega}^{2}(L+x)$...(i)
Also, $\quad \sqrt{\frac{K}{m}}=\omega_{0}$
$\Rightarrow \quad K=m \omega_{0}^{2}$
Substituting the value in Eq. (i)
$$
\begin{aligned}
m \omega_{0}^{2} x &=m \omega^{2}(L+x) \\
\Rightarrow \quad x &=\frac{\omega^{2} L}{\omega_{0}^{2}-\omega^{2}}
\end{aligned}
$$

Topic: Physics, Oscillations, WBJEE, WBJEE 2013
A small metal sphere of radius $a$ is falling with a velocity $v$ through a vertical column of a viscous liquid. If the coefficient of viscosity of the liquid is $\eta$, then the sphere encounters an opposing force of

Options:
A $6 \pi \eta a^{2} \mathrm{v}$
B $\frac{6 \mathrm{\eta v}}{\mathrm{\pi a}}$
C $6\pi \eta av$
D $\frac{\pi \eta V}{6 a^{3}}$

The correct answer is:
$6\pi \eta av$

Solution:
Stokes established that if a sphere of radius a moves with velocity $v$ through a fluid of viscosily
Coefficient $\mu$ the viscous force opposing the motion of thy sphere is $F=6\pi \eta a v$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2014
A small spherical body of radius $r$ and density p moves with the terminal velocity $v$ in a fluid of coefficient of viscosity $\eta$ and density $\bar{\sigma}$. What will be the net force on the body?

Options:
A $\frac{4 \pi}{3} r^{3}(p-\sigma) g$
B $6 \pi \eta rv$
C Zero
D Infinity

The correct answer is:
Zero

Solution:
When the spherical body falls with constant velocity, i.e. terminal velocity then the net force becomes zero, i.e the weight of body is equal to the buoyancy force. Hence, $F_{\text {net }}=0$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2019
A small steel ball bounces on a steel plate held horizontally. On each bounce the speed of the ball arriving at the plate is reduced by a factor
e (coefficient of restitution) in the rebound, so that $V_{upward}=eV_{downward}$
If the ball is initially dropped from a height of $0.4 \mathrm{m}$ above the plate and if 10 seconds later the bouncing ceases, the value of $e$ is

Options:
A $\sqrt{\frac{2}{7}}$
B $\frac{3}{4}$
C $\frac{13}{18}$
D $\frac{17}{18}$

The correct answers are:
$\frac{17}{18}$

Solution:
Given, steel ball bounces on a steel ple held horizontally, so $$
V_{\text {upward }}=e V_{\text {downward }}
$$
According to the relation,
$h_{n}=e^{2 n} h$
$\begin{aligned} \therefore \quad t &=\sqrt{\frac{2 h}{g}}+2 \sqrt{\frac{2 h e^{2}}{g}}+2 \sqrt{\frac{2 h e^{4}}{g}}+\ldots \\ &=\sqrt{\frac{2 h}{g}}\left[1+2 e+2 e^{2}+\ldots\right] \\ &=\sqrt{\frac{2 h}{g}}\left[\frac{1+e}{1-e}\right] \\ \therefore \quad t &=10 \mathrm{sec} \\ \therefore 10 &=\sqrt{\frac{2(0 \cdot 4)}{10}}\left(\frac{1+e}{1-e}\right) \end{aligned}$
Solving above relation $\therefore e=\frac{25 \sqrt{2}-1}{25 \sqrt{2}+1}=\frac{17}{18}$

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2017
A smooth massless string passes over a smooth fixed pulley. Two masses $m_{1}$ and $m_{2},\left(m_{1}>m_{2}\right)$ are tied at the two ends of the string. The masses are allowed to move under gravity starting from rest. The total external force acting on the two masses is

Options:
A $\left(m_{1}+m_{2}\right) g$
B $\frac{\left(m_{1}-m_{2}\right)^{2}}{m_{1}+m_{2}} g$
C $\left(m_{1}-m_{2}\right) g$
D $\frac{\left(m_{1}+m_{2}\right)^{2}}{m_{1}-m_{2}} g$

The correct answer is:
$\frac{\left(m_{1}-m_{2}\right)^{2}}{m_{1}+m_{2}} g$

Solution:
We know that Acceleration, $a_{\mathrm{CM}}=\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)^{2} \times g\left(\because m_{1}>m_{2}\right)$
So, resultant external force.
$$
\begin{aligned}
F &=\left(m_{1}+m_{2}\right) a_{\mathrm{CM}} \\
&=\left(m_{1}+m_{2}\right) \times\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)^{2} \times g \\
&=\frac{\left(m_{1}-m_{2}\right)^{2}}{\left(m_{1}+m_{2}\right)} \times g
\end{aligned}
$$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2014
A solid rectangular sheet has two different coefficients of linear expansion $\alpha_{1}$ and $\alpha_{2}$ along its length and breadth respectively. The coefficient of surface expansion is (for $\left.\alpha_{1} t< < 1, \alpha_{2} t< < 1\right)$

Options:
A $\frac{\alpha_{1}+\alpha_{2}}{2}$
B $2\left(\alpha_{1}+\alpha_{2}\right)$
C $\frac{4 \alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}}$
D $\alpha_{1}+\alpha_{2}$

The correct answer is:
$\alpha_{1}+\alpha_{2}$

Solution:
The coefficient of linear expansion along its length $=\alpha_{1}$

The coefficient of linear expansion along its breadth $=\alpha_{2}$ Increase in length,
$$
L_{i}=l_{0}\left(1+\alpha_{1} \Delta t\right)
$$
Increase in breadth,
$$
B_{t}=b_{0}\left(1+\alpha_{2} \Delta t_{2}\right)
$$
Let coefficient of surface expansion is $\beta$
$$
\begin{aligned}
\text { Area } &=\text { length } \times \text { breadth } \\
&=L_{0}\left(1+\alpha_{1} \Delta t\right) \times b_{0}\left(1+\alpha_{2} \Delta t\right) \\
&=L_{0} b_{0}\left(1+\alpha_{1} \Delta t\right)\left(1+\alpha_{2} \Delta t\right) \\
&=S_{0}\left(1+\alpha_{1} \Delta t+\alpha_{2} \Delta t+\ldots\right) \\
\text { where, } S_{0} &=l_{0} \cdot b_{0}
\end{aligned}
$$
$=$ Initial area of surface In state of expansion,
$$
\begin{aligned}
S_{t} &=L_{r} \times B_{r} \\
&=L_{0} b_{0}\left(1+\alpha_{1} \Delta t\right)\left(1+\alpha_{2} \Delta t\right) \\
&=S_{0}\left(1+\alpha_{1} \Delta t+\alpha_{2} \Delta t+\ldots\right) \\
S_{t} &=S_{0}(1+\beta \Delta t) \\
\therefore S_{0}(1+\beta \Delta t) &=S_{0}\left(1+\alpha_{1} \Delta t+\alpha_{2} \Delta t+\ldots\right) \\
\beta \cdot \Delta t &=\alpha_{1} \Delta t+\alpha_{2} \Delta t \\
\beta &=\alpha_{1}+\alpha_{2}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2017
A solid spherical ball and a hollow spherical ball of two different materials of densities $\rho_{1}$ and $\rho_{2}$ respectively have same outer radii and same mass. What will be the ratio, the moment of inertia (about an axis passing through the centre) of the hollow sphere to that of the solid sphere?

Options:
A $\frac{\rho_{2}}{\rho_{1}}\left(1-\frac{\rho_{2}}{\rho_{1}}\right)^{\frac{5}{3}}$
B $\frac{\rho_{2}}{\rho_{1}}\left[1-\left(1-\frac{\rho_{2}}{\rho_{1}}\right)^{\frac{5}{3}}\right]$
C $\frac{\rho_{2}}{\rho_{1}}\left(1-\frac{\rho_{1}}{\rho_{2}}\right)^{\frac{5}{3}}$
D $\frac{\rho_{2}}{\rho_{1}}\left[1-\left(1-\frac{\rho_{1}}{\rho_{2}}\right)^{\frac{5}{3}}\right]$

The correct answer is:
$\frac{\rho_{2}}{\rho_{1}}\left[1-\left(1-\frac{\rho_{1}}{\rho_{2}}\right)^{\frac{5}{3}}\right]$

Solution:
$\therefore \quad M_{1}=\rho_{1} \frac{4}{3} \pi R^{3}$ and $M_{2}=\rho_{2} \frac{4}{3} \pi\left(R^{3}-r^{3}\right)$
$\therefore \quad \rho_{1} \frac{4}{3} \pi R^{3}=\rho_{2} \frac{4}{3} \pi\left(R^{3}-r^{3}\right)$
$\Rightarrow \quad \rho_{1} R^{3}=\rho_{2}\left(R^{3}-r^{3}\right)$
$\Rightarrow \quad \frac{\rho_{1}}{\rho_{2}}=1-\frac{r^{3}}{R^{3}}$
$\therefore \frac{r}{R}=\left(1-\frac{\rho_{1}}{\rho_{2}}\right)^{\frac{1}{3}}$
Moment of inertia, $\begin{aligned} \frac{I_{M}}{I_{S}} &=\frac{\frac{2}{5} M_{2}\left[\frac{R^{5}-r^{3}}{R^{3}-r^{3}}\right]}{\frac{2}{5} M_{1} R^{2}} \\ &=\frac{\rho_{2} \frac{4}{3} \pi\left(R^{3}-r^{3}\right)\left[\frac{R^{5}-r^{5}}{\left(R^{3}-r^{3}\right)}\right]}{\rho_{1} \frac{4}{3} \pi R^{3} R^{2}} \\ &=\frac{\rho_{2}}{\rho_{1}}\left[\frac{R^{5}-r^{5}}{R^{5}}\right]=\frac{\rho_{2}}{\rho_{1}}\left[1-\left(\frac{r}{R}\right)^{3}\right] \end{aligned}$
$\therefore \quad \frac{I_{H}}{I_{s}}=\frac{\rho_{2}}{\rho_{1}}\left[1-\left(1-\frac{\rho_{1}}{\rho_{2}}\right)^{\frac{5}{3}}\right]$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2018
A solution containing \(4 \mathrm{~g}\) of polymer in 4.0 litre solution at \(27^{\circ} \mathrm{C}\) shows an osmotic pressure of \(3.0 \times 10^{-4} \mathrm{~atm}\). The molar mass of the polymer in \(\mathrm{g} / \mathrm{mol}\) is

Options:
A 820000
B 82000
C 8200
D 820

The correct answer is:
82000

Solution:
Hint : \(\pi=i C(M) R T\)
\(3.0 \times 10^{-4}=1 \times C(M) \times 0.0821 \times 300\)
\(\therefore \mathrm{C}(\mathrm{M})=1.22 \times 10^{-5} \text {, Molarity }=\frac{\text { no. of moles }}{\text { vol. of solution }(\mathrm{L})}\)
\(1.22 \times 10^{-5}=\frac{4 / \mathrm{M}}{4}\). Hence \(M=81967 \approx 82000 \mathrm{~g} / \mathrm{mol}\)

Topic: Chemistry, Solutions, WBJEE, WBJEE 2023
A solution is made by a concentrated solution of $\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}$ with a concentrated solution of $\mathrm{NaNO}_{2}$ in $50 \%$ acetic acid. A solution of a salt containing metal $\mathrm{M}$ is added to the mixture, when a yellow precipitate is formed. Metal ' $\mathrm{M}$ ' is

Options:
A Magnesium
B Sodium
C Potassium
D Zinc

The correct answer is:
Potassium

Solution:
$\begin{array}{l}
\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{3}+\mathrm{NaNO}_{2} \longrightarrow \mathrm{Na}_{3}\left[\mathrm{Co}\left(\mathrm{NO}_{2}\right)_{6}\right] \\
\mathrm{Na}_{3}\left[\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{6}\right]+\mathrm{K}^{+} \longrightarrow \mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{NO}_{2}\right)_{6}\right] \downarrow \text { yellow ppt }
\end{array}$

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2021