datavorous/entrance-exam-dataset huggingface dataset:
Consider the circuit shown. If all the cells have negligible internal resistance, what will be the current through the $2 \Omega$ resistor when steady state is reached?

Options:
A $0.66 \mathrm{~A}$
B $0.29 \mathrm{~A}$
C $0 \mathrm{~A}$
D $0.14 \mathrm{~A}$

The correct answer is:
$0 \mathrm{~A}$

Solution:
Hint:

$2-8 I-8 I-2=0$
$-8 I=0$
$I=0$

Topic: Physics, Alternating Current, WBJEE, WBJEE 2020
Consider the circuit shown in the figure.
The value of the resistance $X$ for which the thermal power generated in it is independent of small variation of its resistance is

Options:
A $X=R$
B $X=\frac{R}{3}$
C $X=\frac{R}{2}$
D $X=2 R$

The correct answer is:
$X=\frac{R}{2}$

Solution:
For above circuit
$$
\begin{array}{l}
\frac{1}{R^{\prime}}=\frac{1}{R}+\frac{1}{X}=\frac{R+X}{R X} \\
R^{\prime}=\frac{R X}{R+X}
\end{array}
$$
The current in the circuit $i=\frac{E}{R+R^{\prime}}=\frac{E}{\left(R+\frac{R X}{R+X}\right)}$
$V_{k=}=\frac{E \cdot \frac{R X}{R+X}}{\left(R+\frac{R X}{R+X}\right)}=\frac{E X}{R+2 X}$
Power dissipated in the circuit
$\begin{aligned} P_{x} &=\frac{V_{B X}^{2}}{X}=\frac{E^{2} \cdot X^{2}}{X(R+2 X)^{2}} \\ &=\frac{E^{2} X}{(R+2 X)^{2}} \\ \frac{d P_{X}}{d X} &=E^{2} \frac{(R-2 X)}{(R+2 X)^{3}} \\ d P_{X} &=\frac{E^{2}(R-2 X)}{(R+2 X)^{3}} \cdot d X \end{aligned}$
$\left(d P_{x}\right)$ will be zero for all $(d X)$ if $X=\frac{R}{2} \quad\left[d P_{x}=\frac{E^{2}\left(R-2 \times \frac{R}{2}\right)}{\left(R+2 \times \frac{R}{2}\right)^{3}}\right]$
$$
=0

Topic: Physics, Current Electricity, WBJEE, WBJEE 2017
Consider the circuit shown in the figure where all the resistances are of magnitude $1 \mathrm{k} \Omega$. If the current in the extreme right resistance $X$ is $1 \mathrm{mA},$ the potential difference between $A$ and $B$ is

Options:
A $34 \mathrm{V}$
B $21 \mathrm{V}$
C $68 \mathrm{V}$
D $55 \mathrm{V}$

The correct answer is:
$34 \mathrm{V}$

Solution:
The potential between points $A$ and $B$ $V_{N}=34$ volt

Topic: Physics, Current Electricity, WBJEE, WBJEE 2017
Consider the curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. The portion of the tangent at any point of the curve intercepted between the point of contact and the directrix subtends at the corresponding focus an angle of

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

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

Solution:
Hint:
Property

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2020
Consider the curve $y=$ be $^{-\frac{x}{ \text { a }}}$ where $a$ and $b$ are non-zero real numbers. Then

Options:
A $\frac{x}{a}+\frac{y}{b}=1$ is tangent to the curve at $(0,0)$
B $\frac{x}{a}+\frac{y}{b}=1$ is tangent to the curve where the curve crosses the axis of $y$
C $\frac{x}{a}+\frac{Y}{b}=1$ is tangent to the curve at $(a, 0)$
D $\frac{x}{a}+\frac{y}{b}=1$ is tangent to the curve at $(2 a, 0)$

The correct answer is:
$\frac{x}{a}+\frac{y}{b}=1$ is tangent to the curve where the curve crosses the axis of $y$

Solution:
Hint:
$y-b=-\frac{b}{a}(x) \Rightarrow \frac{x}{a}+\frac{y}{b}=1$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2020
Consider the equation $y-y_1=m\left(x-x_1\right)$. If $m$ and $x_1$ are fixed and different lines are drawn for different values of $y_1$, then

Options:
A the lines will pass through a fixed point
B there will be a set of parallel lines
C all lines intersect the line $x=x_1$
D all lines will be parallel to the line $y=x_1$

The correct answers are:
there will be a set of parallel lines, all lines intersect the line $x=x_1$

Solution:
$m$ and $x_1$ are fixed and $y_1$ is variable
Hence, (B) There will be a set of parallel lines
(C) All lines intersect the line $x=x_1$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2022
Consider the following reaction for $2 \mathrm{NO}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2} \mathrm{F}(g) . \quad$ The expression for the rate of reaction in terms of the rate of change of partial pressure of reactant and product is/are

Options:
A rate $=-\frac{1}{2}\left[\frac{d p\left(\mathrm{NO}_{2}\right)}{d t}\right]$
B rate $=\frac{1}{2}\left[\frac{d p\left(\mathrm{NO}_{2}\right)}{d t}\right]$
C rate $=-\frac{1}{2}\left[\frac{d p\left(\mathrm{NO}_{2} F\right)}{d t}\right]$
D rate $=\frac{1}{2}\left[\frac{d p\left(\mathrm{NO}_{2} \mathrm{F}\right)}{\mathrm{dt}}\right]$

The correct answers are:
rate $=-\frac{1}{2}\left[\frac{d p\left(\mathrm{NO}_{2}\right)}{d t}\right]$, rate $=\frac{1}{2}\left[\frac{d p\left(\mathrm{NO}_{2} \mathrm{F}\right)}{\mathrm{dt}}\right]$

Solution:
$2 \mathrm{NO}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2} \mathrm{F}(g)$
$$
\begin{aligned}
\text { Rate of reaction } &=-\frac{1}{2}\left[\frac{d p\left(\mathrm{NO}_{2}\right)}{d t}\right] \\
&=-\left[\frac{d p\left(F_{2}\right)}{d t}\right] \\
&=+\frac{1}{2}\left[\frac{d p\left(\mathrm{NO}_{2} F\right)}{d t}\right]
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2013
Consider the following salts: $\mathrm{NaCl}, \mathrm{HgCl}_{2}, \mathrm{Hg}_{2} \mathrm{Cl}_{2}, \mathrm{CuCl}_{2}, \mathrm{CuCl}$ and $\mathrm{AgCl}$. Identify the correct set of insoluble salts in water.

Options:
A $\mathrm{Hg}_{2} \mathrm{Cl}_{2}, \mathrm{CuCl}, \mathrm{AgCl}$
B $\mathrm{Hg} \mathrm{Cl}_{2}, \mathrm{CuCl}, \mathrm{AgCl}$
C $\mathrm{Hg}_{2} \mathrm{Cl}_{2}, \mathrm{CuCl}_{2}, \mathrm{AgCl}$
D $\mathrm{Hg}_{2} \mathrm{Cl}_{2}, \mathrm{CuCl}, \mathrm{NaCl}$

The correct answer is:
$\mathrm{Hg}_{2} \mathrm{Cl}_{2}, \mathrm{CuCl}, \mathrm{AgCl}$

Solution:
Salts such as $\mathrm{NaCl}, \mathrm{HgCl}_{2}, \mathrm{CuCl}_{2}$ are soluble in water while salts such as $\mathrm{Hg}_{2} \mathrm{Cl}_{2}, \mathrm{CuCl}$ and $\mathrm{AgCl}$ are insoluble in water.

Topic: Chemistry, Practical Chemistry, WBJEE, WBJEE 2015
Consider the following two first order reactions occurring at 298 K with same initial concentration of $A$ :(1) $A \rightarrow B ;$ rate constant, $k=0.693 \mathrm{min}^{-1}$ (2) $A \rightarrow C$; half-life, $l_{1 / 2}=0.693$ min Choose the correct option.

Options:
A Reaction (1) is faster than reaction (2).
B Reaction (1) is slower than reaction (2).
C Both reactions proceed at the same rate.
D Since two different products are formed, rates cannot be compared.

The correct answer is:
Reaction (1) is slower than reaction (2).

Solution:
Given, for two first order reactions.
$$
A \rightarrow B k=0.693 \mathrm{min}^{-1}...(1)
$$
and
$$
A \rightarrow C, t_{1 / 2}=0.693 \mathrm{min}...(2)
$$
For first order reaction
$$
t_{1 / 2}=\frac{0.693}{k}
$$
Thus, for the reaction $A \rightarrow B$
$$
t_{1 / 2}=\frac{0.693}{k}=\frac{0.693}{0.693}=1.0 \mathrm{min}
$$
Also, for first order reaction, lower the $t_{1 / 2}$ value, faster is the reaction.
$$
t_{1 / 2} \propto \frac{1}{k}
$$
and $k$ for $A \rightarrow C$ is $\frac{0.693}{0.693}=1 \mathrm{min}^{-1}$
Thus, reaction (2) is faster than reaction
(1) or reaction (1) is slower than reaction (2) Hence, option (b) is the correct answer.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2019
Consider the function $f(x)=\frac{x^{3}}{4}-\sin \pi x+3$

Options:
A $f(x)$ does not attain value within the interval [-2,2]
B $f(x)$ takes on the value $2 \frac{1}{3}$ in the interval [-2,2]
C $f(x)$ takes on the value $3 \frac{1}{4}$ in the interval [-2,2]
D $f(x)$ takes no value $\rho, 1 < p < 5$ in the interval [-2,2]

The correct answers are:
$f(x)$ takes on the value $2 \frac{1}{3}$ in the interval [-2,2], $f(x)$ takes on the value $3 \frac{1}{4}$ in the interval [-2,2]

Solution:
\(f(-2)=1\) and \(f(2)=5\) and \(f\) is continuous also.
So intermediate value theorem, function \(f(x)\) takes all values between 1 to 5 .
\(\Rightarrow \quad 2 \frac{1}{3}\) and \(3 \frac{1}{4}\) lies in 1 to 5 so option \(B, C\) are correct

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2019
Consider the function $f(x)=\cos x^{2}$. Then,

Options:
A $f$ is of period $2 \pi$
B $f$ is of period $\sqrt{2 \pi}$
C $f$ is not periodic
D $f$ is of period $\pi$

The correct answer is:
$f$ is not periodic

Solution:
We have,
$$
f(x)=\cos x^{2}
$$
Let $T$ be the period of $f(x)$. Then
$$
\begin{aligned}
& & f(x+T) &=f(x) \\
\Rightarrow & & \cos (x+T)^{2} &=\cos x^{2}
\end{aligned}
$$
But there is no value of $T$ for which
$$
\cos (x+T)^{2}=\cos x^{2}
$$
$\therefore f(x)$ is not periodic

Topic: Mathematics, Functions, WBJEE, WBJEE 2019
Consider the function $y=\log _{a}\left(x+\sqrt{\left.x^{2}+1\right)}\right.$ $a>0, a \neq 1 .$ The inverse of the function

Options:
A does not exist
B is $x=\log _{1 / a}\left(y+\sqrt{y^{2}+1}\right)$
C is $x=\sinh (y$ loga)
D is $x=\cosh \left(-y \log \frac{1}{a}\right)$

The correct answers are:
is $x=\sinh (y$ loga)

Solution:
Given, $y=\log _{a}\left(x+\sqrt{x^{2}+1}\right), a>0, a \neq 1$
$\Rightarrow \quad a^{y}=\left(x+\sqrt{x^{2}+1}\right)$
$\Rightarrow \begin{aligned} a^{-y} &=\frac{1}{x+\sqrt{x^{2}+1}} \\ &=\sqrt{x^{2}+1}-x \end{aligned}$
$\Rightarrow \quad a^{y}-a^{-y}=2 x \Rightarrow x=\frac{a^{y}-a^{-y}}{2}$
$\Rightarrow \quad f^{-1}(y)=\frac{e^{y \log a}-e^{-y \log a}}{2}$
$\Rightarrow f^{-1}(y)=\sinh (y \log a)$$\quad\left(\because \frac{e^{x}-e^{-x}}{2}=\sinh (x)\right)$

Topic: Mathematics, Functions, WBJEE, WBJEE 2018
Consider the functions $f_{1}(x)=x, f_{2}(x)=2+\log _{e} x, x>0$. The graphs of the functions intersect.

Options:
A once in $(0,1)$ but never in $(1, \infty)$
B once in $(0,1)$ and once in $\left(e^{2}, \infty\right)$
C once in $(0,1)$ and once in $\left(e, e^{2}\right)$
D more than twice in $(0, \infty)$

The correct answer is:
once in $(0,1)$ and once in $\left(e, e^{2}\right)$

Solution:
$\mathrm{f}_{1}(\mathrm{x})=\mathrm{x}, \mathrm{f}_{2}(\mathrm{x})=2+\log _{\mathrm{e}} \mathrm{x}$
Let $g(x)=f_{2}(x)-f_{1}(x)=2+\log _{e} x-x$
$g\left(o^{+}\right) < 0, g(1)>0, g(e)>0, g\left(e^{2}\right) < 0$
and value of $g(x)$ for all $x \geq e^{2}$ is negative.
$\therefore \mathrm{g}(\mathrm{x})=0$ has two roots in $(0,1)$ and $\left(\mathrm{e}, \mathrm{e}^{2}\right)$

Topic: Mathematics, Functions, WBJEE, WBJEE 2021
Consider the logic circuit with inputs A, B, C and output \(Y\). How many combinations of \(A, B\) and \(C\) gives the output \(Y=0\) ?

Options:
A 8
B 5
C 7
D 1

The correct answer is:
7

Solution:
Hint :
\(\begin{aligned}
& \overline{(A+\overline{B})+\overline{C}}=\overline{A+\overline{B}} \cdot C=(\overline{A} \cdot B \cdot C) \\
& \therefore y=1 \\
& \text { if } A=0, B=1, C=1
\end{aligned}\)
For rest of all cases \(Y=O\)
Total cases \(=8\)
\(\therefore\) Ans \((8-1)=7\)

Topic: Physics, Semiconductors, WBJEE, WBJEE 2023
Consider the non-constant differentiable function $f$ of one variable which obeys the relation $\frac{f(x)}{f(y)}=f(x-y) .$ If $f^{\prime}(0)=p$ and $f^{\prime}(5)=q,$ then $f^{\prime}(-5)$ is

Options:
A $\frac{p^{2}}{q}$
B $\frac{q}{p}$
C $\frac{p}{q}$
D $q$

The correct answer is:
$\frac{p^{2}}{q}$

Solution:
We have, $\frac{f(x)}{f(y)}=f(x-y)$
$\Rightarrow \quad f(x)=a^{k x}$
$\therefore \quad f^{\prime}(x)=k a^{b x} \log a$
$\begin{array}{lr}\text { Again, } & f^{\prime}(0)=P \\ \Rightarrow & k a^{\circ} \log a=P \\ \Rightarrow & k \log a=P\end{array}$
Also, $\quad f^{\prime}(5)=q$
$\Rightarrow \quad k a^{5 k} \log a=q$
$a^{5 k} p=q$
$a^{5 k}=\frac{q}{P}$
Now, $\begin{aligned} f^{\prime}(-5) &=k a^{-4} \log a \\ &=\frac{k \log a}{a^{\Delta}} \end{aligned}$ $=\frac{P}{\left(\frac{q}{p}\right)}=\frac{p^{2}}{q}$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2017
Consider the parabola $y^{2}=4 x$. Let $P$ and $Q$ be points on the parabola where $P(4,-4)$ and $Q (9,6)$. Let $R$ be a point on the arc of the parabola between $P$ and $Q$. Then, the area of $\Delta P Q R$ is largest when

Options:
A $\angle P Q A=90^{\circ}$
B $R(4,4)$
C $R\left(\frac{1}{4}, 1\right)$
D $R\left(1, \frac{1}{4}\right)$

The correct answer is:
$R\left(\frac{1}{4}, 1\right)$

Solution:
Let $\quad f(x)=x \log x+x-3$
$\Rightarrow \quad f^{\prime}(x)=x \cdot \frac{1}{x}+\log x+1$
$\Rightarrow \quad f^{\prime}(x)=\log x+2>0$
$\Rightarrow f(1)=-2$ and $f(3)=3 \log 3, f(1) \cdot f(3 < 0)$
Hence, exactly one root in $x \in(1,3)$ as $f(x)>0$

Equation of $P Q$ is $2 x-y=12$
Perpendicular distance $A R=\left|\frac{2 t^{2}-2 t-12}{\sqrt{5}}\right|=\frac{2}{\sqrt{5}}(t-3 \cdot(t+4)$
$A R$ is Maximum, at $t=\frac{1}{2}$
$\therefore R$ is $\left(\frac{1}{4}, 1\right)$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2018
Consider the P-V diagram for 1 mole of an ideal monatomic gas shown in the figure. Which of the following statements is/are true?

Options:
A The change in internal energy for the whole process is zero
B Heat is rejected during the process
C Change in internal energy for process $A \rightarrow B$ is $-\frac{3}{2} P_{0} V_{0}$
D Work done by the gas during the entire process is $2 \mathrm{P}_{0} \mathrm{~V}_{0}$

The correct answers are:
The change in internal energy for the whole process is zero, Heat is rejected during the process, Change in internal energy for process $A \rightarrow B$ is $-\frac{3}{2} P_{0} V_{0}$

Solution:
$$
\begin{array}{l}
\Delta \mathrm{U}=\mathrm{n} \mathrm{Cv} \Delta \mathrm{T} \\
\Delta \mathrm{T}=\frac{\mathrm{P}_{\mathrm{c}} \mathrm{V}_{\mathrm{c}}}{\mathrm{nR}}-\frac{\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}}{\mathrm{nR}}=0
\end{array}
$$
Hence $\Delta \mathrm{U}=0$ (option $\mathrm{A}$ is correct)
$$
\begin{array}{l}
\Delta \mathrm{Q}=\Delta \mathrm{U}+\text { work } \\
\Delta \mathrm{U}=0, \text { work }=(-)
\end{array}
$$
$\Rightarrow \Delta Q=(-)$; Hence heat is rejected during the process (option B is correct) for process $A B$
$\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}}\left(\frac{\mathrm{P}_{\mathrm{C}} \mathrm{V}_{\mathrm{B}}}{\mathrm{nR}}-\frac{\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}}{\mathrm{nR}}\right)=\frac{3 \mathrm{R}}{2}\left(\frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}-\frac{\mathrm{P}_{0} 2 \mathrm{~V}_{0}}{\mathrm{R}}\right)=-\frac{3}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$ (option $\mathrm{C}$ is correct)
during entire process work done by gas is $-P_{0} V_{0} \quad$ (option $D$ is correct)

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2021
Consider the real valued function $\mathrm{h}:\{0,1,2 \ldots \ldots 100\} \rightarrow \mathrm{R}$ such that $\mathrm{h}(0)=5, \mathrm{~h}(100)=20$ and satisfying $h(p)=\frac{1}{2}\{h(p+1)+h(p-1)\}$ for every $p=1,2 \ldots .99 .$ Then the value of $h(1)$ is

Options:
A $5.15$
B $5.5$
C 6
D $6.15$

The correct answer is:
$5.15$

Solution:
$h(p)=\frac{1}{2}(h(p+1)+h(p-1)) \Rightarrow h(p-1), h(p), h(p+1)$ are in A.P.
$h(100)=h(0)+99 d$
$\Rightarrow \frac{20-5}{99}=d \Rightarrow d=\frac{15}{99} \Rightarrow h(1)=h(0)+d=5+\frac{15}{99}=5.15$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2021
Consider the vectors
$\overrightarrow{\mathrm{A}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}, \overrightarrow{\mathrm{B}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{C}}=\frac{1}{\sqrt{5}}(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$. What is the value of $\overrightarrow{\mathrm{C}} .(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}) ?$

Options:
A 1
B 0
C $3 \sqrt{2}$
D $18 \sqrt{5}$

The correct answer is:
0

Solution:
Hint:
$\overrightarrow{\mathrm{C}}(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})=\left|\begin{array}{ccc}1 & 1 & -1 \\ 2 & -1 & 1 \\ 1 & -2 & 2\end{array}\right|=1(-2+2)-1(4-1)-1(-4+1)=-3+3=0$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2020
Consider three vectors $A=i+j-2 k,\mathbf{B}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} \text { and } \mathbf{C}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} . \text { A vector } \mathbf{X} \text { of }$ the form $\alpha A+\beta B(\alpha$ and $\beta$ are numbers) is perpendicular to C. The ratio of $\alpha$ and $\beta$ is

Options:
A $1: 1$
B $2: 1$
C $-1: 1$
D $3: 1$

The correct answer is:
$1: 1$

Solution:
Vector $\mathrm{X}$ of the form $\alpha \mathrm{A}+\beta \mathrm{B}$
$\begin{aligned}
\mathrm{X} &=\alpha \mathrm{A}+\beta \mathrm{B} \\
&=\alpha(\hat{i}+\hat{j}-2 \hat{k})+\beta(\hat{i}-\hat{j}+\hat{k}) \\
X &=\hat{i}(\alpha+\beta)+\hat{j}(\alpha-\beta)+\hat{k}(-2 \alpha+\beta)
\end{aligned}$
A vector $\mathrm{X}$ is perpendicular to $\mathrm{C},$ i.e., $\mathrm{X} \cdot \mathrm{C}=0$
$\text {or } -9 \alpha+9 \beta=0 \text { or } \alpha=\beta$
or $\frac{\alpha}{\beta}=\frac{1}{1} \text { or } \alpha: \beta=1: 1$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2014
Consider two concentric conducting sphere of radii $\mathrm{R}$ and $2 \mathrm{R}$ respectively. The inner sphere is given a charge $+\mathrm{Q}$. The other sphere is grounded. The potential at $r=\frac{3 R}{2}$ is

Options:
A $\frac{1}{4 \pi \varepsilon_0} \frac{Q}{6 R}$
B 0
C $\frac{1}{4 \pi \varepsilon_0} \frac{2 Q}{3 R}$
D $\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}$

The correct answer is:
$\frac{1}{4 \pi \varepsilon_0} \frac{Q}{6 R}$

Solution:
$q \rightarrow$ Charge appearing on shell of radius $2 R$ after earthing
Calculating $\mathrm{q} \quad \mathrm{V}_{\text {outer shell }}=\frac{\mathrm{KQ}}{2 \mathrm{R}}+\frac{\mathrm{Kq}}{2 \mathrm{R}}=0 \quad \mathrm{q}=-\mathrm{Q}$
Calculating potential at $r=\frac{3 R}{2}$
$$
V_{\text {inner shell }}+V_{\text {outer shell }}=\frac{K Q}{r}+\frac{K q}{2 R}=\frac{K Q}{\frac{3 R}{2}}+\frac{K(-Q)}{2 R}=\frac{K Q}{6 R}=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{6 R}
$$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2022
Consider two concentric spherical metal shells of radii $r_{1}$ and $r_{2}\left(r_{2}>r_{1}\right)$. If the outer shell has a charge $q$ and the inner one is grounded, the charge on the inner shell is

Options:
A $\frac{-r_{2}}{5}, q$
B zero
C $\frac{-r_{1}}{r_{2}} q$
D $-q$

The correct answer is:
$\frac{-r_{1}}{r_{2}} q$

Solution:
Let, the charge on the inner shell be $q^{\prime}$
The total charge will be zero
So, $\frac{k q^{\prime}}{r_{1}}+\frac{k q}{r_{2}}=0$
$\left(\because r_{2}>r_{1}\right)$
$\Rightarrow \quad q^{\prime}=-\frac{K q / r_{2}}{K / r_{1}}$
$$
q^{\prime}=-\left(\frac{r_{1}}{r_{2}}\right) q
$$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2014
Consider two infinitely long wires parallel to Z-axis carrying same current I in the positive Z direction. One wire passes through the point $L$ at coordinates $(-1,+1)$ and the other wire passes through the point $M$ at coordinates $(-1,-1)$. The resultant magnetic field at the origin $\mathrm{O}$ will be

Options:
A $\frac{\mu_{0} I}{2 \sqrt{2 \pi}} \hat{\mathrm{j}}$
B $\frac{\mu_{0} I}{2 \pi} \hat{j}$
C $\frac{\mu_{0} \mathrm{I}}{2 \sqrt{2 \pi}} \hat{\mathrm{i}}$
D $\frac{\mu_{0} I}{4 \pi} \hat{j}$

The correct answer is:
$\frac{\mu_{0} I}{2 \pi} \hat{j}$

Solution:
$\mathrm{B}_{\text {net }}=\sqrt{2} \mathrm{~B}$ where $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$
$\mathrm{B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \sqrt{2} \hat{\mathrm{j}}$
$=\frac{\mu_{0} \mathrm{I}}{2 \pi \sqrt{2}} \times \sqrt{2} \hat{\mathrm{j}}$ and $\mathrm{r}=\sqrt{2}$
$\mathrm{B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{2 \pi} \hat{\mathrm{j}}$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2021
Consider two particles of different masses, In which of the following situations the heavier of the two particles will have smaller de-Broglie wavelength?

Options:
A Both have a free fall through the same height
B Both move with the same kinetic energy
C Both move with the same linear momentum
D Both move with the same speed

The correct answers are:
Both have a free fall through the same height, Both move with the same kinetic energy, Both move with the same speed

Solution:
The de-Broglie equation is
$$
\begin{aligned}
\lambda &=\frac{h}{m v}=\frac{h}{\sqrt{2 m k}} \Rightarrow m=\frac{h}{\lambda v} \\
\Rightarrow \quad m \propto \frac{1}{\lambda}, \text { if } \frac{h}{v} &=\mathrm{constant}
\end{aligned}
$$
Again, $\frac{h}{v}$ will be constant only when $v$ is constant
Thus, in present case the de-Broglie wavelength of heavier particle will be smaller. When the velocities of two particles will be same or both move with same kinetic energy(K).

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2015
Correct pair of compounds which gives blue colouration/precipitate and white precipitate, respectively, when their Lassaigne's test is separately done is

Options:
A NH- $\mathrm{NH}_{2}$, HCl and CICH, COOH
B $\mathrm{NH}_{2} \mathrm{CSNH}_{2}$ and $\mathrm{PhCH}_{2} \mathrm{Cl}$
C $\mathrm{NH}_{2} \mathrm{NH}_{2}, \mathrm{COOH}$ and $\mathrm{NH}_{2} \mathrm{CONH}_{2}$
D

The correct answer is:

Solution:
$\begin{array}{ll}\text {Hydrozine, } & \mathrm{NH}_{2} \cdot \mathrm{NH}_{2} \text { does not respond }\end{array}$
Lassaigne's test because it does not contain any carbon and hence, NaCN is not formed. Compound, $\mathrm{NH}_{2}-\mathrm{CS}-\mathrm{NH}_{2}$ contains both $\mathrm{N}$ and $\mathrm{S}$, hence, it will give red colour in Lassaigne test.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2013
Correct statement(s) in cases of $n$ -butanol and $t$ -butanol is (are)

Options:
A both are having equal solubility in water
B $t$ -butanol is more soluble in water than $n$ -butanol
C boiling point of $t$ -butanol is lower than $n$ -butanol
D boiling point of $n$ -butanol is lower than $t$ -butanol

The correct answers are:
$t$ -butanol is more soluble in water than $n$ -butanol, boiling point of $t$ -butanol is lower than $n$ -butanol

Solution:
More branching results high solubility and low boiling point.

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2013
$\cos (2 x+7)=a(2-\sin x)$ can have a real solution for

Options:
A all real values of a
B $a \in[2,6]$
C $\mathrm{a} \in[-\infty, 2] \backslash\{0\}$
D $\mathrm{a} \in(0, \infty)$

The correct answer is:
$a \in[2,6]$

Solution:
Step 1. find the solution of given equation:
\(\begin{aligned}
&\cos 2 x+a \sin x=2 a-7 \\
&\Rightarrow 1-2 \sin ^2 x+a \sin x=2 a-7 ;\left[\because \cos 2 x=1-2 \sin ^2 x\right] \\
&\Rightarrow 2 \sin ^2 x-a \sin x+2 a-8=0
\end{aligned}\)
Step 2. Find the roots of above equation:
\(\begin{aligned}
&\Rightarrow \sin x=\frac{a \pm \sqrt{a^2-16(a-4)}}{4} ;\left[\because \boldsymbol{x}=\frac{-\mathbf{b} \pm \sqrt{\mathbf{b}^2-4 \mathbf{a c}}}{2 \mathbf{a}}\right] \\
&=\frac{a \pm(a-8)}{4} \\
&\Rightarrow \sin x=\frac{a-4}{2} \text { or } \sin x=2(\text { not possible }) \\
&\Rightarrow \sin x=\frac{a-4}{2} \\
&\because-1 \leq \sin x \leq 1 \\
&\therefore-1 \leq \frac{a-4}{2} \leq 1 \\
&\Rightarrow 2 \leq a \leq 6 \\
&\Longrightarrow a \in[2,6]
\end{aligned}\)

Topic: Mathematics, Limits, WBJEE, WBJEE 2020
$\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}$

Options:
A is equal to zero
B lies between 0 and 3
C is a negative number
D lies between 3 and 6

The correct answer is:
is a negative number

Solution:
Let $\frac{2 \pi r}{7}=\theta$
$\Rightarrow \quad 2 \pi r=3 \theta+4 \theta$
$\Rightarrow \quad 4 \theta=2 \pi r-3 \theta$
$\Rightarrow \quad \sin 4 \theta=\sin (2 \pi r-3 \theta)$
$\Rightarrow \quad \sin 4 \theta=-\sin 3 \theta$
$\Rightarrow 2 \sin 2 \theta \cos 2 \theta=-\left[3 \sin \theta-4 \sin ^{3} \theta\right.$
$\Rightarrow 2 \times 2 \sin \theta \cos \theta\left(2 \cos ^{2} \theta-1\right)$
$=-3 \sin \theta+4 \sin ^{3} \theta$
$\Rightarrow \sin \theta\left[8 \cos ^{3} \theta-4 \cos \theta+3-4\left(1-\cos ^{2} \theta\right)\right]=0$
$\Rightarrow 8 \cos ^{3} \theta+4 \cos ^{2} \theta-4 \cos \theta-1=0$ (i)
Thus, $\cos \frac{2 \pi}{7}, \cos \frac{4 \pi}{7}$ and $\cos \frac{6 \pi}{7}$ are the roots of
the above equation. $\therefore \quad \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}=-\frac{1}{2}$

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2014