datavorous/entrance-exam-dataset huggingface dataset:
A particle is projected at $60^{\circ}$ to the horizontal with a kinetic energy $\mathrm{K}$. The kinetic energy at the highest point is

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

The correct answer is:
$\frac{\mathrm{K}}{4}$

Solution:
At highest point kinetic energy $=1 / 2 \mathrm{~m}\left(\mathrm{v} \cos 60^{\circ}\right)^2=1 / 4 \times 1 / 2 \mathrm{~m} \mathrm{v} v^2=\mathrm{K} / 4$

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2009
A particle is projected from the ground with a kinetic energy $E$ at an angle of $60^{\circ}$ with the horizontal. Its kinetic energy at the highest point of its motion will be

Options:
A $E / \sqrt{2}$
B $E / 2$
C $E / 4$
D $E / 8$

The correct answer is:
$E / 4$

Solution:
At ground, $E=\frac{1}{2} m u^{2}$
At highest point, $E^{\prime}=\frac{1}{2} m\left(u \cos 60^{\circ}\right)^{2}$
$$
E^{\prime}=\frac{E}{4}
$$

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2012
A particle is projected vertically upwards. If it has to stay above the ground for 12 seconds, then

Options:
A velocity of projection is $192 \mathrm{ft} / \mathrm{sec}$
B greatest height attained is $600 \mathrm{ft}$
C velocity of projection is $196 \mathrm{ft} / \mathrm{sec}$
D greatest height attained is $576 \mathrm{ft}$

The correct answers are:
velocity of projection is $192 \mathrm{ft} / \mathrm{sec}$, greatest height attained is $576 \mathrm{ft}$

Solution:
Hint:
$V=u-g t$ at $t=6$
$u-g t=0$
$\Rightarrow u=6 g=192 f t /$ sec $\quad\left(g=32 f t / \sec ^{2}\right) \ldots \ldots . .(\mathrm{i})$
$x=u t=\frac{1}{2} g t^{2}$
$=192.6-\frac{1}{2} 32.6^{2}$
$=576 \mathrm{ft}$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2020
A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motion, the phase difference $(\delta)$ between the two motions is

Options:
A $\delta=\frac{\pi}{3}$
B $\delta=\frac{2 \pi}{3}$
C $\delta=\pi$
D $\delta=\frac{\pi}{2}$

The correct answer is:
$\delta=\frac{2 \pi}{3}$

Solution:
$A_{\text {res }}^2=A_1^2+A_2^2+2 A_1 A_2 \cos \delta \quad A^2=A^2+A^2+2 A^2 \cos \delta, \cos \delta=-\frac{1}{2} \delta=120^{\circ}$

Topic: Physics, Oscillations, WBJEE, WBJEE 2022
A particle is travelling along a straight line $O X$. The distance $x$ (in metre) of the particle from 0 at a time t is given by $x=37+27 t-t^{3}$, where t is time in seconds. The distance of the particle from $O$ when it comes to rest is

Options:
A $81 \mathrm{m}$
B $91 \mathrm{m}$
C $101 \mathrm{m}$
D $111 \mathrm{m}$

The correct answer is:
$91 \mathrm{m}$

Solution:
Given, $x=37+27 t-t^{3}$
$$
v=\frac{d x}{d t}=27-3 t^{2}
$$
According to problem,
$$
v=0 \Rightarrow 27-3 t^{2}=0
$$
Here, $t=3 \mathrm{s}$
$$
x=37+27 \times 3-(3)^{2}=91 \mathrm{m}
$$

Topic: Physics, Motion In One Dimension, WBJEE, WBJEE 2012
A particle moves along $x$ -axis and its displacement at any time is given by $x(t)=2 t^{3}-3 t^{2}+4 t$ in SI units. The velocity of the particle when its acceleration is zero, is

Options:
A $2.5 \mathrm{ms}^{-1}$
B $3.5 \mathrm{ms}^{-1}$
C $4.54 \mathrm{ms}^{-1}$
D $8.5 \mathrm{ms}^{-1}$

The correct answer is:
$2.5 \mathrm{ms}^{-1}$

Solution:
Given, $x(t)=2 t^{3}-3 t^{2}+4 t$
So, velocity, $v=\frac{d x}{d t}=\left(6 t^{2}-6 t+4\right)$
and acceleration $a=\frac{d v}{d t}=(12 t-6)$
when acceleration is zero, $i . e . .(12 t-6)=0$
or
$$
t=\frac{6}{12}=\frac{1}{2} \mathrm{s}
$$
$\therefore$ Velocity of the particle at zero acceleration is
$$
v=6\left(\frac{1}{2}\right)^{2}-6\left(\frac{1}{2}\right) 4=2.5 \mathrm{m} / \mathrm{s}
$$

Topic: Physics, Motion In One Dimension, WBJEE, WBJEE 2013
A particle moves with constant acceleration along a straight line starting from rest. The percentage increase in its displacement during the 4th second compared to that in the 3rd second is

Options:
A $33 \%$
B $40 \%$
C $66 \%$
D $77 \%$

The correct answer is:
$40 \%$

Solution:
We know that
$S=u+\frac{1}{2} a(2 n-1)$
$S_{3rd \text { sec}}=0+\frac{1}{2} a(2 \times 3-1)=\frac{5}{2} a$ (for $n=3$ s)
$S_{4th \text { sec}}=0+\frac{1}{2} a(2 \times 4-1)=\frac{7}{2} a \quad($ for $n=4 \mathrm{s})$
So, the percentage increase
$=\frac{S_{4 t h}-S_{3 \mathrm{rd}}}{S_{3 \mathrm{rd}}} \times 100$
$\begin{array}{l} =\frac{\frac{7}{2} a-\frac{5}{2} a}{\frac{5}{2} a} \times 100 \\ =\frac{\frac{2 a}{2}}{\frac{5}{2}} \times 100=2 \times 20=40 \% \end{array}$

Topic: Physics, Motion In One Dimension, WBJEE, WBJEE 2014
A particle moving in a straight line starts from rest and the acceleration at any time $\mathrm{t}$ is $a-k t^2{ }^2$ where a and $k$ are positive constants. The maximum velocity attained by the particle is

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

The correct answer is:
$\frac{2}{3} \sqrt{a^3 / k}$

Solution:
$\frac{d v}{d t}=a-k t^2=(\sqrt{a}-\sqrt{k t})(\sqrt{a}+\sqrt{k t})[\because a, k>0]$
$v$ will be $\max$ at $t=\sqrt{\frac{a}{k}}$
$$
\because v=a t-\frac{k}{3} t^3+C
$$
at $\mathrm{t}=0, \mathrm{v}=0 \Rightarrow \mathrm{C}=0$
$\begin{aligned} \therefore v &=a t-\frac{k}{3} t^3 \\ \therefore v_{\max } &=a \sqrt{\frac{a}{k}}-\frac{k}{3} \frac{a}{k} \sqrt{\frac{a}{k}} \\ &=\frac{2 a}{3} \sqrt{\frac{a}{\mathrm{k}}} \end{aligned}$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2022
A particle of mass $M$ and charge $q$, initially at rest, is accelerated by a uniform electric field $E$ through a distance $D$ and is then allowed to approach a fixed static charge $Q$ of the same sign. The distance of the closest approach of the charge $q$ will then be

Options:
A $\frac{q Q}{4 \pi \varepsilon_{0} D}$
B $\frac{Q}{4 \pi \varepsilon_{0} E D}$
C $\frac{q Q}{2 \pi \varepsilon_{0} D^{2}}$
D $\frac{Q}{4 \pi \varepsilon_{0} E}$

The correct answer is:
$\frac{Q}{4 \pi \varepsilon_{0} E D}$

Solution:
At the distance of closest approach, the entire $\mathrm{KE}$ of a particle is converted into electric potential energy. i.e.. $\mathrm{KE}=\mathrm{PE}$
Here, KE of the particle = work done in moving the particle in uniform electric field (E) through a distance (D)
$$
=q E \times D
$$
PE of the particle = potential $\times$ charge
$$
=\frac{q}{4 \pi c_{0} \gamma_{0}} \times Q
$$
$\therefore \quad q E D=\frac{q Q}{4 \pi \varepsilon_{0} \gamma_{0}} \Rightarrow r_{0}=\frac{Q}{4 \pi \varepsilon_{0} E D}$

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2013
A particle of mass $M$ and charge $q$ is at rest at the mid point between two other fixed similar charges each of magnitude $Q$ placed a distance $2 d$ apart. The system is collinear as shown in the figure. The particle is now displaced by a small amount $x(x< < d)$ along the joining the two charges and is left to itself. It will now oscillate about the mean position with a time period $\left(\varepsilon_{0}=\right.$ permittivity of free space)

Options:
A $2 \sqrt{\frac{\pi^{3} M \varepsilon_{0} d}{Q q}}$
B $2 \sqrt{\frac{\pi^{2} M \varepsilon_{0} d^{3}}{Q q}}$
C $2 \sqrt{\frac{\pi^{3} M \varepsilon_{0} d^{3}}{Q q}}$
D $2 \sqrt{\frac{\pi^{3} M \varepsilon_{0}}{Q q d^{3}}}$

The correct answer is:
$2 \sqrt{\frac{\pi^{3} M \varepsilon_{0} d^{3}}{Q q}}$

Solution:
Restoring force on displacement of $x$
$\begin{aligned} F &=K\left[\frac{q}{(d-x)^{2}}-\frac{Q q}{(d+x)^{2}}\right] \\ &=K Q q\left[\frac{1}{(d-x)^{2}}-\frac{1}{(d+x)^{2}}\right] \\ &=K Q q\left[\frac{4 d x}{\left(d^{2}-x^{2}\right)^{2}}\right] \\ &=K Q q\left[\frac{4 d x}{d^{4}}\right] \text { If }(d>>x) \\ F &=K Q q\left[\frac{4 x}{d^{3}}\right] \end{aligned}$
Acceleration, $a=\frac{F}{m}=\frac{4 K Q q x}{M d^{3}}$
or $\omega^{2}=\frac{4 K Q q}{M d^{3}}$
$\therefore \quad T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{M d^{3}}{4 K O q}}$
$=2 \sqrt{\frac{\pi^{3} M d^{3} \varepsilon_{0}}{Q q}}$

Topic: Physics, Oscillations, WBJEE, WBJEE 2013
A particle of mass $m$ is attached to three identical massless springs of spring constant ' $k$ ' as shown in the figure. The time period of vertical oscillation of the particle is

Options:
A $2 \pi \sqrt{\frac{m}{k}}$
B $2 \pi \sqrt{\frac{m}{2 k}}$
C $2 \pi \sqrt{\frac{m}{3 k}}$
D $\pi \sqrt{\frac{m}{k}}$

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

Solution:
$$
\begin{aligned}
& \text { Hints : } \mathrm{T}=2 \pi \sqrt{\frac{m}{\mathrm{~K}_{\mathrm{eq}}}} \\
& \mathrm{F}=\mathrm{K} x+2 \mathrm{~K} x \cos ^2 45 \\
& \mathrm{~K}_{\mathrm{eq}} x=\mathrm{K} x+\mathrm{K} x \\
& \mathrm{~K}_{\mathrm{eq}}=2 \mathrm{~K}
\end{aligned}
$$

Topic: Physics, Oscillations, WBJEE, WBJEE 2010
A particle of mass \(m\) is located in a one dimensional potential field where potential energy is given by : \(\mathrm{V}(\mathrm{x})=\mathrm{A}(1-\cos \mathrm{px})\), where \(\mathrm{A}\) and \(\mathrm{p}\) are constants. The period of small oscillations of the particle is

Options:
A \(2 \pi \sqrt{\frac{m}{(A p)}}\)
B \(2 \pi \sqrt{\frac{m}{\left(A p^2\right)}}\)
C \(2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{A}}}\)
D \(\frac{1}{2 \pi} \sqrt{\frac{A p}{m}}\)

The correct answer is:
\(2 \pi \sqrt{\frac{m}{\left(A p^2\right)}}\)

Solution:
Hints: \(v_x=A(1-\cos p x)\)
\(F=-\frac{d u}{d x}=-A p \sin p x\)
For small (x)
\(\mathrm{F}=-\mathrm{AP}^2 \mathrm{x}\)
\(a=-\frac{A p^2}{m} x \quad a=-\omega^2 x\)
\(\omega=\sqrt{\frac{A P^2}{\mathrm{~m}}} \quad \therefore \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{A p^2}}\)

Topic: Physics, Oscillations, WBJEE, WBJEE 2011
A particle of mass \(\mathrm{m}\) is projected at a velocity \(\mathrm{u}\), making an angle \(\theta\) with the horizontal ( \(\mathrm{x}\)-axis). If the angle of projection \(\theta\) is varied keeping all other parameters same, then magnitude of angular momentum (L) at its maximum height about the point of projection varies with \(\theta\) as,

Options:
A

B

C

D

The correct answer is:

Solution:
Hint :
\(\begin{aligned}
& \mathrm{L}_0=m u \cos \theta \mathrm{H} \\
& =m u \cos \theta \frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \\
& =\frac{m u^3 \sin ^2 \theta \cos \theta}{2 \mathrm{~g}}
\end{aligned}\)
\(L\) is zero for \(\theta=0^{\circ}\) and \(\theta=\frac{\pi}{2}\)

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2023
A particle starts at the origin and moves l unit horizontally to the right and reaches $P_{1}$, then it moves $\frac{1}{2}$ unit vertically up and 2 reaches $P_{2}$, then it moves $\frac{1}{4}$ unit horizontally to right and reaches $P_{3}$, then it moves $\frac{1}{8}$ unit vertically down and reaches $P_{4}$, then it moves $\frac{1}{16}$ unit horizontally to right and reaches $P_{\mathrm{S}}$ and so on. Let $P_{n}=\left(x_{n}, y_{n}\right)$ and $\lim _{n \rightarrow \infty} x_{n}=\alpha$ and $\lim _{n \rightarrow \infty} y_{n}=\beta .$ Then, $(\alpha, \beta)$ is

Options:
A (2,3)
B $\left(\frac{4}{3}, \frac{2}{5}\right)$
C $\left(\frac{2}{5}, 1\right)$
D $\left(\frac{4}{3}, 3\right)$

The correct answer is:
$\left(\frac{4}{3}, \frac{2}{5}\right)$

Solution:
According to the given information in question, we can draw the situation of particle at different stages as following

Here, $x_{1}=1, x_{2}=1, x_{3}=\frac{5}{4}, x_{4}=\frac{5}{4}$ and $x_{5}=\frac{21}{16}$
$y_{1}=0, y_{2}=\frac{1}{2}, y_{3}=\frac{1}{2}, y_{4}=\frac{3}{8}$ and $y_{5}=\frac{3}{8}$
$\therefore \quad x_{n}=1+\frac{1}{4}+\frac{1}{4^{2}}+\ldots \infty=\frac{1}{1-\frac{1}{4}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$
$\begin{aligned} \text { and } y_{n}=& 0+\frac{1}{2}-\frac{1}{8}+\ldots \infty \\ &=\frac{\frac{1}{2}}{1-\left(-\frac{1}{4}\right)}=\frac{\frac{1}{2}}{\left(1+\frac{1}{4}\right)}=\frac{\frac{1}{2}}{\frac{5}{4}}=\frac{2}{5} \\ \therefore \quad(\alpha, \beta)=\left(\frac{4}{3}, \frac{2}{5}\right) \end{aligned}$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2019
A particle starts moving from rest from a fixed point in a fixed direction. The distance s from the fixed point at a timet is given by $s=t^{2}+a t-b+17,$ where $a$ and $b$ are real numbers. If the particle comes to rest after 5s at a distance of $s=25$ units from the fixed point, then values of $a$ and $b$ are. respectively

Options:
A 10, -33
B -10,-30
C -8.33 .
D -10.33

The correct answer is:
-10.33

Solution:
Given, $s=t^{2}+a t-b+17$
After 5 s. $\left[\frac{d s}{d t}\right]_{t=5}=[2 t+a]_{t=5}=0$
$\Rightarrow \quad a=-10$
At $t=5 \mathrm{s}, s=25$ units
$\therefore \quad 25=5^{2}+5 a-b+17$
$$
\mid \because s=25 \text { and } t=5 \mid
$$
$\Rightarrow \quad 25=25+5(-10)-b+17$
$\Rightarrow \quad 50-17=-b$
$\Rightarrow \quad b=-33$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2015
A particle vibrating simple harmonically hat an acceleration of $16 \mathrm{cms}^{-2}$ when it is at distance of $4 \mathrm{cm}$ from the mean position, lt time period is

Options:
A $1 \mathrm{s}$
B $2.572 \mathrm{s}$
C $3.142 \mathrm{s}$
D $6.028 \mathrm{s}$

The correct answer is:
$3.142 \mathrm{s}$

Solution:
In SHM, the acceleration of vibrating particle is proportional to displacement
$$
\begin{aligned}
|\mathbf{a}| &=\omega^{2} x \\
\text { Here } a=16 \mathrm{cm} / \mathrm{s}^{2} \\
&=16 \times 10^{-2} \mathrm{m} / \mathrm{s}^{2}
\end{aligned}
$$
Displacement, $x=4 \mathrm{cm}=4 \times 10^{-2} \mathrm{m}$
$\therefore \quad 16 \times 10^{-2}=\omega^{2}\left(4 \times 10^{-2}\right)$
$$
\begin{aligned}
\omega^{2} &=4 \\
\omega &=\pm 2 \mathrm{rad} / \mathrm{s} \\
\omega &=\frac{2 \pi}{T} \\
T &=\frac{2 \pi}{\omega}=\frac{2 \pi}{2}=\pi=3.142 \mathrm{s}
\end{aligned}
$$

Topic: Physics, Oscillations, WBJEE, WBJEE 2016
A particle with charge $e$ and mass $m$, moving along the $X$ -axis with a uniform speed $u$, enters a region where a uniform electric field $E$ is acting along the $Y$ -axis. The particle starts to move in a parabola. Its focal length (neglecting any effect of gravity) is

Options:
A $\frac{2 m u^{2}}{e E}$
B $\frac{e E}{2 m u^{2}}$
C $\frac{m u}{2 e E}$
D $\frac{m u^{2}}{2 e E}$

The correct answer is:
$\frac{m u^{2}}{2 e E}$

Solution:
From parabola

$y=\frac{1}{2} \times\left[\frac{E e}{m}\right] \times \frac{x^{2}}{u^{2}}$
$=\frac{E \cdot e}{2 m u^{2}} \cdot X^{2}$
$\begin{array}{ll}\text { As } x=4 a y & \text { (for parabola) }\end{array}$
$\therefore \quad X^{2}=\frac{2 m u^{2}}{E \cdot e} \cdot y$
$a=\frac{2 m u^{2}}{4 E \cdot e}=\frac{m u^{2}}{2 E \cdot e}$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2017
A particle with charge $Q$ coulomb, tied at the end of an inextensible string of length $R$ metre, revolves in a vertical plane. At the centre of the circular trajectory, there is a fixed charge of magnitude $Q$ coulomb. The mass of the moving charge $M$ is such that $M g=\frac{Q^{2}}{4 \pi \in_{0} R^{2}}$. If at the highest position of the particle, the tension of the string just vanishes, the horizontal velocity at the lowest point has to be

Options:
A 0
B $2 \sqrt{g R}$
C $\sqrt{2 g R}$
D $\sqrt{5 g R}$

The correct answer is:
$2 \sqrt{g R}$

Solution:
According to the question, when tension in the string is zero.

As $T=0$ $M g-\frac{K Q^{2}}{R^{2}}=\frac{m v^{2}}{R} \quad\left(k=\frac{1}{4 \pi \varepsilon_{0}}\right)$
$\frac{m v^{2}}{R}=$ centripetal force required to keep the body in a circular path
$\Rightarrow \quad v=0$$\quad\left[\because M g=\frac{k Q^{2}}{R^{2}}\right]$
So the required work done to keep charge in vertical motion
$\therefore \quad W_{g}=\Delta K E=$ change in $\mathrm{K.E.}$
$\Rightarrow \quad m g(2 R)=\frac{1}{2} m v_{0}^{2}$
\begin{array}{l}
v_{0}^{2}=2 g \cdot 2 R=4 g R \\
v_{0}=\sqrt{4 g R}=2 \sqrt{g R}
\end{array}

Topic: Physics, Electrostatics, WBJEE, WBJEE 2017
A particle with charge $q$ moves with a velocity $v$ in a direction perpendicular to the directions of uniform clectric and magnetic ficlds, $E$ and $B$ respectively, which are mutually perpendicular to each other. Which one of the following gives the condition for which the particle moves undeflected in its original trajectory?

Options:
A $v=\frac{E}{B}$
B $v=\frac{B}{E}$
C $v=\sqrt{\frac{E}{B}}$
D $v=q \frac{B}{E}$

The correct answer is:
$v=\frac{E}{B}$

Solution:
According to the question,
Charge on particle is $=q$
Velocity of particle $=v$
Due to uniform electric field,
electric force on particle
$F_{\text {electric }}=q E$
Due to a uniform magnetic field, magnetic force on particle is given by,
$$
F_{\text {magnetic }}=q(v \times B)
$$
When $v$ is perpendicular to $E$ and $B,$ which are mutually perpendicular to each other, $F_{\text {electric }}=F_{\text {magnetic }}$
From Eqs. (i) and (ii)
$$
q E=q v B
$$
$\therefore$
$$
v=\frac{E}{B}
$$
So, $v=\frac{E}{B}$ is the condition for which the particle moves undeflected in its original trajectory.

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2019
A person goes to office by car, scooter, bus and train, probability of which are $1 / 7,3 / 7$. $2 / 7$ and $1 / 7,$ respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is $2 / 9,1 / 9,4 / 9,$ and $1 / 9,$ respectively. Given that he reached office in time, the probability that he travelled by a car, is

Options:
A $1 / 7$
B $2 / 7$
C $3 / 7$
D $4 / 7$

The correct answer is:
$1 / 7$

Solution:
Let the event of person goes to office by a car, scooter, bus or train be $A, B, C$ and $D,$ respectively. We have, $P(A)=\frac{1}{7}, P(B)=\frac{3}{7}, P(C)=\frac{2}{7}$
and $P(D)=\frac{1}{7}$
Let $E=$ He reached office in time
We have
$P\left(\frac{\bar{E}}{A}\right)=\frac{2}{9}, P\left(\frac{\bar{E}}{B}\right)=\frac{1}{9}, P\left(\frac{\bar{E}}{C}\right)=\frac{4}{9}$
and $P\left(\frac{\bar{E}}{D}\right)=\frac{1}{9}$
$\begin{aligned} P\left(\frac{A}{E}\right)=& P(A) \cdot P\left(\frac{E}{A}\right) \\ &+P(C) \cdot P\left(\frac{E}{C}\right)+P(D) \cdot P\left(\frac{E}{D}\right) \\ &=\frac{\frac{1}{7} \cdot \frac{7}{9}}{P(A) \cdot P\left(\frac{E}{A}\right)+P(B) \cdot P\left(\frac{E}{B}\right)} \\ &=\frac{1}{7} \cdot \frac{7}{9}+\frac{3}{7} \cdot \frac{8}{9}+\frac{2}{7} \cdot \frac{5}{9}+\frac{1}{7} \cdot \frac{8}{9} \\=& \frac{7}{7+24+10+8}=\frac{7}{49}=\frac{1}{7} \end{aligned}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2015
A plane meets the co-ordinate axes at the points $A, B, C$ respectively in such a way that the centroid of $\triangle A B C$ is $\left(1, r, r^{2}\right)$ for some real $r$. If the plane passes through the point $(5,5,-12)$ then $r=$

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

The correct answers are:
$\frac{3}{2}$, $-4$

Solution:
Centroid is $\mathrm{G}(\mathrm{a} / 3, \mathrm{~b} / 3, \mathrm{c} / 3) \equiv\left(1, \mathrm{r}, \mathrm{r}^{2}\right)$
$a / 3=1 \quad \Rightarrow a=3, b=3 r, c=3 r^{2}$
Now,
$\begin{array}{l}
\frac{5}{a}+\frac{5}{b}-\frac{12}{c}=1 \\
\Rightarrow \frac{5}{3}+\frac{5}{3 r}-\frac{12}{3 r^{2}}=1 \\
\Rightarrow \frac{5 r^{2}+5 r-12}{3 r^{2}}=1 \\
\Rightarrow 5 r^{2}+5 r-12=3 r^{2} \\
\Rightarrow 2 r^{2}+5 r-12=0 \\
\Rightarrow r=\frac{-5 \pm \sqrt{25+96}}{4} \\
\Rightarrow r=\frac{-5 \pm 11}{4}=-4,3 / 2
\end{array}$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2021
A plane progressive wave is given by \(y=2 \cos 6.284(330 \mathrm{t}-\mathrm{x})\). What is period of the wave?

Options:
A \(\frac{1}{330} \mathrm{~S}\)
B \(2 \pi \times 330 \mathrm{~S}\)
C \((2 \pi \times 330)^{-1} \mathrm{~S}\)
D \(\frac{6.284}{330} \mathrm{~S}\)

The correct answer is:
\(\frac{1}{330} \mathrm{~S}\)

Solution:
Hints: \(y=2 \cos 2 \pi(330 t-x)\)
\(\omega=2 \pi \times 330\)
\(\therefore \mathrm{T}=\frac{1}{330} \mathrm{~s}\)

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2011
A plano-concave lens is made of glass of refractive index 1.5 and the radius of curvature of its curved face is \(100 \mathrm{~cm}\). What is the power of the lens?

Options:
A \(+0.5 \mathrm{D}\)
B \(-0.5 \mathrm{D}\)
C \(-2 \mathrm{D}\)
D \(+2 \mathrm{D}\)

The correct answer is:
\(-0.5 \mathrm{D}\)

Solution:
\(\begin{aligned}
& \text {Hints: } P=\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& =(1.5-1)\left(\frac{1}{\infty}-\frac{1}{1 \mathrm{~m}}\right) \\
& =0.5(-1) \\
& P=-0.5 \mathrm{D}
\end{aligned}\)

Topic: Physics, Ray Optics, WBJEE, WBJEE 2011
A plano-convex lens $(f=20 \mathrm{~cm})$ is silvered at plane surface. Now focal length will be :

Options:
A $20 \mathrm{~cm}$
B $40 \mathrm{~cm}$
C $30 \mathrm{~cm}$
D $10 \mathrm{~cm}$

The correct answer is:
$10 \mathrm{~cm}$

Solution:
$$
\begin{aligned}
& \text { Hints : } \mathrm{P}=2 \mathrm{P}_{\mathrm{L}}+\mathrm{P}_{\mathrm{M}} \\
& \mathrm{P}_{\mathrm{M}}=0 \\
& \mathrm{P}=\frac{1}{f} \times 2=\frac{2}{f} \\
& -\frac{1}{\mathrm{~F}}=\frac{2}{f} \\
& \mathrm{~F}=-\frac{f}{2}
\end{aligned}
$$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2010
A plot of In \(\mathrm{k}\) against \(\frac{1}{\mathrm{~T}}\) (abscissa) is expected to be a straight line with intercept on ordinate axis equal to

Options:
A \(\frac{\Delta \mathrm{S}^{\circ}}{2.303 \mathrm{R}}\)
B \(\frac{\Delta \mathrm{S}^{\circ}}{\mathrm{R}}\)
C \(-\frac{\Delta S^{\circ}}{R}\)
D \(\mathrm{R} \times \Delta \mathrm{S}^{\circ}\)

The correct answer is:
\(\frac{\Delta \mathrm{S}^{\circ}}{\mathrm{R}}\)

Solution:
Hints: \(\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \operatorname{InK}\)
or, \(\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}=-\mathrm{RT} \operatorname{InK}\)
or, \(\operatorname{InK}=\frac{-\Delta \mathrm{H}^{\circ}}{\mathrm{RT}}+\frac{\Delta \mathrm{S}^{\circ}}{\mathrm{R}}\) comparing with \(\mathrm{y}=\mathrm{m} \cdot \mathrm{x}+\mathrm{c}\)
\(\therefore \mathrm{y}\) intercept is \(\frac{\Delta \mathrm{S}^{\circ}}{\mathrm{R}}\)

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2011
A point charge $-q$ is carried from a point $A$ to another point $B$ on the axis of a charged ring of radius $r$ carrying a charge $+q$. If the point $A$ is at a distance $\frac{4}{3} r$ from the centre of the
ring and the point $B$ is $\frac{3}{4} r$ from the centre but
on the opposite side, what is the net work that need to be done for this?

Options:
A $-\frac{7}{5} \cdot \frac{q^{2}}{4 \pi \varepsilon_{0} r}$
B $-\frac{1}{5} \cdot \frac{q^{2}}{4 \pi \varepsilon_{0} I}$
C $\frac{7}{5} \cdot \frac{q^{2}}{4 \pi \varepsilon_{0} r}$
D $\frac{1}{5} \cdot \frac{q^{2}}{4 \pi \varepsilon_{0} r}$

The correct answer is:
$-\frac{1}{5} \cdot \frac{q^{2}}{4 \pi \varepsilon_{0} I}$

Solution:
A charged circular ring of radius $R$ is shown in figure.

From the figure, $d=\sqrt{x^{2}+r^{2}}$
$\therefore$ Potential due to a uniform ring of positive charge $q$ at point $P$
$$
\begin{aligned}
V &=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{d} \\
&=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{\sqrt{x^{2}+r^{2}}} \quad[\text { from Eq. }
\end{aligned}
$$
Now,
$\therefore C B=\sqrt{O C^{2}+O B^{2}}=\sqrt{r^{2}+\left(\frac{3}{4}\right)^{2} r^{2}}=\frac{5}{4} r$
Similarly, $A C=\frac{5}{3} r$
$\therefore$ Work done in bringing a point charge $-q$ from $A$
to $B$,
$\begin{aligned} W &=-q\left(V_{B}-V_{A}\right) \\ &=-q\left[\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{C B}-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{C A}\right] \\=& \frac{-1}{4 \pi \varepsilon_{0}} \cdot q^{2}\left[\frac{1}{\frac{5}{4} r}-\frac{1}{5}\right] \\=& \frac{-1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{5 r}[4-3]=-\frac{1}{5} \cdot \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{r} \end{aligned}$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2018
A point charge $+q$ is placed at the centre of a cube of side $\mathrm{L}$. The electric flux emerging from the cube is

Options:
A $\frac{q}{\varepsilon_0}$
B Zero
C $\frac{6 q L^2}{\varepsilon_0}$
D $\frac{q}{6 L^2 \varepsilon_0}$

The correct answer is:
$\frac{q}{\varepsilon_0}$

Solution:
No Solution Available

Topic: Physics, Electrostatics, WBJEE, WBJEE 2010