datavorous/entrance-exam-dataset huggingface dataset:
A biconvex lens of focal length $f$ and radii of curvature of both the surfaces $R$ is made of a material of refractive index $n_{1}$. This lens is placed in a liquid of refractive index $n_{2}$. Now this lens will behave like

Options:
A either as a convex or as a concave lens depending solely on $\mathrm{R}$
B a convex lens depending on $n_{1}$ and $n_{2}$
C a concave lens depending on $n_{1}$ and $n_{2}$
D a convex lens of same focal length irrespective of $R, n_{1}$ and $n_{2}$

The correct answers are:
a convex lens depending on $n_{1}$ and $n_{2}$, a concave lens depending on $n_{1}$ and $n_{2}$

Solution:
The lens will behave like a convex lens or a concave lens depending upon the value of $n_{1}$ and $n_{2}$.

Topic: Physics, Ray Optics, WBJEE, WBJEE 2013
A block of ice at temperature $-20^{\circ} \mathrm{C}$ is slowly heated and converted to steam at $100^{\circ} \mathrm{C}$. Which of the following diagram is most appropriate?

Options:
A

B

C

D

The correct answer is:

Solution:
Hints:

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2010
A block of mass $1 \mathrm{kg}$ starts from rest at $x=0$ and moves along the $X$ -axis under the action of a force $F=k t,$ where $t$ is time and
$k=1 \mathrm{Ns}^{-1}$. The distance the block will trave] in 6 seconds is

Options:
A $36 \mathrm{m}$
B $72 \mathrm{m}$
C $108 \mathrm{m}$
D $18 \mathrm{m}$

The correct answer is:
$36 \mathrm{m}$

Solution:
From Newton's second law
$$
\begin{array}{l}
F=m a \\
m=1 \mathrm{kg} \\
F=1 \times a=\frac{d v}{d t}
\end{array}
$$
According to the question,
$$
F=k t=\frac{d v}{d t}
$$
$$
d v=k t \cdot d t
$$
Integration gives
$$
\begin{aligned}
v &=\frac{k t^{2}}{2}=\frac{d x}{d t} \\
x &=\frac{k t^{3}}{6}=\frac{1 \times 6 \times 6 \times 6}{6} \\
&=36 \mathrm{m}
\end{aligned}
$$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2017
A block of mass $m(=0.1 \mathrm{kg})$ is hanging over a frictionless light fixed pulley by an inextensible string of negligible mass. The other end of the string is pulled by a constant force $F$ in the vertically downward direction. The linear momentum of the block increases by $2 \mathrm{kgms}^{-1}$ in $1 \mathrm{s}$ after the block starts from rest. Then, (given $g=10 \mathrm{ms}^{-2}$ )

Options:
A The tension in the string is $F$
B The tension in the string is $3 \mathrm{N}$
C The work done by the tension on the block is $20 \mathrm{J}$ during this 1 s
D The work done against the force of gravity is $10 \mathrm{J}$

The correct answers are:
The tension in the string is $F$, The tension in the string is $3 \mathrm{N}$, The work done against the force of gravity is $10 \mathrm{J}$

Solution:
The free body diagram

$\Rightarrow \quad F=2+m g=3 N$
also, $a=\frac{\text { unbalanced force }}{\text { mass }}=\frac{2}{0.1}=20 \mathrm{m} / \mathrm{s}^{2}$
$S=\frac{1}{2} a t^{2}=\frac{1}{2} \times 20 \times 1=10 \mathrm{m}$
Honce, work done by tension
$$
=F \times 10=3 \times 10=30 \mathrm{J}
$$
So, the work done against gravity
$$
=m g \times S=1 \times 10=10 \mathrm{J}
$$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2013
A block of mass $m_{2}$ is placed on a horizontal table and another block of mass $m_{1}$ is placed on top of it. An increasing horizontal force \(F=\alpha t\) is exerted on the upper block but the lower block never moves as a result. If the coefficient of friction between the blocks is $\mu_{i}$ and that between the lower block and the table is $\mu_{2}$, then what is the maximum possible value of $\mu_{1} / \mu_{2} ?$

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

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

Solution:
According to the question.

$\therefore \quad N_{2}=m_{2} g+N_{1}=\left(m_{1}+m_{2}\right) g$
: Lower block never moves.
$\therefore \quad f_{r_{2}} \geq f_{n}$
$\quad \mu_{2} N_{2} \geq \mu_{1} N_{1} \Rightarrow \mu_{2}\left(m_{1}+m_{2}\right) g \geq \mu_{1} m_{1} g$
$\Rightarrow \quad \frac{m_{1}+m_{2}}{m_{1}} \geq \frac{\mu_{1}}{\mu_{2}}$
$\therefore \quad \frac{\mu_{1}}{\mu_{2}} \leq 1+\frac{m_{2}}{m_{1}}$
$\left.\therefore \quad \frac{\mu_{1}}{\mu_{2}}\right|_{\operatorname{man}}=1+\frac{m_{2}}{m_{1}}$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2018
A block of mass m rests on a horizontal table with a co-efficient of static friction $\mu .$ What minimum force must be applied on the block to drag it on the table?

Options:
A $\frac{\mu}{\sqrt{1+\mu^{2}}} \mathrm{mg}$
B $\frac{\mu-1}{\mu+1} \mathrm{mg}$
C $\frac{\mu}{\sqrt{1-\mu^{2}}} \mathrm{mg}$
D $\mu \mathrm{mg}$

The correct answer is:
$\frac{\mu}{\sqrt{1+\mu^{2}}} \mathrm{mg}$

Solution:
Hint:

$F \sin \theta+N=m g \quad N=m g-F \sin \theta$
$F \cos \theta=\mu(m g-F \sin \theta) \quad F=\frac{\mu m g}{\cos \theta+\mu \sin \theta}$,
for $F_{\min }, \frac{d}{d \theta}[\cos \theta+\mu \sin \theta]=0$
$\therefore \tan \theta=\mu$
$\therefore F_{\min }=\frac{\mu \mathrm{mg}}{\sqrt{1+\mu^{2}}}$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2020
A block of mass $m$ slides with speed $v$ on a frictionless table towards another stationary block of mass $m$. A massless spring with spring constant $\mathrm{k}$ is attached to the second block as shown in figure. The maximum distance the spring gets compress through is

Options:
A $\sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \mathrm{v}$
B $\sqrt{\frac{m}{2 k}} v$
C $\sqrt{\frac{\mathrm{k}}{\mathrm{m}}} \mathrm{v}$
D $\sqrt{\frac{\mathrm{k}}{2 \mathrm{~m}}} \mathrm{v}$

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

Solution:
$\frac{1}{2} \mathrm{kx}^{2}=\frac{1}{2} \times \frac{\mathrm{m} \times \mathrm{m}}{2 \mathrm{~m}} \mathrm{v}^{2} \Rightarrow \mathrm{x}=\sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}} \mathrm{v}$

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2021
A body is projected from the ground with a velocity $\mathbf{v}=(3 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}) \mathrm{ms}^{-1} .$ The maximum height attained and the range of the body respectively are (given $g=10 \mathrm{ms}^{-2}$ )

Options:
A $5 \mathrm{m}$ and $6 \mathrm{m}$
B $3 \mathrm{m}$ and $10 \mathrm{m}$
C $6 \mathrm{m}$ and $5 \mathrm{m}$
D $3 \mathrm{m}$ and $5 \mathrm{m}$

The correct answer is:
$5 \mathrm{m}$ and $6 \mathrm{m}$

Solution:
(a) Given, $v=(3 i+10 j) m / s$
$\Rightarrow \quad v_{x}=3$ and $v_{y}=10$
$\therefore$ Maximum height attained, $H=\frac{v_{y}^{2}}{2 g}$
$$
\begin{aligned}
&=\frac{10 \times 10}{2 \times 10}=5 \mathrm{m} \\
\text { Range }=v_{x} \times T=V_{x} \times \frac{2 v_{y}}{g} &=\frac{3 \times 2 \times 10}{10}=6 \mathrm{m}
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2013
A body is projected with a speed $u \mathrm{~m} / \mathrm{s}$ at an angle $\beta$ with the horizontal. The kinetic energy at the highest point is $3 / 4$ th of the initial kinetic energy. The value of $\beta$ is :

Options:
A $30^{\circ}$
B $45^{\circ}$
C $60^{\circ}$
D $120^{\circ}$

The correct answer is:
$30^{\circ}$

Solution:
Hints: (K.E.) at maximum height $=\frac{1}{2} m\left(u^2 \cos ^2 \beta\right)$
$\mathrm{K} . \mathrm{E} .=\mathrm{K} \cos ^2 \beta$
Here, $K \cos ^2 \beta=\frac{3}{4} \mathrm{~K}$
$$
\begin{aligned}
& \cos \beta=\frac{\sqrt{3}}{2} \\
& \beta=30^{\circ}
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2010
A body of density $1.2 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ is dropped from rest from a height $1 \mathrm{~m}$ into a liquid of density $2.4 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$. Neglecting all dissipative effects, the maximum depth to which the body sinks before returning to float on the surface is

Options:
A $0.1 \mathrm{~m}$
B $1 \mathrm{~m}$
C $0.01 \mathrm{~m}$
D $2 \mathrm{~m}$

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

Solution:
$\mathrm{d}=1.2 \times 10^{3}, \mathrm{~h}=1 \mathrm{~m}, \rho=2.4 \times 10^{3}$
By work Energy Theorem,
$$
\begin{array}{l}
\mathrm{Mg}(\mathrm{h}+\mathrm{d})=\mathrm{Bd} \\
\mathrm{V}\left(1.2 \times 10^{-3}\right) \mathrm{g}(1+\mathrm{d})=\mathrm{V}\left(2.4 \times 10^{-3}\right) \mathrm{g} \mathrm{d} \\
\therefore \mathrm{d}=1 \mathrm{~m}
\end{array}
$$

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2021
A body of mass \(2 \mathrm{~kg}\) moves in a horizontal circular path of radius \(5 \mathrm{~m}\). At an instant, its speed is \(2 \sqrt{5} \mathrm{~m} / \mathrm{s}\) and is increasing at the rate of \(3 \mathrm{~m} / \mathrm{s}^2\). The magnitude of force acting on the body at the instant is,

Options:
A \(6 \mathrm{~N}\)
B \(8 \mathrm{~N}\)
C \(14 \mathrm{~N}\)
D \(10 \mathrm{~N}\)

The correct answer is:
\(10 \mathrm{~N}\)

Solution:
Hint: \(F=\) ma
\(\begin{aligned}
& =m \sqrt{a_c^2+a_T^2} \\
& =m \sqrt{\left(\frac{20}{5}\right)^2+9} \\
& =2 \sqrt{16+9}=2 \times 5=10 \mathrm{~N}
\end{aligned}\)

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2023
A body of mass \(6 \mathrm{~kg}\) is acted upon by a force which causes a displacement in it given by \(x=\frac{t^2}{4}\) metre where \(t\) is the time in second. The work done by the force in 2 seconds is

Options:
A \(12 \mathrm{~J}\)
B 9J
C \(6 \mathrm{~J}\)
D \(3 \mathrm{~J}\)

The correct answer is:
\(3 \mathrm{~J}\)

Solution:
\(\begin{aligned}
& \text {Hints : } \mathrm{m}=6 \mathrm{~kg} \quad \mathrm{x}=\frac{\mathrm{t}^2}{4} \\
& \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}=\frac{\mathrm{t}}{2} \quad \mathrm{v}(0)=0 ; \mathrm{v}(2)=\frac{2}{2}=1 \\
& \mathrm{~K}_{\mathrm{i}}=\frac{1}{2} \mathrm{~m}(0)^2=0 ; \mathrm{K}_{\mathrm{f}}=\frac{1}{2} \mathrm{~m}(1)^2=\frac{1}{2} \times 6 \times 1=3 ; \mathrm{W}=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}=3-0=3 \mathrm{~J}
\end{aligned}\)

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2011
A body of mass $m$ is thrown vertically upward with speed $\sqrt{3} v_e$, where $v_e$ is the escape velocity of a body from earth surface. The final velocity of the body is

Options:
A 0
B $2 v_e$
C $\sqrt{3} v_e$
D $\sqrt{2} \mathrm{v}_{\mathrm{e}}$

The correct answer is:
$\sqrt{2} \mathrm{v}_{\mathrm{e}}$

Solution:
$-\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{~m}\left(\sqrt{3} \mathrm{v}_{\mathrm{e}}\right)^2=0+\frac{1}{2} \mathrm{mv}^2, \quad \mathrm{v}=\sqrt{2} \mathrm{v}_{\mathrm{e}}$

Topic: Physics, Gravitation, WBJEE, WBJEE 2022
A body of mass $m$ is thrown with velocity u from the origin of a co-ordinate axes at an angle $\theta$ with the horizon. The magnitude of the angular momentum of the particle about the origin at time $\mathrm{t}$ when it is at the maximum height of the trajectory is proportional to

Options:
A $\mathrm{u}$
B $u^2$
C $\mathrm{u}^3$
D independent of $u$

The correct answer is:
$\mathrm{u}^3$

Solution:
$\vec{L}=m u \cos \theta h_{\max }=m u \cos \theta \frac{u^2 \sin ^2 \theta}{2 g} \Rightarrow L=\frac{m u^3 \sin ^2 \theta \cos \theta}{2 g} \quad L \propto u^3$

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2022
A body starts from rest, under the action of an engine working at a constant power and moves along a straight line. The displacement s is given as a function of time ( $f$ as

Options:
A $s=a t+b t^{2}, a$ and $b$ are constants
B $s=b t^{2}, b$ is a constant
C $s=a t^{3 / 2},$ a is a constant
D $s=a t,$ a is a constant

The correct answer is:
$s=a t^{3 / 2},$ a is a constant

Solution:
Given, Power $(P)=$ constant
Kinetic Energy (KE) $=\frac{1}{2} m v^{2}$
We know that, $P=\frac{K E}{\Delta t} \Rightarrow P=\frac{m v^{2}}{\Delta t}$
$\because P=$ constant.
Hence, velocity of the body $v \propto \sqrt{t}$...(i)
As, Velocity $v=\frac{d s}{d t}$...(ii)
From Eqs. (i) and (ii), we get
So, $\quad \frac{d s}{d t} \propto \sqrt{t}$
Integrating the above cquation w.r.t. time $(t)$
$$
\int \frac{d s}{d t} \propto \int \sqrt{t}
$$
we get, displacement of the body $s \propto t^{3 / 2}$
Displacement $s=a t^{3 / 2}$, where $a$ is constant.

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2019
A body when fully immersed in a liquid of specific gravity 1.2 weighs 44 gwt. The same body when fully immersed in water weighs 50 gwt. The mass of the body is

Options:
A 36 \mathrm{g}$
B 48 \mathrm{g}$
C 64 \mathrm{g}$
D $80 \mathrm{g}$

The correct answer is:
$80 \mathrm{g}$

Solution:
Balancing equation
$$
W=m g-V \rho g
$$
Hence,
$$
\begin{array}{l}
44=m-1.2 \mathrm{V} \\
50=m-V
\end{array}
$$
On solving Eqs. (i) and (ii) we get
$$
m=80 \mathrm{g}
$$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2012
A box is moved along a straight line by a machine delivering constant power. The distance moved by the body in time \(\mathrm{t}\) is proportional to

Options:
A \(\mathrm{t}^{\frac{1}{2}}\)
B \(t^{\frac{3}{4}}\)
C \(\mathrm{t}^{\frac{3}{2}}\)
D \(\mathrm{t}^2\)

The correct answer is:
\(\mathrm{t}^{\frac{3}{2}}\)

Solution:
\(\begin{aligned}
& \text {Hints: } \mathrm{P}=\mathrm{Fv}=\mathrm{m} \cdot \frac{\mathrm{dv}}{\mathrm{dt}} \cdot \mathrm{v} \\
& \int \mathrm{vdv}=\int \mathrm{P} / \mathrm{mdt} ; \quad \frac{\mathrm{v}^2}{2}=\frac{\mathrm{Pt}}{\mathrm{m}} \\
& V=\sqrt{\frac{2 p}{m}} \mathrm{t}^{\frac{1}{2}} ; \frac{\mathrm{dx}}{\mathrm{dt}}=\sqrt{\frac{2 \mathrm{p}}{\mathrm{m}}} \mathrm{t}^{\frac{1}{2}} \\
& \int d x=\sqrt{\frac{2 p}{m}} \int t^{\frac{1}{2}} d t ; \quad x=\sqrt{\frac{2 p}{m}} \frac{t^{\frac{3}{2}}}{\frac{3}{2}}=\frac{2}{3} \sqrt{\frac{2 p}{m}} \mathrm{t}^{\frac{3}{2}} \\
& \mathrm{x} \alpha \mathrm{t}^{\frac{3}{2}} \\
&
\end{aligned}\)

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2011
A boy of mass $40 \mathrm{~kg}$ is climbing a vertical pole at a constant speed. If the coefficient of friction between his palms and the pole is 0.8 and $g=10 \mathrm{~m} / \mathrm{s}^2$, the horizontal force that he is applying on the pole is

Options:
A $300 \mathrm{~N}$
B $400 \mathrm{~N}$
C $500 \mathrm{~N}$
D $600 \mathrm{~N}$

The correct answer is:
$500 \mathrm{~N}$

Solution:
Hints: Here $\mu=0.8$
Frictional force $=\mu \mathrm{N}_1=m g$
$$
\mathrm{N}_1=\frac{m g}{\mu}=\frac{400}{0.8}=500 \mathrm{~N}
$$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2010
A bulb is placed at the centre of a circular track of radius $10 \mathrm{~m}$. A vertical wall is erected touching the track at a point P. A man is running along the track with a speed of $10 \mathrm{~m} / \mathrm{sec}$. Starting from $\mathrm{P}$ the speed with which his shadow is running along the wall when he is at an angular distance of $60^{\circ}$ from $P$ is

Options:
A $30 \mathrm{~m} / \mathrm{sec}$
B $40 \mathrm{~m} / \mathrm{sec}$
C $60 \mathrm{~m} / \mathrm{sec}$
D $80 \mathrm{~m} / \mathrm{sec}$

The correct answer is:
$40 \mathrm{~m} / \mathrm{sec}$

Solution:
$\overrightarrow{\mathrm{v}}=\mathrm{r} \frac{\mathrm{d} \theta}{\mathrm{dt}}=10$
$\tan \theta=\frac{\mathrm{y}}{\mathrm{r}}$ or, $\mathrm{y}=\mathrm{r} \tan \theta$ or, $\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{r} \sec ^{2} \theta \cdot \frac{\mathrm{d} \theta}{\mathrm{dt}}=10 \times \sec ^{2}\left(60^{\circ}\right)=40 \mathrm{~m} / \mathrm{sec}$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2021
A bullet is fired with a velocity u making an angle of $60^{\circ}$ with the horizontal plane. The horizontal component o the velocity of the bullet when it reaches the maximum height is

Options:
A $\mathrm{u}$
B $0$
C $\frac{\sqrt{3 u}}{2}$
D $\frac{\mathrm{u}}{2}$

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

Solution:
Horizontal velocity would be constant so the value of velocity at the highest point will be $u / 2$

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2009
A bullet of mass $4.2 \times 10^{-2} \mathrm{kg},$ moving at a speed of $300 \mathrm{ms}^{-1}$, gets stuck into a block with a mass 9 times that of the bullet. If the block is free to move without any kind of friction, the heat generated in the process will be

Options:
A $45 \mathrm{cal}$
B $405 \mathrm{cal}$
C $450 \mathrm{cal}$
D $1701 \mathrm{cal}$

The correct answer is:
$405 \mathrm{cal}$

Solution:
Let mass of bullet = $m$
Mass of block $=\mathrm{M}$
Velocity of bullet $=v=300 \mathrm{m} / \mathrm{s}$
Velocity of combined system $M+m=V$
Here, fhom momentum conservation $\frac{M+m}{m} V=v$
$v=\frac{300 \times 42 \times 10^{-4}}{4.2 \times 10^{-4}+9\left(4.2 \times 10^{-4}\right)}$
$30 \mathrm{ms}$
Now, heat produced = Loss in kinetic energy of bullet
$$
\begin{array}{l}
=\frac{1}{2} m^{2}-\frac{1}{2}(M+m) V^{2} \\
=\frac{1}{2} \times 4.2 \times 10^{-4}(300)^{2}-\frac{1}{2}\left(4.2 \times 10^{-2}\right. \\
+9 \times 4.2 \times 1)(30)^{2} \\
=63 \times 270 \\
=1701 \mathrm{J} \\
=\frac{1701}{4.2} \mathrm{Cal}
\end{array}
$$
$=405 \mathrm{Cal}$

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2017
A bullet of mass $m$ travelling with a speed $v$ hits a block of mass $M$ initially at rest and gets embedded in it. The combined system is free to move and there is no other force acting on the system. The heat generated in the process will be

Options:
A Zero
B $\frac{m v^{2}}{2}$
C $\frac{M m v^{2}}{2(M-m)}$
D $\frac{m M v^{2}}{2(M+m)}$

The correct answer is:
$\frac{m M v^{2}}{2(M+m)}$

Solution:
Mass of bullet $=m_{1}=m$
Initial speed of bullet $=u_{1}=v$
Mass of block $=m_{2}=M$
Initial speed of block $=u_{2}=0$
Let the common velocity of the bodies after collision $=V$ According to conservation of linear momentum
$$
\begin{aligned}
m \times v+M \times 0 &=(m+M) V \\
V &=\frac{m v}{(m+M)}
\end{aligned}
$$
: Heat generated = loss in $\mathrm{KE}$
$=$ Initial KE - Final KE
$=\frac{1}{2} m v^{2}-\frac{1}{2}(m+M) v^{2}$
$=\frac{1}{2} m v^{2}-\frac{1}{2}(m+M)\left(\frac{m v}{m+M}\right)^{2}$
$=\frac{1}{2} m v^{2}-\frac{1}{2} \frac{m^{2} v^{2}}{(m+M)}$
$=\frac{1}{2} m v^{2}\left(1-\frac{m}{m+M}\right)$
$=\frac{1}{2} m v^{2}\left(\frac{m+M-m}{m+M}\right)$
$=\frac{1}{2} \frac{m M}{(m+M)} v^{2}$

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2013
A bullet on penetrating $30 \mathrm{cm}$ into its target loses its velocity by $50 \%$. What additional distance will it penetrate into the target before it comes to rest?

Options:
A $30 \mathrm{cm}$
B $20 \mathrm{cm}$
C $10 \mathrm{cm}$
D $5 \mathrm{cm}$

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

Solution:
Given $S=30 \mathrm{cm}=30 \times 10^{-2} \mathrm{m}$ and loss in
velocity is $50 \%$ Equation of motion
$$
v^{2}=u^{2}+2 a s
$$
Here, $\frac{u^{2}}{4}=u^{2}+2 a \times 30 \times 10^{-2}$
$$
0=\frac{u^{2}}{4}+3 a \times x
$$
On solving Eqs. (i) and (ii), we get
$$
x=10 \mathrm{cm}
$$

Topic: Physics, Motion In One Dimension, WBJEE, WBJEE 2012
A candidate is required to answer 6 out of 12 questions which are divided into two parts A and $B$, each containing 6 questions and he/she is not permitted to attempt more than 4 questions from any part. In how many different ways can he/she make up hisher choice of 6 questions?

Options:
A 850
B 800
C 750
D 700

The correct answer is:
850

Solution:
$$
\begin{array}{|c|c|}
\hline A & B \\
\hline 4 & 2 \\
\hline 3 & 3 \\
\hline 2 & 4 \\
\hline
\end{array}
$$
$\therefore$ Total number of ways $={ }^{6} C_{4} \times{ }^{6} C_{2}+{ }^{6} C_{3} \times{ }^{6} C_{3}+{ }^{6} C_{2} \times{ }^{6} C_{4}$
$=\frac{6 \times 5}{2 \times 1} \times \frac{6 \times 5}{2 \times 1}+\frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1}+\frac{6 \times 5}{2 \times 1} \times \frac{6 \times 5}{2 \times 1}$
$=15 \times 15+20 \times 20+15 \times 15$
$=225+400+225$
$=850$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2019
A capacitor of capacitance $C$ is connected in series with a resistance $R$ and $D C$ source of emf $E$ through a key. The capacitor starts charging when the key is closed. By the time the capacitor has been fully charged. what amount of energy is dissipated in the resistance $R ?$

Options:
A $\frac{1}{2} C E^{2}$
B 0
C $\mathrm{CE}^{2}$
D $\frac{E^{2}}{R}$

The correct answer is:
$\frac{1}{2} C E^{2}$

Solution:
We know that, energy stored in the capacitor
$$
=\frac{1}{2} C E^{2}
$$
and energy supplied by the source of emf
$$
=C E^{2}
$$
$\therefore$ Energy dissipated in resistance $R$ = Energy supplied by the source of emf $E$ - Energy stored in the capacitor
$$
\begin{array}{l}
=C E^{2}-\frac{1}{2} C E^{2} \\
=\left(1-\frac{1}{2}\right) C E^{2}=\frac{1}{2} C E^{2}
\end{array}
$$

Topic: Physics, Capacitance, WBJEE, WBJEE 2019
A car is moving with a speed of $72 \mathrm{km}\mathrm{h}^{-1}$ towards a roadside source that emits sound at a frequency of $850 \mathrm{Hz}$. The car driver listens to the sound while approaching the source and again while moving away from the source after crossing it. If the velocity of sound is $340 \mathrm{ms}^{-1}$, the difference of the two frequencies, the driver hears is

Options:
A $50 \mathrm{Hz}$
B $85 \mathrm{Hz}$
C $100 \mathrm{Hz}$
D $150 \mathrm{Hz}$

The correct answer is:
$100 \mathrm{Hz}$

Solution:
By Doppler's Effect When observer is moving with velocity $v_{0}$ towards a source at rest then approach frequency
$$
N_{\text {Approach }}=N\left(\frac{v+v_{0}}{v}\right)
$$
where $N=850 \mathrm{Hz}, v=340 \mathrm{ms}^{-1}, v_{0}=72 \mathrm{kmh}^{-1}$
$$
=20 \mathrm{ms}^{-1}=850\left(\frac{340+20}{340}\right)
$$
Similarly. when observer is moving away from source, then
$$
N_{\text {Soparation }}=N\left(\frac{v-v_{0}}{v}\right)=850\left(\frac{340-20}{340}\right)
$$
The different of the two frequency.
$$
\begin{array}{l}
N_{\text {Approuch }}-N_{\text {Soparation }} \\
\qquad \begin{aligned}
&=850\left(\frac{360}{340}\right)-850\left(\frac{320}{340}\right) \\
&=\frac{850}{340} \times 40=\frac{850}{8.5}=100 \mathrm{Hz}
\end{aligned}
\end{array}
$$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2014
A car moving at a velocity of $17 \mathrm{ms}^{-1}$ towards an approaching bus that blows a horn at a frequency of $640 \mathrm{Hz}$ on a straight track. The frequency of this horn appears to be $680 \mathrm{Hz}$ to the car driver. If the velocity of sound in air is $340 \mathrm{ms}^{-1}$, then the velocity of the approaching bus is

Options:
A $2 \mathrm{ms}^{-1}$
B $4 \mathrm{ms}^{-1}$
C $8 \mathrm{ms}^{-1}$
D $10 \mathrm{ms}^{-1}$

The correct answer is:
$4 \mathrm{ms}^{-1}$

Solution:
Given, velocity of sound, $v=340 \mathrm{m} / \mathrm{s}$
Velocity of listner, $v_{L}=17 \mathrm{m} / \mathrm{s}$
Velocity of source $=v_{s}$
Frequency of hom emitted
$$
v=640 \mathrm{Hz}
$$

The apparent frequency
$$
\begin{aligned}
v^{\prime} &=v \frac{\left(v+v_{l}\right)}{v-v_{s}} \\
680 &=640\left(\frac{340+17}{340-v_{s}}\right)
\end{aligned}
$$
On solving we get $v_{s}=4 \mathrm{m} / \mathrm{s}$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2013
A carbon resistor with colour code is shown in the figure. There is no fourth band in the resistor. The value of the resistance is

Options:
A $24 \mathrm{M} \Omega \pm 20 \%$
B $14 \mathrm{k} \Omega \pm 5 \%$
C $24 \mathrm{k} \Omega \pm 20 \%$
D $34 \mathrm{k} \Omega \pm 10 \%$

The correct answer is:
$24 \mathrm{k} \Omega \pm 20 \%$

Solution:
RYO
By colour coding of resistors.
\(R=24 \times 10^3 \Omega=24 \mathrm{k} \Omega\)

Topic: Physics, Current Electricity, WBJEE, WBJEE 2021
A cell of emf $E$ is connected to a resistance $R_{1}$ for time $t$ and the amount of heat generated in it is $H$. If the resistance $R_{1}$ is replaced by another resistance $R_{2}$ and is connected to the cell at the same time $t,$ the amount of heat generated in $R_{2}$ is $4 \mathrm{H}$. Then internal resistance of the cell is

Options:
A $\frac{2 R_{1}+R_{2}}{2}$
B $\sqrt{R_{1} R_{2}} \frac{2 \sqrt{R_{2}}-\sqrt{R_{1}}}{\sqrt{R_{2}}-2 \sqrt{R_{1}}}$
C $\sqrt{R_{1} R_{2}} \frac{\sqrt{R_{2}}-2 \sqrt{R_{1}}}{2 \sqrt{R_{2}}-\sqrt{R_{1}}}$
D $\sqrt{R_{1} R_{2}} \frac{\sqrt{R_{2}}-\sqrt{R_{1}}}{\sqrt{R_{2}}+\sqrt{R_{1}}}$

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

Solution:
We have, $H=I^{2} R$ According to given condition, $I_{1}^{2} R_{1}=H$
and
$$
I_{2}^{2} R_{2}=4 H
$$
$$
\Rightarrow \quad \frac{E^{2}}{\left(R_{1}+r\right)^{2}} R_{1}=H \text { and } \frac{E^{2}}{\left(R_{2}+r\right)^{2}} R_{2}=4 H
$$
$\therefore \quad \frac{R_{2}}{\left(R_{2}+r\right)^{2}}=\frac{4 R_{1}}{\left(R_{1}+r\right)^{2}}$
$\Rightarrow \quad \sqrt{R_{2}}\left(R_{1}+r\right)=2 \sqrt{R_{1}}\left(R_{2}+r\right)$
On solving. $r=\frac{\sqrt{R_{1} R_{2}}\left[\sqrt{R_{1}}-2 \sqrt{R_{2}}\right]}{2 \sqrt{R_{1}}-\sqrt{R_{2}}}$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2013