datavorous/entrance-exam-dataset huggingface dataset:
Amongst the following compounds, the one which would not respond to iodoform test is

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

The correct answer is:
$\mathrm{CH}_{3} \mathrm{COOH}$

Solution:
Acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\) would not respond to iodoform test. In the mechanism of haloform reaction, in the first step, base deprotonates most acidic alpha \(\mathrm{H}\) atom (of methyl keto group).
In acetic acid, the most acidic proton is attached to \(\mathrm{O}\) atom. So deprotonation of \(\alpha-\) hydrogen does not occur and hence no haloform reaction.

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2016
Amongst the following compounds, the one(s) which readily react with ethanolic KCN.

Options:
A Ethyl chioride
B Chiorobenzene
C Benzaldehyde
D Salicylic acid

The correct answers are:
Ethyl chioride, Benzaldehyde

Solution:
(a) Ethyl chioride gives ethyl cyanide, when it reacts with KCN (ethanolic).
$$
\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{KCN} \stackrel{\text { EThande }}{\longrightarrow} \mathrm{C}_{2}\mathrm{H}_{\mathrm{s}}-\mathrm{CN}
$$
(b) It does not react with KCN.
(c) Benzaldehyde : KCN gives HCN in alcoholic medium and HCN gives cyanohydrin with benzaldehyde.

Topic: Chemistry, Carboxylic Acid Derivatives, WBJEE, WBJEE 2016
Amongst the following, the one which can exist in free state as a stable compound is

Options:
A $C_{7}H_{9}O$
B $\mathrm{C}_{8} \mathrm{H}_{12} \mathrm{O}$
C $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}$
D $\mathrm{C}_{10} \mathrm{H}_{17} \mathrm{O}$

The correct answer is:
$\mathrm{C}_{8} \mathrm{H}_{12} \mathrm{O}$

Solution:
Molecules having whole number for degree of unsaturation can exist in free state as stable compounds.
Degree of unsaturation $=\frac{\Sigma n(v-2)}{2}+1$
where, $n=$ number of atoms of a particular type $v=$ valency of the atom.
(a) For $\mathrm{C}_{7} \mathrm{H}_{9} \mathrm{O}$,
$D U=\frac{7(4-2)+9(1-2)+1(2-2)}{2}+1=3 \cdot 5$
(b) For $\mathrm{C}_{8} \mathrm{H}_{12} \mathrm{O}$
$$
\mathrm{DU}=\frac{8(4-2)+9(1-2)+1(2-2)}{2}+1=3 \cdot 0
$$
(c) For $\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{O}$
$$
\mathrm{DU}=\frac{6(4-2)+11(1-2)+1(2-2)}{2}+1=1 \cdot 5
$$
(d) For $\mathrm{C}_{10} \mathrm{H}_{17} \mathrm{O}_{2}$
$D U=\frac{10(4-2)+17(1-2)+2(2-2)}{2}+1=2 \cdot 5$
$\therefore \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}$ will exist in free state or a stable compound.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2013
Amongst the following, which should have the highest rms speed at the same temperature?

Options:
A $\mathrm{SO}_{2}$
B $\mathrm{CO}_{2}$
C $\mathrm{O}_{2}$
D $\mathrm{H}_{2}$

The correct answer is:
$\mathrm{H}_{2}$

Solution:
$\because$ rms velocity $(v_{rms}) \propto \sqrt{\frac{1}{M}}$. Thus, smaller the mass, more will be the rms speed. $\mathrm{H}_{2}$ has minimum molar mass, So it has highest value of rms speed.

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2016
An alpha particle $\left({ }^{4} \mathrm{He}\right)$ has a mass of 4.00300 amu. A proton has a mass of 1.00783 amu and a neutron has a mass of 1.00867 amu respectively. The binding energy of alpha estimated from these data is the closest to

Options:
A $27.9 \mathrm{MeV}$
B $22.3 \mathrm{MeV}$
C $35.0 \mathrm{MeV}$
D $20.4 \mathrm{MeV}$

The correct answer is:
$27.9 \mathrm{MeV}$

Solution:
The mass defect, $\Delta m=2\left(m_{p}+m_{n}\right)-m_{\text {He }}$
$=2(1.00783+1.00867)-4.00300$
$=0.0300$ amu
So, the binding energy, $E=\Delta m c^{2}$
$=0.03 \times 931 \mathrm{MeV}=27.9 \mathrm{MeV}$

Topic: Physics, Nuclear Physics, WBJEE, WBJEE 2013
An altermating current is flowing through a series $L-C - R$ circuit. It is found that the current reaches a value of $1 \mathrm{mA}$ at both $200 \mathrm{Hz}$ and $800 \mathrm{Hz}$ frequency. What is the resonance frequency of the circuit?

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

The correct answer is:
$400 \mathrm{Hz}$

Solution:
$\because$ Resonance frequency, $f_{0}=\sqrt{f_{1} f_{2}}$
$$
=\sqrt{200 \times 800}=400 \mathrm{Hz}
$$

Topic: Physics, Alternating Current, WBJEE, WBJEE 2018
An alternating current in a circuit is given by $I=20 \sin (100 \pi t+0.05 \pi)$ A. The r.m.s. value and the frequency of current respectively are

Options:
A $10 \mathrm{A}$ and $100 \mathrm{Hz}$
B $10 \mathrm{A}$ and $50 \mathrm{Hz}$
C $10 \sqrt{2} \mathrm{A}$ and $50 \mathrm{Hz}$
D $10 \sqrt{2}$ A and $100 \mathrm{Hz}$

The correct answer is:
$10 \sqrt{2} \mathrm{A}$ and $50 \mathrm{Hz}$

Solution:
Given, $I=20 \sin (100 \pi t+0.05 \pi)$
The root mean square value of alternating
$$
\text { current }=l_{\mathrm{ms}}=\frac{l_{0}}{\sqrt{2}}=\frac{20}{\sqrt{2}}=10 \sqrt{2} \mathrm{A}
$$
Also, $\quad \omega=100 \pi$
$\Rightarrow \quad 2 \pi f=100 \pi$ or $f=50 \mathrm{Hz}$

Topic: Physics, Alternating Current, WBJEE, WBJEE 2013
An alternating current of rms value $10 \mathrm{~A}$ is passed through a $12 \Omega$ resistor. The maximum potential difference across the resistor is

Options:
A $20 \mathrm{~V}$
B $90 \mathrm{~V}$
C $1969.68 \mathrm{~V}$
D none

The correct answer is:
$1969.68 \mathrm{~V}$

Solution:
$I_{\mathrm{rms}}=10 \mathrm{~A}$
$\begin{aligned} & I_{r m s}=\frac{I_0}{\sqrt{2}} \Rightarrow I_0=\sqrt{2} \times 10=10 \sqrt{2} \\ & \text { Max. P.D. }=\sqrt{2} \times 10 \times 12=120 \times 1.414=169.68 \mathrm{~V}\end{aligned}$

Topic: Physics, Alternating Current, WBJEE, WBJEE 2009
An amount of charge Q passes through a coil of resistance R. If the current in the coil decreases to zero at a uniform rate during time \(T\), then the amount of heat generated in the coil will be,

Options:
A \(\frac{4 Q^2 R}{3 T}\)
B \(\frac{2 Q^2 R}{3 T}\)
C \(\frac{Q^2 R}{4 R}\)
D \(Q^2 R T\)

The correct answer is:
\(\frac{4 Q^2 R}{3 T}\)

Solution:
Hint :
Given, \(\frac{1}{2} \mathrm{I}_0 T=Q \Rightarrow \mathrm{I}_0=\frac{2 \mathrm{Q}}{\mathrm{T}}\)
Equation of \(\mathrm{I}(\mathrm{t}) \Rightarrow \frac{\mathrm{I}}{\mathrm{I}_0}+\frac{\mathrm{t}}{\mathrm{T}}=1\)
\(\begin{aligned}
& I=I_0\left(1-\frac{t}{T}\right)=\frac{2 Q}{T}\left(1-\frac{t}{T}\right) \\
& \text { Heat }=\int_0^T I^2 R d f \\
& =R \int_0^T \frac{4 Q^2}{T^2}\left(1-\frac{t}{T}\right)^2 \\
& =\frac{4 Q^2 R}{T^2}\left[\int_0^T d t+\frac{1}{T^2} \int_0^T t^2 d t-\frac{2}{T} \int_0^T t d t\right]=\frac{4 Q^2 R}{T^2}\left[\not T+\frac{T}{3}-\not T\right]=\frac{4 Q^2 R}{3 T}
\end{aligned}\)

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2023
An artificial satellite moves in a circular orbit around the earth. Total energy of the satellite is given by $E$. The potential energy of the satellite is

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

The correct answer is:
$-2 E$

Solution:
We know that
The potential energy of the satellite
$$
U=-\frac{G M_{e} m}{R_{0}}
$$
The kinetic energy of the satellite
$$
K=\frac{1}{2} \frac{G M_{e} m}{R_{e}}
$$
The total energy
$$
\begin{aligned}
E=U+K &=-\frac{G M_{e} m}{R}+\frac{1}{2} \frac{G M_{e} m}{R_{0}} \\
E &=-\frac{1}{2} \frac{G M_{e} m}{R} \\
2 E &=-\frac{G M_{e} m}{R_{0}} \Rightarrow-2 E=U
\end{aligned}
$$
So. $\mathrm{PE}=-2(\mathrm{TE})$
$$
\mathrm{PE}=-2 E
$$

Topic: Physics, Gravitation, WBJEE, WBJEE 2014
An atomic nucleus having low $n / p$ ratio tries to find stability by

Options:
A the emission of an $\alpha$ -particle
B the emission of a positron
C capturing an orbital electron (K-electron capture)
D emission of a $\beta$ -particle

The correct answer is:
the emission of a positron

Solution:
Emission of a positron results in the conversion of a proton into neutron, along with the release of neutrino. Thus, the number of protons decreases and of neutron increases, i.e. $n / p$ ratio increases and the nuclei becomes stable.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2014
An earth's satellite nejar the surface of the earth takes about 90 min per revolution. A satellite orbiting the moon also takes about 90 min per revolution. Then which of the following is true?

Options:
A \(\rho_m < \rho_e\)
B \(\rho_m > \rho_e\)
C \(\rho_{\mathrm{m}}=\rho_{\mathrm{e}}\)
D No conclusion can be made about the densities

The correct answer is:
\(\rho_{\mathrm{m}}=\rho_{\mathrm{e}}\)

Solution:
Hint :
\(m\left(E_g\right)=m \omega^2 R\)
\(\begin{aligned}
& m \cdot 4 \pi \frac{G \rho R}{3}=m \omega^2 R \\
& \Rightarrow \omega^2 \propto \rho \\
& \Rightarrow T \propto \frac{1}{\sqrt{\rho}}
\end{aligned}\)
If \(\mathrm{T}\) are equal, so will be \(\rho\)

Topic: Physics, Gravitation, WBJEE, WBJEE 2023
An electric bulb, a capacitor, a battery and a switch are all in series in a circuit. How does the intensity of light vary when the switch is
turn on?

Options:
A Continues to increase gradually
B Gradually increases for sometime and then becomes steady
C Sharply rises initially and then gradually decreases
D Gradually increases for sometime and then gradually decreases

The correct answer is:
Sharply rises initially and then gradually decreases

Solution:
Initially, there will be no voltage drop across capacitor, so intensity of bulb will rise sharply and gradually voltage drop across capacitor will increase as a result voltage drop across bulb decreases, so intensity of bulb will decreases.

Topic: Physics, Capacitance, WBJEE, WBJEE 2018
An electric bulb marked as 50W-200V is connected across a $100 \mathrm{V}$ supply. The present power of the bulb is

Options:
A $37.5 \mathrm{W}$
B $25 \mathrm{W}$
C $12.5 \mathrm{W}$
D $10 \mathrm{W}$

The correct answer is:
$12.5 \mathrm{W}$

Solution:
We know that
$$
R=\frac{V^{2}}{P}
$$
Given $\quad V=200 \mathrm{V}, P=40 \mathrm{W}$
and $\quad V=100 \mathrm{V}$
Hence, $\quad R=\frac{V^{2}}{P}=\frac{200 \times 200}{50}$
$$
\begin{array}{l}
P=\frac{V^{2}}{R}=\frac{100 \times 100 \times 50}{200 \times 200} \\
P^{\prime}=12.5 \mathrm{W}
\end{array}
$$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2012
An electric cell of emf $E$ is connected across a copper wire of diameter $d$ and length $l$. The drift velocity of electrons in the wire is $v_{d}$. If the length of the wire is changed to $2 l,$ the new drift velocity of electrons in the copper wire will be

Options:
A $v_{d}$
B $2 v_{d}$
C $v_{d} / 2$
D $v_{d} / 4$

The correct answer is:
$v_{d} / 2$

Solution:
The drift velocity is given as
$$
\begin{aligned}
v_{d} &=\frac{i}{n e A}=\frac{E}{R \times n e A} \\
&=\frac{E \times A}{\rho l \times n e A}=\frac{E}{ln e}
\end{aligned}
$$
When length of wire changed to 2l , the new drift velocity.
$$
\begin{aligned}
v_{d}^{\prime} &=\frac{E}{\rho \times 2 l \times n e} \\
\therefore \quad & \frac{v_{d}^{\prime}}{v_{d}}=\frac{E / \rho 2 \text { lne }}{E / \rho \text { lne }}=\frac{1}{2} \\
\Rightarrow \quad v_{d}^{\prime} &=\frac{v_{d}}{2}
\end{aligned}
$$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2013
An electric current $^{\prime} I^{\prime}$ enters and leaves a uniform circular wire of radius $r$ through diametrically opposite points. A particle carrying a charge $q$ moves along the axis of the circular wire with speed $v$. What is the magnetic force experienced by the particle when it passes through the centre of the circle?

Options:
A $q v \frac{\mu_{0} i}{a}$
B $q v \frac{\mu_{0} i}{2 a}$
C $q v \frac{\mu_{0} i}{2 \pi a}$
D Zero

The correct answer is:
Zero

Solution:
Given, electric current in circular wire $=I$
Radius of wire $=r$
Charge on particle $=q$
Speed of the particle $=v$
The given loop can be as shown in the figure below. $1 / 2$

$\because$ Net magnetic field at the centre $=0$
So, magnetic force $\mathbf{F}_{\text {magnetic }}=q(\mathbf{v} \times \mathbf{B})=0$ So, the magnetic force experienced by the particle when it passes through the centre is $F=0$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2019
An electric dipole of dipole moment \(\vec{p}\) is placed at the origin of the co-ordinate system along the \(z\)-axis. The amount of work required to move a charge ' \(q\) ' from the point \((a, 0,0)\) to the point \((0,0, a)\) is,

Options:
A \(\frac{p q}{4 \pi \varepsilon_0 a}\)
B 0
C \(\frac{-p q}{4 \pi \varepsilon_0 a^2}\)
D \(\frac{p q}{4 \pi \varepsilon_0 a^2}\)

The correct answer is:
\(\frac{p q}{4 \pi \varepsilon_0 a^2}\)

Solution:
Hint :
\(\begin{aligned} W & =q\left(V_B-V_A\right) \\ & =q\left(\frac{p}{4 \pi \epsilon_0} \cdot \frac{\cos 0}{a^2}-\frac{p \cos 90^{\circ}}{4 \pi \epsilon_0 a^2}\right) \\ & =\frac{p q}{4 \pi \varepsilon_0 a^2}\end{aligned}\)

Topic: Physics, Electrostatics, WBJEE, WBJEE 2023
An electrical cable having a resistance of $0.2 \Omega$ delivers $10 \mathrm{kw}$ at $200 \mathrm{~V}$ D.C. to a factory. What is the efficiency of transmission?

Options:
A $65 \%$
B $75 \%$
C $85 \%$
D $95 \%$

The correct answer is:
$95 \%$

Solution:
$P=V I \Rightarrow I=\frac{10 \times 10^3}{200}=50 A$, Power loss $=(50)^2(0.2)=500 \mathrm{~W}$
Efficiency $=\frac{10000 \times 100}{10000+500}=95.23 \%$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2009
An electron accelerated through a potential of $10000 \mathrm{V}$ from rest has a de-Broglie wave length $\lambda$. What should be the accelerating potential, so that the wavelength is doubled?

Options:
A $20000 \mathrm{V}$
B $40000 \mathrm{V}$
C $5000 \mathrm{V}$
D $2500 \mathrm{V}$

The correct answer is:
$2500 \mathrm{V}$

Solution:
$\because$ Kinetic energy of a electron due to accelerated by a potential $V, \mathrm{KE}=e V$
$$
\frac{1}{2} m_{r} v^{2}=c V
$$
$\Rightarrow \quad \frac{1 \times p^{2}}{2 m_{e}}=e V \quad[\because p=m v]$
$\therefore$
$$
p=\sqrt{2 e V m_{e}}
$$
$\because$ de-Broglie wavelength of a particle having momentum $p$
$$
\lambda=\frac{h}{p}
$$
According to question, $\frac{\lambda_{1}}{\lambda_{2}}=\frac{p_{2}}{p_{1}}=\frac{\sqrt{2 e V_{2} m_{c}}}{\sqrt{2 e V_{1} m_{e}}}=\sqrt{\frac{V_{2}}{V_{1}}}$
$\because$
$$
\lambda_{2}=2 \lambda_{1}
$$
$\Rightarrow \quad \frac{\lambda_{1}}{2 \lambda_{1}}=\sqrt{\frac{V_{2}}{10000}}$
$\therefore \quad$ Potential $V_{2}=\frac{10^{4}}{4}=2500 \mathrm{V}$

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2018
An electron enters an electric field having
intensity $\mathbf{E}=3 \overrightarrow{\mathbf{i}}+6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \cdot \mathbf{V} \mathbf{m}^{-1}$ and magnetic
field having induction $\mathrm{B}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ T with a velocity $\mathbf{v}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ ms $^{-1}$. The magnitude of the force acting on the electron is (Given, $\left.e=-1.6 \times 10^{-19} \mathrm{C}\right)$

Options:
A $2.02 \times 10^{-18} \mathrm{N}$
B $5.16 \times 10^{-16} \mathrm{N}$
C $3.72 \times 10^{-17} \mathrm{N}$
D None of the above

The correct answer is:
None of the above

Solution:
Given, $E=3 i+6 j+2 \hat{k} \vee / m$
$$
\begin{array}{l}
B=2 \hat{i}+3 \hat{j} T \\
v=2 \hat{i}+3 \hat{j} m / s \\
e=-1.6 \times 10^{-10} C
\end{array}
$$
The magnitude of electric field
$$
\begin{aligned}
|E| &=\sqrt{(3)^{2}+(6)^{2}+(2)^{2}} \\
&=\sqrt{9+36+4} \\
&=\sqrt{49}=7
\end{aligned}
$$
The force acting on electron due to electric field
$$
F=q E
$$
$$
\begin{aligned}
|F| &=|q||\mathbf{E}| \\
&=1.6 \times 10^{-19} \times 7 \mathrm{N} \\
&=11.2 \times 10^{-19} \mathrm{N} \\
&=1.12 \times 10^{-18} \mathrm{N}
\end{aligned}
$$
Alternate Method
B $\|$ v Magnetic field is parallel to velocily of electron. $\mathrm{v} \| \mathrm{B}$ Magnetic force
$$
\begin{aligned}
F_{m} &=q(\mathbf{v} \times \mathbf{B}) \\
&=q v B \sin \theta \\
&=q v B \sin 0 \\
&=0
\end{aligned}
$$
$\therefore$ Net force $\quad F=F_{e}+F_{m}$
$$
\begin{array}{l}
=\left(1.12 \times 10^{-18}+0\right) \mathrm{N} \\
=1.12 \times 10^{-18} \mathrm{N}
\end{array}
$$
Hence, none of the given options is correct.

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2016
An electron in a circular orbit of radius $0.05 \mathrm{nm}$ performs $10^{16}$ revolutions per second. The maghetic moment due to this rotation of electron is (in $\mathrm{Am}^{2}$ )

Options:
A $2.16 \times 10^{-23}$
B $3.21 \times 10^{-22}$
C $3.21 \times 10^{-24}$
D $1.26 \times 10^{-23}$

The correct answer is:
$1.26 \times 10^{-23}$

Solution:
Given, $r=0.05 \mathrm{nm}=0.05 \times 10^{-9} \mathrm{m}$
$$
\begin{array}{l}
\qquad\left(\because 1 \mathrm{nm}=10^{-9} \mathrm{m}\right) \\
n=10^{16} \text { revolutions } \\
e=1.6 \times 10^{-19} \mathrm{C}
\end{array}
$$
We know that
The magnetic moment $M=A L$
$$
\begin{aligned}
M &=\pi r^{2} \times n e \\
M &=3.14 \times\left(0.05 \times 10^{-9}\right)^{2} \times 10^{16} \times 1.6 \times 10^{-19} \\
M &=0.1256 \times 10^{-18} \times 10^{16} \times 10^{-19} \\
\text {or } M &=1.26 \times 10^{-23} \mathrm{Am}^{2}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2014
An electron in the \(5 \mathrm{~d}\) orbital can be represented by the following \(\left(n, 1, m_0\right)\) values

Options:
A \((5,2,1)\)
B \((5,1,-1)\)
C \((5,0,1)\)
D \((5,2,-1)\)

The correct answers are:
\((5,2,1)\), \((5,2,-1)\)

Solution:
Hint: \(5 \mathrm{~d} \quad \therefore \mathrm{n}=5 \quad \mathrm{I}=2\)
and \(m\) can be -2 to 2

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2023
An electron of charge $e$ and mass $m$ is moving in a circular of radius $r$ with a uniform angular speed $\omega .$ Then which of the following statements are correct?

Options:
A The equivalent current flowing in the circular path is proportional to $r^{2}$
B The magnetic moment due to circular current loop is independent of m
C The magnetic moment due to circular current loop equal to $2 e / m$ times the angular momentum of the electron,
D The angular momentum of the particle is proportional to the areal velocity of electron.

The correct answers are:
The magnetic moment due to circular current loop is independent of m, The angular momentum of the particle is proportional to the areal velocity of electron.

Solution:
Magnetic moment, $m=IA$ $=\frac{e v}{2 \pi f} \times \pi r^{2}=\frac{e v r}{2}$
Angular momentum $=2 m \frac{d A}{d t}$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2013
An electron revolves around the nucleus in a circular path with angular momentum $\overrightarrow{\mathrm{L}} . \mathrm{A}$ uniform magnetic field $\overrightarrow{\mathrm{B}}$ is applied perpendicular to the plane of its orbit. If the electron experiences a torque $\overrightarrow{\mathrm{T}}$, then

Options:
A $\overrightarrow{\mathrm{T}} \| \overrightarrow{\mathrm{L}}$
B $\overrightarrow{\mathrm{T}}$ is anti-parallel to $\overrightarrow{\mathrm{L}}$
C $\overrightarrow{\mathrm{T}} \cdot \overrightarrow{\mathrm{L}}=0$
D Angle between $\vec{T}$ and $\vec{L}$ is $45^{\circ}$

The correct answer is:
$\overrightarrow{\mathrm{T}} \cdot \overrightarrow{\mathrm{L}}=0$

Solution:
$\vec{\tau}=0 \Rightarrow \vec{\tau} \cdot \vec{L}=0$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2022
An element belongs to Group 15 and third period of the periodic table. Its electonic configuration will be

Options:
A \(1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^3\)
B \(1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^4\)
C \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^3\)
D \(1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^2\)

The correct answer is:
\(1 s^2 2 s^2 2 p^6 3 s^2 3 p^3\)

Solution:
Hints: General valence shell electronic configuration of 15 group elements is \(n s^2 n p^3\). where \(n=\) period number.

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2011
An element crystallizes in a body centred cubic lattice. The edge length of the unit cell is $200 \mathrm{pm}$ and the density of the element is $5.0 \mathrm{~g} \mathrm{~cm}^{-3}$. Calculate the number of atoms in $100 \mathrm{~g}$ of this element.

Options:
A $2.5 \times 10^{23}$
B $2.5 \times 10^{24}$
C $5.0 \times 10^{23}$
D $5.0 \times 10^{24}$

The correct answer is:
$5.0 \times 10^{24}$

Solution:
$\rho_{\text {unit cell }}(\mathrm{g} / \mathrm{cc})=\frac{\mathrm{Z} \times \mathrm{M} \cdot \mathrm{Mass}(\mathrm{g} / \mathrm{mol})}{\mathrm{N}_{\mathrm{A}}\left(\mathrm{mol}^{-1}\right) \times(\mathrm{a} \mathrm{pm})^{3} \times 10^{-30}} \quad \mathrm{Z}_{\text {B.c.c }}=2$
or $5=\frac{2 \times \text { M.Mass }}{\mathrm{N}_{\mathrm{A}} \times(200)^{3} \times 10^{-30}}$
or M.Mass $=\frac{5 \times 8 \times \mathrm{N}_{\mathrm{A}} \times 10^{-24}}{2}=2 \mathrm{~N}_{\mathrm{A}} \times 10^{-23}$
Now, no. of moles in $100 \mathrm{~g}=\frac{100}{2 \mathrm{~N}_{\mathrm{A}} \times 10^{-23}}$
$\therefore$ no. of atoms $=\frac{100}{2 \mathrm{~N}_{\mathrm{A}} \times 10^{-23}} \times \mathrm{N}_{\mathrm{A}}=50 \times 10^{23}=5.0 \times 10^{24}$

Topic: Chemistry, Solid State, WBJEE, WBJEE 2021
An element $E$ loses one $\alpha$ and two $\beta$ particle in three successive stages. The resultibg element will be

Options:
A an isobar of $E$
B an isotone of $E$
C an isotope of $E$
D $E$ itself

The correct answer is:
an isotope of $E$

Solution:
Let $Z$ = atomic number
$M=$ atornic mass

Hence, ${ }_{Z} E^{M}$ and ${ }_{2} E^{M-4}$ has same atomic number with different atomic mass thus, they are isotopes.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2016
An element $X$ belongs to fourth period and fifteenth group of the periodic table. Which of the following statements is true?

Options:
A It has completely filled s-orbital and a partialy filled $d$ -orbital
B it has completely filled s-and $p$ -orbitals and a partially filled $d$ -orbital
C It has completely filled $s$ - and $p$ -orbitals and a hat filled $d$ -orbital
D It has a half filled $p$ -orbital and completely fled s-and $d$ -orbitals

The correct answer is:
It has a half filled $p$ -orbital and completely fled s-and $d$ -orbitals

Solution:
The element belonging to 4 th period and 15 th group is arsenic (As). The outer electronic configuration of 15 th group element is $n s^{2} n \rho^{3}$. Since arsenic belongs to 4 th period, therefore it has filled 'd' orbital too. Thus, the electronic configuration of As is $[A r] 4 s^{2} 3 d^{10} 4 \rho^{3} .$
So, it has filled's' and'd orbital and half-fliled porbital.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2016
An equilateral triangle is made by uniform wires $A B, B C, C A .$ A current $I$ enters at $A$ and leaves from the mid point of $B C$. If the lengths of each side of the triangle is $L,$ the magnetic field $B$ at the centroid $O$ of the triangle is

Options:
A $\frac{\mu_{0}}{4 \pi}\left(\frac{4 L}{L}\right)$
B $\frac{\mu_{0}}{2 \pi}\left(\frac{4 L}{L}\right)$
C $\frac{\mu_{0}}{4 \pi}\left(\frac{2 L}{L}\right)$
D zero

The correct answer is:
zero

Solution:
The given situation can be shown as

The magnetic field at centroid $O$ = Magnetic field due to left part + Magnetic field due to right part
$\therefore$
$$
\mathrm{B}=\mathrm{B}_{1}+\mathrm{B}_{2}=0
$$
(: direction of $B$ due to both parts is different of $O$ )

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2013
An equimolar mixture of toluene and chlorobenzene is treated with a mixture of conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ and conc. $\mathrm{HNO}_{3}$. Indicate the correct statement from the following.

Options:
A $p$ -nitrotoluene is formed in excess
B equimolar amounts of $p$ -nitrotoluene and p-nitrochlorobenzene are formed
C $p$ -nitrochlorobenzene is formed in excess
D $m$ -nitrochlorobenzene is formed in excess

The correct answer is:
$p$ -nitrotoluene is formed in excess

Solution:
As toluene and chlorobenzene, both can react with mixture of conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ and conc. HNO $_{3}$. but being more reactive, toluene will yield p-nitrotoluene in excess.

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2012
An ideal gas expands adiabatically against vaccum. Which of the following is correct for the given process?

Options:
A $\Delta S=0$
B $\Delta \mathrm{T}=-\mathrm{ve}$
C $\Delta \mathrm{U}=0$
D $\Delta \mathrm{P}=0$

The correct answer is:
$\Delta \mathrm{U}=0$

Solution:
Hint:
For adiabatic free expansion of an ideal gas
$\mathrm{dq}=0$
$\mathrm{dU}=0$
$\mathrm{dH}=0$
$\mathrm{dT}=0$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2020