datavorous/entrance-exam-dataset huggingface dataset:
Two particles $A$ and $B$ move from rest along a straight line with constant accelerations $f$ and $h,$ respectively. If $A$ takes $m$ seconds more than $B$ and describes $n$ units more than that of $B$ acquiring the same speed, then

Options:
A $(f+h) m^{2}=h n$
B $(f-f h) m^{2}=f h n$
C $(h-f) n=\frac{1}{2}$ $fhm^{2}$
D $\frac{1}{2}(f+h) n=$ $fhm^{2}$

The correct answers are:
$(h-f) n=\frac{1}{2}$ $fhm^{2}$

Solution:
Let $B$ travels $x$ units, $v=u+a t$
According to problem, $h t=f(t+m)$ $\begin{aligned} h t=& f t+f m \\ h t-f t &=f m \\ t(h-f) &=f m \\ \frac{h-f}{f} &=\frac{m}{t} \end{aligned}$
$t=m\left(\frac{f}{h-f}\right)$
$\Rightarrow \quad t^{2}=m^{2}\left(\frac{f}{h-f}\right)^{2}$
Again, $n+x=\frac{1}{2} f(t+m)^{2}$
$\Rightarrow \quad h+\frac{1}{2} h t^{2}=\frac{1}{2} f(t+m)^{2}$
From Eqs. (i) and (ii) $\frac{m}{n}=\frac{2(h-f)}{m f h}$
$\Rightarrow \quad(h-f) n=\frac{m^{2} f h}{2}$

Topic: Mathematics, Functions, WBJEE, WBJEE 2019
Two particles are simultaneously projected in the horizontal direction from a point $P$ at a certain height. The initial velocities of the particles are oppositely directed to each other and have magnitude $v$ each. The separation between the particles at a time when their position vectors (drawn from the point $P$ ) are mutually perpendicular, is

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

The correct answer is:
$\frac{4 v^{2}}{g}$

Solution:
According to the question,
Representation of position vectors of two particles (drawn from the point $P$

In two dimension, the position vectors $\mathbf{r}_{1}$ and $\mathbf{r}_{2}$ represented as
$$
\begin{array}{l}
\mathbf{r}_{1}=v t \hat{\mathbf{i}}-\frac{1}{2} g t^{2} \mathbf{j} ...(i) \\
\mathbf{r}_{2}=v t(-\hat{\mathbf{i}})-\frac{1}{2} g t^{2}(\mathbf{j})...(ii)
\end{array}
$$
$\because$ We know that, when the two vectors are mutually perpendicular, i.e.
$$
\theta=90^{\circ}
$$
So,
$$
\begin{array}{l}
\mathbf{r}_{1} \cdot \mathbf{r}_{2}=r_{1} r_{2} \cos 90^{\circ} \\
\mathbf{r}_{1} \cdot \mathbf{r}_{2}=0
\end{array}
$$
Substituting the values $\mathbf{r}_{1}$ and $\mathbf{r}_{2}$ in the above relation, we get
$$
\left((v t) \hat{\mathbf{i}}-\frac{1}{2} g t^{2} \mathbf{j}\right) \cdot\left(v t(-\hat{\mathbf{i}})-\frac{1}{2} g t^{2}(\mathbf{j})\right)=0
$$
$-v^{2} t^{2}+\frac{1}{4} 4 g^{2} t^{4}=0($ where, $\hat{\mathbf{i}} \cdot \hat{\mathbf{i}}=\hat{\mathbf{j}} \cdot \hat{\mathbf{j}}=\mathbf{k} \cdot \mathbf{k}=1)$
$$
\begin{aligned}
v^{2} t^{2} &=\frac{1}{4} g^{2} t^{4} \\
\Rightarrow \quad v^{2} &=\frac{1}{4} g^{2} t^{2}
\end{aligned}
$$
$\therefore$ Magnitude of velocity of the particles, $v=\frac{1}{2} g t$
We know that, separation distance between particles at a time $t$
$$
\begin{array}{c}
\Delta x=2 v t \\
\Delta x=2 \times v \times \frac{2 v}{g} \Rightarrow \dot{\Delta} x=\frac{4 v^{2}}{g}
\end{array}
$$

Topic: Physics, Motion In One Dimension, WBJEE, WBJEE 2019
Two particles have masses $\mathrm{m} \& 4 \mathrm{~m}$ and their kinetic energies are in the ratio $2: 1$. What is the ratio of their linear momenta?

Options:
A $\frac{1}{\sqrt{2}}$
B $\frac{1}{2}$
C $\frac{1}{4}$
D $\frac{1}{16}$

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

Solution:
$\frac{K E_1}{K E_2}=\frac{\frac{p_1^2}{2 m}}{\frac{p_2^2}{2 \times 4 m}}=\frac{2}{1} \Rightarrow \frac{p_1}{p_2}=\frac{1}{\sqrt{2}}$

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2009
Two particles move in the same straight line starting at the same moment from the same point in the same direction. The first moves with constant velocity $u$ and the second starts from rest with constant acceleration $f$. Then,

Options:
A they will be at the greatest distance at the end of time $\frac{u}{2 f}$ from the start
B they will be at the greatest distance at the end of time $\frac{u}{f}$ from the start
C their greatest distance is $\frac{u^{2}}{2 f}$
D their greatest distance is $\frac{u^{2}}{f}$

The correct answers are:
they will be at the greatest distance at the end of time $\frac{u}{f}$ from the start, their greatest distance is $\frac{u^{2}}{2 f}$

Solution:
We know, $S=u t+\frac{1}{2} a t^{2}$
Condition for first, As, velocity is constant $u=u, v=u$ (constant)
$\therefore \quad a=0, s=u t \ldots(i)$
[from Eq. (i) Condition for second $u=0, a=f$ (constant)
$\therefore$
$$
s=\frac{1}{2}a t^{2}
$$
$s=\frac{1}{2} f t^{2}$
(ui) [from Eq. (i))
On differentiating Eqs. (ii) and (iui), we get $\frac{d S}{d t}=u$
and $\frac{d S}{d t}=\frac{1}{2} f(2t)$
$\frac{d S}{d t}=f t$
$u=f t$
[from Eq.
$t=\frac{u}{f}$
Thus, they will be at the greatest distance at the end of the $\frac{u}{f}$ from the start. For greatest distance
$$
S=\frac{1}{2} f t^{2}
$$
Put $\quad t=\frac{u}{f}$
$$
\begin{array}{l}
S=\frac{1}{2} f\left(\frac{u^{2}}{f^{2}}\right) \\
S=\frac{u^{2}}{2 f}
\end{array}
$$
Hence, option (b) and (c) are correct.

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2017
Two particles of mass $m_{1}$ and $m_{2}$; approach each other due to their mutual gravitational attraction only. Then,

Options:
A accelerations of both the particles are equal
B acceleration of the particle of mass $m_{1}$ is proportional to $\overline{m_{1}}$
C acceleration of the particle of mass $m_{1}$ is proportional to $m_{2}$
D acceleration of the particle of mass $m_{1}$ is inversely proportional to $m_{1}$

The correct answer is:
acceleration of the particle of mass $m_{1}$ is proportional to $m_{2}$

Solution:
The gravitational force acting between the two masses $m_{1}$ and $m_{2}$ is given by
$$
F_{O}=\frac{G m_{1} m_{2}}{r^{2}}
$$
Force on mass $m_{1}$.
$$
F_{1}=\frac{G m_{1} m_{2}}{r^{2}}=m_{1} a_{1}
$$
where, $a_{1}=$ acceleration
$$
\Rightarrow \quad a_{1}=\frac{G m_{2}}{I^{2}} \Rightarrow a_{1} \propto m_{2}
$$
and similarly, $a_{2} \propto m_{1}$

Topic: Physics, Gravitation, WBJEE, WBJEE 2015
Two pith balls, each carrying charge $+q$ are hung from a hook by two strings. It is found that when each charge is tripled, angle between the strings double. What was the initial angle between the strings?

Options:
A $30^{\circ}$
B $60^{\circ}$
C $45^{\circ}$
D $90^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
Hint:
$\mathrm{T} \sin \theta=\mathrm{Fe}, \mathrm{T} \cos \theta=\mathrm{mg}$
$\tan \theta=\frac{\mathrm{Fe}}{\mathrm{mg}}=\frac{\mathrm{k} \cdot \mathrm{q}^{2}}{(2 \mathrm{~L} \sin \theta)^{2} \mathrm{mg}}$
In 2 nd case
$\begin{array}{l}
\tan 2 \theta=\frac{\mathrm{kq} \mathrm{q}^{2}}{(2 \mathrm{~L} \sin 2 \theta)^{2} \cdot \mathrm{mg}} \\
\frac{\tan 2 \theta}{\tan \theta}=\frac{9 \sin ^{2} \theta}{\sin ^{2} 2 \theta} \\
\frac{2}{1-\tan ^{2} \theta}=\frac{9}{4 \cos ^{2} \theta}=\frac{9}{4} \sec ^{2} \theta \\
\frac{2}{1-\tan ^{2} \theta}=\frac{9}{4}\left(1-\tan ^{2} \theta\right)
\end{array}$
$\tan ^{2} \theta=x, \frac{2}{1-x}=\frac{9}{4}(1+x)$
$\Rightarrow 8=9-9 x^{2} \Rightarrow x=\frac{1}{3} \Rightarrow \tan ^{2} \theta=\frac{1}{3} \Rightarrow \theta=30^{\circ}$
Angle between the strings $=2 \theta=60^{\circ}$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2020
Two positive charges $Q$ and $4 Q$ are placed at points $A$ and $B$ respectively, where $B$ is at a distance $d$ units to the right of $A$. The total electric potential due to these charges is minimum at $P$ on the line through $A$ and $B$. What is (are) the distance (s) of $P$ from $A ?$

Options:
A $\frac{d}{3}$ units to the right of $A$
B $\frac{d}{3}$ units to the left of $A$
C $\frac{d}{5}$ units to the right of $A$
D $d$ units to the left of $A$

The correct answers are:
$\frac{d}{3}$ units to the right of $A$

Solution:
$\because V_{P}$ is minimum. $\therefore$
$E_{p}=0 \Rightarrow E_{A}=E_{B}$
$\Rightarrow \quad \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{r^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4 Q}{(d-r)^{2}}$
$\Rightarrow \quad \frac{1}{r}=\frac{2}{d-r} \Rightarrow d-r=2 r$
$\therefore$ Distance of $P$ from the $A r=\frac{d}{3}$ units to the right of $A .$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2018
Two radionctive substances $A$ and $B$ have decay constants $5 \lambda$ and $\lambda$ respectively. At $t=0,$ they have the same number of nuclei. The ratio of number of nuclei of $A$ to that of $B$ will be $(1 / e)^{2}$
after a time interval of

Options:
A $\frac{1}{\lambda}$
B $\frac{1}{2 \lambda}$
C $\frac{1}{3 \lambda}$
D $\frac{1}{4 \lambda}$

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

Solution:
We know that
$N=N_{0} e^{-\lambda t}$
For $A$,
$N_{A}=N_{0} e^{-5 \lambda_{L}}$
For $B$,
$N_{B}=N_{0} e^{-\lambda t}$
Given,
$\begin{aligned}
\frac{N_{A}}{N_{B}} &=\frac{1}{e^{2}} \\
\frac{N}{N_{B}} &=\frac{1}{e^{4 \lambda t}} \\
t &=\frac{1}{2 \lambda}
\end{aligned}$

Topic: Physics, Nuclear Physics, WBJEE, WBJEE 2012
Two rods of equal length and diameter have thermal conductivities 3 and 4 units respectively. If they are joined in series, the thermal conductivity of the combination would be

Options:
A $3.43$
B $3.5$
C $3.4$
D $3.34$

The correct answer is:
$3.43$

Solution:
In series $R=R_1+R_2$
$\begin{aligned} & \frac{2 \ell}{K_{e f f} A}=\frac{\ell}{K_1 A}+\frac{\ell}{K_2 A} \\ & K_{e f f}=\frac{24}{7}=3.43\end{aligned}$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2009
Two simple harmonic motions are given by $x_{1}=a \sin \omega t+a \cos \omega t$ and $x_{2}=a \sin \omega t+\frac{a}{\sqrt{3}} \cos \omega t$ The ratio of the amplitudes of first and second motion and the phase difference between them are respectively

Options:
A $\sqrt{\frac{3}{2}}$ and $\frac{\pi}{12}$
B $\frac{\sqrt{3}}{2}$ and $\frac{\pi}{12}$
C $\frac{2}{\sqrt{3}}$ and $\frac{\pi}{12}$
D $\sqrt{\frac{3}{2}}$ and $\frac{\pi}{6}$

The correct answer is:
$\sqrt{\frac{3}{2}}$ and $\frac{\pi}{12}$

Solution:
The given situation can be shown as

Fon second SHM
Ratio of amplitude $=\frac{a_{1}}{a_{2}}=\frac{\sqrt{3}}{\sqrt{2}}$
and phase difference, $\frac{\pi}{4}-\frac{\pi}{6}=\frac{\pi}{12}$

Topic: Physics, Oscillations, WBJEE, WBJEE 2013
Two soap bubbles of radii \(\mathrm{x}\) and \(\mathrm{y}\) coalesee to constitute a bubble of radius \(\mathrm{z}\). Then \(\mathrm{z}\) is requal to

Options:
A \(\sqrt{x^2+y^2}\)
B \(\sqrt{x+y}\)
C \(x+y\)
D \(\frac{x+y}{2}\)

The correct answer is:
\(\sqrt{x^2+y^2}\)

Solution:
Hints: \(\mathrm{n}=\mathrm{n}_1+\mathrm{n}_2\)
\(\mathrm{pv}=\mathrm{p}_1 \mathrm{v}_1+\mathrm{p}_2 \mathrm{v}_2\)
\(\mathrm{p}_1=\mathrm{p}_0+\frac{4 \mathrm{~T}}{\mathrm{x}}, \mathrm{p}_2=\mathrm{p}_0+\frac{4 \mathrm{~T}}{\mathrm{y}}, \mathrm{p}=\mathrm{p}_0+\frac{4 \mathrm{~T}}{\mathrm{z}}\)
If the process takes place is vaccume then \(\mathrm{p}_0=0\)
\(\mathrm{p}_1=\frac{4 \mathrm{~T}}{\mathrm{x}}, \mathrm{p}_2=\frac{4 \mathrm{~T}}{\mathrm{y}}, \mathrm{p}=\frac{4 \mathrm{~T}}{\mathrm{z}}\)
If process is isothermal
\(\begin{aligned}
& \therefore \mathrm{p}_1 \mathrm{v}_1+\mathrm{p}_2 \mathrm{v}_2=\mathrm{pv} \\
& \therefore \mathrm{z}=\sqrt{\mathrm{x}^2+\mathrm{y}^2}
\end{aligned}\)

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2011
Two solid spheres of same metal but of mass $\mathrm{M}$ and $8 \mathrm{M}$ fall simultaneously on a viscous liquid and their terminal velocities are $v$ and $n v$ then value of $n$ is

Options:
A 16
B 8
C 4
D 2

The correct answer is:
4

Solution:
$$
\begin{aligned}
& \text { Hints: } m=\frac{4}{3} \pi r^3 \times \rho \\
& m \propto r^3 \\
& \left(\frac{r_1}{r_2}\right)^3=\frac{1}{8} \\
& \frac{r_1}{r_2}=\frac{1}{2} \\
& 6 \pi n r \mathrm{~V}=\frac{4}{3} \pi r^3(d=\rho) \\
& \mathrm{V} \propto r^2, \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{1}{4} \\
& n=4
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2010
Two solid spheres $S_{1}$ and $S_{2}$ of same uniform density fall from rest under gravity in a viscous medium and after some time, reach terminal velocities $v_{1}$ and $v_{2}$ respectively. If ratio of masses $\frac{m_{1}}{m_{2}}=8$, then $\frac{v_{1}}{v_{2}}$ will be equal to

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

The correct answer is:
4

Solution:
$\mathrm{v}=\frac{2 \mathrm{r}^{2}}{9 \mathrm{n}}(\mathrm{d}-\rho) \mathrm{y}$
$\mathrm{m}=\frac{4}{3} \pi \mathrm{r}^{3} \cdot \mathrm{d}$
$\mathrm{r} \propto \mathrm{m}^{1 / 3}$
$\Rightarrow \mathrm{v}_{\mathrm{T}} \propto \mathrm{r}^{2} \propto \mathrm{m}^{2 / 3}$
$\therefore \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\left[\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}\right]^{2 / 3}=(8)^{2 / 3}=4$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2021
Two spheres of equal masses but radii $r_1$ and $r_2$ are allowed to fall in a liquid of infinite column. The ratio of their terminal velocities is

Options:
A $1$
B $r_1: r_2$
C $\mathrm{r}_2: \mathrm{r}_1$
D None of the above

The correct answer is:
None of the above

Solution:
We have $\mathrm{v}_{\mathrm{T}}=\frac{2 r^2(\sigma-\rho) g}{9 \eta}$
$\frac{\mathrm{v}_1}{\mathrm{v}_2}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^2 \frac{\left(\sigma_1-\rho\right)}{\left(\sigma_2-\rho\right)} ;$ given $\mathrm{m}_1=\mathrm{m}_2 \Rightarrow\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^3=\frac{\sigma_2}{\sigma_1}$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2009
Two spheres of the same material, but of radii $R$ and $3 R$ are allowed to fall vertically downwards through a liquid of density $\rho$. The ratio of their terminal velocities is

Options:
A $1: 3$
B $1: 6$
C $1: 9$
D $1: 1$

The correct answer is:
$1: 9$

Solution:
The terminal velocity, $v=\frac{2 r^{2}(p-\sigma) g}{9 \eta}$
As, all other parameters are constant, therefore $v\propto r^{2}$
$\Rightarrow \quad \frac{v_{R}}{v_{3 R}}=\frac{(R)^{2}}{(3 R)^{2}}=\frac{1}{9}$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2013
Two substance \(A\) and \(B\) of same mass are heated at constant rate. The variation of temperature \(\theta\) of the substance with time \(\mathrm{t}\) is shown in the figure. Choose the correct statement

Options:
A Specific heat of \(A\) is greater than that of \(B\)
B Specific heat of \(B\) is greater than that of \(A\)
C Both have same specific heat
D None of the above is true

The correct answer is:
Specific heat of \(A\) is greater than that of \(B\)

Solution:
Hint :
\(\begin{aligned} & \Delta \mathrm{H}=\mathrm{mC} \Delta \theta \\ & \frac{\mathrm{dH}}{\mathrm{dt}}=\mathrm{mC} \frac{\mathrm{d} \theta}{\mathrm{dt}} \quad \frac{\mathrm{dH}}{\mathrm{dt}}=\mathrm{a} \text { constant } \\ & \therefore \frac{d \theta}{d t} \propto \frac{1}{C} \\ & \text { i.e. slope } \propto \frac{1}{C} \quad \therefore C_B < C_A \\ & \end{aligned}\)

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2023
Two tangents to the circle $x^{2}+y^{2}=4$ at the points $A$ and $B$ meet at $M(-4,0)$. The area of the quadrilateral MAOB, where $O$ is the origin is

Options:
A $4 \sqrt{3}$ sq. units
B $2 \sqrt{3}$ sq. units
C $\sqrt{3}$ sq. units
D $3 \sqrt{3}$ sq. units

The correct answer is:
$4 \sqrt{3}$ sq. units

Solution:
$\begin{aligned} & O M=4, O A=2 \\ \therefore & M A=\sqrt{16-4}=\sqrt{12}=2 \sqrt{3} \\ \therefore & \text { Area }(M A O B) \\=& 2 \times \text { area } \Delta M A O \\=& 2 \times \frac{1}{2} \times M A \times O A \\=& 2 \sqrt{3} \times 2=4 \sqrt{3} \text { sq. units } \end{aligned}$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2021
Two temperature scales \(\mathrm{A}\) and \(\mathrm{B}\) are related by \(\frac{\mathrm{A}-42}{110}=\frac{\mathrm{B}-72}{220}\). At which temperature two scales have the same reading ?

Options:
A \(-42^0\)
B \(-72^0\)
C \(+12^0\)
D \(-40^{\circ}\)

The correct answer is:
\(+12^0\)

Solution:
Hints: \(\frac{\mathrm{A}-42}{110}=\frac{\mathrm{B}-72}{220}, \mathrm{~A}=\mathrm{B}\)
\(\frac{A-42}{110}=\frac{A-72}{220}\)
\(\begin{aligned}
& 2 \mathrm{A}-84=\mathrm{A}-72 \\
& \mathrm{~A}=12
\end{aligned}\)

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2011
Two thin lenses of focal lengths \(20 \mathrm{~cm}\) and \(25 \mathrm{~cm}\) are placed in cotact. The effective power of the combination is

Options:
A 9D
B \(2 \mathrm{D}\)
C 3D
D \(\mathrm{TD}\)

The correct answer is:
9D

Solution:
Hints: \(P=P_1+P_2\)
\(=\frac{1}{f_1}+\frac{1}{f_2}=\frac{100}{20}+\frac{100}{25}=5+4=9 D\)

Topic: Physics, Ray Optics, WBJEE, WBJEE 2011
Two weights of the mass $m_{1}$ and $m_{2}\left(>m_{1}\right)$ are joined by an inextensible string of negligible mass passing over a fixed frictionless pulley. The magnitude of the acceleration of the loads is

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

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

Solution:
According to the question, we can draw the following diagram

Here, $T$ is the tension in the string.
In equilibrium condition.
For mass $m_{1}, \quad T-m_{1} g=m_{1} a$....(i)
For mass $m_{2}, \quad m_{2} g-T=m_{2} a$....(ii)
After adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
m_{2} g-m_{1} g &=m_{1} a+m_{2} a \\
\left(m_{2}-m_{1}\right) g &=\left(m_{1}+m_{2}\right) a \\
a &=\frac{\left(m_{2}-m_{1}\right)}{\left(m_{1}+m_{2}\right)} g
\end{aligned}
$$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2019
Two wires $A$ and $B$ of same length are made of same material. Load $(F)$ vs. elongation $(x)$ graph for these two wires is shown in the figure. Then which of the following statement(s) is / are true?

Options:
A The cross-section area of $A$ is greater than that of $B$.
B Young's modulus of $A$ is greater than Young's modulus of $B$.
C The cross-sectional area of $B$ is greater than that of $A$.
D Young's modulus of both A and B are same.

The correct answers are:
The cross-section area of $A$ is greater than that of $B$., Young's modulus of both A and B are same.

Solution:
same material have identical Young's Modulus.
For a given strain and Young's Modulus.
stress in same
$$
\begin{aligned}
&\frac{F_A}{A_A}=\frac{F_B}{A_B} \Rightarrow \frac{F_A}{F_B}=\frac{A_A}{A_B} \\
&\text { as } F_A>F_B, A_A>A_B
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, WBJEE, WBJEE 2022
Two wires of same radius having lengths $l_{1}$
and $I_{2}$ and resistivities $\rho_{1}$ and $\rho_{2}$ are connected in series. The equivalent resistivity will be

Options:
A $\frac{\rho_{1} l_{2}+\rho_{2} I_{1}}{\rho_{1}+\rho_{2}}$
B $\frac{\rho_{1} l_{1}+\rho_{2} /_{2}}{l_{1}+l_{2}}$
C $\frac{\rho_{1} l_{1}-\rho_{2} /_{2}}{l_{1}-l_{2}}$
D $\frac{\rho_{1} l_{2}+\rho_{2} l_{1}}{l_{1}+l_{2}}$

The correct answer is:
$\frac{\rho_{1} l_{1}+\rho_{2} /_{2}}{l_{1}+l_{2}}$

Solution:
$R=\frac{\rho l}{A}$
where, $\rho=$ resistivity $l=$ length of the resistance wire $A=$ area of cross-section
When the wires are connected in senes, then
$$
\begin{aligned}
R &=R_{1}+R_{2} \\
R &=\frac{\rho_{1} l_{1}}{A_{1}}+\frac{\rho_{2}l_{2}}{A_{2}} \\
R &=\frac{\rho_{1} l_{1}}{A}+\frac{\rho_{2}l_{2}}{A} \\
\end{aligned}
$$
$\mathrm{\rho}_{\mathrm{eq}}=$ resistivity of combination
$$
\begin{array}{l}
I=I_{1}+I_{2} \\
A=A
\end{array}
$$
Thus, $\rho_{\mathrm{eq}} \frac{I_{1}+I_{2}}{A}=\frac{\rho _{1}l_{1}+\rho_{2} l_{2}}{A}$
$$
\rho_{e q}=\frac{\rho_{1}l_{1}+\rho_{2} l_{2}}{I_{1}+I_{2}}
$$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2016
Under identical conditions, the $\mathrm{S}_{\mathrm{N}} 1$ reaction will occur most efficiently with

Options:
A tert-butyl chloride
B 1 -chlorobutane
C 2 -methyl- 1 -chloropropane
D 2 -chlorobutane

The correct answer is:
tert-butyl chloride

Solution:
$\mathrm{S}_{\mathrm{N}^{1}}$ reaction is most effective in $3^{\circ}$ carbon.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2012
Under isothermal conditions, two soap bubbles of radii a and b coalesce to form a single bubble of radius c. If the external pressure is $\mathrm{P}$, then surface tension of the bubbles is

Options:
A $\frac{\mathrm{P}\left(\mathrm{c}^{3}-\mathrm{a}^{3}+\mathrm{b}^{3}\right)}{4\left(\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{c}^{2}\right)}$
B $\frac{\mathrm{P}\left(\mathrm{c}^{3}-\mathrm{a}^{3}-\mathrm{b}^{3}\right)}{4\left(\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{c}^{2}\right)}$
C $\frac{\mathrm{P}\left(\mathrm{c}^{2}+\mathrm{a}^{2}-\mathrm{b}^{2}\right)}{4\left(\mathrm{a}^{3}+\mathrm{b}^{3}-\mathrm{c}^{3}\right)}$
D $\frac{\mathrm{P}\left(\mathrm{c}^{3}+\mathrm{b}^{3}-\mathrm{c}^{3}\right)}{4\left(\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{c}^{2}\right)}$

The correct answer is:
$\frac{\mathrm{P}\left(\mathrm{c}^{3}-\mathrm{a}^{3}-\mathrm{b}^{3}\right)}{4\left(\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{c}^{2}\right)}$

Solution:
Total number of moles $=$ constant
$$
\begin{array}{l}
\frac{P_{1} V_{1}}{R T}+\frac{P_{2} V_{2}}{R T}=\frac{P V}{R T} \Rightarrow P_{1} V_{1}+P_{2} V_{2}=P V \Rightarrow\left(P+\frac{4 T}{a}\right) \frac{4}{3} \pi a^{3}+\left(P+\frac{4 T}{b}\right) \frac{4}{3} \pi b^{3}=\left(P+\frac{4 T}{c}\right) \frac{4}{3} \pi c^{3} \\
\Rightarrow P\left(a^{3}+b^{3}-c^{3}\right)=4 T\left(c^{2}-a^{2}-b^{2}\right) \Rightarrow T=\frac{P\left(a^{3}+b^{3}-c^{3}\right)}{4\left(c^{2}-a^{2}-b^{2}\right)} \Rightarrow T=\frac{P}{4} \frac{\left(c^{3}-a^{3}-b^{3}\right)}{\left(a^{2}+b^{2}-c^{2}\right)}
\end{array}
$$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2021
Under the same reaction conditions, initial concentration of $1.386 \mathrm{~mol} \mathrm{dm}^{-3}$ of a substance becomes half in $40 \mathrm{~s}$ and 20 s through first-order and zero-order kinetics respectively. Ratio $\left(\frac{k_{1}}{k_{0}}\right)$ of the rate constants for first-order $\left(k_{1}\right)$ and zero-order $\left(\mathrm{k}_{0}\right)$ of the reactions is

Options:
A $0.5 \mathrm{~mol}^{-1} \mathrm{dm}^{3}$
B $0.5 \mathrm{~mol} \mathrm{dm}^{-3}$
C $1.0 \mathrm{~mol} \mathrm{dm}^{-3}$
D $2.0 \mathrm{~mol}^{-1} \mathrm{dm}^{3}$

The correct answer is:
$0.5 \mathrm{~mol}^{-1} \mathrm{dm}^{3}$

Solution:
For First order reaction
$\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{k}_{1}} \Rightarrow \mathrm{k}_{1}=\frac{0.693}{40} \mathrm{~s}^{-1}\quad_{-} \text {(i) }$
For Zero order reaction
$\mathrm{t}_{1 / 2}=\frac{[\mathrm{R}]_{0}}{2 \mathrm{k}_{0}} \Rightarrow \mathrm{k}_{0}=\frac{1.386}{2 \times 20} \mathrm{~mol} / \mathrm{dm}^{3} / \mathrm{s} \quad_{-} \text {(ii) }$
$\underline{\text { Using (i) and (ii) }}$
$\frac{\mathrm{k}_{1}}{\mathrm{k}_{0}}=\frac{0.693 / 40}{1.386 / 2 \times 20}=\frac{0.693}{1.386}=0.5 \mathrm{~mol}^{-1}-\mathrm{dm}^{3}$

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2021
Under which of the following condition(s) does(do) the system of equations $\left(\begin{array}{ccc}1 & 2 & 4 \\ 2 & 1 & 2 \\ 1 & 2 & (a-4)\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}6 \\ 4 \\ a\end{array}\right)$ possesses (possess) unique solution?

Options:
A $\forall \mathrm{a} \in \mathrm{R}$
B $a=8$
C for all integral values of a
D $a \neq 8$

The correct answer is:
$a \neq 8$

Solution:
$\Delta \neq 0$
$\Rightarrow\left|\begin{array}{ccc}
1 & 2 & 4 \\
2 & 1 & 2 \\
1 & 2 & \mathrm{a}-4
\end{array}\right| \neq 0 \Rightarrow \mathrm{a} \neq 8$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2022
' \(\underline{\mathrm{G}}\) ' in the above sequence of reactions is

Options:
A \(\left(\mathrm{CH}_3\right)_2 \mathrm{CHCOOCH}_2 \mathrm{CH}_3\)
B \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOCH}_2 \mathrm{CH}_3\)
C \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOCH}_2 \mathrm{CH}_2 \mathrm{CH}_3\)
D \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOCH}\left(\mathrm{CH}_3\right)_2\)

The correct answer is:
\(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOCH}_2 \mathrm{CH}_2 \mathrm{CH}_3\)

Solution:
Hint :

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2023
Upon treatment with Al(OEt) $_{3}$ followed by usual reactions (work up), CH $_{3} \mathrm{CHO}$ will produce

Options:
A only $\mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}$
B a mixture of $\mathrm{CH}_{3} \mathrm{COOH}$ and ErOH
C only $\mathrm{CH}_{3} \mathrm{COOH}$
D only EtOH

The correct answer is:
only $\mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}$

Solution:

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2012