datavorous/entrance-exam-dataset huggingface dataset:
The velocity of a particle executing a simple harmonic motion is $13 \mathrm{ms}^{-1}$, when its distance from the equilibrium position $(Q)$ is $3 \mathrm{m}$ and its velocity is $12 \mathrm{ms}^{-1},$ when it is $5 \mathrm{m}$ away from $Q .$ The frequency of the simple harmonic motion is

Options:
A $\frac{5 \pi}{8}$
B $\frac{5}{8 \pi}$
C $\frac{8 \pi}{5}$
D $\frac{8}{5 \pi}$

The correct answer is:
$\frac{5}{8 \pi}$

Solution:
The speed of a particle executing simple
harmonic motion is $v=\omega \sqrt{a^{2}-x^{2}}$
where, $a=$ Amplitude $\omega=$ Angular frequency $x=$ Displacement
or
$$
v^{2}=\omega^{2}\left(a^{2}-x^{2}\right)
$$
According to the question,
Here
$$
\begin{aligned}
v_{1}^{2} &=\omega^{2}\left(a^{2}-x_{1}^{2}\right) \\
v_{2}^{2} &=\omega^{2}\left(a^{2}-x_{2}^{2}\right) \\
v_{1}^{2}-v_{2}^{2} &=\omega^{2}\left(x_{2}^{2}-x_{1}^{2}\right) \\
\omega &=\sqrt{\frac{v_{1}^{2}-v_{2}^{2}}{x_{2}^{2}-x_{1}^{2}}} \\
v_{1}=& 13 \mathrm{m} / \mathrm{s}
\end{aligned}
$$
$$
\begin{aligned}
v_{2} &=12 \mathrm{m} / \mathrm{s} \\
x_{1} &=3 \mathrm{m} \\
x_{2} &=5 \mathrm{m} \\
\omega &=\sqrt{\frac{(13)^{2}-(12)^{2}}{(5)^{2}-(3)^{2}}} \\
&=\sqrt{\frac{169-144}{25-9}}=\sqrt{\frac{25}{16}}=\frac{5}{4} \\
\omega &=2 \pi r \\
v &=\frac{\omega}{2 \pi}=\frac{5 / 4}{2 \pi} \\
v &=\frac{5}{8 \pi}
\end{aligned}
$$

Topic: Physics, Oscillations, WBJEE, WBJEE 2017
The velocity of sound in air at $20^{\circ} \mathrm{C}$ and 1 atm pressure is $344.2 \mathrm{m} / \mathrm{s}$. At $40^{\circ} \mathrm{C}$ and $2 \mathrm{atm}$
pressure, the velocity of sound in air is approximately

Options:
A $350 \mathrm{m} / \mathrm{s}$
B $356 \mathrm{m} / \mathrm{s}$
C $363 \mathrm{m} / \mathrm{s}$
D $370 \mathrm{m} / \mathrm{s}$

The correct answer is:
$356 \mathrm{m} / \mathrm{s}$

Solution:
$\quad T_{1}=273+20=293 \mathrm{K}$
$$
T_{2}=273+40=313 \mathrm{K}
$$
The velocity of sound wave in gases or air is given by
$$
v=\sqrt{\frac{\gamma R T_{1}}{M}}
$$
where, $\gamma=$ Ratio of $C_{p} / C_{V}$
$$
\begin{array}{l}R=\text { gas constant } \\ \qquad \begin{array}{l}v_{1}=\sqrt{\frac{\gamma R T_{1}}{M}} \\ v_{2}=\sqrt{\frac{\gamma R T_{2}}{M}}\end{array} \\ \frac{v_{1}}{v_{2}}=\sqrt{\frac{T_{1}}{T_{2}}}=\sqrt{\frac{293}{313}}\end{array}
$$
$\begin{aligned} \text { Given, } v_{1}=344.2 \mathrm{m} / \mathrm{s} & \\ \begin{aligned} \therefore \quad v_{2} &=v_{1} \sqrt{\frac{313}{293}} \\ &=344.2 \times \sqrt{1068} \\ &=344.2 \times 103 \mathrm{m} / \mathrm{s} \\ &=35575 \mathrm{m} / \mathrm{s} \\ &=356 \mathrm{m} / \mathrm{s} \end{aligned} \end{aligned}$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2016
The velocity $(v)$ of a particle (under a force $F$ ) depends on its distance $(x)$ from the origin (with $x>0) v \propto\frac{1}{\sqrt{x}}$. Find how the magnitude
of the force $(F)$ on the particle depends on $x ?$

Options:
A $F \propto \frac{1}{x^{3 / 2}}$
B $F \propto \frac{1}{x}$
C $F \propto \frac{1}{x^{2}}$
D $F \times x$

The correct answer is:
$F \propto \frac{1}{x^{2}}$

Solution:
According to the question.
$$
v \propto \frac{1}{\sqrt{x}}
$$
or $v=\frac{k}{\sqrt{x}}$
$$
\frac{d v}{d t}=\frac{d}{d t} \cdot \frac{K}{\sqrt{x}}=k \cdot \frac{-1}{2} \cdot x^{-\frac{1}{2}-1} \cdot \frac{d x}{d t}
$$
$a=-\frac{k}{2} \cdot x^{-\frac{3}{2}} \cdot \frac{k}{\sqrt{x}}\left[\because \frac{d x}{d t}=v\right.$ and $v=\frac{k}{\sqrt{x}}$ from $\left.\mathrm{Eq}(i)\right]$
$$
=\frac{-k^{2}}{2} \cdot \frac{1}{x^{2}}
$$
$=m \frac{k^{2}}{2} \cdot \frac{1}{x^{2}}$
(magnitue
$F=\frac{k^{2}}{2} \cdot \frac{m}{x^{2}}$
$\therefore$
$F x \frac{1}{x^{2}}$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2018
The volume of ethyl alcohol (density $1.15 \mathrm{g} / \mathrm{cc}$ ) that has to be added to prepare $100 \mathrm{cc}$ of $0.5 \mathrm{M}$ ethyl alcohol solution in water is

Options:
A $1.15 \mathrm{cc}$
B $2 \mathrm{cc}$
C $2.15 \mathrm{cc}$
D $2.30 \mathrm{cc}$

The correct answer is:
$2 \mathrm{cc}$

Solution:
Molarity $=\frac{\text { mass } \times 1000}{\text { molecular weight } \times \text { volume of solution }}$
Molecular weight of ethyl alcohol,
$$
\begin{aligned}
\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} &=12 \times 2+5+16+1 \\
&=24+22=46 \mathrm{g} \mathrm{mol}^{-1}
\end{aligned}
$$
On substituting the values, we get
$$
\begin{array}{l}
0.5 \mathrm{M}=\frac{\text { mass } \times 1000}{46 \times 100} \\
\text { Mass }=\frac{0.5 \times 46}{10}=2.3 \mathrm{g}
\end{array}
$$
$\therefore$ Volume of ethyl alcohol required.
$$
V=\frac{\text { mass of ethyl alcohol }}{\text { density of ethyl alcohol }}=\frac{2.3 \mathrm{g}}{1.15 \mathrm{g} / \mathrm{cc}}=2 \mathrm{cc}
$$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2014
The water equivalent of a calorimeter is $10 \mathrm{g}$ and it contains $50 \mathrm{g}$ of water at $15^{\circ} \mathrm{C}$. Some amount of ice, initially at $-10^{\circ} \mathrm{C}$ is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (when specific heat of ice $=0.5 \mathrm{cal} \mathrm{gm}^{-1}{ }^{\circ} \mathrm{C}^{-1},$ specific heat of water $=1.0$
cal $\mathrm{gm}^{-1 \circ} \mathrm{C}^{-1}$ and latent heat of melting of ice $=80$ cal $\mathrm{gm}^{-1}$ )?

Options:
A $10 \mathrm{g}$
B $18 \mathrm{g}$
C $20 \mathrm{g}$
D $30 \mathrm{g}$

The correct answer is:
$20 \mathrm{g}$

Solution:
Let the mass of ice $=m$
Applying calorimetry principle, heat given $=$ heat taken $\left(m_{1}+m_{2}\right) s_{1}\left(t_{1}-t\right)=\frac{m L}{2}+m s_{2}\left(t-t_{2}\right)$
Putting values, we get $(10+50) \times 1 \times(15-0)=\frac{m}{2} \times 80+m \times 0.5[0-(-10)]$
$\Rightarrow 60 \times 15=40 m+\frac{m}{2} \times 10=45 m$
$\therefore \quad m=\frac{60 \times 15}{45}=20 \mathrm{g}$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2018
The wavelength of second Balmer line in hydrogen &spectrum is 600 nm The wavelength for its third line in Lyman series is

Options:
A $800 \mathrm{nm}$
B $600 \mathrm{nm}$
C $400 \mathrm{nm}$
D None of the above

The correct answer is:
None of the above

Solution:
$\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$
For Balmer series,
$\begin{array}{l}
\therefore n_{1}=2, n_{2}=3,4,5 \ldots \\
\frac{1}{\lambda_{B}}=R\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right]
\end{array}$
$\begin{array}{l}
=R\left[\frac{1}{4}-\frac{1}{16}\right] \\
\frac{1}{\lambda_{B}}=R \cdot \frac{3}{16}
\end{array}$
Lyman series is obtained when $n_{1}=1, n_{2}=2,3,4, \ldots$ For third line in Lyman series,
$\begin{aligned}
\frac{1}{\lambda_{L}} &=R\left[\frac{1}{1^{2}}-\frac{1}{4^{2}}\right]=R\left[1-\frac{1}{16}\right] \\
\frac{1}{\lambda_{L}} &=R \cdot \frac{15}{16} \\
\frac{1 / \lambda_{B}}{\frac{1}{\lambda_{L}}} &=\frac{R \cdot \frac{3}{16}}{R \cdot \frac{15}{16}} \\
\frac{\lambda_{L}}{\lambda_{B}} &=\frac{3}{15} \\
\lambda_{L} &=\frac{3}{15} \times \lambda_{\theta} \\
\lambda_{B} &=600 \mathrm{nm}=600 \times 10^{-9} \mathrm{m} \\
\lambda_{L} &=\frac{3}{15} \times 600 \times 10^{-9} \\
&=120 \times 10^{-9} \mathrm{m}=120 \mathrm{nm}
\end{aligned}$
Hence, none of the given options is correct.

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2016
The weight of oxalic acid that will be required to prepare a $1000 \mathrm{mL}(\mathrm{N} / 20)$ solution is

Options:
A $126 / 100 \mathrm{g}$
B $63 / 40 \mathrm{g}$
C $63 / 20 \mathrm{g}$
D $126 / 20 \mathrm{g}$

The correct answer is:
$63 / 20 \mathrm{g}$

Solution:
$w=\frac{N E V}{1000}=\frac{1}{20} \times \frac{63 \times 1000}{1000}=\frac{63}{20} g$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2012
The well known compounds, $(+)-$ lactic acid and $(-)-$ lactic acid, have the same molecular formula, $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}$. The correct relationship between them is

Options:
A constitutional isomerism
B geometrical isomerism
C identicalness
D optical isomerism

The correct answer is:
optical isomerism

Solution:
Both are optical isomers because they rotate the plane of polarised light in opposite directions.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2012
The white precipitate $(Y)$, obtained on passing colourless and odourless gas $(X)$ through an ammoniacal solution of $\mathrm{NaCl}$, loses about $37 \%$ of its weight on heating and a white residue $(\mathrm{Z})$ of basic nature is left. Identify $(\mathrm{X}),(\mathrm{Y})$ and $(\mathrm{Z})$ from following sets.

Options:
A $$\mathrm{N}_{2} \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} \mathrm{NH}_{4} \mathrm{Cl}
$$
B $$\mathrm{O}_{2} \mathrm{NaNH}_{4} \mathrm{CO}_{3} \mathrm{NaHCO}_{3}
$$
C $$\mathrm{CO}_{2} \mathrm{NH}_{4} \mathrm{HCO}_{3} \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}
$$
D $$\mathrm{CO}_{2} \mathrm{NaHCO}_{3} \mathrm{Na}_{2} \mathrm{CO}_{3}
$$

The correct answer is:
$$\mathrm{CO}_{2} \mathrm{NaHCO}_{3} \mathrm{Na}_{2} \mathrm{CO}_{3}
$$

Solution:
$\begin{array}{c}
\mathrm{NH}_{3}+\underset{(\mathrm{x})}{\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow} \longrightarrow \mathrm{NH}_{4} \mathrm{HCO}_{3} \\
\mathrm{NH}_{4} \mathrm{HCO}_{3}+\mathrm{NaCl} \longrightarrow \mathrm{NaHCO}_{3} \downarrow+\mathrm{NH}_{4} \mathrm{Cl} \\
2 \mathrm{NaHCO}_{3} \stackrel{\Delta}{\longrightarrow} \underset{(\mathrm{y})}{\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}} \\
{(\mathrm{z})}
\end{array}$

Topic: Chemistry, s Block Elements, WBJEE, WBJEE 2021
The work done in carrying a charge $\mathrm{Q}$ once around a circle of radius $\mathrm{r}$ about a charge $\mathrm{q}$ at the centre is

Options:
A $\frac{q Q}{4 \pi \varepsilon_0 r}$
B $\frac{q Q}{4 \pi \varepsilon_0} \frac{1}{\pi r}$
C $\frac{q Q}{4 \pi \varepsilon_0}\left(\frac{1}{2 \pi r}\right)$
D $0$

The correct answer is:
$0$

Solution:
Work done by conservative force in a round trip is zero

Topic: Physics, Electrostatics, WBJEE, WBJEE 2009
The work function of Cesium is $2.27 \mathrm{eV}$. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of $600 \mathrm{nm}$ wavelength is

Options:
A $0.5 \mathrm{V}$
B $-0.2 \mathrm{V}$
C $-0.5 \mathrm{V}$
D None of the above

The correct answer is:
None of the above

Solution:
Work function of cesium $=2.27 \mathrm{eV}$
The wavelength of incident light
$$
\lambda=600 \mathrm{nm}=600 \times 10^{-9} \mathrm{m}
$$
The energy of incident photon $=h v$
$$
\begin{aligned}
E &=\frac{h c}{\lambda} \\
&=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{600 \times 10^{-9} \times 1.6 \times 10^{-19}} e \mathrm{V} \\
&=2.06 \mathrm{eV} \\
\frac{h c}{\lambda} < \phi
\end{aligned}
$$
$2.06 \mathrm{eV} < 2.27 \mathrm{eV}$
So, no emission of electrons will take place. Hence, none of the given options is correct.

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2016
The work function of metals is in the range of $2 \mathrm{eV}$ to $5 \mathrm{eV}$. Find which of the following wavelength of light cannot be used for photoelectric effect? (Consider, Plank constant $=4 \times 10^{-15} \mathrm{eVs},$ velocity of light $\left.=3 \times 10^{8} \mathrm{m} / \mathrm{s}\right)$

Options:
A $510 \mathrm{nm}$
B $650 \mathrm{nm}$
C $400 \mathrm{nm}$
D $570 \mathrm{nm}$

The correct answer is:
$650 \mathrm{nm}$

Solution:
Given that, $2 \mathrm{eV} \leq \phi \leq 5 \mathrm{eV}$
Wavelength corresponding to minimum and maximum values of work function are
$$
\begin{aligned}
\lambda_{\max } &=\frac{h C}{E}=\frac{4 \times 10^{-15} \times 3 \times 10^{8}}{2} \\
&=6 \times 10^{-7} \mathrm{m}=600 \mathrm{nm} \\
\text { and } \lambda_{\min } &=\frac{h C}{E}=\frac{4 \times 10^{-15} \times 3 \times 10^{8}}{5} \\
&=2.4 \times 10^{-7} \mathrm{m}=240 \mathrm{nm}
\end{aligned}
$$
Clearly, wavelength range would be
$$
240 \leq \lambda \leq 600
$$
Thus, wavelength $650 \mathrm{nm}$ is not suitable for photoelectric effect.

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2015
The yield of a acetanilide in the reaction (100\% conversion) of 2 moles of aniline with 1 mole of acetic anhydride is

Options:
A 2708
B $135 \mathrm{g}$
C $67.5 \mathrm{g}$
D 177 g

The correct answer is:
$135 \mathrm{g}$

Solution:
The reaction between aniline and acetic anhydride is as follows :

Topic: Chemistry, Amines, WBJEE, WBJEE 2017
There are 7 greeting cards, each of a different colour and 7 envelopes of same 7 colours as that of the cards. The number of ways in which the cards can be put in envelopes, so that exactly 4 of the cards go into envelopes of respective colour is,

Options:
A ${ }^{7} C_{3}$
B $2^{7} C_{3}$
C $3!^{4} C_{4}$
D $3!^{7} C_{3}^{4} C_{3}$

The correct answer is:
$2^{7} C_{3}$

Solution:
Required number of ways
$$
\begin{array}{l}
={ }^{7} C_{4} \times D(3) \\
={ }^{7} C_{3} \times\left[3 \cdot\left(1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}\right)\right] \\
{\left[\because{ }^{n} C_{r}={ }^{n} C_{n-r, a n d}\right.} \\
\left.D(n)=n\left[1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\ldots\right]\right] \\
={ }^{7} C_{3} \times 2=2\left({ }^{7} C_{3}\right)
\end{array}
$$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2019
There are 100 students in a class. In an examination, 50 of them failed in Mathematics, 45 failed in Physics, 40 failed in Biology and 32 failed in exactly two of the three subjects. Only one student passed in all the subjects. Then, the number of students failing in all the three subjects

Options:
A is 12
B is 4
C is 2
D cannot be determined from the given information

The correct answer is:
is 2

Solution:
According to question,
$\begin{array}{c}
a+b+c+d+e+f+g+h=100 \\
h=1 \\
a+b+d+e=50
\end{array}$
Venn diagram shows failed students $b+c+d+f=45$
$\begin{array}{l}
d+e+f+g=40 \\
b+e+f=32
\end{array}$
On solving, the above equations, we get $d=2$

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2012
There are \(n\) elastic balls placed on a smooth horizontal plane. The masses of the balls are \(m, \frac{m}{2}, \frac{m}{2^2}, \ldots . \frac{m}{2^{n-1}}\) respectively. If the first ball hits the second ball with velocity \(\mathrm{v}_0\), then the velocity of the \(\mathrm{n}^{\text {th }}\) ball will be,

Options:
A \(\frac{4}{3} v_0\)
B \(\left(\frac{4}{3}\right)^n v_0\)
C \(\left(\frac{4}{3}\right)^{n-1} v_0\)
D \(v_0\)

The correct answer is:
\(\left(\frac{4}{3}\right)^{n-1} v_0\)

Solution:
Hint : 1st Collision
\(\mathrm{V}_1=\frac{2 \times \mathrm{m}}{\frac{\mathrm{m}}{2}+\mathrm{m}} \mathrm{V}_0=\frac{4}{3} \mathrm{~V}_0\)
2nd Collision
\(V_2=\frac{2 \times \frac{m}{2}}{\frac{m}{2}+\frac{m}{2^2}} \times V_0=\left(\frac{4}{3}\right)^2 V_0\)
3rd Collision
\(\left(\frac{4}{3}\right)^3 \mathrm{~V}_0\)
\(\ldots(n-1)\) collision, \(\therefore V_{n-1}=\left(\frac{4}{3}\right)^{n-1} V_0\)

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2023
There are $n$ white and $n$ black balls marked $1,2,3, \ldots . n$. The number of ways in which we can arrange these balls in a row so that neighbouring balls are of different colours is

Options:
A $(n !)^2$
B (2n)!
C $2(n !)^2$
D $\frac{(2 n) !}{(n !)^2}$

The correct answer is:
$2(n !)^2$

Solution:
BW BW ..... $=n ! \times n !$
or
WB WB ..... $=2(n !)^2$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2022
There are two coins, one unbiased with probaility $\frac{1}{2}$ or getting heads and the other one is biased with probability $\frac{3}{4}$ of getting heads. A coin is selected at random and tossed. It shows heads up. Then, the probability that the unbiased coin wa selected is

Options:
A $\frac{2}{3}$
B $\frac{3}{5}$
C $\frac{1}{2}$
D $\frac{2}{5}$

The correct answer is:
$\frac{2}{5}$

Solution:
Let $E \rightarrow$ Event of head showing up
$E_{1} \rightarrow$ Event of biased coin chosen
$E_{2} \rightarrow$ Event of unbiased coin chosen Now, $P\left(E_{2}\right)=\frac{1}{2}$ and $P\left(E_{1}\right)=\frac{1}{2}$
Also, $P\left(\frac{E}{E_{2}}\right)=\frac{1}{2}$ and $P\left(\frac{E}{E_{1}}\right)=\frac{3}{4}$
(by conditional probability) By Baye's theorem
$$
\begin{aligned}
P\left(\frac{E_{2}}{E}\right)=& \frac{P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}{P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)+P\left(E_{1}\right) \cdot P\left(\frac{E}{E_{1}}\right)} \\
&=\frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{3}{4}}=\frac{2}{5}
\end{aligned}
$$

Topic: Mathematics, Probability, WBJEE, WBJEE 2013
There is a group of 265 persons who like either singing or dancing or painting. In this group 200 like singing, 110 like dancing and 55 like painting. If 60 persons like both singing and dancing, 30 like both singing and painting and 10 hike all three activities, then the number of persons who like only dancing and painting is

Options:
A 10
B 20
C 30
D 40

The correct answer is:
10

Solution:
Let $D$ denotes dancing. $P$ denotes painting and $S$ denotes singing.
$\begin{array}{l}
\therefore \quad n(D \cup P \cup S)=265 \\
n(S)=200, n(D)=110, n(P)=55 \\
n(S \cap D)=60, n(S \cap P)=30 \\
\text { and } n(D \cap P \cap S)=10 \\
\because n(D \cup P \cup S)=n(D)+n(P)+n(S)-n(D \cap P) \\
-n(P \cap S)-n(S \cap D)+n(D \cap P \cap S) \\
\therefore \quad 265=110+55+200-n(D \cap P) \\
\Rightarrow \quad 265=285-n(D \cap P) \\
\Rightarrow \quad n(D \cap P)=20
\end{array}$
$\therefore$ Only person who like dancing and painting
$\begin{array}{l}
=n(D \cap P)-n(D \cap P \cap S) \\
=20-10=10
\end{array}$

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2014
There is a small air bubble at the centre of a solid glass sphere of radius $r$ and refractive index $\mu .$ What will be the apparent distance of the bubble from the centre of the sphere, when viewed from outside?

Options:
A r
B $\frac{r}{\mu}$
C $r\left(1-\frac{1}{\mu}\right)$
D Zero

The correct answer is:
Zero

Solution:
'
As the object is at centre of the sphere.
$\therefore$ All rays will fall normally on surface, hence they do not deflect. Thus, a virtual image is formed at the
centre $O$.
$\therefore$ Apparent depth $=$ Real depth

Topic: Physics, Ray Optics, WBJEE, WBJEE 2018
This alcohol is

Options:
A
B
C
D

The correct answer is:

Solution:
Hint:
Hydride shift, here $\mathrm{D}^{-}$, shift decides the product

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2020
This product in the above reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
Hint:

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2020
Three blocks are pushed with a force $\mathrm{F}$ across a frictionless table as shown in figure. Let $\mathrm{N}_{1}$ be the contact force between the left two blocks and $\mathrm{N}_{2}$ be the contact force between the right two blocks. Then

Options:
A $F>N_{1}>N_{2}$
B $F>N_{2}>N_{1}$
C $\quad F>N_{1}=N_{2}$
D $F=N_{1}=N_{2}$

The correct answer is:
$F>N_{1}>N_{2}$

Solution:
$\mathrm{F}=6 \mathrm{a}, \mathrm{N}_{1}=5 \mathrm{a}, \mathrm{N}_{2}=3 \mathrm{a}$
$\mathrm{F}>\mathrm{N}_{1}>\mathrm{N}_{2}$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2021
Three blocks of masses 4 kg. 2 kg. 1 kg respectively are in contact on a frictionless table as shown in the figure. If a force of $14 \mathrm{N}$ is applied on the 4 kg block, the contact force between the $4 \mathrm{kg}$ and the $2 \mathrm{kg}$ block will be

Options:
A $2 \mathrm{N}$
B $6 \mathrm{N}$
C $8 \mathrm{N}$
D $14 \mathrm{N}$

The correct answer is:
$6 \mathrm{N}$

Solution:
We know that, $F=m a$
$$
a=\frac{F}{m}=\frac{14}{7}=2 \mathrm{ms}^{-2}
$$

Hence, from the figure
$$
14-N=4 a
$$
$$
\begin{array}{r}
14-N=8 \\
N=6 \mathrm{N}
\end{array}
$$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2012
Three bodies of the same material and having masses $m, m$ and $3 m$ are at temperatures $40^{\circ} \mathrm{C}, \quad 50^{\circ} \mathrm{C} \quad$ and $\quad 60^{\circ} \mathrm{C}$ respectively. If the bodies are brought in thermal contact, the final temperature will be

Options:
A $45^{\circ} \mathrm{C}$
B $54^{\circ} \mathrm{C}$
C $52^{\circ} \mathrm{C}$
D $48^{\circ} \mathrm{C}$

The correct answer is:
$54^{\circ} \mathrm{C}$

Solution:
Let the final temperature after the masses in thermal contact is $\theta$, then from the principle of calonimetry.
Heat lost $=$ Heat gained
$\begin{aligned} \Rightarrow 3 m s(60-\theta) &=m s(\theta-50)+m s(\theta-40) \\ \Rightarrow \quad 3(60-\theta) &=\theta-50+\theta-40 \\ \Rightarrow \quad 180-3 \theta &=2 \theta-90 \\ \theta &=\frac{270}{5}=54^{\circ} \mathrm{C} \end{aligned}$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2015
Three capacitors $3 \mu \mathrm{F}, 6 \mu \mathrm{F}$ and $6 \mu \mathrm{F}$ are connected in aeries to a source of $120 \mathrm{V}$. The potential difference, in volt, across the $3 \mu \mathrm{F}$ capacitor will be

Options:
A 24
B 30
C 40
D 60

The correct answer is:
60

Solution:
The combination of three charges in series
$$
\begin{array}{l}
\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} \\
\frac{1}{C}=\frac{1}{3}+\frac{1}{6}+\frac{1}{6} \Rightarrow C=\frac{6}{4}=1.5 \mu \mathrm{F}
\end{array}
$$
The charge of this circuit
$$
\begin{array}{l}
q=C V=1.5 \times 120 \\
q=180 \mu \mathrm{C}
\end{array}
$$
The potential difference across the $3 \mu \mathrm{F}$
$$
q=C V
$$
$$
V=\frac{q}{C}=\frac{180}{3}=60 \mathrm{V}
$$

Topic: Physics, Capacitance, WBJEE, WBJEE 2014
Three capacitors of capacitance 1.0,2.0 and $5.0 \mu \mathrm{F}$ are connected in series to a $10 \mathrm{V}$ source. The potential difference across the $2.0 \mu \mathrm{F}$ capacitor is

Options:
A $\frac{100}{17} \mathrm{V}$
B $\frac{20}{17} \mathrm{v}$
C $\frac{50}{17} \mathrm{V}$
D $10 \mathrm{V}$

The correct answer is:
$\frac{50}{17} \mathrm{V}$

Solution:
When the capacitors are connected in series, the resultant capacitance of combination
$$
\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}
$$
$$
\begin{array}{l}
=\frac{1}{1}+\frac{1}{2}+\frac{1}{5}=\frac{17}{10} \\
C=\frac{10}{17} \mu \mathrm{F}
\end{array}
$$
The charge will be same on all the capacitors in series
$$
Q=C V=\frac{10}{17} \times 10=\frac{100}{17}
$$
The potential difference across $2.0 \mu \mathrm{F}$ capacitor
$$
V^{*}=\frac{Q}{C}=\frac{100 / 17}{2}=\frac{50}{17} \text { volt}
$$

Topic: Physics, Capacitance, WBJEE, WBJEE 2017
Three concentric metallic shells A, B and C of radii $a, b$ and $c(a < b < c)$ have surface charge densities $+\sigma,-\sigma$ and $+\sigma$ respectively. The potential of shell B is

Options:
A $\quad(a+b+c) \frac{\sigma}{\varepsilon_0}$
B $\frac{\sigma \mathrm{c}}{\varepsilon_0}$
C $\left(\frac{a^2}{c}-\frac{b^2}{c}+c\right) \frac{\sigma}{e_0}$
D $\left(\frac{a^2}{b}-b+c\right) \frac{\sigma}{e_0}$

The correct answer is:
$\left(\frac{a^2}{b}-b+c\right) \frac{\sigma}{e_0}$

Solution:
$V_b=V_{A, b}+V_{B, b}+V_{C, b}$
$=\frac{K \sigma 4 \pi a^2}{b}-\frac{K \sigma 4 \pi b^2}{b}+\frac{K \sigma 4 \pi c^2}{c}$
$=\frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{b}-b+c\right)$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2022