datavorous/entrance-exam-dataset huggingface dataset:
A charge of 0.8 coulomb is divided into two charges $Q_{1}$ and $Q_{2}$. These are kept at a separation of $30 \mathrm{cm}$. The force on $Q_{1}$ is maximum when

Options:
A $Q_{1}=Q_{2}=0.4 \mathrm{C}$
B $Q_{1}=0.8 \mathrm{C}, Q_{2}$ negligible
C $Q_{1}$ negligible, $Q_{2}=0.8 \mathrm{C}$
D $Q_{1}=0.2 \mathrm{C}, Q_{2}=0.6 \mathrm{C}$

The correct answer is:
$Q_{1}=Q_{2}=0.4 \mathrm{C}$

Solution:
$$
\frac{Q_{1}}{30 c m} \stackrel{Q_{2}}{6}
$$
Checking the given options one by one
(i) $Q_{1}=Q_{2}=0 \cdot 4 \mathrm{C}$
The force on $Q_{1}$ due to $Q_{2}$
$$
\begin{aligned}
F &=k_{p} \frac{Q_{1} \times Q_{2}}{30 \times 10^{-2}} \\
&=k \frac{Q_{1} Q_{2} \times 100}{30} \\
&=k \frac{0-4 \times 0.4 \times 100}{30} \\
&=\frac{k \times 0 \cdot 16 \times 100}{30}
\end{aligned}
$$
(ii) When $Q_{1}=0 \cdot 8 \mathrm{C} Q_{2}=0$
$$
F=\frac{k \times Q_{1} Q_{2}}{30 \times 10^{-2}}=0
$$
(iii) when $Q_{1}=0, Q_{2}=0-8 \mathrm{C}$
$$
F=0
$$
(iv) $Q_{1}=0 \cdot 2 \mathrm{C} \mathrm{Q}_{2}=0 \cdot 6 \mathrm{C}$
$$
\begin{aligned}
F &=\frac{k \cdot 0 \cdot 2 \times 0 \cdot 6}{30 \times 10^{-2}} \\
&=\frac{k \times 0 \cdot 12 \times 10^{2}}{30}
\end{aligned}
$$
Thus we find that, in option(a), the force on
$Q_{1}$ will be maximum.
$\therefore$ correct answer is $Q_{1}=Q_{2}=0 \cdot 4 \mathrm{C}$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2017
A charge $q$ is placed at one corner of a cube. The electric flux through any of the three faces adjacent to the charge is zero. The flux through any one of the other three faces is

Options:
A $q / 3 \varepsilon_{0}$
B $q / 6 \varepsilon_{0}$
C $q / 12 \varepsilon_{0}$
D $q / 24 \varepsilon_{0}$

The correct answer is:
$q / 24 \varepsilon_{0}$

Solution:
Consider the diagram

In above the flux coverage of three face is shown in figure. Which is like a quadrant in any plane. So, flux will be
$$
\phi=\frac{1}{4} \times \frac{q}{6 \varepsilon_{0}}=\frac{q}{24 \varepsilon_{0}}
$$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2015
A charge $+q$ is placed at the origin $O$ of $X-Y$ axes as shown in the figure. The work done in taking a charge $Q$ from $A$ to $B$ along the straight line $A B$ is

Options:
A $\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{a-b}{a b}\right)$
B $\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{b-a}{a b}\right)$
C $\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{b}{a^{2}}-\frac{1}{b}\right)$
D $\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{a}{b^{2}}-\frac{1}{b}\right)$

The correct answer is:
$\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{a-b}{a b}\right)$

Solution:
We know that

and $\quad V=\frac{Q}{4 \pi \varepsilon_{0} r}$
$\begin{aligned} \text { Hence, } \quad W &=q\left[\frac{Q}{4 \pi \varepsilon_{0} b}-\frac{Q}{4 \pi \varepsilon_{0} a}\right] \\ &=\frac{Q q}{4 \pi \varepsilon_{0}}\left[\frac{a-b}{a b}\right] \end{aligned}$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2012
A charged particle in a uniform magnetic field \(B=B_0 \hat{k}\) starts moving from the origin with velocity \(v=3 \hat{i}+4 \hat{k}\) m/s. The trajectory of the particle and the time \(t\) at which it reaches \(2 \mathrm{~m}\) above \(x-y\) plane are,

Options:
A Circular path, \(\frac{1}{2}\) sec.
B Helical path, \(\frac{1}{2}\) sec.
C Circular path, \(\frac{2}{3}\) sec.
D Helical path, \(\frac{2}{3}\) sec.

The correct answer is:
Helical path, \(\frac{1}{2}\) sec.

Solution:
Hint :
Velocity along z-direction will be const.
\(\mathrm{t}=\frac{\mathrm{s}}{V_z}=\frac{2}{4}=\frac{1}{2} \mathrm{sec}\)
Path will be helical.

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2023
A charged particle of charge \(q\) and mass \(m\) is placed at a distance \(2 R\) from the centre of a vertical cylindrical region of radius \(R\) where magnetic field varies as \(\vec{B}=\left(4 t^2-2 t+6\right) \hat{k}\) where \(t\) is time. Then which of the following statement(s) is/are true?

Options:
A Induced electric field lines form closed loops
B Electric field varies linearly with \(r\) if \(r < R\), where \(r\) is the radial distance from the centerline of the cylinder
C The charged particle will move in clockwise direction when viewed from top
D Acceleration of the charged particle is \(\frac{7 q}{2 m}\) when \(t=2 \mathrm{sec}\)

The correct answers are:
Induced electric field lines form closed loops, Electric field varies linearly with \(r\) if \(r < R\), where \(r\) is the radial distance from the centerline of the cylinder

Solution:
Hint :
\(r < R\)
\(E \times 2 \pi r=\frac{d \phi}{d t}=\frac{d}{d t}\left(4 t^2-2 t+6\right) \times \pi r^2\)
\(\begin{aligned}
& E \times 2 \pi r=(8 t-2) \pi r^2 \\
& E=\frac{(8 t-2) r}{2} \\
& E=(4 t-1) r
\end{aligned}\)
\(E \propto r\)
For \(r > R\)
\(\begin{aligned}
& E \times 2 \pi \cdot(2 R)=\frac{d}{d t}\left[4 t^2-2 t+6\right] \times \pi[R]^2 \\
& E .4 \pi R=[8 t-2] \pi R^2 \\
& E=\frac{[8 t-2] R}{4} \quad \text { at } t=2, E=\frac{14}{4} R=\frac{7 R}{2} \\
& \text { acceleration }=\frac{E q}{m}=\frac{7 R q}{2 m}=\frac{7 q R}{2 m}
\end{aligned}\)

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2023
A charged particle of mass $m_{1}$ and charge $q_{1}$ is revolving in a circle of radius $r .$ Another charged particle of charge $q_{2}$ and mass $m_{2}$ is situated at the centre of the circle. If the velocity and time period of the revolving particle be $v$ and $T$ respectively, then

Options:
A $v=\sqrt{\frac{q \cdot q_{2} r}{4 \pi \varepsilon_{0} m_{1}}}$
B $v=\frac{1}{m_{1}} \sqrt{\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r}}$
C $T=\sqrt{\frac{16 \pi^{3} \varepsilon_{0} m_{1}^{2} r^{3}}{q_{1} q_{2}}}$
D None of the above

The correct answers are:
None of the above

Solution:
The necessary centripetal force, to move in a circular path is provided by the Coulomb's force.
$$
\begin{array}{r}
\frac{m_{1} v^{2}}{\gamma}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}} \\
v=\sqrt{\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q_{1} q_{2}}{r m_{1}}}
\end{array}
$$
We know, $v=r\omega$
$$
\omega=\frac{2 \pi}{T}
$$
$$
\begin{array}{l}
T=\frac{2 \pi}{\omega} \\
=\frac{2 \pi \cdot r}{v}
\end{array}
$$
Substitution the value of $v$
$T=\frac{2 \pi r}{\sqrt{\frac{q_{1} \cdot q_{2}}{4 \pi \varepsilon_{0} R \cdot m_{1}}}}$
$T=\sqrt{\frac{4 \pi^{2} r^{2} \times 4 \pi \varepsilon_{0} \cdot r \cdot m_{1}}{q q_{2}}}$
$T=\sqrt{\frac{16 \pi^{3} \varepsilon_{0} r^{3} \cdot m_{1}}{q_{1} \cdot q_{2}}}$
None of the given options is correct.
Both the charges are positive or both negative. (Sino answers have the product $q A_{2}$ inside square loy Therefore, circular motion is not possible. The statement of question is wrong. Either $q_{1}$ or should be negative.

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2016
A chord $A B$ is drawn from the point $A(0,3)$ on
the circle $x^{2}+4 x+(y-3)^{2}=0,$ and is extended to $M$ such that $A M=2 A B$. The locus
of $M$ is

Options:
A $x^{2}+y^{2}-8 x-6 y+9=0$
B $x^{2}+y^{2}+8 x+6 y+9=0$
C $x^{2}+y^{2}+8 x-6 y+9=0$
D $x^{2}+y^{2}-8 x+6 y+9=0$

The correct answer is:
$x^{2}+y^{2}+8 x-6 y+9=0$

Solution:
Given, $A M=2 A B$
$\Rightarrow B$ is mid-point of $A M$
$\therefore$ Coordinate of $B$ is $\left(\frac{0+x_{1}}{2}, \frac{3+y_{1}}{2}\right)$
$$
=\left(\frac{x_{1}}{2}, \frac{y_{1}+3}{2}\right)
$$
since, $B$ lies on the circle $x^{2}+4 x+(y-3)^{2}=0$
$\therefore\left(\frac{x_{1}}{2}\right)^{2}+4\left(\frac{x_{1}}{2}\right)+\left(\frac{y_{1}+3}{2}-3\right)^{2}=0$
$\Rightarrow \quad \frac{x_{1}^{2}}{4}+2 x_{1}+\left(\frac{y_{1}-3}{2}\right)^{2}=0$
$\Rightarrow \quad \frac{x_{1}^{2}}{4}+2 x_{1}+\frac{y_{1}^{2}+9-6 y_{1}}{4}=0$
$\Rightarrow \quad x_{1}^{2}+y_{1}^{2}+8 x_{1}-6 y_{1}+9=0$
Hence, locus of a point is $x^{2}+y^{2}+8 x-6 y+9=0$

Topic: Mathematics, Circle, WBJEE, WBJEE 2018
A circle passing through (0,0),(2,6) (6,2) cut the $x$ -axis at the point $P \neq(0,0)$ Then, the lenght of $O P,$ where $O$ is the origin, is

Options:
A $\frac{5}{2}$
B $\frac{5}{\sqrt{2}}$
C 5
D 10

The correct answer is:
5

Solution:
Let the equation of circle is
$$
x^{2}+y^{2}+2 g x+2 f y+c=0
$$
When, circle (i) passes through the onigin
Then, $\quad c=0$
When, circle (i) passes through the point (2,6).
Then $4+36+4 g+12 f+0=0$
$\Rightarrow \quad 4 g+12 f+40=0$
$\Rightarrow g+3 f=-10$
When, circle (1) passes through the point (6,2) Then, $36+4+12 g+4 t+0=0$ [from Eq. $\Rightarrow \quad 12 g+4 f+40=0$
$\Rightarrow \quad 3 g+t+10=0$
On solving Eqs. (iii) and (iv), we get $g=\frac{-5}{2}$ and $f=\frac{-5}{2}$
$\therefore$ Equation of circle becomes
$$
x^{2}+y^{2}-5 x-5 y=0
$$
Circle cut the $x$ -axis.
So, put $y=0$ in Eq. $(v)$, we get
$$
x^{2}-5 x=0 \Rightarrow x(x-5)=0
$$
$\Rightarrow \quad x=5$
So, the circle cut the $x$ -axis at point $P(5,0)$
$\therefore$ The length of $O P=5$

Topic: Mathematics, Circle, WBJEE, WBJEE 2013
A circuit consists of three batteries of emf $E_{1}=1 \mathrm{V}, E_{2}=2 \mathrm{V}$ and $E_{3}=3 \mathrm{V}$ and internal resistances $1 \Omega, 2 \Omega$ and $1 \Omega$ respectively which are connected in parallel as shown in the figure. The potential difference between points $P$ and $Q$ is

Options:
A $1.0 \mathrm{V}$
B $2.0 \mathrm{V}$
C $2.2 \mathrm{V}$
D $3.0 \mathrm{V}$

The correct answer is:
$2.0 \mathrm{V}$

Solution:
$1 \Omega .2 \Omega$ and $1 \Omega$ are in parallel

So, the required internal resistance
$$
\begin{array}{l}
\frac{1}{r}=\frac{1}{r_{1}}+\frac{1}{r-{2}}+\frac{1}{r-{3}} \\
\frac{1}{r}=\frac{1}{1}+\frac{1}{2}+\frac{1}{1} \\
\frac{1}{r}=\frac{2+1+2}{2} \Rightarrow r=\frac{2}{5} \Omega
\end{array}
$$
The potential difference between points $P$ and $Q$
$\begin{aligned} E_{\text {dif }} &=\frac{\frac{E_{1}}{r_{1}}+\frac{E_{2}}{5_{2}}+\frac{E_{3}}{5_{3}}}{1 / r}=\frac{1}{1}+\frac{2}{2}+\frac{3}{1} \\ &=\frac{2+2+6}{5 / 2} \\=& \frac{2}{5 / 2}=\frac{10 / 2}{5 / 2}=\frac{5}{5} \times 2=2 \mathrm{V} \end{aligned}$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2014
A circular coil is placed near a current carrying conductor, both lying on the plane of the paper. The current is flowing through the conductor in such a way that the induced current in the loop is clockwise as shown in the figure. The current in the wire is,

Options:
A time dependent and downward.
B steady and upward.
C An alternating current.
D None of these

The correct answer is:
time dependent and downward.

Solution:
Hint : If current is increasing in upward direction, so magnetic field is increasing in out of plane in order to oppose it induced current will be in clockwise direction
Similarly if current is decreasing in downward direction, so magnetic field is decreasing into the plane in order to support it induced current will be in clockwise direction.

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2023
A circular disc rolls on a horizontal Iloor without slipping and the centre of the disc moves with a uniform velocity $v$. Which of the following values of the velocily at a point on the rim of the disc can have?

Options:
A $v$
B $-v$
C $2 v$
D Zero

The correct answers are:
$v$, $2 v$, Zero

Solution:
Consider the rolling motion of a circular disc as shown in the adjacent diagram. Velocity of point of contact $(P)$ is $v_{P}=v-\omega R=v-v=0$
$$
[\because v=\omega R]
$$
velocity of point ${ }^{\prime} S^{\prime}, v_{S}=v+\omega R=2 V$
velocity of centre ' $O^{\prime}, v_{0}=v-\omega(0)=V$ Velocity of a point ' $T$ at distance $R$ from $P$ is $V_{T}=\omega R=\mathrm{O}_{V}$
Thus possible velocities are $v, 2 v$ and $0 .$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2015
A circular loop of radius $r$ of conducting wire connected with a voltage source of zero internal resistance produces a magnetic field
$B$ at its centre. If instead, a circular loep of radius $2 r,$ made of same material, having the same cross-section is connected to the same voltage source, what will be the magnetic ficld at its centre?

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

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

Solution:
$$
B=\frac{\mu_{0}}{2} \cdot \frac{I}{r}
$$
$\because$
$$
I_{1}=\frac{V}{R_{1}}
$$
Case (I) $l_{1}=2 \pi r$ (length of the loop).
$$
R_{1}=\rho \frac{l_{1}}{A} \text { and } B_{1}=\frac{\mu_{0}}{2} \cdot \frac{I_{1}}{r}
$$
Case (II) $I_{2}=2 \pi \cdot 2 r=2 l_{1}$
$$
\begin{array}{l}
R_{2}=\rho \cdot \frac{l_{2}}{A}=\rho \cdot \frac{2 l_{1}}{A}=2 R_{1} \\
I_{2}=\frac{V}{R_{2}}=\frac{V}{2 R_{1}}=\frac{I_{1}}{2} \\
B_{2}=\frac{\mu_{0}}{2} \cdot \frac{I_{2}}{2 r}=\frac{\mu_{0}}{2} \cdot \frac{I_{1}}{2 \cdot 2 r}
\end{array}
$$
From Eqs. (i) and (ii),
$$
\frac{B_{2}}{B_{1}}=\frac{\frac{\mu_{0}}{2} \cdot \frac{I_{1}}{2 \cdot 2 r}}{\frac{\mu_{0}}{2} \cdot \frac{I_{1}}{r}}=\frac{1}{4}
$$
Magnetic field $B_{2}=\frac{B_{1}}{4}$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2018
A coin is tossed again and again. If tail appears on first three tosses, then the chance that head appears on fourth toss is

Options:
A \(\frac{1}{16}\)
B \(\frac{1}{2}\)
C \(\frac{1}{8}\)
D \(\frac{1}{4}\)

The correct answer is:
\(\frac{1}{2}\)

Solution:
Hints: \(p=1.1 .1 \cdot \frac{1}{2}=\frac{1}{2}\)

Topic: Mathematics, Probability, WBJEE, WBJEE 2011
A common emitter transistor amplifier is connected with a load resistance of $6 \mathrm{k} \Omega$. When a small a.c. signal of 15 $\mathrm{mV}$ is added to the base emitter voltage, the alternating base current is $20 \mu \mathrm{A}$ and the alternating collector current is $1.8 \mathrm{~mA}$. What is the voltage gain of the amplifier?

Options:
A 90
B 640
C 900
D 720

The correct answer is:
720

Solution:
Hint:
$R_{C}=6 \times 10^{3} \Omega$
$\Delta \mathrm{l}_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}=15 \times 10^{-3} \mathrm{~V}$
$\Delta \mathrm{I}_{\mathrm{B}}=20 \times 10^{-6} \mathrm{~A}$
$\Delta \mathrm{l}_{\mathrm{C}}=1.8 \times 10^{-3} \mathrm{~A}$
$\mathrm{~A}_{\mathrm{v}}=\beta \times \frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\Delta \mathrm{l}_{\mathrm{C}}}{\Delta_{\mathrm{B}}} \times \frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}_{\mathrm{B}}}=\frac{1.8 \times 10^{-3}}{20 \times 10^{-6}} \times \frac{6 \times 10^{3} \times 20 \times 10^{-6}}{15 \times 10^{-3}}=720$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2020
A compound formed by elements $X$ and $Y$ crystallises in the cubic structure, where $X$ atoms are at the corners of a cube and $Y$ atoms are at the centre of the body. The formula of the compounds is

Options:
A $XY$
B $XY_{2}$
C $X_{2} Y_{3}$
D $X Y_{3}$

The correct answer is:
$XY$

Solution:
Number of $X$ atoms at the corners $=8$ Number of $X$ atoms per unit cell $=8 \times \frac{1}{8}=1$ atom Number of $Y$ atoms at the centre of the body $=1$ atom Hence, the formula of the compound is $X Y$.

Topic: Chemistry, Solid State, WBJEE, WBJEE 2018
A compound is formed by substitution of two chlorine for two hydrogens in propane. The number of possible isomeric compounds is

Options:
A $4$
B $3$
C $5$
D $2$

The correct answer is:
$5$

Solution:
$\mathrm{C}_3 \mathrm{H}_8 \stackrel{-2 \mathrm{H}}{+2 \mathrm{Cl}}>\mathrm{C}_3 \mathrm{H}_6 \mathrm{Cl}_2$, following isomers of $\mathrm{C}_3 \mathrm{H}_6 \mathrm{Cl}_2$ is possible
Due to presence of chiral carbon compound (IV) is optically active and forms an enantiomer. So total no of isomers $=5$

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2009
A compressive force is applied to a uniform rod of rectangular cross-section so that its length decreases by $1 \%$. If the Poisson's ratio for the material of the rod be $0.2,$ which of the following statements is correct? "The volume approximately ...........

Options:
A decreases by $1 \%$
B decreases by $0.8 \%$
C decreases by $0.6 \%$
D increases by $0.2 \%$

The correct answer is:
decreases by $0.6 \%$

Solution:
Given, Decrement in the length $=1 \%$ Poisson's ratio for material of the rod, $\sigma=0.2$
As we know that, Volume, $v=\pi r^{2} l,[$ where $, l$ is the length of the rod $]$
$$
\frac{\Delta V}{V}=\frac{2 \Delta r}{r}+\frac{\Delta l}{l}...(i)
$$
since, Poisson's ratio, $\sigma=\frac{-\Delta D / D}{\Delta L / L}$
$$
=\frac{-\Delta r / r}{\Delta L / L}
$$
So, Eq. (i) can be written as,
$$
\begin{aligned}
\frac{\Delta V}{V}=\frac{\Delta l}{l}(1-2 \sigma) \\
\frac{\Delta V}{V} \times 100 &=\left(\frac{\Delta l}{l} \times 100\right) \times(1-2 \sigma) \\
&=-1 \times[1-2 \times(0.2)] \\
&=-1 \times[1-0.4] \\
&=-0.6 \%
\end{aligned}
$$
Here, negative sign shows the decrement in the volume.

Topic: Physics, Mechanical Properties of Solids, WBJEE, WBJEE 2019
A conducting circular loop of resistance $20 \Omega$ and cross-sectional area $20 \times 10^{-2} \mathrm{~m}^{2}$ is placed perpendicular to a spatially uniform magnetic field B, which varies with time tas $\mathrm{B}=2 \sin (50 \pi t)$ T. Find the net charge flowing through the loop in $20 \mathrm{~ms}$ starting from $\mathrm{t}=0$

Options:
A $0.5 \mathrm{C}$
B $0.2 \mathrm{C}$
C $0 \mathrm{C}$
D $0.14 \mathrm{C}$

The correct answer is:
$0 \mathrm{C}$

Solution:
Hint : $\Delta \mathrm{q}=\frac{\Delta \varphi}{\mathrm{R}}=\frac{\varphi_{2}-\varphi_{1}}{\mathrm{R}}=0, \Delta \mathrm{q}=0\left[\varphi_{1}=\varphi_{2}=0\right.$ at $\mathrm{t}=0$ and $\left.\mathrm{t}=20 \mathrm{~ms}\right]$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2020
A conductivity cell has been calibrated with
$0.01 \mathrm{M} \quad 1: 1$ electrolyte solution (specific conductance, $k=1.25 \times 10^{-3} \mathrm{S} \mathrm{cm}^{-1}$ ) in the cell and the measured resistance was $800 \Omega$ at $25^{\circ} \mathrm{C} .$ The cell constant will be

Options:
A $1.02 \mathrm{cm}^{-1}$
B $0.102 \mathrm{cm}^{-1}$
C $1.00 \mathrm{cm}^{-1}$
D $0.5 \mathrm{cm}^{-1}$

The correct answer is:
$1.00 \mathrm{cm}^{-1}$

Solution:
Given, $\mathrm{K}=1.25 \times 10^{-3} \mathrm{S} \mathrm{cm}^{-1}$
$$
\rho=\frac{1}{\mathrm{x}}=\frac{1}{1.25 \times 10^{-3}} \mathrm{S}^{-1} \mathrm{cm}
$$
From
$$
\begin{aligned}
R &=\rho \frac{1}{A} \\
800 &=\frac{1}{1.25 \times 10^{-3}} \times \frac{1}{A}
\end{aligned}
$$
(where, $\frac{1}{A}=$ cell constant)
$\therefore$
$$
\begin{aligned}
\frac{1}{A} &=800 \times 1.25 \times 10^{-3} \\
&=1 \mathrm{cm}^{-1}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2013
A container $\mathrm{AB}$ in the shape of a rectangular parallelopiped of length $5 \mathrm{~m}$ is divided internally by a movable partition $\mathrm{P}$ as shown in the figure. The left compartment is filled with a given mass of an ideal gas of molar mass 32 while the right compartment is filled with an equal mass of another ideal gas of molar mass 18 at same temperature. What will be the distance of P from the left wall A when equilibrium is established?

Options:
A $2.5 \mathrm{~m}$
B $1.8 \mathrm{~m}$
C $3.2 \mathrm{~m}$
D $2.1 \mathrm{~m}$

The correct answer is:
$1.8 \mathrm{~m}$

Solution:
Hint:

At equilibrium $\quad P_{1}=P_{2}, T_{1}=T_{2}$ (assuming wall to be conducting)
$\frac{P_{1} V_{1}}{\mu_{1}}=\frac{P_{2} V_{2}}{\mu_{2}}=R T$
$\frac{(5-x) 32}{m}=\frac{x \times 18}{m}$
So, $x=3.2 \mathrm{~m}$
Hence the distance of the piston from $\mathrm{A}=5-3.2=1.8 \mathrm{~m}$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2020
A convex lens of focal length \(30 \mathrm{~cm}\) produces 5 times magnified real image of an object. What is the object distance ?

Options:
A \(36 \mathrm{~cm}\)
B \(25 \mathrm{~cm}\)
C \(30 \mathrm{~cm}\)
D \(150 \mathrm{~cm}\)

The correct answer is:
\(36 \mathrm{~cm}\)

Solution:
Hints: \(\frac{1}{5 \mathrm{u}}-\left(\frac{1}{-\mathrm{u}}\right)=\frac{1}{30}\)
\(\begin{aligned}
& \frac{1}{5 \mathrm{u}}+\frac{1}{\mathrm{u}}=\frac{1}{30}, \frac{5+1}{5 \mathrm{u}}=\frac{1}{30} \\
& \mathrm{u}=36 \mathrm{~cm} .
\end{aligned}\)

Topic: Physics, Ray Optics, WBJEE, WBJEE 2011
A copper coin was electroplated with $\mathrm{Zn}$ and then heated at high temperature until there is a change in colour. What will be the resulting colour?

Options:
A white
B black
C silver
D golden

The correct answer is:
golden

Solution:
A copper coin was electroplated with zinc
(Zn) and then heated at high temperature until there is a change in colour. The resulting colour will be golden. The golden colour is due to the zinc migrating through the copper to form the alpha-form of brass alloy (percentage of $\mathrm{Cu}>65 \%$ and that of $\mathrm{Zn} < $ $35 \%)$
$$
\mathrm{Zn}+\mathrm{Cu} \stackrel{\Delta}{\longrightarrow} \text { Brass }
$$
Thus, the option (d) is correct answer.

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2019
A cricket ball of mass \(0.25 \mathrm{~kg}\) with speed \(10 \mathrm{~m} / \mathrm{s}\) collides with a bat and returns with same speed within \(0.01 \mathrm{~S}\). The force acted on bat is

Options:
A \(25 \mathrm{~N}\)
B \(50 \mathrm{~N}\)
C \(250 \mathrm{~N}\)
D \(500 \mathrm{~N}\)

The correct answer is:
\(500 \mathrm{~N}\)

Solution:
Hints: \(\Delta \mathrm{P}=2 \mathrm{mV}=2 \times 0.25 \times 10=5 \frac{\mathrm{kgm}}{\mathrm{s}}\)
\(\mathrm{F}=\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}=\frac{5}{0.01}=500 \mathrm{~N}\)

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2011
A cricket ball thrown across a field is at heights $h_{1}$ and $h_{2}$ from the point of projection at times $t_{1}$ and $t_{2}$ respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is

Options:
A $\frac{h_{1} t_{2}^{2}-h_{2} t_{1}^{2}}{h_{1} t_{2}-h_{2} t_{1}}$
B $\frac{h_{1} t_{1}^{2}+h_{2} t_{2}^{2}}{h_{2} t_{1}+h_{1} t_{2}}$
C $\frac{h_{1} t_{2}^{2}+h_{2} t_{1}^{2}}{h_{1} t_{2}+h_{2} t_{1}}$
D $\frac{h t_{1}^{2}-h t_{2}^{2}}{h_{1}-h_{2} t_{2}}$

The correct answer is:
$\frac{h_{1} t_{2}^{2}-h_{2} t_{1}^{2}}{h_{1} t_{2}-h_{2} t_{1}}$

Solution:
For vertically moment (for $\left.h_{1}\right)$
or
$\begin{array}{l}
h_{1}=u \sin \theta t_{1}-\frac{1}{2} g t_{1}^{2} \\
t_{1}=\frac{h_{1}+\frac{1}{2} g t_{1}^{2}}{u \sin \theta}
\end{array}$
$h_{2}=u \sin \theta t_{2}-\frac{1}{2} g t_{2}^{2}$
or $t_{2}=\frac{h_{2}+\frac{1}{2} g t_{2}^{2}}{u \sin \theta}$
Divide Eq. (i) by Eq. (ia)
$\begin{array}{l}
\frac{t_{1}}{t_{2}}=\frac{h_{1}+\frac{1}{2} g t_{1}^{2} / u \sin \theta}{h_{2}+\frac{1}{2} g t_{2}^{2} / u \sin \theta} \\
\Rightarrow h_{1} t_{2}-h_{2} t_{1}=\frac{g}{2}\left(t_{1} t_{2}^{2}-t_{1}^{2} t_{2}\right)
\end{array}$
The time of flight of the ball $T=\frac{2 u \sin \theta}{g}=\frac{2}{g}(u \sin \theta) \quad$
$=\frac{2}{g}\left[\frac{h_{1}+\frac{1}{2} g t_{1}^{2}}{t_{1}}\right]=\frac{2}{t_{1}}\left[\frac{h_{1}}{g}+\frac{t_{1}^{2}}{2}\right]$
$=\frac{h_{1}}{t_{1}} \times \frac{2}{g}+t_{1}=\frac{h_{1}}{t_{1}} \times\left(\frac{t_{1} t_{2}^{2}-t_{1}^{2} t_{2}}{h_{1} t_{2}-h_{2} t_{1}}\right)+t_{1}$
$=\frac{h_{1} t_{1} t_{2}^{2}-h_{1} t_{1}^{2} t_{2}+h_{1} t_{1}^{2} t_{2}-h_{2} t_{1}^{3}}{t_{1}\left(h_{1} t_{2}-h_{2} t_{1}\right)}$
$=\left(\frac{h_{1} t_{2}^{2}-h_{2} t_{1}^{2}}{h_{1} t_{2}-h_{2} t_{1}}\right)$

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2014
A cubical vessel of height \(1 \mathrm{~m}\) is full of water. What is the amount of work done in pumping water out of the vessel? (Take \(\mathrm{g}=\) \(10 \mathrm{~m} \mathrm{~s}^{-2}\) )

Options:
A \(1250 \mathrm{~J}\)
B \(5000 \mathrm{~J}\)
C \(1000 \mathrm{~J}\)
D \(2500 \mathrm{~J}\)

The correct answer is:
\(5000 \mathrm{~J}\)

Solution:
Hints:
\(\begin{aligned}
& \mathrm{V}=\ell^3=1 \mathrm{~m}^3 \\
& \mathrm{~m}=1 \times 1000=1000 \mathrm{~kg} ; \mathrm{W}=\mathrm{mgh}=1000 \times 10 \times \frac{1}{2}=5000 \mathrm{~J}
\end{aligned}\)

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2011
A current $I=I_{0} e^{-\lambda t}$ is flowing in a circuit consisting of a parallel combination of resistance $R$ and capacitance $C$. The total charge over the entire pulse period is

Options:
A $\frac{I_{0}}{\lambda}$
B $\frac{2 I_{0}}{\lambda}$
C $I_{0} \lambda$
D $e^{I_{0}\lambda}$

The correct answer is:
$\frac{I_{0}}{\lambda}$

Solution:
The current $1=1_{0} e^{-x}$ is of exponential nature.
$I=\frac{d Q}{d t}=I_{0} e^{-\lambda t}$
$d Q=I_{0} e^{-\lambda t} d t$
$\int_{0}^{Q} d Q=I_{0} \int_{t=0}^{t=-\infty} e^{-\lambda t} \cdot d t$
$Q=I_{0} \times \frac{1}{\lambda}=\frac{I_{0}}{\lambda}$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2016