datavorous/entrance-exam-dataset huggingface dataset:
Bond order of $\mathrm{He}_{2}, \mathrm{He}_{2}^{+}$ and $\mathrm{He}_{2}^{2+}$ are respectively:

Options:
A $1, \frac{1}{2}, 0$
B $0, \frac{1}{2}, 1$
C $\frac{1}{2}, 1,0$
D $1,0, \frac{1}{2}$

The correct answer is:
$0, \frac{1}{2}, 1$

Solution:
Hint:
$\mathrm{He}_{2}\left(4 \mathrm{e}^{-}\right)=\sigma(1 \mathrm{~s})^{2}, \sigma^{*}(1 \mathrm{~s})^{2} ; \mathrm{B} . \mathrm{O}=\frac{2-2}{2}=0$
$\mathrm{He}_{2}^{+}\left(3 \mathrm{e}^{-}\right)=\sigma(1 \mathrm{~s})^{2} \sigma^{*}(1 \mathrm{~s})^{1} ; \mathrm{B} . \mathrm{O}=\frac{2-1}{2}=\frac{1}{2}$
$\mathrm{He}_{2}^{+2}\left(2 \mathrm{e}^{-}\right)=\sigma(\mathrm{1s})^{2} ; \quad$ B.O. $=\frac{2-0}{2}=1$

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2020
By passing excess $\mathrm{Cl}_{2}(\mathrm{g})$ in boiling toluene, which one of the following compounds is exclusively formed?

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2012
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{F}^{18}$ is a $\mathrm{F}^{18}$ radio-isotope labelled organic compound. $\mathrm{F}^{18}$ decays by positron emission. The product resulting on decay is

Options:
A $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{18}$
B $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Ar}^{10}$
C $\mathrm{B}^{12} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{F}$
D $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{16}$

The correct answer is:
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{18}$

Solution:
${ }_{9} \mathrm{F}^{18} \longrightarrow{ }_{x} \mathrm{E}^{y}+{ }_{+1} e^{0}$
$$
x=8, y=18
$$
$\therefore \quad \mathrm{E}^{y}={ }_{8} \mathrm{O}^{18}$
$\because$ Position has one unit of the charge and
zero mass. Thus, $\mathrm{F}^{18}$ changes to $\mathrm{O}^{18}$ therefore resulting decay product is $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{18}$

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2017
$\mathrm{C}_6 \mathrm{H}_6(\mathrm{liq})+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{liq})$
Benzene burns in oxygen according to the above equation. What is the volume of oxygen (at STP) needed for complete combustion of 39 gram of liquid benzene?

Options:
A $11.2$ litre
B $22.4$ litre
C 84 litre
D 168 litre

The correct answer is:
84 litre

Solution:
$\mathrm{n}_{\mathrm{C}_6 \mathrm{H}_6}=\frac{39}{78}=\frac{1}{2}$ mole
From Balanced equation,
$\begin{aligned}
&\mathrm{n}_{\mathrm{C}_6 \mathrm{H}_{\mathrm{G}}}=\frac{2}{15} \mathrm{n}_{\mathrm{O}_2} \\
&\frac{1}{2}=\frac{2}{15} \times \frac{\mathrm{xL}}{22.4 \mathrm{~L}} \\
&\therefore \mathrm{x}=84 \mathrm{~L}
\end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2022
Calculate $\mathrm{K}_{\mathrm{C}}$ for the reversible process given below if $\mathrm{K}_{\mathrm{P}}=167$ and $\mathrm{T}=800^{\circ} \mathrm{C}$
$\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$

Options:
A 1.95
B 1.85
C 1.89
D 1.60

The correct answer is:
1.89

Solution:
Hints: $K_p=K_C(R T)^{\Delta n}$
for $\mathrm{eq}^{\mathrm{n}} \quad \mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{n}=1$
$\mathrm{K}_{\mathrm{C}}=\frac{\mathrm{K}_{\mathrm{P}}}{(\mathrm{RT})^{\Delta \mathrm{n}}}=\frac{167}{(0.0821 \times 1073)^1}=1.89$

Topic: Chemistry, Chemical Equilibrium, WBJEE, WBJEE 2010
Cards are drawn one-by-one without replacement from a well shuffled pack of 52 cards. Then, the probability that a face card (jack, queen or king) will appear for the first time on the third turn is equal to

Options:
A $\frac{300}{2197}$
B $\frac{36}{85}$
C $\frac{12}{85}$
D $\frac{4}{51}$

The correct answer is:
$\frac{12}{85}$

Solution:
Ist turn Total number of face card $=12$
Total number of elements in sample space, $n(s)=52$
$\therefore P_{1}$ (no face card in first turn) $=\frac{52-12}{52}=\frac{40}{52}$
IInd turn $P_{2}$ (no face card in second turn)
$$
=\frac{(52-13)}{(52-1)}=\frac{39}{51}
$$
IIIrd turn $P_{3}$ (face card in third turn)
$$
=\frac{(52-40)}{(51-1)}=\frac{12}{50}
$$
$\therefore$ Required $P$ (face card on third turn)
$$
\begin{array}{l}
=P_{1} \times P_{2} \times P_{3} \\
=\frac{40}{52} \times \frac{39}{51} \times \frac{12}{50}=\frac{12}{85}
\end{array}
$$

Topic: Mathematics, Probability, WBJEE, WBJEE 2013
Case \(-1:\) An ideal gas of molecular weight \(M\) at temperature \(T\).
Case \(-2:\) Another ideal gas of molecular weight \(2 \mathrm{M}\) at temperature \(\mathrm{T} / 2\)
Identify the correct statement in context of above two cases.

Options:
A Average kinetic energy and avereage speed will be the same in the two cases
B Both the averages are halved
C Both the averages are doubled
D Only average speed is halved in the second case

The correct answer is:
Both the averages are halved

Solution:
Hint : As temperature is halved, average \(K E\) is halved.
Average speed \((C) \propto \sqrt{\frac{T}{M}}\)
In case \(-I,(C) \propto \sqrt{\frac{T}{M}}\)
\(\text { In case }-II ~(C) \propto \sqrt{\frac{T}{2 \times 2 M}}=\frac{1}{2} \sqrt{\frac{T}{M}}\)
So average speed is also halved.

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2023
Centre of mass (C.M.) of three particles of masses $1 \mathrm{~kg}, 2 \mathrm{~kg}$ and $3 \mathrm{~kg}$ lies at the point $(1,2,3)$ and C.M. of another system of particles of $3 \mathrm{~kg}$ and $2 \mathrm{~kg}$ lies at the point $(-1,3,-2)$. Where should we put a particle of mass $5 \mathrm{~kg}$ so that the C.M. of entire system lies at the C.M. of the first system ?

Options:
A $(3,1,8)$
B $(0,0,0)$
C $(1,3,2)$
D $(-1,2,3)$

The correct answer is:
$(3,1,8)$

Solution:
$\left.\begin{array}{l}6 \mathrm{~kg}[1,2,3] \\ 5 \mathrm{~kg}[-1,3,-2] \\ 5 \mathrm{~kg}[\mathrm{x}, \mathrm{y}, \mathrm{z}]\end{array}\right) \rightarrow(\overline{\mathrm{x}}, \overline{\mathrm{y}}, \overline{\mathrm{z}})=(1,2,3)$
$\bar{x}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}$
$1=\frac{5(-1)+5 \times x}{10}$
$-5+5 x=10 \Rightarrow x=3$
$\bar{y}=2=\frac{5 \times 3+5 y}{10}, 15+5 y=20, y=1$
$\bar{z}=3=\frac{5(-2)+5 . z}{10},-10+5 z=30, z=8$

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2021
Certain amount of an ideal gas is taken from its initial state 1 to final state 4 through the paths $1 \rightarrow 2 \rightarrow 3 \rightarrow 4$ as shown in figure. $A B, C D, E F$ are all isotherms. If $v_p$ is the most probable speed of the molecules, then

Options:
A $v_p$ at $3=v_p$ at $4>v_p$ at $2>v_p$ at 1
B $v_p$ at $3>v_p$ at $1>v_p$ at $2>v_p$ at 4
C $v_p$ at $3>v_p$ at $2>v_p$ at $4>v_p$ at 1
D $v_p$ at $2=v_p$ at $3>v_p$ at $1>v_p$ at 4

The correct answer is:
$v_p$ at $3=v_p$ at $4>v_p$ at $2>v_p$ at 1

Solution:
$v_p \propto \sqrt{T}$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2022
$\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C} \mathrm{MgBr}$ can be prepared by the
reaction of

Options:
A $\mathrm{CH}_{3}-\mathrm{C}=\mathrm{C}-\mathrm{Br}$ with $\mathrm{MgBr}_{2}$
B $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}$ with $\mathrm{MgBr}_{2}$
C $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}$ with $\mathrm{KBr}$ and $\mathrm{Mg}$ metal
D $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}$ with $\mathrm{CH}_{3} \mathrm{MgBr}$

The correct answer is:
$\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}$ with $\mathrm{CH}_{3} \mathrm{MgBr}$

Solution:
$\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CMg}$ Br can be prepared by the reaction of $\mathrm{CH}_{3}-\mathrm{C}=\mathrm{CH}$ with $\mathrm{CH}_{3} \mathrm{MgBr}$. This
method is used in preparation of higher alkynes from lower alkynes. The chemical equation for the formation of $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CMgBr}$ is given below
$$
\begin{array}{r}
\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}+\mathrm{CH}_{3} \mathrm{MgBr} \longrightarrow \\
\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CMg} X
\end{array}
$$

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2018
Chlorine gas reacts with red hot calcium oxide to give

Options:
A bleaching powder and dichlorine monoxide
B bleaching powder and water
C calcium chioride and chiorine dioxide
D calcium chioride and oxygen

The correct answer is:
calcium chioride and oxygen

Solution:
$2 \mathrm{CaO}+\mathrm{Cl}_{2} \stackrel{\text { Red hot }}{\longrightarrow} \mathrm{CaCl}_{2}+\mathrm{O}_{2} \uparrow$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2013
Choose the correct statement for the $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ complex ion (Atomic no. of $\mathrm{Ni}=28$ )

Options:
A The complex is square planar and paramagnetic
B The complex is tetrahedral and diamagnetic
C The complex is square planar and diamagnetic
D The complex is tetrahedral and paramagnetic

The correct answer is:
The complex is square planar and diamagnetic

Solution:

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2022
Choose the correct statement(s) among the following.

Options:
A
B $\mathrm{CH}_{3} \mathrm{CHO}$ on reaction with HCN gives racemic mixture
C
D $\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{NOH}$ shows geometrical isomerism

The correct answers are:
$\mathrm{CH}_{3} \mathrm{CHO}$ on reaction with HCN gives racemic mixture, $\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{NOH}$ shows geometrical isomerism

Solution:
These are cis and trans isomers not enantiomers (Thus, false).
(b) $CH_{3}CHO$ on reaction with HCN gives racemic mixture of cyanohydrin (Thus, true).

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2016
Chords of an ellipse are drawn through the positive end of the minor axis. Their midpoint lies on

Options:
A a circle
B a parabola
C an ellipse
D a hyperbola

The correct answers are:
an ellipse

Solution:
$\begin{array}{ll}
\frac{x_1+0}{2}=h & \frac{y_1+b}{2}=k \\
\Rightarrow x_1=2 h & \Rightarrow y_1=2 k-b
\end{array}$
$\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1 \Rightarrow \frac{4 h^2}{a^2}+\frac{(2 k-b)^2}{b^2}=1 \Rightarrow \frac{h^2}{a^2 / 4}+\frac{\left(k-\frac{b}{2}\right)^2}{b^2 / 4}=1$
$\therefore \frac{x^2}{a^2 / 4}+\frac{\left(y-\frac{b}{2}\right)^2}{b^2 / 4}=1 \rightarrow$ Locus of midpoint which is an ellipse

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2022
$\mathrm{Cl}_{2} \mathrm{O}_{7}$ is the anhydride of

Options:
A $\mathrm{HOCl}$
B $\mathrm{HClO}_{2}$
C $\mathrm{HClO}_{3}$
D $\mathrm{HClO}_{4}$

The correct answer is:
$\mathrm{HClO}_{4}$

Solution:
$\mathrm{Cl}_{2} \mathrm{O}_{7}$ is the anhydride of $\mathrm{HClO}_{4}$. It reacts with water slowly to give perchloric acid.
$\mathrm{Cl}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \underset{\text { Perchloric acid }}{2 \mathrm{HClO}_{4}}$
Its structure is as follows:

It is less reactive than other oxides of chlorine and does not react with $P, S,$ coal or paper at room temperature.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2018
CO is practically non-polar since

Options:
A the $\sigma$ -electron drift from $C$ to $O$ is almost nullified by the $\pi$ -electron drift from $\mathrm{O}$ to $\mathrm{C}$
B the $\sigma$-clectron drift from O to C is almost nullified by the $\pi$ -electron drift from $\mathrm{C}$ to $\mathrm{O}$
C the bond moment is low
D there is a triple bond berween $\mathrm{C}$ and $\mathrm{O}$

The correct answer is:
the $\sigma$ -electron drift from $C$ to $O$ is almost nullified by the $\pi$ -electron drift from $\mathrm{O}$ to $\mathrm{C}$

Solution:
$\mathrm{C} \frac{\pi}{\sigma} \mathrm{O}$
In $\mathrm{CO}$ molecule, $\mathrm{C}-\mathrm{O} \sigma$ moment and $\mathrm{O}-\mathrm{C}$ $\pi$ -moment cancels each other. So that it is non-polar.

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2012
Cold ferrous sulphate solution on absorptive of NO develops brown colour due to the formation of

Options:
A paramagnetic $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{NO})\right] \mathrm{SO}_{4}$
B diamagnetic $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\left(\mathrm{N}_{3}\right)\right] \mathrm{SO}_{4}$
C paramagnetic $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\left(\mathrm{NO}_{3}\right)\right]\left[\mathrm{SO}_{4} \mathrm{l}_{2}\right.$
D diamagnetic $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{SO}_{4}\right)\right] \mathrm{NO}_{3}$

The correct answer is:
paramagnetic $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{NO})\right] \mathrm{SO}_{4}$

Solution:
Cold FeSO, solution on absorption of NO develops brown colour due to the formation of $[Fe(H_{2}O)_{5}(NO)]SO_4$
$\mathrm{FeSO}_{4}+\mathrm{NO}+5 \mathrm{H}_{2} \mathrm{O} \rightarrow [Fe(H_{2}O)_{5}(NO)]SO_4$
This complex has 3.89 BM magnetic moment shows that it has 3 unpaired electrons

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2015
Commercial sample of $\mathrm{H}_{2} \mathrm{O}_{2}$ is labeled as $10 \mathrm{V}$. Its % strength is nearly

Options:
A 3
B 6
C 9
D 12

The correct answer is:
3

Solution:
10 V $\mathrm{H}_{2} \mathrm{O}_{2}$ means $1 \mathrm{L}$ of this solution will produce $10 \mathrm{LO}_{2}$ at STP.
$$
\begin{array}{c}
2 \mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\quad \mathrm{O}_{2} \\
68 \mathrm{g} \quad \quad \quad \quad 22.4 \text{L at STP}
\end{array}
$$
$\therefore 22.4 \mathrm{L}$ of $\mathrm{O}_{2}$ is obtained from $\mathrm{H}_{2} \mathrm{O}_{2}=68 \mathrm{g}$
10 L of $O_2$ will be obtained from
$$
\mathrm{H}_{2} \mathrm{O}_{2}=\frac{68}{22.4} \times 10=30.36 \mathrm{g}
$$
$1000 \mathrm{mL}$ of the given solution contains $30.36 \mathrm{g}$ $\mathrm{H}_{2} \mathrm{O}_{2}$ and $100 \mathrm{mL}$ of the given solution contains
$$
\frac{30.36 \times 100}{1000}=3.03 \mathrm{g} \mathrm{H}_{2} \mathrm{O}_{2}
$$
Thus, $\%$ strength of $\mathrm{H}_{2} \mathrm{O}_{2}$ is $3.03(\approx 3)$

Topic: Chemistry, Hydrogen, WBJEE, WBJEE 2014
Compound $X$ is tested and the results are shown in the table.
$\begin{array}{l|l}\hline \text{ Test } & \text{ Result } \\\hline \text{ Aqueous sodium } & \text{ Gas given off which turns }\\\text{ hydroxide is added, }& \text{ damp red litmus paper }\\\text{ then heated gently. }& \text{ blue. }\\\text{ Dilute hydrochioric }& \text{ Effervescence, gas given }\\\text{ acid is added. }& \text{ off which turns lime water }\\& \text{ milky and acidified }\\& \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \text{ paper green. }\\\hline\end{array}$
Which ions are present in compound $X ?$

Options:
A Ammonium ions and sulphite ions
B Ammonium ions and carbonate ions
C Sodium ions and carbonate ions
D Ammonium ions and sulphate ions

The correct answer is:
Ammonium ions and sulphate ions

Solution:
It is a test for ammonia

(ii) it is a test for $\mathrm{SO}_{4}^{2-}$ ions (Orange)

Topic: Chemistry, Practical Chemistry, WBJEE, WBJEE 2016
Compounds with spin-only magnetic moment equivalent to five unpaired electrons are

Options:
A $\mathrm{K}_{4}\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]$
B $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$
C $\mathrm{K}_{3}\left[\mathrm{FeF}_{6}\right]$
D $\mathrm{K}_{4}\left[\mathrm{MnF}_{6}\right]$

The correct answers are:
$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}$, $\mathrm{K}_{3}\left[\mathrm{FeF}_{6}\right]$, $\mathrm{K}_{4}\left[\mathrm{MnF}_{6}\right]$

Solution:
Among the given species
$\mathrm{H}_{2} \mathrm{O}$ and $\mathrm{F}$ are weak field ligands and do not cause pairing of electrons in the central metal atom. As CN is a strong field ligand, it causes pairing of electrons in the central metal atom.
Now,
Oxidation state of :
(b) Fe in $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{3}=(+) 3$ and electronic
configuration of $\mathrm{Fe}^{3 +}$ ions is $\mathrm{3d}^{5} 4 \mathrm{s}^{0}$
i.e. have 5 unpaired electrons
(c) Fe in $\mathrm{K}_{3}\left[\mathrm{FeF}_{6}\right]$
Oxidation no. of $\mathrm{Fe}=(+) 3$ electronic configuration of $\mathrm{Fe}^{3}$ is $\mathrm{3d}^{4} 4 \mathrm{s}^{0}$
i.e. have 5 unpaired electrons.
(d) Oxidation state of $\mathrm{Mn}$ in $\mathrm{K}_{4}\left[\mathrm{MnF}_{6}\right]=(+) 2$ and electronic configuration of $\mathrm{Mn}^{2}+$ is $\mathrm{3d}^{5} 4 \mathrm{s}^{0}$
i.e., have 5 unpaired electrons. Hence, options (b) (c) and (d) are the correct options.

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2019
Consider a conducting wire of length $\mathrm{L}$ bent in the form of a circle of radius $\mathrm{R}$ and another conductor of length 'a' (a < < R) is bent in the form of a square. The two loops are then placed in same plane such that the square loop is exactly at the centre of the circular loop. What will be the mutual inductance between the two loops?

Options:
A $\mu_{0} \frac{\pi a^{2}}{L}$
B $\mu_{0} \frac{\pi \mathrm{a}^{2}}{16 \mathrm{~L}}$
C $\mu_{0} \frac{\pi a^{2}}{4 L}$
D $\mu_{0} \frac{\mathrm{a}^{2}}{4 \pi \mathrm{L}}$

The correct answer is:
$\mu_{0} \frac{\pi \mathrm{a}^{2}}{16 \mathrm{~L}}$

Solution:
Hint:

$2 \pi R=L_{\cdots} \ldots \ldots \ldots .(1)$
$4 s=a \ldots \ldots \ldots . .(2)$
$\frac{\mu_{0} I \times S^{2}}{2 R}=\phi, M=\frac{\mu_{0} S^{2}}{2 R}$
$=\frac{\mu_{0}}{2} \times \frac{a^{2} \times 2 \pi}{16 \times L}=\frac{\mu_{0} a^{2} \pi}{16 L}$

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2020
Consider a positively charged infinite cylinder with uniform volume charge density \(r > 0\). An electric dipole consisting of \(+Q\) and \(-Q\) charges attached to opposite ends of a massless rod is oriented as shown in the figure. At the instant as shown in the figure, the dipole will experience,

Options:
A a force to the left and no torque
B a force to the right and a clockwise torque
C a force to the right and a counter clockwise torque
D non froce but only a clockwise torque

The correct answer is:
a force to the right and a clockwise torque

Solution:
Hint :
\(E \propto \frac{1}{r}\)
\(\begin{aligned} & \mathrm{F}_1 > \mathrm{F}_2 \\ & \therefore \quad \mathrm{F}_{\text {net }} \text { towards right } \tau_{\text {net }} \text { clockwise. }\end{aligned}\)

Topic: Physics, Electrostatics, WBJEE, WBJEE 2023
Consider a pure inductive A.C. circuit as shown in the figure. If the average powr consumed is $P$, then

Options:
A $P>0$
B $\quad P < 0$
C $P=0$
D $\mathrm{P}$ is infinite

The correct answer is:
$P=0$

Solution:
Pure inductor consumer no power as I and $\mathrm{V}$ are in $90^{\circ}$ phase difference.

Topic: Physics, Alternating Current, WBJEE, WBJEE 2021
Consider a quadratic equation \(a x^2+2 b x+c=0\) where \(a, b, c\) are positive real numbers. If the equation has no real root, then which of the following is true?

Options:
A a, b, c cannot be in A.P. or H.P. but can be in G.P.
B a, b, c cannot be in G.P. or H.P. but can be in A.P.
C \(a, b, c\) cannot be in A.P. or G.P. but can be in H.P.
D a, b, c cannot be in A.P., G.P. or H.P.

The correct answer is:
\(a, b, c\) cannot be in A.P. or G.P. but can be in H.P.

Solution:
Hint : For no real root
\(\begin{aligned}
& (2 b)^2-4 a c < 0 \\
& b^2 < a c \\
& b < \sqrt{a c} \because a, b, c > 0
\end{aligned}\)
If in \(A P\), then \(a+c=2 b\)
i.e., \(\mathrm{x}=-1\) a solution
\(\begin{aligned}
& \therefore a, b, c \text { not in AP } \\
& \because b^2 < a c
\end{aligned}\)
\(\therefore a, b, c\) not in GP
For HP \(\quad b=\frac{2 a c}{a+c}=\frac{2 \sqrt{a c}}{\sqrt{\frac{a}{c}}+\sqrt{\frac{c}{a}}}\)
\(\sqrt{\frac{a}{c}}+\sqrt{\frac{c}{a}} \geq 2\)
\(\therefore \frac{2 \sqrt{a c}}{\sqrt{\frac{a}{c}}+\sqrt{\frac{c}{a}}} \leq 2 \therefore a, b, c\) in HP is possible

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2023
Consider a region in free space bounded by the surfaces of an imaginary cube having sides of length $a$ as shown in the figure. A charge $+Q$ is placed at the centre $O$ of the cube. $P$ is such a point out side the cube that the line $O P$ perpendicularly intersects the surface $A B C D$ at $R$ and also $O R=R P=a / 2$. A charge $+Q$ is placed at point $P$ also. What is the total electric flux through the five faces of the cube other than $A B C D ?$

Options:
A $\frac{Q}{\varepsilon_{0}}$
B $\frac{5 Q}{6 \varepsilon_{0}}$
C $\frac{10 Q}{6 \varepsilon_{0}}$
D zero

The correct answer is:
$\frac{Q}{\varepsilon_{0}}$

Solution:
For surface $A B C D$ clectric flux is zero. Because at surface $A B C D$ net electric field is zero. Using Gauss's law,
$$
\oint \mathbf{E} \cdot d \mathbf{S}=\frac{Q_{\mathrm{in}}}{\varepsilon_{0}}
$$
Electric flux through the five faces of the cube,
$$
\phi_{E}=\frac{Q}{\varepsilon_{0}}
$$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2018
Consider a tangent to the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{1}=1$ at any point. The locus of the midpoint of the portion intercepted between the axes is

Options:
A $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1$
B $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$
C $\frac{1}{3 x^{2}}+\frac{1}{4 y^{2}}=1$
D $\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$

The correct answers are:
$\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$

Solution:
Hint:
Tangent at $\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$
$\frac{x x_{1}}{2}+\frac{y y_{1}}{1}=1$
Let mid point of intercept be $\mathrm{P}(\mathrm{h}, \mathrm{k})$
$\mathrm{h}=\frac{1}{\mathrm{x}_{1}}, \mathrm{k}=\frac{1}{2 \mathrm{y}_{1}}$ or $\mathrm{x}_{1}=\frac{1}{\mathrm{~h}}, \mathrm{y}_{1}=\frac{1}{2 \mathrm{k}}$
locus is $\frac{1}{2 x^{2}}+\frac{1}{4 y^{2}}=1$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2020
Consider a thermodynamic process where internal energy $U=A P^2 \mathrm{~V}(\mathrm{~A}=$ constant). If the process is performed adiabatically, then

Options:
A $A P^2(V+1)=$ constant
B $(\mathrm{AP}+1)^2 \mathrm{~V}=\mathrm{constant}$
C $(\mathrm{AP}+1) \mathrm{V}^2=$ constant
D $\frac{V}{(A P+1)^2}=$ constant

The correct answer is:
$(\mathrm{AP}+1)^2 \mathrm{~V}=\mathrm{constant}$

Solution:
$U=A P^2 V$
$\frac{d U}{d V}=A P^2+2 P A V \frac{d P}{d V}$ [For adiabatic process $\left.d Q=0, d U=-d w=-P d V\right]$
$-\frac{P d V}{d V}=A P^2+2 P A V \frac{d P}{d V}$
$-1=A P+2 A V \frac{d P}{d V} \Rightarrow-1-A P=2 A V \frac{d P}{d V}$
$-\int \frac{d V}{V}=\int \frac{2 A d P}{A P+1}$
$-\ell n V+C=\frac{2 A}{A} \ln (A P+1)$
$C=\ell n(A P+1)^2+\ell n V \Rightarrow C=(A P+1)^2 V$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2022
Consider an engine that absorbs 130 cal of heat from a hot reservoir and delivers 30 cal heat to a cold reservoir in each cycle. The engine also consumes 2 J energy in each cycle to overcome friction. If the engine works at 90 cycles per minute, what will be the maximum power delivered to the load? [Assume the thermal equivalent of heat is $4.2 \mathrm{~J} / \mathrm{cal}]$

Options:
A $816 \mathrm{~W}$
B $819 \mathrm{~W}$
C $627 \mathrm{~W}$
D $630 \mathrm{~W}$

The correct answer is:
$627 \mathrm{~W}$

Solution:
Hint:
work done per cycle $=100 \times 4.2-2=418 \mathrm{~J}$
Total work done $=418 \times 90$
Power $=\frac{418 \times 90}{60}=627 \mathrm{~W}$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2020
Consider the circuit given here. The potential difference $V_{B C}$ between the points $B$ and $C$ is

Options:
A $1 \mathrm{V}$
B $0.5 \mathrm{V}$
C $0 \mathrm{V}$
D $-1 \mathrm{V}$

The correct answer is:
$0.5 \mathrm{V}$

Solution:
From circuit the current,
$$
I=\frac{E}{R}
$$
The equivalent resistance of above circuit
$$
\begin{array}{l}
=3+2+1 \\
=6 \mathrm{k} \Omega \\
=6 \times 10^{3} \Omega \\
E=3 \text { volt }
\end{array}
$$
$$
\begin{aligned}
I &=\frac{3}{6 \times 10^{3}} \\
&=0.5 \times 10^{-3} \mathrm{A}
\end{aligned}
$$
Porential, $\begin{aligned} V_{A D} &=i R=0.5 \times 10^{-3} \times 3 \times 10^{3} \\ &=1.5 \mathrm{V} \end{aligned}$
Charge,
$$
\begin{array}{l}
\frac{1}{C}=\frac{1}{1}+\frac{1}{2}=\frac{3}{2} \\
C=\frac{2}{3} \\
Q=C \cdot V_{A D} \\
=\frac{2}{3} \times 1 \cdot 5 \\
=1 \mu \mathrm{C}
\end{array}
$$
Applying KVL (Kirchhoffs Voltage Law) from $B$ to $C$
$$
\begin{array}{c}
v_{s}-0 \cdot 5 \times 10^{-3} \times 2 \times 10^{3}+\frac{1}{2}=V_{c} \\
V_{n}-V_{c}=1-\frac{1}{2}=0.5 \mathrm{V}
\end{array}
$$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2017