datavorous/entrance-exam-dataset huggingface dataset:
Three identical convex lenses each of focal length \(f\) are placed in a straight line separated by a distance \(f\) from each other. An object is located at \(\mathrm{f} / 2\) in front of the leftmost lens. Then,

Options:
A Final image will be at \(f / 2\) behind the rightmost lens and its magnification will be -1 .
B Final image will be at \(f / 2\) behind the rightmost lens and its magnification will be +1 .
C Final image will be at \(f\) behind the rightmost lens and its magnification will be -1 .
D Final image will be at \(f\) behind the rightmost lens and its magnification will be +1 .

The correct answer is:
Final image will be at \(f / 2\) behind the rightmost lens and its magnification will be -1 .

Solution:
Hint :
For first lens
\(\begin{aligned}
& u=-\frac{f}{2} \\
& \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{f}=\frac{1}{v}-\frac{1}{\frac{-f}{2}} \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{2}{f} \Rightarrow \frac{1}{v}=\frac{-1}{f} \\
& v=-f \\
& m_1=\frac{v}{u}=\frac{-f}{\frac{-f}{2}}=2
\end{aligned}\)
For second lens
\(u=-(f+f)=-2 f\)
\(\begin{aligned}
& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}-\frac{1}{-2 f}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{2 f} \Rightarrow \frac{1}{v}=\frac{1}{2 f} \\
& v=2 f \\
& m_1=\frac{v}{u}=\frac{2 f}{-2 f}=-1
\end{aligned}\)
For third lens
\(\begin{aligned}
& u=f \\
& f=f \\
& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{2}{f} \Rightarrow v=\frac{f}{2}
\end{aligned}\)
\(m_3=\frac{v}{u}=\frac{\frac{f}{2}}{f}=\frac{1}{2}\)
Total magnification \(=m_1 m_2 m_3=2 \times(-1) \times \frac{1}{2}=-1\)

Topic: Physics, Ray Optics, WBJEE, WBJEE 2023
Three identical square plates rotate about the axes shown in the figure in such a way that their kinetic energies are equal. Each of the rotation axes passes through the centre of the square. Then the ratio of angular speeds $\omega_{1}: \omega_{2}: \omega_{3}$ is

Options:
A 01:01:01
B $\sqrt{2}: \sqrt{2}: 1$
C $1: \sqrt{2}: 1$
D $1: 2: \sqrt{2}$

The correct answer is:
$\sqrt{2}: \sqrt{2}: 1$

Solution:
We know that
$$
K=\frac{1}{2} \mathrm{l} \omega^{2}
$$
where $K=$ kinetic energy
$$
\begin{array}{l}
I=\text { moment of inertia } \\
\omega=\text { angular speed }
\end{array}
$$
So, $\quad \omega \propto \frac{1}{\sqrt{I}}$
$$
\begin{aligned}
\omega_{1}: \omega_{2}: \omega_{3} &=\frac{1}{\sqrt{I_{1}}}: \frac{1}{\sqrt{I_{2}}}: \frac{1}{\sqrt{I_{3}}} \\
&=\frac{1}{\sqrt{1}}: \frac{1}{\sqrt{1}}: \frac{1}{\sqrt{2}}=\sqrt{2}: \sqrt{2}: 1
\end{aligned}
$$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2014
Three infinite plane sheets carrying uniform charge densities $-\sigma, 2 \sigma, 4 \sigma$ are placed parallel to $X Z$ plane at $Y=a, 3 a$, 4a respectively. The electric field at the point $(0,2 a, 0)$ is

Options:
A $\frac{5 \sigma}{2 \varepsilon_{0}} \hat{j}$
B $-\frac{7 \sigma}{2 \varepsilon_{0}} \hat{j}$
C $\frac{\sigma}{2 \varepsilon_{0}} \hat{j}$
D $\frac{5 \sigma}{-2 \varepsilon_{0}} \hat{\mathrm{j}}$

The correct answer is:
$-\frac{7 \sigma}{2 \varepsilon_{0}} \hat{j}$

Solution:
$\begin{aligned} \mathrm{E}_{\text {net }} &=7 \mathrm{E} \\ &=7 \frac{\sigma}{2 \varepsilon_{0}} \text { alone }-\text { ve } \mathrm{Y} \text {-axis } \\ & \therefore \overrightarrow{\mathrm{E}}=-\frac{7 \sigma}{2 \varepsilon_{0}} \hat{\mathrm{j}} \end{aligned}$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2021
Three lines are drawn from the origin $O$ with direction cosines proportional to (1,-1,1) (2,-3,0) and $(1,0,3) .$ The three lines are

Options:
A not coplanar
B coplanar
C perpendicular to each other
D coincident

The correct answer is:
coplanar

Solution:
$\quad$ $\Delta=\left|\begin{array}{ccc}1 & -1 & 1 \\ 2 & -3 & 0 \\ 1 & 0 & 3\end{array}\right|$
$=1(-9-0)-(-1) \cdot(6-0)+1(0-(-3))$
$=-9+6+3=0$
since, $\Delta=0$
$\therefore$ Lines are coplanar.

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2017
Three numbers are chosen at random from 1 to 20 . The probability that they are consecutive is

Options:
A $\frac{1}{190}$
B $\frac{1}{120}$
C $\frac{3}{190}$
D $\frac{5}{190}$

The correct answer is:
$\frac{3}{190}$

Solution:
Hints: Total number of cases ; ${ }^{20} \mathrm{C}_3=\frac{20 \times 19 \times 18}{2 \times 3}=20 \times 19 \times 3=1140$
Total number of favourable cases $=18$
$\therefore$ Required probability $=\frac{18}{1140}=\frac{3}{190}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2009
Three particles, each of mass ' $m$ ' grams situated at the vertices of an equilateral $\triangle A B C$ of side ' $a$ ' $\mathrm{cm}$ (as shown in the figure). The moment of inertia of the system about a line $\mathrm{AX}$ perpendicular to $\mathrm{AB}$ and in the plane of $\mathrm{ABC}$ in $\mathrm{g}-\mathrm{cm}^2$ units will be

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

The correct answer is:
$\frac{5}{4} \mathrm{ma}^2$

Solution:
$\mathrm{I}=\mathrm{~ma}^2+\mathrm{m}\left(\mathrm{a} \cos 60^{\circ}\right)^2=m \mathrm{a}^2+\frac{\mathrm{ma}^2}{4}=\frac{5 \mathrm{ma}^2}{4}$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2022
Three unequal positive numbers a, b, c are such that a, b, c are in G.P. while $\log \left(\frac{5 c}{2 a}\right), \log \left(\frac{7 b}{5 c}\right), \log \left(\frac{2 a}{7 b}\right)$ are in A.P. Then $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are the lengths of the sides of

Options:
A an isosceles triangle
B an equilateral triangle
C a scalene triangle
D a right-angled triangle

The correct answer is:
a scalene triangle

Solution:
$\log \frac{5 c}{2 a}+\log \frac{2 a}{7 b}=2 \log \frac{7 b}{5 c} \Rightarrow \log \frac{5 c}{7 b}=\log \frac{49 b^{2}}{25 c^{2}}$
$\begin{array}{l}
\Rightarrow 5^{3} \mathrm{c}^{3}=7 \mathrm{~b}^{3} \Rightarrow 5 \mathrm{c}=7 \mathrm{~b} \Rightarrow \mathrm{c}=\frac{7}{5} \mathrm{~b} \\
\because \mathrm{b}^{2}=\mathrm{ac}=\mathrm{a} \cdot \frac{7}{5} \mathrm{~b} \Rightarrow \mathrm{a}=\frac{5 \mathrm{~b}}{7}
\end{array}$
Sides are $\frac{5 b}{7}, b, \frac{7}{5} b$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2021
Three vectors $\mathbf{A}=a \mathbf{i}+\mathbf{j}+\mathbf{k} ; \mathbf{B}=\mathbf{i}+b \mathbf{j}+\mathbf{k}$
and $\mathbf{C}=\mathbf{i}+\mathbf{j}+c \mathbf{k}$ are mutually perpendicular (i, $\mathbf{j}$ and $\mathbf{k}$ are unit vectors along $X, Y$ and $Z$ axes respectively). The respective values of $a$, $b$ and $c$ are

Options:
A 0,0,0
B $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}$
C 1,-1,1
D $\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$

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

Solution:
Three vectors are mutually perpendicular, so the scalar or dot product of two vectors will zero. So,
$\mathbf{A} \cdot \mathbf{B}=0, \mathbf{B} \mathbf{C}=0,$
So, $\mathbf{A} \cdot \mathbf{B}=(a \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+b \hat{\mathbf{j}}+\hat{\mathbf{k}})=0$
$\Rightarrow a+b+1=0$
$\mathbf{B} \cdot \mathbf{C}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \cdot \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}=$
$\Rightarrow 1+b+c=0$
$\mathbf{A} \cdot \mathbf{C}=(a \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+c \hat{\mathbf{k}})=0$
$\mathbf{A} \cdot \mathbf{C}=0$
$\Rightarrow \quad a+1+c=0$
(iii Addition of Eqs. (i), (ii) and (iii),
$\begin{array}{l}
2(a+b+c)+3=0 \\
a+b+c=-\frac{3}{2}
\end{array}$
$\Rightarrow \quad-1+c=-\frac{3}{2}$
$\begin{aligned}
c &=-\frac{3}{2}+1 \\
&=-\frac{1}{2}=0.5 \\
a &=-\frac{1}{2} \\
b &=-\frac{1}{2} \\
c &=-\frac{1}{2}
\end{aligned}$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2017
Time period $T$ of a simple pendulum of length $l$ is given by $T=2 \pi \sqrt{\frac{l}{g}}$. If the length is increased by $2 \%$, then an approximate change in the time period is

Options:
A $2 \%$
B $1 \%$
C $\frac{1}{2} \%$
D None of these

The correct answer is:
$1 \%$

Solution:
Given, $T=2 \pi \sqrt{\frac{1}{g}}$
On differentiating w.r.t. $l,$ we get
$\begin{aligned} \frac{d T}{d l} &=\frac{2 \pi}{\sqrt{g}} \cdot \frac{1}{2 \sqrt{I}} \\ \Delta T &=\frac{d T}{d l} \Delta l=\frac{\pi}{\sqrt{g} l} \cdot\left(\frac{2 I}{100}\right) \\ &=2 \pi \sqrt{\frac{1}{g}} \cdot \frac{1}{100}=\frac{T}{100} \\ \frac{\Delta T}{T} &=\frac{1}{100}=1 \% \end{aligned}$
Hence, approximate change in the time period is $1 \%$.

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2016
To a solution of a colourless efflorescent sodium salt, when dilute acid is added, a colourless gas is evolved along with formation of a white precipitate. Acidified dichromate solution turns green when the colourless gas is passed through it. The sodium salt is

Options:
A $\mathrm{Na}_{2} \mathrm{SO}_{3}$
B $\mathrm{Na}_{2} \mathrm{~S}$
C $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$
D $\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}$

The correct answer is:
$\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$

Solution:
Hint:
$\underset{\text { (colouriess, efflorescent) }}{\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(\mathrm{~s})}+2 \mathrm{H}^{+}(\mathrm{dil} .) \rightarrow \mathrm{S}(\mathrm{s})+2 \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{SO}_{2} \uparrow+\mathrm{H}_{2} \mathrm{O}(\ell)$
$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}($ aq. $)+3 \mathrm{SO}_{2}(\mathrm{~g})+2 \mathrm{H}^{+}(\mathrm{aq} .) \longrightarrow 2 \mathrm{~K}^{+}($ aq. $)+\underset{(\text { green })}{2 \mathrm{Cr}^{+3}}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+3 \mathrm{SO}_{4}^{-2}(\mathrm{aq})$

Topic: Chemistry, Practical Chemistry, WBJEE, WBJEE 2020
To a solution of colourless sodium salt, a solution of lead nitrate was added to have a white precipitate which dissolves in warm water and reprecipitates on cooling. Which of the following acid radical is present in the salt?

Options:
A $\mathrm{Cl}^{-}$
B $\mathrm{SO}_4{ }^{2-}$
C $\mathrm{S}^{2-}$
D $\mathrm{NO}_3^{-}$

The correct answer is:
$\mathrm{Cl}^{-}$

Solution:
$2 \mathrm{NaCl}+\mathrm{Pb}\left(\mathrm{NO}_3\right)_2 \rightarrow \mathrm{PbCl}_2(\mathrm{ppt})+2 \mathrm{NaNO}_3$

Topic: Chemistry, Practical Chemistry, WBJEE, WBJEE 2022
To determine the coefficient of friction between a rough surface and a block, the surface is kept inclined at $45^{\circ}$ and the block is released from rest. The block takes a time $t$ in moving a distance $d$. The rough surface is then replaced by a smooth surface and the same experiment is repeated. The block now takes a time $t / 2$ in moving down the same distance d. The coefficient of friction is

Options:
A $3 / 4$
B $5 / 4$
C $1 / 2$
D $1 / \sqrt{2}$

The correct answer is:
$3 / 4$

Solution:
If the same wedge is made rough then time taken by it to come down becomes $n$ times more. The coefficient of friction
$$
\begin{array}{l}
\mu=\left[1-\frac{1}{n^{2}}\right] \tan \theta \Rightarrow \mu=\left[1-\frac{1}{2^{2}}\right] \tan 45^{\circ} \\
\mu=\frac{4-1}{4} \Rightarrow \mu=\frac{3}{4}
\end{array}
$$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2014
To determine the composition of a bimetallic alloy, a sample is first weighed in air and then in water. These weights are found to be $w_{1}$ and $w_{2}$ respectively. If the densities of the two constituents metals are $\rho_{1}$ and $\rho_{2}$ respectively. then the weight of the first metal in the sample is (where $\rho_{w}$ is the density of water)

Options:
A $\frac{\rho_{1}}{\rho_{w}\left(\rho_{2}-\rho_{1}\right)}\left[w_{1}\left(\rho_{2}-\rho_{w}\right)-w_{2} \rho_{2}\right]$
B $\frac{\rho_{1}}{\rho_{w}\left(\rho_{2}+\rho_{1}\right)}\left[w_{1}\left(\rho_{2}-\rho_{w}\right)+w_{2} \rho_{2}\right]$
C $\frac{\rho_{1}}{\rho_{w}\left(\rho_{2}-\rho_{1}\right)}\left[w_{1}\left(\rho_{2}+\rho_{w}\right)-w_{2} \rho_{1}\right]$
D $\frac{\rho_{1}}{\rho_{w}\left(\rho_{2}-\rho_{1}\right)}\left[w_{1}\left(\rho_{1}-\rho_{w}\right)-w_{z} \rho_{1}\right]$

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

Solution:
By Archimedes' Principle
$$
F=v \rho_{w} g \Rightarrow\left(w_{1}-w_{2}\right) g=v \rho_{w} g
$$
Let, the total volume be $v$ and first metal weight be $x$
$$
\begin{array}{l}
w_{1}-w_{2}=\left(v_{1}+v_{2}\right) \rho_{w} \\
w_{1}-w_{2}=v_{1} \rho_{w}+v_{2} g_{w} \\
w_{1}-w_{2}=\left(\frac{x}{\rho_{1}} \rho_{w}+\frac{w_{1}-x}{\rho_{2}} \rho_{w}\right)
w_{1}-w_{2}=\frac{x \rho_{\mathcal{P}} \rho_{w}+\left(w_{1}-x\right) \rho_{w} \rho_{1}}{\rho_{\jmath} \rho_{2}}
\end{array} \quad\left(\because v=\frac{m}{\rho}\right)
$$
$$
\begin{aligned} w_{1} \rho_{1} \rho_{2}-w_{2} \rho_{1} \rho_{2} &=x \rho_{2} \rho_{w}+w_{1} \rho_{w} \rho_{1}-x \rho_{w} \rho_{1} \\ x\left(\rho_{2}-\rho_{1}\right) \rho_{w} &=\rho_{1}\left[w_{1}\left(\rho_{2}-\rho_{w}\right)-w_{2} \rho_{2}\right] \end{aligned}\\
x=\frac{\rho_{1}}{\rho_{w}\left(\rho_{2}-\rho_{1}\right)}\left[w_{1}\left(\rho_{2}-\rho_{w}\right)-w_{z} \rho_{2}\right]
$$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2014
To observe an elevation of boiling point of $0.05^{\circ} \mathrm{C},$ the amount of a solute (mol. wt. $=100$ ) to be added to $100 \mathrm{g}$ of water $\left(K_{b}=0.5\right)$ is

Options:
A $2 \mathrm{g}$
B $0.5 \mathrm{g}$
C $1 \mathrm{g}$
D $0.75 \mathrm{g}$

The correct answer is:
$1 \mathrm{g}$

Solution:
Elevation of boiling point,
$$
\Delta T_{b}=\frac{w \times K_{b} \times 1000}{M \times W}
$$
(Here, w and $W=$ weights of solute and solvent respectively.
$M=$ molecular weight of solute and $K_{b}=$ constant
On substituting values, we get
of
$$
\begin{aligned}
0.05 &=\frac{W \times 0.5 \times 1000}{100 \times 100} \\
w &=\frac{0.05 \times 100 \times 100}{0.5 \times 1000}=1 \mathrm{g}
\end{aligned}
$$

Topic: Chemistry, Solutions, WBJEE, WBJEE 2014
To which of the following quantities, the radius of the circular path of a charged particle moving at right angles to a uniform magnetic field is directly proportional?

Options:
A energy of the particle
B magnetic field
C charge of the particle
D momentum of the particle

The correct answer is:
momentum of the particle

Solution:
If velocity of particle, $\mathrm{v}$ is perpendicular to magnetic field, $\mathbf{B}$ i.e. $\theta=90^{\circ},$ then particle will experience maximum magnetic force, i.e. $F_{\max }=q v B .$ This force acts in a direction perpendicular to the motion of charged particle. Therefore the trajectory of the particle is a circle. In this case path of charged particle is circular and magnetic force provides the necessary centripetal force,
i.e.
$$
q v B=\frac{m v^{2}}{r}
$$
$\Rightarrow$ Radius of path, $r=\frac{m v}{q B}$
$$
r=\frac{p}{q B} \quad[\because p=m v]
$$
where, $p=$ momentum of the particle
$\therefore r$ momentum
Hence, option (d) is correct

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2019
To which of the following the angular velocity of the electron in the $n$ -th Bohr orbit is proportional?

Options:
A $n^{2}$
B $\frac{1}{n^{2}}$
C $\frac{1}{n^{3 / 2}}$
D $\frac{1}{n^{3}}$

The correct answer is:
$\frac{1}{n^{3}}$

Solution:
According to the Bohr's atomic model.
$\text {Angular momentum, } L=m v=\frac{n h}{2 \pi}....(i)$
since, angular velocity $\omega=\frac{v}{r}$
$\Rightarrow v=r \omega$
From Eq. (i), we get
$\begin{aligned}
m(r \omega) r &=\frac{n h}{2 \pi} \\
m \omega r^{2} &=\frac{n h}{2 \pi} \\
\omega &=\frac{n h}{2 \pi m r^{2}}....(ii)
\end{aligned}$
since, the radius of the electron in $n^{t h}$ orbit of
Bohr's atomic model is given as, $r=\frac{n^{2} h^{2} \varepsilon_{0}}{\pi m z e^{2}} \ldots$ (iii)
Squaring the Eq. (iii) and substituting its value in Eq. (ii), we get
$\omega=\frac{n h(\pi n z e)^{2}}{2 \pi m\left(n^{4} h^{4} \varepsilon_{0}^{2}\right)} \Rightarrow \omega \propto \frac{1}{n^{3}}$

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2019
To write the decimal number 37 in binary, how many binary digits are required?

Options:
A $5$
B $6$
C $7$
D $4$

The correct answer is:
$6$

Solution:

Topic: Physics, Semiconductors, WBJEE, WBJEE 2009
Transforming to parallel axes through a point $(p, q), \quad$ the $\quad$ equation $2 x^{2}+3 x y+4 y^{2}+x+18 y+25=0$ becomes
$2 x^{2}+3 x y+4 y^{2}=1 .$ Then

Options:
A $p=-2, q=3$
B $p=2, q=-3$
C $p=3, q=-4$
D $p=-4, q=3$

The correct answer is:
$p=2, q=-3$

Solution:
Given equations are
$$
\begin{array}{r}
2 x^{2}+3 x y+4 y^{2}+x+18 y+25=0 ...(i)\\
2 x^{2}+3 x y+4 y^{2}+1=0 ....(ii)
\end{array}
$$
Let the origin be transferred to $(p, q)$ axes being parallel to the previous axes; then the equation
(i) becomes.
$$
\begin{array}{l}
2\left(x^{\prime}+p\right)^{2}+3\left(x^{\prime}+p\right)\left(y^{\prime}+q\right) \\
+4\left(y^{\prime}+q\right)^{2}+\left(x^{\prime}+p\right)+18\left(y^{\prime}+q\right)+25=0 \\
\Rightarrow 2 x^{2}+2 p^{2}+4 x^{\prime} p+3 x^{\prime} y^{\prime}+3 x^{\prime} q+3 p y^{\prime} \\
+3 p q+4 y^{2}+4 q^{2}+8 y^{\prime} q+x^{\prime}+p \\
+18 y^{\prime}+18 q+25=0 \\
\Rightarrow 2 x^{2}+4 y^{a}+3 x^{\prime} y^{\prime}+(4 p+3 q+1) x^{\prime} \\
1+(3 p+8 q+18) y^{\prime}+2 p^{2}+3 p q \\
+4 q^{2}+p+25=0
\end{array}
$$
From Eq. (ii) coefficient of $x^{\prime}$ and $y^{\prime}$ must be
zero.
$\therefore$
$$
4 p+3 q+1=0 ...(iii)
$$
$$
3 p+8 q+18=0 ...(iv)
$$
By solving Eqs. (iii) and (iv), we get
$$
p=2, q=-3
$$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2017
Treatment of $\text { with } \mathrm{NaNH}_{2} \text { /liq. } \mathrm{NH}_{3}$ gives

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2013
Twenty metres of wire is available to fence off a flower bed in the form of a circular sector. What must the radius of the circle be, if the area of the flower bed be greatest?

Options:
A $10 \mathrm{~m}$
B $4 \mathrm{~m}$
C $5 \mathrm{~m}$
D $6 m$

The correct answers are:
$5 \mathrm{~m}$

Solution:
Let, radius $=r$, arc length $=\ell$
$2 \mathrm{r}+\ell=20 \Rightarrow \ell=20-2 \mathrm{r}$
$\Rightarrow \mathrm{A}=\frac{1}{2}\mathrm{r}^{2} \theta = \frac{1}{2} \ell \mathrm{r} \Rightarrow \frac{\mathrm{dA}}{\mathrm{dr}}=\frac{1}{2}(20-4 \mathrm{r})=0 \Rightarrow \mathrm{r}=5 \quad \therefore \mathrm{r}=5 \mathrm{~m}\right.$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2022
Two aromatic compounds having formula \(\mathrm{C}_7 \mathrm{H}_8 \mathrm{O}\) which are easily identifiable by \(\mathrm{FeCl}_3\) solution test (violet colouration) are

Options:
A o- cresol and benzyl alcohol
B m-cresol and p-cresol
C o-cresol and p-cresol
D methyl phenyl ether and benzyl alcohol

The correct answer is:
o- cresol and benzyl alcohol

Solution:
Hints: \(\mathrm{O}\) - cresol contains phenolic group, thus it gives violet coloration with \(\mathrm{FeCl}_3\) where as benzylalchol donot contains phenolic group, hence no coloration with \(\mathrm{FeCl}_3\). Hence Identifiable

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2011
Two base balls (masses: \(m_1=100 \mathrm{~g}\), and \(m_2=50 \mathrm{~g}\) ) are thrown. Both of them move with uniform velocity, but the velocity of \(m_2\) is 1.5 times that of \(m_1\). The ratio of de Broglie wavelengths \(\lambda\left(m_1\right): \lambda\left(m_2\right)\) is given by

Options:
A \(4: 3\)
B \(3: 4\)
C \(2: 1\)
D \(1: 2\)

The correct answer is:
\(3: 4\)

Solution:
Hint : \(\frac{\lambda_1}{\lambda_2}=\frac{m_2 \mathrm{~V}_2}{m_1 \mathrm{~V}_1}=\frac{50 \times 1.5 \mathrm{~V}_1}{100 \times \mathrm{V}_1}=\frac{1.5}{2}=\frac{3}{4}\)

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2023
Two black bodies at temperatures $327^{\circ} \mathrm{C}$ and $427^{\circ} \mathrm{C}$ are kept in an evacuated chamber at $27^{\circ} \mathrm{C}$. The ratio of their rates of loss of heat are :

Options:
A $\frac{6}{7}$
B $\left(\frac{6}{7}\right)^2$
C $\left(\frac{6}{7}\right)^3$
D $\frac{243}{464}$

The correct answer is:
$\frac{243}{464}$

Solution:
Hints: Rate ofloss of heat $\propto\left(\mathrm{T}^4-\mathrm{T}_0^4\right)$
$$
\frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{T}_1^4-\mathrm{T}_0^4}{\mathrm{~T}_2^4-\mathrm{T}_0^4}=\frac{(600)^4-(300)^4}{(700)^4-(300)^4}=\frac{6^4-3^4}{7^4-3^4}
$$
$$
\frac{E_1}{E_2}=\frac{243}{464}
$$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2010
Two blocks of $2 \mathrm{~kg}$ and $1 \mathrm{~kg}$ are in contact on a frictionless table. If a force of $3 \mathrm{~N}$ is applied on $2 \mathrm{~kg}$ block, then the force of contact between the two blocks will be :

Options:
A $0 \mathrm{~N}$
B $1 \mathrm{~N}$
C $2 \mathrm{~N}$
D $3 \mathrm{~N}$

The correct answer is:
$1 \mathrm{~N}$

Solution:
Hints: Common acceleration $=\frac{3}{3}=1 \mathrm{~m} / \mathrm{sec}^2$
$$
\mathrm{N}_1=1 \mathrm{~N}
$$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2010
Two bodies of masses $m_{1}$ and $m_{2}$ are separated by a distance $R$. The distance of the centre of mass of the bodies from the mass $m_{1}$ is

Options:
A $\frac{m_{2} R}{m_{1}+m_{2}}$
B $\frac{m_{1} R}{m_{1}+m_{2}}$
C $\frac{m_{1} m_{2}}{m_{1}+m_{2}} R$
D $\frac{m_{1}+m_{2}}{m_{1}} R$

The correct answer is:
$\frac{m_{2} R}{m_{1}+m_{2}}$

Solution:
The bodies are separated by a distance $R$. so the coordinates of $m_{1}$ and $m_{2}$ will be (0,0) and $(R, 0)$

From the formula of centre of mass,
$$
\begin{aligned}
X_{\mathrm{cm}} &=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}} \\
&=\frac{m_{1} \times 0+m_{2} \times R}{m_{1}+m_{2}} \\
X_{\mathrm{cm}} &=\frac{m_{2} \cdot R}{m_{1}+m_{2}}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2016
Two cells $A$ and $B$ of emf $2 \mathrm{V}$ and $1.5 \mathrm{V}$ respectively, are connected as shown in figure through an external resistance $10 \Omega$ The internal resistance of each cell is $5 \Omega$. The potential difference $E_{A}$ and $E_{B}$ across the terminals of the cells $A$ and $B$ respectively are

Options:
A $E_{A}=20 \mathrm{V}, E_{B}=1.5 \mathrm{V}$
B $E_{A}=2.125 \mathrm{V}, E_{B}=1.375 \mathrm{V}$
C $E_{A}=1.875 \mathrm{V}, E_{B}=1.625 \mathrm{V}$
D $E_{A}=1.875 \mathrm{V}, E_{B}=1.375 \mathrm{V}$

The correct answer is:
$E_{A}=1.875 \mathrm{V}, E_{B}=1.625 \mathrm{V}$

Solution:
The figure can be redrawn as,

The current through the circuit
$$
\begin{aligned}
i &=\frac{\text { net emf }}{\text { effective resistance }} \\
&=\frac{2-1.5}{5+5+10}=\frac{0.5}{20} \\
&=\frac{1}{40}=0.025 \mathrm{A}
\end{aligned}
$$
The terminal potential difference of the batteries
and
$$
\begin{aligned}
V_{A} &=\varepsilon_{A}-i I_{A}=2-0.025 \times 5 \\
&=2-0.0125=1.875 \mathrm{V} \\
V_{B} &=\varepsilon_{B}+i I_{B}
\end{aligned}
$$
$$
=1.5+0.025 \times 5
$$
$$
=1,5+0.0125=1.625 \mathrm{V}
$$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2015
Two cells with the same e.m.f. \(E\) and different internal resistances \(r_1\) and \(r_2\) are connected in series to an external resistance \(R\). The value of \(R\) so that the potential difference across the first cell be zero is

Options:
A \(\sqrt{r_1 r_2}\)
B \(r_1+r_2\)
C \(\mathrm{r}_1-\mathrm{r}_2\)
D \(\frac{r_1+r_2}{2}\)

The correct answer is:
\(\mathrm{r}_1-\mathrm{r}_2\)

Solution:
Hints:
\(I=\frac{2 E}{R+r_1+r_2}\)
Potential difference across first cell \(\mathrm{V}=\mathrm{E}-\mathrm{Ir}_1=0\)
\(\begin{aligned}
& E-\frac{2 E r_1}{R+r_1+r_2}=0 \\
& {\left[\frac{R+r_1+r_2-2 r_1}{R+r_1+r_2}\right]=0} \\
& \Rightarrow R+r_2-r_1=0 \\
& R=r_1-r_2
\end{aligned}\)

Topic: Physics, Current Electricity, WBJEE, WBJEE 2011
Two charges, each equal to $-q$ are kept at $(-a, 0)$ and $(a, 0)$. A charge $q$ is placed at the origin. If $q$ is given a small displacement along y direction, the force acting on q is proportional to

Options:
A $y$
B -y
C $\frac{1}{y}$
D $-\frac{1}{y}$

The correct answer is:
-y

Solution:
$\vec{F}=+q$ (Electric field due to the two negative charges $=\vec{E}$ )
$$
\vec{F}=+q\left[\frac{k(-2 q) \vec{y}}{a^3}\right] \Rightarrow \vec{F}=-\frac{2 k q^2 y}{a^3} \Rightarrow \vec{F} \propto-\vec{y}
$$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2022