datavorous/entrance-exam-dataset huggingface dataset:
A metal $M$ (specific heat 0.16 ) forms a metal chloride with $65 \%$ chlorine present in it. The formula of the metal chloride will be

Options:
A MCI
B $M C l_{2}$
C $M \mathrm{Cl}_{3}$
D $M C l_{4}$

The correct answer is:
$M C l_{2}$

Solution:
Given, specific heat $=0.16$
Let metal chloride be $M \mathrm{Cl}_{x}$ then. $\frac{6.4}{\text { specific heat }}=$ Atomic weight of metal $\frac{6.4}{0.16}=$ Atomic weight of metal
Atomic weight $=40$
40 is the atomic weight of calcium. According to question, metal chloride $\left(M C l_{x}\right)$ have $\approx 65 \%$ chlorine present in it. $\frac{x \times \text { Atomic weight of chlorine }}{40+x \times \text { Atomic weight of chlorine }} \times 100=65$
$$
\begin{array}{l}
\frac{x \times 355}{40+x \times 355} \times 100=65 \\
x=209 \approx 2(a p p r o x .)
\end{array}
$$
So, the formula of metal chloride will be $M \mathrm{Cl}_{2}$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2018
A metal rod is fixed rigidly at two ends so as to prevent its thermal expansion. If $L, \alpha$ and $Y$ preventively denote the length of the rod, respective of linear thermal expansion and coefficient of Young's modulus of its material, then for an Young's modulus of its material, then for an increase in temperature of the rod by $\Delta T$, the longitudinal stress developed in the rod is

Options:
A inversely proportional to $\alpha$
B inversely proportional to Y
C directly proportional to $\Delta T / Y$
D independent of $L$

The correct answer is:
directly proportional to $\Delta T / Y$

Solution:
Strain $=\alpha \cdot \Delta T$
Stress $Y \propto \Delta T$
where $L=$ length of the rod $\boldsymbol{\alpha}=$ coefficient of linear thermal expansion
$Y=$ Young's modulus of its material So, the longitudinal stress developed in the rod is directly proportional to $\frac{\Delta T}{Y}$

Topic: Physics, Mechanical Properties of Solids, WBJEE, WBJEE 2014
A metal sphere of radius $R$ carrying charge $q$ is surrounded by a thick concentric metal shell of inner and outer radii a and $\mathrm{b}$ respectively. The net charge on the shell is zero. The potential at the centre of the sphere, when the outer surface of the shell is grounded will be

Options:
A $\frac{\mathrm{q}}{4 \pi \epsilon_{0}}\left(\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{~b}}\right)$
B $\frac{\mathrm{q}}{4 \pi \epsilon_{0}} \frac{1}{\mathrm{a}}$
C $\frac{\mathrm{q}}{4 \pi \epsilon_{0}}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{a}}\right)$
D $\frac{\mathrm{q}}{4 \pi \epsilon_{0}} \frac{1}{\mathrm{R}}$

The correct answer is:
$\frac{\mathrm{q}}{4 \pi \epsilon_{0}}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{a}}\right)$

Solution:
\(V_{A}=V_{B}+V_{C}+V_{c}\)
\(V_{c}=\frac{1}{4 x \varepsilon_{0}}\left(\frac{-q}{a}\right)+\frac{1}{4 x \varepsilon_{0}}\left(\frac{q}{R}\right)+0\)
\(=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q}{R}-\frac{q}{a}\right)\)
\(=\frac{2}{4 \pi \varepsilon 0}\left[\frac{1}{R}-\frac{1}{a}\right]\)

Topic: Physics, Electrostatics, WBJEE, WBJEE 2021
A metallic block of mass $20 \mathrm{~kg}$ is dragged with a uniform velocity of $0.5 \mathrm{~ms}^{-1}$ on a horizontal table for $2.1 \mathrm{~s}$. The co-efficient of static friction between the block and the table is $0.10$. What will be the maximum possible rise in temperature of the metal block if the specific heat of the block is $0.1$ C.G.S unit ? Assume $g=10 \mathrm{~ms}^{-2}$ uniform rise in temperature throughout the whole block. [lgnore absorption of heat by the table]

Options:
A $0.0025^{\circ} \mathrm{C}$
B $0.025^{\circ} \mathrm{C}$
C $0.001^{\circ} \mathrm{C}$
D $0.05^{\circ} \mathrm{C}$

The correct answer is:
$0.0025^{\circ} \mathrm{C}$

Solution:
Hint:
friction $f=\mu \mathrm{mg}=0.1 \times 20 \times 10=20$
$\mathrm{~W}_{\mathrm{f}}=$ heat
$\Rightarrow(\mathrm{f.u}) \mathrm{t}=\mathrm{mc} \Delta \mathrm{T}$
$\Rightarrow 20 \times 0.5 \times 2.1=20 \times 0.1 \times 4.2 \times 10^{3} . \Delta \mathrm{T}$
$\Rightarrow \Delta \mathrm{T}=0.0025^{\circ} \mathrm{C}$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2020
A missile is fired from the ground level rises \(x\) meters vertically upwards in \(t \sec\), where \(x=100 t-\frac{25}{2} t^2\). The maximum height reached is

Options:
A \(100 \mathrm{~m}\)
B \(300 \mathrm{~m}\)
C \(200 \mathrm{~m}\)
D \(125 \mathrm{~m}\)

The correct answer is:
\(200 \mathrm{~m}\)

Solution:
Hint : \(x=100 t-\frac{25}{2} t^2\)
\(\frac{d x}{d t}=100-25 \mathrm{t}\) for maximum and minimum \(\frac{d x}{d t}=0 \Rightarrow t=4 \mathrm{sec}\)
\(\frac{d^2 x}{d t^2}=-25 < 0\)
\(\therefore\) at \(\mathrm{x}=4\) it will have maixmum height
\(x_{\max }=100 \times 4-\frac{25}{2} \times 16=400-200=200 \mathrm{~m}\)

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2023
A modified gravitational potential is given by \(V=-\frac{G M}{r}+\frac{A}{r^2}\). If the constant \(A\) is expressed in terms of gravitational constant (G), mass ( \(M\) ) and velocity of light (c), then from dimensional analysis, \(A\) is,

Options:
A \(\frac{\mathrm{G}^2 \mathrm{M}^2}{\mathrm{c}^2}\)
B \(\frac{\mathrm{GM}}{\mathrm{c}^2}\)
C \(\frac{1}{\mathrm{c}^2}\)
D Dimensionless

The correct answer is:
\(\frac{\mathrm{G}^2 \mathrm{M}^2}{\mathrm{c}^2}\)

Solution:
Hint : \(V=-\frac{G M}{r}+\frac{A}{r^2}\)
\([A]=\frac{[G M]}{[r]}\left[r^2\right]=[G M][r]\)
now, we know, \(\frac{G M}{r}\) gives dimension of \(c^2\)
\(\frac{[\mathrm{GM}]}{[\mathrm{r}]}=\left[\mathrm{c}^2\right] \Rightarrow[\mathrm{r}]=\frac{[\mathrm{GM}]}{\left[\mathrm{c}^2\right]}\)
\(\Rightarrow[\mathrm{A}]=\frac{[\mathrm{GM}][\mathrm{GM}]}{\left[\mathrm{c}^2\right]}\)
\([\mathrm{A}]=\frac{\mathrm{G}^2 \mathrm{M}^2}{\mathrm{c}^2}\)

Topic: Physics, Gravitation, WBJEE, WBJEE 2023
A mouse of mass \(\mathrm{m}\) jumps on the outside edge of a rotating ceiling fan of moment of inertia I and radius \(\mathrm{R}\). The fractional loss of angular velocity of the fan as a result is

Options:
A \(\frac{m R^2}{\mathrm{I}+\mathrm{mR}^2}\)
B \(\frac{I}{I+m R^2}\)
C \(\frac{\mathrm{I}-\mathrm{mR}^2}{\mathrm{I}}\)
D \(\frac{I-m R^2}{I+m R^2}\)

The correct answer is:
\(\frac{m R^2}{\mathrm{I}+\mathrm{mR}^2}\)

Solution:
Hint : \(I \omega_0=\left(\mathrm{I}+\mathrm{mR}^2\right) \omega\)
\(\omega_0 \rightarrow\) Initial angular velocity
\(\omega=\frac{I \omega_0}{I+m R^2}\)
\(\omega \rightarrow\) Final angular velocity
So, \(\frac{\omega_0-\omega}{\omega_0}=I-\frac{I}{I+m R^2}=\frac{m R^2}{I+m R^2}\)

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2023
A moving line intersects the lines $x+y=0$ and $x-y=0$ at the points A, B respectively such that the area of the triangle with vertices $(0,0), A \& B$ has a constant area $C$. The locus of the mid-point $A B$ is given by the equation

Options:
A $\left(x^{2}+y^{2}\right)^{2}=C^{2}$
B $\left(x^{2}-y^{2}\right)^{2}=C^{2}$
C $(x+y)^{2}=C^{2}$
D $(x-y)^{2}=C^{2}$

The correct answer is:
$\left(x^{2}-y^{2}\right)^{2}=C^{2}$

Solution:
Let mid point of $\mathrm{AB}=(\mathrm{h}, \mathrm{k})$
Let, $A=(\alpha,-\alpha), B=(\beta, \beta)$
$\therefore \beta+\alpha=2 \mathrm{~h}, \quad \beta-\alpha=2 \mathrm{k}$
Area $\Delta \mathrm{AOB}=\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}=\frac{1}{2} \sqrt{2 \alpha^{2}} \sqrt{2 \beta^{2}}=|\alpha \beta|=\mathrm{c}$
$\begin{array}{l}
\therefore \alpha^{2} \beta^{2}=\mathrm{c}^{2} \Rightarrow 16 \alpha^{2} \beta^{2}=16 \mathrm{c}^{2} \Rightarrow\left\{(\beta+\alpha)^{2}-(\beta-\alpha)^{2}\right\}^{2}=16 \mathrm{c}^{2} \\
\Rightarrow\left(4 \mathrm{~h}^{2}-4 \mathrm{k}^{2}\right)^{2}=16 \mathrm{c}^{2} \rightarrow \text { locus }:\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)^{2}=\mathrm{c}^{2}
\end{array}$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2021
A negative charge is placed at the midpoint between two fixed equal positive charges, separated by a distance $2 d$. If the negative charge is given a small displacement $x(x< < a)$ perpendicular to the line joining the positive charges, how the force $(F)$ developed on it will approximately depend on $x ?$

Options:
A $F \propto x$
B $F \propto \frac{1}{x}$
C $F \propto x^{2}$
D $F \propto \frac{1}{x^{2}}$

The correct answer is:
$F \propto x$

Solution:
According to the question,

Force experienced by the charge $-q$ due to charge $Q$
$$
\begin{array}{l}
\qquad F=-\frac{2 k Q q}{r^{2}} \cos \theta \ldots(\mathrm{i})\left[\text { where, } k=\frac{1}{4 \pi \varepsilon_{0}}\right] \\
\text { From diagram, } \cos \theta=\frac{x}{r}
\end{array}
$$
By substituting the value of $\cos \theta$ in Eq. (i)
$$
F=-\frac{2 k Q q}{r^{2}} \cdot \frac{x}{r}
$$
or
$$
F=-\frac{2 k Q x}{r^{3}}
$$
or
$$
F=-\frac{2 k Q x}{\left(x^{2}+d^{2}\right)^{3 / 2}} \quad\left[\begin{array}{l}
\because r^{2}=x^{2}+d^{2} \\
r=\sqrt{x^{2}+d^{2}}
\end{array}\right]
$$
For, $x< < d,$ so $x^{2}$ can be neglected
$$
F=-\frac{2 k Q x}{d^{3}}
$$
So, the force developed by negative charge $(-q)$ due to the system of the charges as shown in the figure is,
$$
F=\frac{-4 k Q q x}{d^{3}}
$$
$\Rightarrow \quad F \propto x$
So, the force developed by negative charge is directly proportional to the distance $x$.

Topic: Physics, Electrostatics, WBJEE, WBJEE 2019
A neutral conducting solid sphere of radius $\mathrm{R}$ has two spherical cavities of radius a and $b$ as shown in the figure. Centre to centre distance between two cavities is $c . q_a$ and $q_b$ charges are placed at the centres of cavities respectively. The force between $q_a$ and $q_b$ is

Options:
A $\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_{\mathrm{a}} \mathrm{q}_{\mathrm{b}}}{\mathrm{c}^2}$
B $\frac{1}{4 \pi \varepsilon_0} q_a q_b\left(\frac{1}{a^2}+\frac{1}{b^2}\right)$
C zero
D insufficient data

The correct answer is:
$\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_{\mathrm{a}} \mathrm{q}_{\mathrm{b}}}{\mathrm{c}^2}$

Solution:
Considering only interaction between charge $q_a$ and $q_b$.

Topic: Physics, Electrostatics, WBJEE, WBJEE 2022
A non-zero current passes through the galvanometer $G$ shown in the circuit when the key $K$ is closed and its value does not change when the key is opened. Then, which of the following statement(s) is/are true?

Options:
A The galvanometer resistance is infinite.
B The current through the galvanometer is $40 \mathrm{mA}$.
C After the key is closed, the current through the $200 \Omega$ resistor is same as the current through the $300 \Omega$ resistor.
D The galvanometer resistance is 150$\Omega$

The correct answers are:
The current through the galvanometer is $40 \mathrm{mA}$., After the key is closed, the current through the $200 \Omega$ resistor is same as the current through the $300 \Omega$ resistor., The galvanometer resistance is 150$\Omega$

Solution:
$\therefore I_{C}=\frac{10}{100+G}$
$300 I_{3}=\frac{10}{100+G} G$
$I_{3}=\frac{G}{30}\left[\frac{1}{100+G}\right]$
$\therefore \quad I_{3}+I_{G}=\frac{1}{(100+G)}\left[10+\frac{G}{30}\right]=\frac{(300+G)}{(100+G) \times 30}$
According to the problem, $10=\left(\frac{200}{3}+\frac{300 G}{300+G}\right)\left[\frac{(300+G)}{(100+6) 30}\right]$
$Y=R C$
$10=\frac{60000+1100 G}{3} \times \frac{1}{100+G} \times \frac{1}{30}$
$\Rightarrow 900(100+G)=60000+1100 G$
$\Rightarrow \quad 30000=200 G$
$\therefore$
$G=150 \Omega$
$I_{G}=\frac{10}{100+150}=\frac{10}{250}=40 \mathrm{mA}$
As $\quad \frac{200}{100}=\frac{300}{150}$
So, $\quad I_{200}=I_{300}$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2018
A NOR gate and a NAND gate are connected as shown in the figure. Two different sets of inputs are given to this setup. In the first case, the inputs to the gates are $A=0, B=0$, $C=0 .$ In the second case, the inputs are $A=1, B=0, C=1 .$ The output $D$ in the first case and second case respectively are

Options:
A 0 and 0
B 0 and 1
C 1 and 0
D 1 and 1

The correct answer is:
1 and 1

Solution:
In first case $A=0, B=0$
$\therefore$ Output of NOR gate, $Y=\overline{A+B}=1$
This output is the input for NAND gate, $i . e ., Y=1$
and $C=0$
$\therefore \quad D=\overline{Y \cdot C}=1$
In second case $A=1, B=0$
$\therefore$ Output of NOR gate, $Y=\overline{A+B}=0$
This output is the input for NAND gate $i . e . . Y=0$
and $C=1$
$\therefore \quad D=\overline{Y \cdot C}=1$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2013
A nucleus X emits a beta particle to produce a nucleus Y. If their atomic masses are $\mathrm{M}_{\mathrm{x}}$ and $\mathrm{M}_{\mathrm{y}}$ respectively, the maximum energy of the beta particle emitted is (where $m_{e}$ is the mass of an electron and $c$ is the velocity of light)

Options:
A $\left(\mathrm{M}_{\mathrm{x}}-\mathrm{M}_{\mathrm{y}}-\mathrm{m}_{\mathrm{e}}\right) \mathrm{c}^{2}$
B $\left(M_{x}-M_{y}+m_{e}\right) \mathrm{c}^{2}$
C $\left(M_{x}-M_{y}\right) \mathrm{c}^{2}$
D $\left(M_{x}-M_{y}-2 m_{e}\right) \mathrm{c}^{2}$

The correct answer is:
$\left(M_{x}-M_{y}\right) \mathrm{c}^{2}$

Solution:
Hint:

$\mathrm{m}_{\mathrm{x}}=\mathrm{M}_{\mathrm{x}}-\mathrm{zm}_{\mathrm{e}}$
$\frac{\mathrm{m}_{\mathrm{y}}=\mathrm{M}_{\mathrm{y}}-(\mathrm{z}+1) \mathrm{m}_{\mathrm{e}}}{\mathrm{m}_{\mathrm{x}}-\mathrm{m}_{\mathrm{y}}=\mathrm{M}_{\mathrm{x}}-\mathrm{M}_{\mathrm{y}}+\mathrm{m}_{\mathrm{e}}}$
$\mathrm{E}=\Delta \mathrm{mC}^{2}=\left(\mathrm{m}_{\mathrm{x}}-\mathrm{m}_{\mathrm{y}}-\mathrm{m}_{\mathrm{e}}\right) \mathrm{C}^{2}=\left(\mathrm{M}_{\mathrm{x}}-\mathrm{M}_{\mathrm{y}}\right) \mathrm{C}^{2}$

Topic: Physics, Nuclear Physics, WBJEE, WBJEE 2020
A pair of parallel metal plates are kept with a separation 'd'. One plate is at a potential $+\mathrm{V}$ and the other is at ground potential. A narrow beam of electrons enters the space between the plates with a velocity $\mathrm{v}_{0}$ and in a direction parallel to the plates. What will be the angle of the beam with the plates after it travels an axial distance L?

Options:
A $\tan ^{-1}\left(\frac{\mathrm{eVL}}{\mathrm{mdv}_{0}}\right)$
B $\tan ^{-1}\left(\frac{\mathrm{eVL}}{\mathrm{mdv}_{0}^{2}}\right)$
C $\sin ^{-1}\left(\frac{\mathrm{eVL}}{\mathrm{mdv}_{0}}\right)$
D $\cos ^{-1}\left(\frac{\mathrm{eVL}}{\mathrm{mdv}_{0}^{2}}\right)$

The correct answer is:
$\tan ^{-1}\left(\frac{\mathrm{eVL}}{\mathrm{mdv}_{0}^{2}}\right)$

Solution:
Hint:

$t=\frac{L}{v_{0}}$
$v_{y}=\frac{e}{m} \cdot \frac{v}{d} \times \frac{L}{v_{0}}=\frac{e v L}{m d v_{0}}$
$\tan \theta=\frac{v_{y}}{v_{x}}=\frac{e v L}{m d v_{0} \cdot v_{0}}$
$\Rightarrow \theta=\tan ^{-1}\left(\frac{e v L}{m d v_{0}^{2}}\right)$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2020
A parallel beam of light is incident on a glass prism in the shape of a quarter cylinder of radius $R=0.05 \mathrm{m}$ and refractive index $n=1.5,$ placed on a horizontal table as shown in the figure. Beyond the cylinder, a patch of light is found whose the nearest distance $x$ from the cylinder is

Options:
A $(3 \sqrt{3}-4) \times 10^{-2} \mathrm{m}$
B $(2 \sqrt{3}-2) \times 10^{-2} \mathrm{m}$
C $(3 \sqrt{5}-5) \times 10^{-2} \mathrm{m}$
D $(3 \sqrt{2}-3) \times 10^{-2} \mathrm{m}$

The correct answer is:
$(3 \sqrt{5}-5) \times 10^{-2} \mathrm{m}$

Solution:
Given, radius $R=0.05 \mathrm{m}$
Refractive index $n=1 \cdot 5$

The critical angle, $\sin c=\frac{1}{n}=\frac{1}{1 \cdot 5}$
$$
=\frac{10}{15}=\frac{2}{3}
$$
From above figure, $\cos c=\frac{R}{R+X}$
$\sqrt{1-\sin ^{2} c}=\frac{R}{R+X}$
$\sqrt{1-\frac{4}{9}}=\frac{R}{R+X}$
$\therefore$
$$
\begin{aligned}
\frac{\sqrt{5}}{3} &=\frac{R}{R+X} \\
\sqrt{5}(R+X) &=3 R \\
\sqrt{5} R+\sqrt{5} X &=3 R \\
X &=(3 \sqrt{5}-5) \times 10^{-2} \mathrm{m}
\end{aligned}
$$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2017
A parallel plate capacitor in series with a resistance of $100 \Omega$, an inductor of $20 \mathrm{mH}$ and an AC voltage source of variable frequency shows resonance at a frequency of $\frac{1250}{\pi}$ Hz If this capacitor is charged by a DC voltage source to a voltage 25 v, what amount of charge will be stored in cach plate of the capacitor?

Options:
A $0.2 \mu C$
B $2 \mathrm{mC}$
C $0.2 \mathrm{mC}$
D $0.2 \mathrm{C}$

The correct answer is:
$0.2 \mathrm{mC}$

Solution:
In given, Series $R-L-C$ circuit Resistance $R=100 \Omega$
Inductance of Inductor, $L=20 \mathrm{mH}=20 \times 10^{-3} \mathrm{H}$
Resonance frequency $f=\frac{1250}{\pi} \mathrm{Hz}$
Source voltage $V_{D C}=25 \mathrm{V}$
According to resonant frequency.
$$
\omega_{0}=2 \pi f_{0}=\frac{1}{\sqrt{L C}}
$$
or $\left(2 \pi f_{0}\right)^{2}=\frac{1}{L C}$
or $\quad 4 \pi^{2} \frac{1250 \times 1250}{\pi \times \pi}=\frac{1}{L C}$
Or
$$
\begin{aligned}
C=& \frac{1000}{1250 \times 1250 \times 4 \times 20} \\
&\left(\because \mathrm{By} \text { substituting } L=20 \times 10^{-3}\right) \\
C=& 8 \times 10^{-6} \mathrm{F}
\end{aligned}
$$
or
We know that, Charge $Q_{s}=C V$
$$
Q_{s}=8 \times 10^{-6} \times 25=0.2 \mathrm{mC}
$$
So, the amount of charge stored in each plate of capacitor is $0.2 \mathrm{mC}$.

Topic: Physics, Alternating Current, WBJEE, WBJEE 2019
A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved farther apart by pulling at them by means of insulating handles, then

Options:
A the energy stored in the capacitor decreases
B the capacitance of the capacitor increases
C the charge on the capacitor decreases
D the voltage across the capacitor increases

The correct answer is:
the voltage across the capacitor increases

Solution:
We know that
The parallel plate capacitor $C=\frac{\varepsilon_{0} \cdot A}{d}$
When $d$ (distance between two parallel plate) increase, then $C$ will be decrease
$\because$
$Q=C V \quad$ where $Q=$ constant
$\therefore$ The voltage across the capacitor increases.

Topic: Physics, Capacitance, WBJEE, WBJEE 2014
A parent nucleus $X$ undergoes $\alpha$ -decay with a half-life of 75000 yrs. The daughter nucleus $Y$ undergoes $\beta$ -decay with a half-life of 9 months. In a particular sample, it is found that the rate of emission of $\beta$ -particles is nearly constant (over several months) at $10^{7} / \mathrm{h} .$ What will be the number of $\alpha$ -particles emitted in an hour?

Options:
A $10^{2}$
B $10^{7}$
C $10^{12}$
D $10^{14}$

The correct answer is:
$10^{7}$

Solution:
According to the question,
$X \stackrel{\text { a-particle }}{\longrightarrow} Y \longrightarrow Z(\text { say) }$
According to Rutherford Soddy law of radioactive decay, the rate of decay of radioactive along at any instant is proportional to the number of atoms present at that instant
$\frac{d N}{d t}=-\lambda N$
since, $\lambda=$ constant (decay constant) and $N=$ number of atoms
Rate of decay of $Y$ particle is given as,
$\lambda_{X} N_{X}-\lambda_{\gamma} N_{Y}=\frac{d N_{Y}}{d t}=0$
[ $\because$ Decay rate for $\beta$ -particle become constant after some time]
Given, rate of emission of $\beta$ -particle $=10^{7} \mathrm{h}$ $\lambda_{X} N_{X}=\lambda_{\gamma} N_{Y}=10^{7} / \mathrm{h}$

Topic: Physics, Nuclear Physics, WBJEE, WBJEE 2019
A particle is executing linear simple harmonic motion of amplitude A. At what displacement is the energy of the particle half potential and half kinetic?

Options:
A $\frac{\mathrm{A}}{4}$
B $\frac{\mathrm{A}}{2}$
C $\frac{\mathrm{A}}{\sqrt{2}}$
D $\frac{\mathrm{A}}{\sqrt{3}}$

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

Solution:
Hints : Total Energy $(\mathrm{E})=\frac{1}{2} m \omega^2 \mathrm{~A}^2$
P.E. $=\frac{1}{2} m \omega^2 x^2$
As P.E. $=\frac{E}{2}$
Then, $\frac{1}{2} m \omega^2 \mathrm{~A}^2 \times \frac{1}{2}=\frac{1}{2} m \omega^2 x^2$
$$
x^2=\frac{\mathrm{A}^2}{2} \Rightarrow x=\frac{\mathrm{A}}{\sqrt{2}}
$$

Topic: Physics, Oscillations, WBJEE, WBJEE 2010
A particle is in motion along a curve $12 y=x^{3}$. The rate of change of its ordinate exceeds that of abscissa in

Options:
A $-2 < x < 2$
B $x=\pm 2$
C $x < -2$
D $x>2$

The correct answers are:
$x < -2$, $x>2$

Solution:
Given, curve, $12 y=x^{3}$
and $\frac{d y}{d t}>\frac{d x}{d t}$ ....(i)
Now, $12 \frac{d y}{d t}=3 x^{2} \frac{d x}{d t}$....(ii)
From Eqs. (i) and (ii), we get
$\Rightarrow$
$$
\begin{aligned}
3 x^{2} \frac{d x}{d t} &>12 \frac{d x}{d t} \\
3 x^{2} &>12
\end{aligned}
$$
$\Rightarrow \quad x^{2}-4>0$
$\begin{array}{ll}\Rightarrow & (x-2)(x+2)>0 \\ \Rightarrow & x \in(-\infty,-2) \cup(2, \infty)\end{array}$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2018
A particle is moving in a straight line. At time $t$, the distance between the particle from its starting point is given by $x=t-6 t+z$. Its acceleration will be zero at

Options:
A $t=1$ unit time
B $t=2$ unit time
C $t=3$ unit time
D $t=4$ unit time

The correct answer is:
$t=2$ unit time

Solution:
Hints : $x=t-6 t^2+t^3$
$$
\begin{aligned}
& \frac{d x}{d t}=1-12 t+3 t^2 \\
& \frac{d^2 x}{d t^2}=-12+6 t
\end{aligned}
$$
Acceleration $=\frac{d^2 x}{d t^2}$
$\therefore$ Acceleration $=0 \Rightarrow 6 t-12=0 \Rightarrow t=2$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2009
A particle is moving in $x-y$ plane according to $\vec{r}=b \cos \omega t \hat{i}+b \sin \omega t \hat{j}$. Where $\omega$ is constant. Which of the following statement(s) is / are true?

Options:
A $\frac{E}{\omega}$ is a constant where $E$ is the total energy of the particle
B The trajectory of the particle in $x-y$ plane is a circle
C In $a_x-a_y$ plane, trajectory of the particle is an ellipse $\left(a_x, a_y\right.$ denotes the components of acceleration)
D $\vec{a}=\omega^2 \vec{v}$

The correct answers are:
$\frac{E}{\omega}$ is a constant where $E$ is the total energy of the particle, The trajectory of the particle in $x-y$ plane is a circle

Solution:
$\vec{r}=b \cos \omega t i+b \sin \omega t j$
$\vec{v}=\frac{d \vec{r}}{d t}=-b \omega \sin \omega t+b \omega \cos \omega t j$
$(\vec{v})=\sqrt{b^2 \omega^2 \sin ^2 \omega t+b^2 \omega^2 \cos ^2 \omega t}$
$(\vec{v})=b \omega \rightarrow$ constant $\quad \therefore E=\frac{1}{2} m v^2=$ constant
$E=\frac{1}{2} m\left[b^2 \omega^2\right] \quad \Rightarrow \frac{E}{\omega}=\frac{1}{2} m b^2 \cdot \omega=$ constant
$x=b \cos \omega t$
$y=b \sin \omega t$
$x^2+y^2=b^2 \rightarrow$ equation of circle
$\therefore \vec{a} \perp \vec{v} \quad$ So $\vec{a} \neq \omega^2 \vec{v}$

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2022
A particle is moving uniformly in a circular path of radius $r$. When it moves through an angular displacement $\theta$, then the magnitude of the corresponding linear displacement will be

Options:
A $2 r \cos \left(\frac{\theta}{2}\right)$
B $2 r \cot \left(\frac{\theta}{2}\right)$
C $2 r \tan \left(\frac{\theta}{2}\right)$
D $2 r \sin \left(\frac{\theta}{2}\right)$

The correct answer is:
$2 r \sin \left(\frac{\theta}{2}\right)$

Solution:
$\ln \Delta A O B \quad \sin \frac{\theta}{2}=\frac{A B}{A O}$
$\Leftrightarrow A O=r)$
$\begin{aligned}
A B &=A O \sin \frac{\theta}{2} \Rightarrow A B=r \sin \frac{\theta}{2} \\
A C &=A B+B C \quad(\because A B=B C) \\
&=r \sin \frac{\theta}{2}+r \sin \frac{\theta}{2} \\
A C &=2 r \sin \frac{\theta}{2}
\end{aligned}$
So, the magnitude of the corresponding linear displacement will be $2 r \sin \frac{\theta}{2}$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2014
A particle is moving with a constant speed \(v\) in a circle. What is the magnitude of average velocity after half rotation?

Options:
A \(2 v\)
B \(2 \frac{v}{\pi}\)
C \(\frac{v}{2}\)
D \(\frac{v}{2 \pi}\)

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

Solution:
Hints:
\(\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{V}} ; \mathrm{t}_{\mathrm{o}}=\frac{\mathrm{T}}{2}=\frac{\pi \mathrm{r}}{\mathrm{V}} ; \quad \mathrm{V}_{\mathrm{av}}=\frac{2 \mathrm{r}}{\frac{\pi \mathrm{r}}{\mathrm{v}}}=\frac{2 \mathrm{v}}{\pi}\)

Topic: Physics, Motion In One Dimension, WBJEE, WBJEE 2011
A particle is moving with a uniform speed $v$ in a circular path of radius $r$ with the centre at $O$. When the particle moves from a point $P$ to $Q$ on the circle such that $\angle P O Q=\theta$, then the magnitude of the change in velocity is

Options:
A $2 \mathrm{v} \sin (2 \theta)$
B zero
C $2 v \sin \left(\frac{\theta}{2}\right)$
D $2 v \cos \left(\frac{\theta}{2}\right)$

The correct answer is:
$2 v \sin \left(\frac{\theta}{2}\right)$

Solution:
$\therefore$ Change in magnitude of velocity
$\begin{aligned}
|\Delta v| &=\sqrt{v^{2}+v^{2}-2 v^{2} \cos \theta} \\
&=2 v \sin \frac{\theta}{2}
\end{aligned}$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2013