datavorous/entrance-exam-dataset huggingface dataset:
Upon treatment with $I_{2}$ and aqueous $\mathrm{NaOH}$, which of the following compounds will form iodoform?

Options:
A $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}$
B $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}$
C $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$
D $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \cdot \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}$

The correct answer is:
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \cdot \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}$

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2012
Using binomial theorem, the value of $(0.999)^3$ correct to 3 decimal places is

Options:
A 0.999
B 0.998
C 0.997
D 0.995

The correct answer is:
0.997

Solution:
$$
\begin{aligned}
& \text { Hints : }{ }^3 C_0-{ }^3 C_1(.001)+{ }^3 C_2(.001)^2-{ }^3 C_3(.001)^3 \\
& =1-.003+3(.000001)-(.000000001)=0.997
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2009
Value of $\tan ^{-1}\left(\frac{\sin 2-1}{\cos 2}\right)$ is

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

The correct answer is:
$1-\frac{\pi}{4}$

Solution:
Hints : $\tan ^{-1}\left(\frac{\sin 2-1}{\cos 2}\right)=\tan ^{-1}\left(\frac{-(\sin 1-\cos 1)^2}{(\cos 1-\sin 1)(\cos 1+\sin 1)}\right) \quad=-\tan ^{-1}\left(\frac{\cos 1-\sin 1}{\cos 1+\sin 1}\right)=1-\frac{\pi}{4}$

Topic: Mathematics, Inverse Trigonometric Functions, WBJEE, WBJEE 2010
Vitamin $\mathrm{C}$ is

Options:
A Citric acid
B Lactic acid
C Paracetamol
D Ascorbic acid

The correct answer is:
Ascorbic acid

Solution:
No Solution Available

Topic: Chemistry, Biomolecules, WBJEE, WBJEE 2010
Water is flowing in streamline motion through a horizontal tube. The pressure at a point in the tube is $p$ where the velocity of flow is $v$. At another point, where the pressure is $p / 2,$ the velocity of flow is [density of water $=\rho]$

Options:
A $\sqrt{v^{2}+\frac{p}{\rho}}$
B $\sqrt{v^{2}-\frac{p}{\rho}}$
C $\sqrt{v^{2}+\frac{2 p}{\rho}}$
D $\sqrt{v^{2}-\frac{2 p}{\rho}}$

The correct answer is:
$\sqrt{v^{2}+\frac{p}{\rho}}$

Solution:
As the water is flowing through the horizontal tube, so in the streamline flow of water, the sun of static pressure and dynamic pressure is constant $i, e$
$$
P+\frac{1}{2} \rho v^{2}=\frac{P}{2}+\frac{1}{2} \rho v_{1}^{2}
$$
$\Rightarrow \quad v_{1}=\sqrt{\frac{P}{\rho}+v^{2}}$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2013
We define a binary relation - on the set of all $3 \times 3$ real matrices as $A \sim B,$ if and only if there exist invertible matrices $P$ and $Q$ such that $B=P A Q^{-1} .$ The binary relation $\sim$ is

Options:
A neither reflexive nor symmetric
B reflexive and symmetric but not transitive
C symmetric and transitive but not reflexive
D an equivalence relation

The correct answer is:
an equivalence relation

Solution:
Let the relation defined as $R=\left\{(A, B): B=P A Q^{-1}\right\}$
For reflexive
$A=I A I^{-1}$
$\Rightarrow(A, A) \in R$
$\Rightarrow R$ is reflexive
For symmetric
$\operatorname{Let}(A, B) \in R$
$\therefore B=P A Q^{-1}$
$\Rightarrow \quad B Q=P A\left(Q^{-1} Q\right)$
$\Rightarrow \quad B Q=P A$
$\Rightarrow \quad P^{-1}(B Q)=P^{-1} P(A)$
$\Rightarrow \quad P^{-1} B Q=A$
$\Rightarrow(B, A) \in R$
$\Rightarrow R$ is symmetric.
For transitive Let $(A, B) \in R,(B, C) \in R$
Then, $\quad A=P B Q^{-1}$ and $B=R C S^{-1}$
$\Rightarrow \quad A=P R C S^{-1} Q^{-1}$
$\begin{array}{ll}\Rightarrow & A=(P R) C(Q S)^{-1} \\ \Rightarrow & (A, C) \in R\end{array}$
$\Rightarrow R$ is transitive. Hence, $R$ is an equivalence relation.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2014
What amount of clectricity can deposit
I mole of Al metal at cathode when passed through molten $\mathrm{AlCl}_{3} ?$

Options:
A $0.3 \mathrm{F}$
B $1 \mathrm{F}$
C $3 \mathrm{F}$
D $1 / 3 \mathrm{F}$

The correct answer is:
$3 \mathrm{F}$

Solution:
The number of electrons involved in the reaction are three as shown below
$$
\mathrm{Al}^{3+}+3 e^{-} \longrightarrow \mathrm{Al}
$$
It means the conversion of every aluminium ion to aluminium atom requires three electrons. Therefore, the amount of electricity required for one mole of $\mathrm{Al}^{3+}$ ions $=3 \mathrm{F}$.

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2018
What current will flow through the $2 \mathrm{k} \Omega$ resistor in the circuit shown in the figure?

Options:
A $3 \mathrm{mA}$
B $6 \mathrm{mA}$
C $12 \mathrm{mA}$
D $36 \mathrm{mA}$

The correct answer is:
$3 \mathrm{mA}$

Solution:
From KCL we have

$i=\frac{72}{8 \times 10^{3}}=9 \times 10^{-3} \mathrm{A}$
$$
\begin{aligned}
i \times 6 &=(9-i) \times 3 \\
i &=3 \mathrm{mA}
\end{aligned}
$$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2012
What is obtained when nitrobenzene is treated sequentially with (i) $\mathrm{NH}_4 \mathrm{Cl} / \mathrm{Zn}$ dust and (ii) $\mathrm{H}_2 \mathrm{SO}_4 / \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$ ?

Options:
A meta-chloronitrobenzene
B para-chloronitrobenzene
C nitrosobenzene
D benzene

The correct answer is:
nitrosobenzene

Solution:
Hints:

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2010
What is the correct order of acidity of salicylic acid, 4-hydroxybenzoic acid, and $2,6-$ dihydroxybenzoic acid?

Options:
A 2,6-dihydroxybenzoic acid > salicylic acid > 4-hydroxybenzoic acid
B 2,6-dihydroxybenzoic acid > 4-hydroxybenzoic acid > salicylic acid
C salicylic acid > 2, 6-dihydroxybenzoic acid > 4-hydroxybenzoic acid
D salicylic acid > 4-hydroxybenzoic acid > 2,6-dihydroxybenzoic acid

The correct answer is:
2,6-dihydroxybenzoic acid > salicylic acid > 4-hydroxybenzoic acid

Solution:
The conjugate base of 2, 6-dihydroxybenzoic acid is highly stabilised by $\mathrm{H}$-bonding. The effect is lesser in salicyclic acid.

Topic: Chemistry, Carboxylic Acid Derivatives, WBJEE, WBJEE 2022
What is the current $I$ shown in the given circuit?

Options:
A $\frac{V}{2 R}$
B $\frac{V}{R}$
C $\frac{V}{16 R}$
D $\frac{V}{8 R}$

The correct answer is:
$\frac{V}{16 R}$

Solution:
According to the question,

Total resistance of the given circuit
$$
R_{\mathrm{eq}}=2 R
$$
$\therefore \quad I=\frac{V / 2 R}{8} \quad\left(\because I_{1}=V / 2 R\right)$
So, current $I$ in the circuit $I=\frac{V}{16 R} \quad\left(\because I=\frac{I_{1}}{8}\right)$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2019
What is the edge length of the unit cell of a body centred cubic crystal of an element whose atomic radius is \(75 \mathrm{pm}\) ?

Options:
A \(170 \mathrm{pm}\)
B \(175 \mathrm{pm}\)
C \(178 \mathrm{pm}\)
D \(173.2 \mathrm{pm}\)

The correct answer is:
\(173.2 \mathrm{pm}\)

Solution:
Hint : In BCC, \(4 r=\sqrt{3} a\)
\(\therefore \mathrm{a}=\frac{4 \mathrm{r}}{\sqrt{3}}=\frac{4 \times 75}{\sqrt{3}}=\frac{300}{\sqrt{3}}=\sqrt{3} \times 100 \mathrm{pm}=173.2 \mathrm{pm}\)

Topic: Chemistry, Solid State, WBJEE, WBJEE 2023
What is the major product of the following reaction?

Options:
A
B
C
D

The correct answer is:

Solution:
Hint:
It's perkin reaction. The reaction starts with generation of nucleophile from propanoic anhydride as follows:

Then onwards, the nucleophile attacks the-CHO group as shown:

Finally, it goes on to give:

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2020
What is the number of ways in which an examiner can assign 10 marks to 4 questions, giving not less than 2 marks to any questions?

Options:
A 4
B 6
C 10
D 16

The correct answer is:
10

Solution:
$x_{1}+x_{2}+x_{3}+x_{4}=10 \quad x_{i} \geq 2$
$\Rightarrow\left(x_{1}-2\right)+\left(x_{2}-2\right)+\left(x_{3}-2\right)+\left(x_{4}-2\right)=2$
$\begin{array}{llll}\mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} & \mathrm{y}_{4}\end{array}$
$\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}+\mathrm{y}_{4}=2 ; \mathrm{y}_{i} \geq 0$
$\therefore$ N.O.W $={ }^{2+4-1} \mathrm{C}_{4-1}={ }^{5} \mathrm{C}_{3}=10$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2021
What is the particle $\mathrm{x}$ in the following nuclear reaction :
${ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \rightarrow{ }_6^{12} \mathrm{C}+\mathrm{x}$

Options:
A electron
B proton
C Photon
D Neutron

The correct answer is:
Neutron

Solution:
${ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \rightarrow{ }_6^{12} \mathrm{C}+{ }_0^1 \mathrm{X}$
Hence X represents neutron $\left({ }_0^1 n\right)$

Topic: Physics, Nuclear Physics, WBJEE, WBJEE 2009
What is the phase difference between two simple harmonic motions represented by $x_{1}=A \sin \left(\omega t+\frac{\pi}{6}\right)$ and $x_{2}=A \cos (\omega t) ?$

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

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

Solution:
Given,
$$
\begin{array}{l}
x_{1}=A \sin \left(\omega t+\frac{\pi}{6}\right) \\
x_{2}=A \cos (\omega t) \\
x_{2}=A \operatorname{sin}\left(\omega t+\frac{\pi}{2}\right)
\end{array}
$$
Phase difference
$$
\Delta \phi=\phi_{2}-\phi_{1}
$$
$$
\begin{array}{l}
\Delta \phi=\frac{\pi}{2}-\frac{\pi}{6} \\
\Delta \phi=\frac{3 \pi-\pi}{6}=\frac{\pi}{3}
\end{array}
$$

Topic: Physics, Oscillations, WBJEE, WBJEE 2012
What is the value of current through the diode in the circuit given?

Options:
A $0 \mathrm{~mA}$
B $1 \mathrm{~mA}$
C $19 \mathrm{~mA}$
D $9 \mathrm{~mA}$

The correct answer is:
$0 \mathrm{~mA}$

Solution:
Since diode is reverse biased

Topic: Physics, Semiconductors, WBJEE, WBJEE 2021
What type of orbital hybridisation is considered on $\mathrm{P}$ in $\mathrm{PCl}_5$ ?

Options:
A $\mathrm{sp}^3 \mathrm{~d}$
B $\mathrm{dsp}^3$
C $\mathrm{sp}^3 \mathrm{~d}^2$
D $\mathrm{d}^2 \mathrm{sp}^3$

The correct answer is:
$\mathrm{sp}^3 \mathrm{~d}$

Solution:
No Solution Available

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2010
What will be current through the $200 \Omega$
resistor in the given circuit, a long time after the switch $K$ is made on?

Options:
A Zero
B $100 \mathrm{mA}$
C $10 \mathrm{mA}$
D $1 \mathrm{mA}$

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

Solution:
In steady state, arm having capacitors does not flow current. So, we can neglect them.
$\therefore$ The given circuit reduces to

$$
R_{\text {net }}=200+400=600 \Omega
$$
$\therefore$ Current in circuit,
$$
\begin{aligned}
I=\frac{V}{R} &=\frac{6}{600}=0.01 \mathrm{A} \\
&=10 \mathrm{mA}
\end{aligned}
$$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2018
What will be the approximate terminal velocity of a rain drop of diameter $1.8 \times 10^{-3} \mathrm{m}$, when density of rain water $\approx 10^{3} \mathrm{kgm}^{-3}$ and the coefficient of viscosity of air $\approx 1.8 \times 10^{-5} \mathrm{N}-sm^{-2}$ ? (Neglect buoyancy of air)

Options:
A $49 \mathrm{ms}^{-1}$
B $98 \mathrm{ms}^{-1}$
C $392 \mathrm{ms}^{-1}$
D $980 \mathrm{ms}^{-1}$

The correct answer is:
$98 \mathrm{ms}^{-1}$

Solution:
Terminal velocity, $v=\frac{2}{9} r^{2} \frac{(\rho-\sigma)}{\eta}$ g
Neglecting buoyancy effect of the fluid, $v=\frac{2}{9} \cdot \frac{\rho}{\eta} r^{2} g$
Putting the given values, we get $v=\frac{2}{9} \times \frac{10^{3} \times\left(0.9 \times 10^{-3}\right)^{2}}{1.8 \times 10^{-5}} \times 9.8=98 \mathrm{ms}^{-1}$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2018
What will be the current flowing through the $6 \mathrm{k} \Omega$ resistor in the circuit shown, where the breakdown voltage of the Zener is $6 \mathrm{V} ?$

Options:
A $\frac{2}{3} m A$
B $1 \mathrm{mA}$
C $10 \mathrm{mA}$
D $\frac{3}{2} m A$

The correct answer is:
$\frac{2}{3} m A$

Solution:
In the given circuit, the zener diode is used as a voltage regulating device. The voltage across $6 \mathrm{k} \Omega$ resistance is $(10-6) \mathrm{V}=4 \mathrm{V}$
Current through $6 \mathrm{k} \Omega$ resistor, $I=\frac{4 \mathrm{V}}{6 \mathrm{k} \Omega}=\frac{4}{6 \times 10^{3}}=\frac{2}{3} \times 10^{-3}=\frac{2}{3} \mathrm{mA}$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2018
What will be the equivalent resistance between the terminals A and B of the infinite resistive network shown in the figure?

Options:
A $\frac{(\sqrt{3}+1) \mathrm{R}}{2}$
B $\frac{(\sqrt{3}-1) R}{2}$
C $3 \frac{R}{2}$
D $(\sqrt{3}+1) \mathrm{R}$

The correct answer is:
$(\sqrt{3}+1) \mathrm{R}$

Solution:
Hint:

$x=2 R+\frac{R x}{R+x}$
$x^{2}-2 R x-2 R^{2}=0$
$x=(1+\sqrt{3}) R$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2020
What will be the mass of one atom of ${ }^{12} \mathrm{C}$ ?

Options:
A 1 a.m.u.
B $1.9923 \times 10^{-23} \mathrm{~g}$
C $1.6603 \times 10^{-22} \mathrm{~g}$
D 6 a.m.u.

The correct answer is:
$1.9923 \times 10^{-23} \mathrm{~g}$

Solution:
Hint:
Mass of 1 atom of ${ }^{12} \mathrm{C}=12$ a.m.u. $=12 \times 1.66 \times 10^{-24} \mathrm{~g}=1.9923 \times 10^{-23} \mathrm{~g}$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2020
What will be the molar specific heat at constant volume of an ideal gas consisting of rigid diatomic molecules?

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

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

Solution:
For a gas at temperature $T$, the internal energy,
$$
U=\frac{f}{2} \mu R T
$$
where, $f=$ degree of freedom
$\Rightarrow$ Change in energy, $\Delta U=\frac{f}{2} \mu R \Delta T$
Also, as we know for any gas heat supplied at constant volume
$$
\left(\Delta Q_{V}\right)=\mu C_{V} \Delta T=\Delta U
$$
where, $C_{V}=$ molar specific heat at constant volume From Eqs. (i) and (ii), we get
$$
C_{V}=\frac{1}{2} f R
$$
For diatomic gas, degree of freedom, $f=5$
$$
C_{V}=\frac{5}{2} R
$$

Topic: Physics, Kinetic Theory of Gases, WBJEE, WBJEE 2019
What will be the normality of the salt solution obtained by neutralising $x$ mL. $y$ (N) HCl with $y \quad$ mL $x$ (N) $\mathrm{NaOH}$, and finally adding $(x+y) \mathrm{mL}$ distilled water

Options:
A $\frac{2(x+y)}{x y} N$
B $\frac{x y}{2(x+y)} N$
C $\left(\frac{2 x y}{x+y}\right) N$
D $\left(\frac{x+y}{x y}\right) N$

The correct answer is:
$\frac{x y}{2(x+y)} N$

Solution:
The reaction for given reaction is $\mathrm{HCl}+\mathrm{NaOH} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$
Number of moles	xy	xy	0	0
Added	$\quad N V \quad N V$
Number of moles left 0	0	xy	xy
$\therefore N($ solution $)=\frac{\text { Number of milliequivalent }}{\text { vol. of solution in mililetre }}$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2017
When $20 \mathrm{~g}$ of naphthoic acid $\left(\mathrm{C}_{11} \mathrm{H}_{8} \mathrm{O}_{2}\right)$ is dissolved in $50 \mathrm{~g}$ of benzene, a freezing point depression of $2 \mathrm{~K}$ is observed. The vant Hoff factor (i) is $\left[\mathrm{K}_{\mathrm{f}}=1.72^{\circ} \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}\right]$

Options:
A $0.5$
B $1.0$
C $2.0$
D $3.0$

The correct answer is:
$0.5$

Solution:
$\begin{array}{l}
\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K}_{\mathrm{f}} \mathrm{m} \\
2=\mathrm{i} \times 1.72 \times \frac{20 \times 1000}{172 \times 50} \\
\therefore \mathrm{i}=\frac{2}{0.01 \times 2 \times 200}=\frac{1}{2}=0.5
\end{array}$

Topic: Chemistry, Solutions, WBJEE, WBJEE 2021
When $100 \mathrm{~g}$ of boiling water at $100^{\circ} \mathrm{C}$ is added into a calorimeter containing $300 \mathrm{~g}$ of cold water at $10^{\circ} \mathrm{C}$, temperature of the mixture becomes $20^{\circ} \mathrm{C}$. Then a metallic block of mass $1 \mathrm{~kg}$ at $10^{\circ} \mathrm{C}$ is dipped into the mixture in the calorimeter. After reaching thermal equilibrium, the final temperature becomes $19^{\circ} \mathrm{C}$. What is the specific heat of the metal in C.G.S. unit?

Options:
A $0.01$
B $0.3$
C $0.09$
D $0.1$

The correct answer is:
$0.1$

Solution:
Hint:
Let, heat capacity of calorimeter = ms
$\begin{aligned}
& 100 \times 1 \times 80=300 \times 1 \times 10+\mathrm{ms} \times 10 \\
\therefore \mathrm{ms}=500 \mathrm{Cal} /{ }^{\circ} \mathrm{C}
\end{aligned}$
Again, $(100+300) \times 1 \times 1+500 \times 1=1000 \times S_{b} \times 9$
$\therefore \mathrm{S}_{\mathrm{b}}=0.1 \mathrm{Cal} / \mathrm{g}^{\circ} \mathrm{C}$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2020
When a $60 \mathrm{mH}$ inductor and a resistor are connected in series with an AC voltage source, the voltage leads the current by $60^{\circ}$. If the inductor is replaced by a $0.5 \mu \mathrm{F}$ capacitor, the voltage lags behind the current by $30^{\circ} .$ What is the frequency of the AC supply?

Options:
A $\frac{1}{2 \pi} \times 10^{4} \mathrm{Hz}$
B $\frac{1}{\pi} \times 10^{4} \mathrm{Hz}$
C $\frac{3}{2 \pi} \times 10^{4} \mathrm{Hz}$
D $\frac{1}{2 \pi} \times 10^{8} \mathrm{Hz}$

The correct answer is:
$\frac{1}{2 \pi} \times 10^{4} \mathrm{Hz}$

Solution:
Given, inductance of inductor, $L=60 \mathrm{mH}$
$$
=60 \times 10^{-3} \mathrm{H}
$$
Phase difference between voltage and current in $L-R$ circuit, $\theta_{1}=60^{\circ}$
Capacitance of capacitor, $C=0.5 \mu \mathrm{F}=0.5 \times 10^{-6} \mathrm{F}$ Phase difference between voltage and current in $\vec{R}-C$ circuit, $\theta_{2}=30^{\circ}$
For $L-R$ circuit.
$$
\begin{array}{l}
\tan \theta_{1}=\frac{X_{L}}{R} \\
\tan \theta_{1}=\frac{\omega L}{R} \quad \ldots(\mathrm{i})\left[\because X_{L}=\omega L\right]
\end{array}
$$
Similarly, for $R$ - $C$ circuit,
$$
\tan \theta_{2}=\frac{X_{C}}{R}
$$
$$
\tan \theta_{2}=\frac{\frac{1}{{\omega C}}}{R} \quad \ldots(\mathrm{ii})\left[\because X_{C}=\frac{1}{\omega C}\right]
$$
From Eqs. (i) and (ii), we get
$$
\begin{array}{l}
\frac{\tan \theta_{1}}{\tan \theta_{2}}=\frac{\omega L}{R \times \frac{1}{\omega C R}}=\omega^{2} L C \\
\frac{\tan 60^{\circ}}{\tan 30^{\circ}}=\omega^{2} L C \\
\frac{\sqrt{3}}{1 / \sqrt{3}}=\omega^{2} L C \\
\omega^{2} L C=3 \\
\omega^{2}=\frac{3}{L C} \\
\omega^{2}=\frac{3}{60 \times 10^{-3} \times 0.5 \times 10^{-6}} \\
\omega^{2}=\frac{3}{30 \times 10^{-9}} \\
\omega=\sqrt{10^{8}} \\
\omega=10^{4}
\end{array}
$$
As we know that,
$$
\omega=2 \pi f
$$
$$
f=\frac{\omega}{2 \pi}=\frac{10^{4}}{2 \pi} \mathrm{Hz}=\frac{1}{2 \pi} \times 10^{4} \mathrm{Hz}
$$

Topic: Physics, Alternating Current, WBJEE, WBJEE 2019
When a certain metal surface is illuminated with light of frequency $v$, the stopping potential for photoelectric current is $V_{0}$. When the same surface is illuminated by light of frequency $\frac{v}{2}$ the stopping potential is $\frac{V_{0}}{4} .$ The threshold
frequency for photoelectric, emission is

Options:
A $\frac{v}{6}$
B $\frac{v}{3}$
C $\frac{2 v}{3}$
D $\frac{4 v}{3}$

The correct answer is:
$\frac{v}{3}$

Solution:
Ist case
$$
\begin{array}{l}
h v=h v_{0}+e V_{0} \\
\frac{h v}{2}=h v_{0}+\frac{e V_{0}}{4}
\end{array}
$$
Solving Eqs. (i) and (ii), we get
$$
v_{0}=\frac{v}{3}
$$

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2012
When a manganous salt is fused with a mixture of $\mathrm{KNO}_3$ and solid $\mathrm{NaOH}$ the oxidation number of Mn changes from +2 to

Options:
A +4
B +3
C +6
D +7

The correct answer is:
+6

Solution:
$$
\text { Hints : } \mathrm{Mn}^{+2}+\mathrm{NO}_3^{-}+\mathrm{O} \overline{\mathrm{H}} \rightarrow \stackrel{(+6)}{\mathrm{M} \mathrm{nO}_4^{-2}}+\mathrm{H}_2 \mathrm{O}
$$

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2010
When a semiconducting device is connected in series with a battery and a resistance, a current is found to flow in the circuit. If however, the polarity of the battery is reversed, practically no current flows in the circuit. The device may be

Options:
A a $p$ -type semiconductor
B a $n$ -type semiconductor
C an intrinsic semiconductor
D a $p-n$ junction

The correct answer is:
a $p-n$ junction

Solution:
When the battery and resistance is connected with $p n$ diode,

In case of forward bias that $p$ is connected to + ve terminal of battery, current flows. If one reverse the polarity i.e., $p$ -end is made negative, no current flows in semiconductor diode.

Topic: Physics, Semiconductors, WBJEE, WBJEE 2017
When a spring is stretched by $10 \mathrm{cm}$, the potential energy stored is $E$. When the spring is stretched by $10 \mathrm{cm}$ more, the potential energy stored in the spring becomes

Options:
A $2 E$
B $4 E$
C $6 E$
D $10 E$

The correct answer is:
$4 E$

Solution:
For 1st situation,
$$
E=\frac{1}{2} k\left(10 \times 10^{-2}\right)^{2}
$$
For 2nd situation,
$$
E^{\prime}=\frac{1}{2} k\left(20 \times 10^{-2}\right)^{2}
$$
On comparing Eqs. (i) and (ii), we get
$$
E^{\prime}=4 E
$$

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2012
When $\mathrm{AgCl}$ is treated with $\mathrm{KCN}$

Options:
A $\mathrm{Ag}$ is precipitated
B a complex ion is formed
C double decomposition takes place
D no reaction takes place

The correct answer is:
a complex ion is formed

Solution:
Hints: $\mathrm{AgCl}+2 \mathrm{KCN} \rightarrow \mathrm{K}\left[\mathrm{Ag}(\mathrm{CN})_2\right]+\mathrm{KCl}$

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2010
When an $A C$ source of emf $E$ with frequency $\omega=100 \mathrm{~Hz}$ is connected across a circuit, the phase difference between
$E$ and current I in the circuit is observed to be $\frac{\pi}{4}$ as shown in the figure. If the circuit consist of only $\mathrm{RC}$ or $\mathrm{RL}$ in series, then

Options:
A $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=5 \mu \mathrm{F}$
B $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{L}=10 \mathrm{H}$
C $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{L}=1 \mathrm{H}$
D $\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=10 \mu \mathrm{F}$

The correct answer is:
$\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=10 \mu \mathrm{F}$

Solution:
.. Since supply voltage lags the current $\operatorname{By}(\phi=\pi / 4)$
$$
\begin{aligned}
&\tan \phi=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}=1, \mathrm{X}_{\mathrm{C}}=\mathrm{R} \ldots \ldots(1) \\
&\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \times 10 \times 10^{-6}}=1 \mathrm{k} \Omega=\mathrm{R}
\end{aligned}
$$

Topic: Physics, Alternating Current, WBJEE, WBJEE 2022
When $\mathrm{BaCl}_{2}$ is added to an aqueous salt solution, a white precipitate is obtained. The anion among $\mathrm{CO}_{3}^{2-}, \mathrm{SO}_{3}^{2-}$ and $\mathrm{SO}_{4}^{2-}$ that was present in the solution can be

Options:
A $\mathrm{CO}_{3}^{2-}$ but not any of the other two
B $S O_{3}^{2-}$ but not any of the other two
C $\mathrm{SO}_{4}^{2-}$ but not any of the other two
D Any of them

The correct answer is:
Any of them

Solution:
$\because$ If $\mathrm{CO}_{3}^{2-}, \mathrm{SO}_{3}^{2-}$ and $\mathrm{SO}_{4}^{2-}$ are present alongwith $\mathrm{BaCl}_{2}$, these can also show white precipitate (as precipitate of all these are also white ).

Topic: Chemistry, Practical Chemistry, WBJEE, WBJEE 2017
When $\mathrm{H}_{2} \mathrm{O}_{2}$ is shaken with an acidified solution of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ in the presence of ether, the cthereal layer turns blue due to the formation of

Options:
A $\mathrm{Cr}_{2} \mathrm{O}_{3}$
B $\mathrm{CrO}_{4}^{2-}$
C $\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$
D $\mathrm{CrO}_{5}$

The correct answer is:
$\mathrm{CrO}_{5}$

Solution:
$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{SO}_{4}+4 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow$ $\mathrm{K}_{2} \mathrm{SO}_{4}+5 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{CrO}_{5}$

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2012