datavorous/entrance-exam-dataset huggingface dataset:
The locus of the vertices of the family of parabolas $6 y=2 a^{3} x^{2}+3 a^{2} x-12 a$ is

Options:
A $x y=\frac{105}{64}$
B $x y=\frac{64}{105}$
C $x y=\frac{35}{16}$
D $x y=\frac{16}{35}$

The correct answer is:
$x y=\frac{105}{64}$

Solution:
$\begin{aligned}
6 y+12 a &=2 a\left(a^{2} x^{2}+\frac{3}{2} a x\right) \\
&=2 a\left(a^{2} x^{2}+2 \cdot \frac{3}{4} a x+\frac{9}{16}\right)-\frac{9}{8} a \\
\Rightarrow 6 y+\frac{105}{8} a &=2 a\left(a x+\frac{3}{4}\right)^{2}
\end{aligned}$
Let Vertices be ( $h, k), \quad h=-\frac{3}{4 a}, k=-\frac{35 a}{16}$
$\therefore \mathrm{hk}=\frac{105}{64}$
Hence, locus is $x y=\frac{105}{64}$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2021
The magnetic field at the point of intersection of diagonals of a square wire loop of side \(\mathrm{L}\) carrying a current \(\mathrm{I}\) is

Options:
A \(\frac{\mu_0 \mathrm{I}}{\pi \mathrm{L}}\)
B \(\frac{2 \mu_0 \mathrm{I}}{\pi \mathrm{L}}\)
C \(\frac{\sqrt{2} \mu_0 I}{\pi \mathrm{L}}\)
D \(\frac{2 \sqrt{2} \mu_{\mathrm{o}} \mathrm{I}}{\pi \mathrm{L}}\)

The correct answer is:
\(\frac{2 \sqrt{2} \mu_{\mathrm{o}} \mathrm{I}}{\pi \mathrm{L}}\)

Solution:
\(\begin{aligned} & \text {Hints: } B=4\left[\frac{\mu_0}{4 \pi} \frac{I}{\left(\frac{\mathrm{L}}{2}\right)}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)\right] \\ & =\frac{\mu_{\mathrm{o}}}{\pi} \frac{2 \mathrm{I}}{\mathrm{L}} \cdot \frac{2}{\sqrt{2}} ; \quad B=\frac{\mu_{\mathrm{o}}}{\pi} \frac{2 \sqrt{2} \mathrm{I}}{\mathrm{L}}\end{aligned}\)

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2011
The magnetic field due to a current in a straight wire segment of length $L$ at a point on
its perpendicular bisector at a distano $r(r>>L)$

Options:
A decreases as $\frac{1}{r}$
B decreases as $\frac{1}{r^{2}}$
C decreases as $\frac{1}{r^{3}}$
D approaches a finite limit as $r \rightarrow \infty$

The correct answer is:
decreases as $\frac{1}{r^{2}}$

Solution:
By Biot-Savart's law, the magnetic field due to a current carrying wire is
$$
B=\frac{\mu_{0}}{4 \pi} \cdot \frac{I d l \sin \theta}{r^{2}}
$$
So,
$$
B=\frac{1}{r^{2}}
$$
So, the magnetic field due to current decreases as inverse square of distance of the point of observation.

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2017
The magnetic flux linked with a coil satisfies the relation $\phi=4 t^{2}+6 t+9$ Wb, where $t$ is the
time in second. The emf induced in the coil at $t=2 \mathrm{s}$ is

Options:
A $22 \mathrm{V}$
B $18 \mathrm{V}$
C $16 \mathrm{V}$
D $40 \mathrm{V}$

The correct answer is:
$22 \mathrm{V}$

Solution:
Given $\phi=4 t^{2}+6 x+9$ and $t=2 \mathrm{s}$ We know that $\varepsilon=\frac{d \phi}{d t}$
Here, $\frac{d \phi}{d t}=8 t+6$
$$
\begin{aligned}
\left(\frac{d \phi}{d t}\right)_{t=2} &=8 \times 2+6 \\
\varepsilon &=22 \mathrm{V}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2012
The magnetic flux through a loop of resistance $10 \Omega$ is given by $\phi=5 t^2-4 t+1$ Weber. How much current is induced in the loop after $0.2 \mathrm{sec}$ ?

Options:
A $0.4 \mathrm{~A}$
B $0.2 \mathrm{~A}$
C $0.04 \mathrm{~A}$
D $0.02 \mathrm{~A}$

The correct answer is:
$0.2 \mathrm{~A}$

Solution:
Hints : $\phi=5 t^2-4 t+1$
$$
\frac{d \phi}{d t}=10 t-4
$$
$$
\mathrm{I}=\frac{e}{\mathrm{R}}=\frac{-d \phi / d t}{\mathrm{R}}=-\frac{10 t-4}{10}
$$
At $t=0.2 \mathrm{sec}$
$$
\mathrm{I}=\frac{-(10 \times 0.2-4)}{10}=-\frac{(2-4)}{10}=+\frac{2}{10}=+0.2 \mathrm{~A}=0.2 \mathrm{~A}
$$

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2010
The magnets of two suspended coll galvanometers are of the same strength so that they produce identical uniform magnetic fields in the region of the coils. The coil of the first one is in the shape of a square of side a and that of the second one is circular of radius $\frac{a}{\sqrt{\pi}},$ When the same current is passed through the coils, the ratio of the torque experienced by the first coil to that experienced by the second one is

Options:
A $1: \frac{1}{\sqrt{\pi}}$
B 1: 1
C $\pi: 1$
D $1: \pi$

The correct answer is:
1: 1

Solution:
In galvanometer $I \sim \theta=G \theta$ Where, $\quad G=\frac{k}{N A B}$
When coil is in square of shape,
$$
G=\frac{k}{N A B}=\frac{k}{N a^{2} B}
$$
When coil is in circular shape of radius $a / \sqrt{\pi}$
$$
\begin{aligned}
G &=\frac{k}{N A B}=\frac{k}{N \pi r^{2} \cdot B} \\
&=\frac{k}{N \pi\left(\frac{a}{\sqrt{\pi}}\right)^{2} \cdot B}=\frac{k}{N a^{2} B}
\end{aligned}
$$
So, value of $G$ is same for both type of shape of coil. So, the torque $e=N B L A \sin \theta$ will be same in both the case
$$
e: e=1: 1
$$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2017
The major product of the above reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
$\mathrm{HBr} \rightleftharpoons \mathrm{H}^{+}+\mathrm{Br}^{-}$
For 1,3 -butadiene at room temperature, there is formation of 1,4 -addition product.

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2015
The major product of the following reaction is $\mathrm{F}_3 \mathrm{C}-\mathrm{CH}=\mathrm{CH}_2+\mathrm{HBr} \rightarrow$

Options:
A $\mathrm{F}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$
B $\mathrm{F}_3 \mathrm{C}-\mathrm{CH}(\mathrm{Br})-\mathrm{CH}_3$
C
D

The correct answer is:
$\mathrm{F}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$

Solution:

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2022
The major products obtained during ozonolysis of 2,3 -dimethyl- 1 -butene and subsequent reductions with $Z \mathrm{n}$ and $\mathrm{H}_{2} \mathrm{O}$ àte

Options:
A methanoic acid and 2 -methyl-2-butanone
B methanal and 3 -methy -2 -butanone
C methanol and 2,2 -dimethyl-3-butanone
D methanoic acid and 2 -methyl-3-butanone

The correct answer is:
methanal and 3 -methy -2 -butanone

Solution:

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2016
The major product(s) obtained form the following reaction of 1 mole of hexadeuteriobenzene is/are

Options:
A
B
C
D

The correct answers are:

Solution:
The reaction for 1 mole of hexadeuterisobenzene occurs as follows :

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2017
The major products obtained in the following reaction is/are

Options:
A
B
C
D

The correct answers are:
,

Solution:
The major product $($ s) obtained in the given reaction are

The mechanism for the above given reaction is as follows. When the $\pi$ -electrons of the alkene approach a molecule of $\mathrm{Br}_{2}$, one of the bromine atom accepts them and releases the shared electrons to the other
bromine atom. A cyclic bromonium ion is formed.
In next step $\mathrm{Br}^{-}$ attacks a carbon atom of the bromonium ion. This release the strain in the three membered ring and form a vicinal dibromide.

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2018
The \(\mathrm{pH}\) of an aqueous solution of \(\mathrm{CH}_3 \mathrm{COONa}\) of concentrated \(\mathrm{C}(\mathrm{M})\) is given by

Options:
A \(7-\frac{1}{2} \mathrm{pK}_{\mathrm{a}}-\frac{1}{2} \log \mathrm{C}\)
B \(\frac{1}{2} \mathrm{pK}_{\mathrm{w}}+\frac{1}{2} \mathrm{pK}_{\mathrm{b}}+\frac{1}{2} \log \mathrm{C}\)
C \(\frac{1}{2} \mathrm{pK}_{\mathrm{w}}-\frac{1}{2} \mathrm{pK}_{\mathrm{b}}-\frac{1}{2} \log \mathrm{C}\)
D \(\frac{1}{2} \mathrm{pK}_{\mathrm{w}}+\frac{1}{2} \mathrm{pK}_{\mathrm{a}}+\frac{1}{2} \log \mathrm{C}\)

The correct answer is:
\(\frac{1}{2} \mathrm{pK}_{\mathrm{w}}+\frac{1}{2} \mathrm{pK}_{\mathrm{a}}+\frac{1}{2} \log \mathrm{C}\)

Solution:
Hints : In case of Hydrolysis of salt of weak acid and strong base, the \(\mathrm{pH}\) is given by \(\frac{1}{2} \mathrm{pK}_{\mathrm{w}}+\frac{1}{2} \mathrm{pK}_{\mathrm{a}}+\frac{1}{2} \log \mathrm{c}\)

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2011
The \(\mathrm{sp}^3 \mathrm{~d}^2\) hybridization of central atom of a molecule would lead to

Options:
A square planar geometry
B Tetrahedral geometry
C Trigonal bipyramidal geometry
D Octahedral geometry

The correct answer is:
Octahedral geometry

Solution:
Hints : Fact

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2011
The maximum and minimum values of $\cos ^{6} \theta+\sin ^{6} \theta$ are respectively

Options:
A 1 and $\frac{1}{4}$
B 1 and 0
C 2 and 0
D 1 and $\frac{1}{2}$

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

Solution:
Let $f(\theta)=\sin ^{6} \theta+\cos ^{6} \theta$
$\Rightarrow \quad f(\theta)=\left(\sin ^{2} \theta\right)^{3}+\left(\cos ^{2} \theta\right)^{3}$
$=\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$
$\left(\sin ^{4} \theta+\cos ^{4} \theta-\sin ^{2} \theta-\cos ^{2} \theta\right)$
$\left[\because a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right]\right.$
$=1 \cdot\left\{\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}-3 \sin ^{2} \theta \cdot \cos ^{2} \theta\right\}$
$=1 \cdot\left(1-\frac{3}{4} \cdot 4 \sin ^{2} \theta \cdot \cos ^{2} \theta\right)$
$=1-\frac{3}{4}(\sin 2 \theta)^{2}(\because \sin 2 A=2 \sin A \cos A)$
$=1-\frac{3}{8}(1-\cos 4 \theta)$
$=1-\frac{3}{8}+\frac{3}{8} \cos 4 \theta$
$f(\theta)=\frac{5}{8}+\frac{3}{8} \cdot \cos 4 \theta$
$-1 \leq \cos 4 \theta \leq 1$
$\Rightarrow \quad \frac{-3}{8} \leq \frac{3}{8} \cos 4 \theta \leq \frac{3}{8}$
$\Rightarrow \quad \frac{5}{8}-\frac{3}{8} \leq \frac{5}{8}+\frac{3}{8} \cos 48 \leq \frac{5}{8}+\frac{3}{8}$
$\Rightarrow \quad \frac{1}{4} \leq f(\theta) \leq 1$
So, the maximum value is 1 and minimum value is $\frac{1}{4}$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2013
The maximum number of atoms that can be in one plane in the molecule p-nitrobenzonitrile are

Options:
A 6
B 12
C 13
D 15

The correct answer is:
15

Solution:
All atoms are on the same plane as all central atoms are either $\mathrm{sp}^{2}$ or sp hybridised $\therefore$ total number of atoms on same plane $\Rightarrow \mathrm{C}_{6} \mathrm{H}_{4}+\mathrm{NO}_{2}+\mathrm{CN}=15$

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2021
The maximum number of electrons in an atom in which the last electron filled has the quantum numbers $n=3, \ell=2$ and $m=-1$ is

Options:
A 17
B 27
C 28
D 30

The correct answer is:
30

Solution:
Hint:
$\mathrm{n}=3$ and $\mathrm{I}=2$ means $3 \mathrm{~d}$ -orbital. Since five orientations of $\mathrm{d}$ -orbitals are degenerate, $\mathrm{m}=-1$ can be assigned to any one. For maximum number of electrons as asked, answer should be atomic number 30.

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2020
The maximum value of $f(x)=e^{\sin x}+e^{\cos x} ; x \in R$ is

Options:
A $2 \mathrm{e}$
B $2 \sqrt{\mathrm{e}}$
C $2 \mathrm{e}^{1 / \sqrt{2}}$
D $2 e^{-1 / \sqrt{2}}$

The correct answer is:
$2 \mathrm{e}^{1 / \sqrt{2}}$

Solution:
$f(x)=e^{\sin x}+e^{\cos x}$
$\begin{aligned}
&f^{\prime}(x)=e^{\sin x} \cdot \cos x-e^{\cos x} \cdot \sin x \\
&\begin{aligned}
f^{\prime \prime}(x) &=e^{\sin x} \cos ^2 x+e^{\cos x} \cdot \sin ^2 x-\sin x \cdot e^{\sin x}-\cos x \cdot e^{\cos x} \\
&=e^{\sin x}\left(1-\sin x-\sin ^2 x\right)+e^{\cos x}\left(1-\cos ^2 x-\cos x\right) \\
f^{\prime}(\pi / 4) &=0 \text { and } f^{\prime \prime}(\pi / 4) < 0 \\
f^{\prime}(x) &=0 \text { at } x=\pi / 4+2 n \pi \text { or } 5 \pi / 4+2 n \pi(n \in z)
\end{aligned}
\end{aligned}$
$f^{\prime}(\pi / 4+2 n \pi)=0$ and $f^{\prime \prime}(\pi / 4+2 n \pi) < 0$
$\therefore f_{\max }=f(\pi / 4+2 n \pi), n \in Z=2 e^{1 / \sqrt{2}}$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2022
The maximum value of $|x|$, when the complex number $z$ satisfies the condition $\left|z+\frac{2}{z}\right|=2$ is

Options:
A $\sqrt{3}$
B $\sqrt{3}+\sqrt{2}$
C $\sqrt{3}+1$
D $\sqrt{3}-1$

The correct answer is:
$\sqrt{3}+1$

Solution:
We have the identity,
$$
\begin{aligned}|z|=\left|z+\frac{2}{z}-\frac{2}{z}\right| \leq\left|z+\frac{2}{z}\right|+\left|\frac{2}{z}\right| \\ \Rightarrow \quad|z| \leq 2+\frac{2}{|z|} \end{aligned}
$$
$\Rightarrow \quad|z|^{2} \leq 2|z|+2$
$\Rightarrow \quad|z|^{2}-2|z|+1 \leq 3$
\begin{array}{ll}
\Rightarrow & (|z|-1)^{2} \leq 3 \\
\Rightarrow & -\sqrt{3} \leq|z|-1 \leq \sqrt{3} \\
\hline
\end{array}
$\Rightarrow \quad-\sqrt{3}+1 \leq|z| \leq \sqrt{3}+1$
Hence, the maximum value $=\sqrt{3}+1$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2012
The mean and variance of a binomial distribution are 4 and 2 respectively. Then the probability of exactly two succsses is

Options:
A $\frac{7}{64}$
B $\frac{21}{128}$
C $\frac{7}{32}$
D $\frac{9}{32}$

The correct answer is:
$\frac{7}{64}$

Solution:
$\mathrm{np}=4, \mathrm{npq}=2$
$\therefore \mathrm{q}=\frac{2}{4}=\frac{1}{2} \Rightarrow \mathrm{p}=\frac{1}{2} \therefore \mathrm{n}=8$
$\mathrm{p}(\mathrm{x}=2)={ }^{8} \mathrm{C}_{2}\left(\frac{1}{2}\right)^{8}=\frac{28}{256}=\frac{7}{64}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2021
The measured freezing point depression for a $0.1 \mathrm{m}$ aqueous $\mathrm{CH}_{3} \mathrm{COOH}$ solution is $0.19^{\circ} \mathrm{C}$. The acid dissociation constant $K_{a}$ at this concentration will be (Given, $K_{f}$ the molal cryoscopic constant $=1.86 \mathrm{K}$ $\mathrm{kg} \mathrm{mol}^{-1}$ )

Options:
A $4.76 \times 10^{-5}$
B $4 \times 10^{-5}$
C $8 \times 10^{-5}$
D $2 \times 10^{-5}$

The correct answer is:
$4 \times 10^{-5}$

Solution:
$\begin{aligned} \Delta T_{f} &=i \times K_{t} \times m \\ \therefore & i=\frac{\Delta T_{f}}{K_{t} \times m}=\frac{0.19}{1.86 \times 0.1}=1.02 \end{aligned}$
Again from, $\alpha=\frac{i-1}{n-1}=\frac{1.02-1}{2-1}$ $=0.02=2.0 \times 10^{-2}$
$$
\begin{aligned}
\text { for } \mathrm{CH}_{3} \mathrm{COOH} & \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+} \\
K \mathrm{a} &=\mathrm{C} \alpha^{2} \\
&=0.1 \times\left(2 \times 10^{-2}\right)^{2} \\
&=4 \times 10^{-5}
\end{aligned}
$$

Topic: Chemistry, Solutions, WBJEE, WBJEE 2013
The melting points of (i) BeCl$_{2}$ (ii) $\mathrm{CaCl}_{2}$ and (iii) $\mathrm{HgCl}_{2}$ follows the order

Options:
A $i < ii < iii$
B $iii < i < ii$
C $i < iii < ii$
D $ii < i < iii$

The correct answer is:
$iii < i < ii$

Solution:
Key Point A compound having more ionic character, has more melting point.
Be and Ca belong to the same group and ionic character increases down the group.
Thus, $\mathrm{BeCl}_{2}$ is less ionic than $\mathrm{CaCl}_{2}$. This means that melting point of $\mathrm{CaCl}_{2}$ is higher than that of $\mathrm{BeCl}_{2}$
Also, $\mathrm{Hg}$ is a transition metal, its compound is less ionic than $\mathrm{BeCl}_{2}$. Therefore, order of melting point will be
or
$$
\begin{array}{c}
\mathrm{CaCl}_{2}>\mathrm{BeCl}_{2}>\mathrm{HgCl}_{2} \\
(\mathrm{ii})>(\mathrm{i})>(\mathrm{iii})
\end{array}
$$Hence, option (b) is the correct answer.

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2019
The metal which can be used to obtain metallic Cu from aqueous CuSO $_{4}$ solution is

Options:
A Na
B Ag
C Hg
D Fe

The correct answer is:
Fe

Solution:
Iron (Fe) is used to obtain Curmetal from CuSO, (aq) because Fe is more reactive than Cu (and also easly available and is cheaper) thus, replace the copper from CuSO $_{4}$ to give copper (Cu)
$\mathrm{CuSO}_{4}(\mathrm{aq})+\mathrm{Fe}(\mathrm{s}) \longrightarrow \mathrm{FeSO}_{4}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2016
The metal-pair that can produce nascent hydrogen in alkaline medium is

Options:
A $\mathrm{Zn}, \mathrm{Al}$
B $\mathrm{Fe}, \mathrm{Ni}$
C Al, Mg
D $\mathrm{Mg}, \mathrm{Zn}$

The correct answer is:
$\mathrm{Zn}, \mathrm{Al}$

Solution:
Fact

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2022
The minimum value of $2^{\sin x}+2^{\cos x}$ is

Options:
A $2^{1-1 / \sqrt{2}}$
B $2^{1+1 / \sqrt{2}}$
C $2^{\sqrt{2}}$
D 2

The correct answer is:
$2^{1-1 / \sqrt{2}}$

Solution:
We know that
$\mathrm{AM} \geq \mathrm{GM}$
$\therefore \quad \frac{2^{\sin x}+2^{\cos x}}{2} \geq \sqrt{2^{\sin x} 2^{\cos x}}$
$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2 \sqrt{2^{\sin x+\cos x}}$
$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2 \times 2^{\frac{\sin x+\cos x}{2}}$
$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2^{1+\frac{\sin x+\cos x}{2}}$
But $\sin x+\cos x=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right) \geq-\sqrt{2}$
$\therefore \quad 2^{\sin x}+2^{\cos x} \geq 2^{1-\frac{\sqrt{2}}{2}}$
$\Rightarrow 2^{\sin x}+2^{\cos x} \geq 2^{1-\frac{1}{\sqrt{2}}}, \forall x \in R$
Hence, minimum value is $2^{1-\frac{1}{\sqrt{2}}}$.

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2014
The minimum value of $\cos \theta+\sin \theta+\frac{2}{\sin 2 \theta}$
for $\theta \in(0, \pi / 2),$ is

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

The correct answer is:
$2+\sqrt{2}$

Solution:
Here, $\cos \theta+\sin \theta+\frac{2}{\sin 2 \theta}, \theta \in\left(0, \frac{\pi}{2}\right)$
For minimum value, $\sin 2 \theta$ must be maximum
$\therefore \quad 2 \theta=\frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{4}$
Hence, $\cos \frac{\pi}{4}+\sin \frac{\pi}{4}+\frac{2}{\sin \frac{\pi}{2}}=\sqrt{2}+2$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2015
The minimum value of $f(x)=e^{\left(x^4-x^3+x^2\right)}$ is

Options:
A e
B $-\mathrm{e}$
C 1
D -1

The correct answer is:
1

Solution:
Hints: $f(x)=e^{\left(x^4-x^3+x^2\right)}, f^{\prime}(x)=e^{x^4-x^3+x^2}$
$e^{x^4-x^3+x^2}\left(4 x^3-3 x^2+2 x\right) x\left(4 x^2-3 x+2\right)$
$\Rightarrow f(x)$ is decreasing for $x < 0$, increasing for $x>0$
$\therefore$ Minimum is at $\mathrm{x}=0 \quad \therefore \mathrm{f}(0)=\mathrm{e}^0=1$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2010
The minimum value of the function $f(x)=2|x-1|+|x-2|$ is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
1

Solution:
Given,
$$
\begin{array}{l}
\quad f(x)=2|x-1|+|x-2| \\
\Rightarrow \quad f(x)=\left\{\begin{array}{c}
-2(x-1)-(x-2), x < 1 \\
2(x-1)-(x-2), 1 \leq x < 2 \\
2(x-1)+(x-2), x \geq 2 \\
-3 x+4, \quad x < 1 \\
x_{1} \\
3 x-4, \quad 1 \leq x < 2 \\
x \geq 2
\end{array}\right.
\end{array}
$$
$f^{\prime}(x)=\left\{\begin{array}{cc}-3 & x < 1 \\ 1 & 1 \leq x < 2 \\ 3 & x \geq 2\end{array}\right.$
So, $f(x)$ will be minimum at $x=1$ and the minimum value is 1 .

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2013
The modulus of $\frac{1-i}{3+i}+\frac{4 i}{5}$ is

Options:
A $\sqrt{5}$ unit
B $\frac{\sqrt{11}}{5}$ unit
C $\frac{\sqrt{5}}{5}$ unit
D $\frac{\sqrt{12}}{5}$ unit

The correct answer is:
$\frac{\sqrt{5}}{5}$ unit

Solution:
$$
\text { Hints: } \frac{1-i}{3+i}+\frac{4 i}{5}=\frac{5-5 i+4 i(3+i)}{5(3+i)}=\frac{5-5 i+12 i-4}{5(3+i)}=\frac{1+7 i}{5(3+i)}=\frac{(1+7 i)(3-i)}{5(9+1)}
$$
$$
=\frac{3+21 i-i+7}{5 \times 10}=\frac{10+20 i}{5 \times 10}=\frac{1+2 i}{5}
$$
$$
\therefore \text { Modulus }=\sqrt{\left(\frac{1}{5}\right)^2+\left(\frac{2}{5}\right)^2}=\sqrt{\frac{1}{25}+\frac{4}{25}}=\sqrt{\frac{1}{5}}=\frac{\sqrt{5}}{5} \text { unit }
$$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2009
The molar solubility (in mol $\mathrm{L}^{-1}$ ) of a sparingly soluble salt $M X_{4}$ is ${ }^{\prime} S$. The corresponding solubility product is $K_{\mathrm{sp}} . S$ in terms of $'K_{\text {ip }}$' s given by the relation

Options:
A $S=\left(\frac{K_{\mathrm{s} \rho}}{128}\right)^{1 / 4}$
B $S=\left(\frac{K_{s p}}{256}\right)^{1 / 5}$
C $S=\left(256 K_{s p}\right)^{1 / 5}$
D $S=\left(128 K_{\mathrm{sp}}\right)^{1 / 4}$

The correct answer is:
$S=\left(\frac{K_{s p}}{256}\right)^{1 / 5}$

Solution:
$\because$ For a compound $A_{x}B_{y}$
$$
K_{s p}=x^{x} \cdot y^{y} \cdot S^{x+y} . \text { (where, } S=\text { solubility })
$$
For $M x_{4}$
$$
\begin{array}{l}
K_{\mathrm{sp}}=1 \times(4)^{4} \cdot \mathrm{S}^{1+4} \\
K_{\mathrm{sp}}=256 \times \mathrm{S}^{5}
\end{array}
$$
$\therefore S=\left(\frac{K_{s}}{256}\right)^{1 / 5}$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2016
The molarity of a $\mathrm{NaOH}$ solution by dissolving $4 \mathrm{~g}$ of it in $250 \mathrm{ml}$ water is

Options:
A $0.4 \mathrm{M}$
B $0.8 \mathrm{M}$
C $0.2 \mathrm{M}$
D $0.1 \mathrm{M}$

The correct answer is:
$0.4 \mathrm{M}$

Solution:
Hints : Molarity $=\frac{4 / 40}{250 / 1000}=0.4$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2010
The mole fraction of a solute in a binary solution is $0.1 .$ At $298 \mathrm{~K}$, molarity of this solution is same as as its molality. Density of this solution at $298 \mathrm{~K}$ is $2.0 \mathrm{~g} \mathrm{~cm}^{-3}$. The ratio of molecular weights of the solute and the solvent $\left(\mathrm{M}_{\text {solute }} / \mathrm{M}_{\text {solvent }}\right)$ is

Options:
A 9
B $\frac{1}{9}$
C $4.5$
D $\frac{1}{4.5}$

The correct answer is:
9

Solution:
$\mathrm{X}_{2}=0.1 \Rightarrow \frac{\mathrm{n}_{2}}{\mathrm{n}_{2}+\mathrm{n}_{1}}=0.1$ $\ldots(1)$
$\mathrm{M}=\frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}} \times \mathrm{d}$ $\ldots(2)$
$\mathrm{m}=\frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}_{1}(\mathrm{~g})}$ $\ldots(3)$
$\mathrm{d}=2 \mathrm{~g} / \mathrm{cc}$ $\ldots(4)$
Given M = m (numerically)
$\begin{array}{l}
\Rightarrow \frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times\left(\mathrm{W}_{2}+\mathrm{W}_{1}\right)} \times \mathrm{d}=\frac{\mathrm{W}_{2} \times 1000}{\mathrm{M}_{2} \times \mathrm{W}_{1}(\mathrm{~g})} \\
\Rightarrow \mathrm{W}_{2}+\mathrm{W}_{1}=\mathrm{W}_{1} \times \mathrm{d}=2 \mathrm{~W}_{1} \\
\Rightarrow \mathrm{W}_{2}=\mathrm{W}_{1}
\end{array}$
Now, $\frac{n_{2}}{n_{2}+n_{1}}=0.1 \Rightarrow \frac{\frac{w_{2}}{M_{2}}}{\frac{W_{2}}{M_{2}}+\frac{W_{1}}{M_{1}}}=0.1 \Rightarrow \frac{\frac{W_{2}}{M_{2}}}{W_{2}\left(\frac{1}{M_{2}}+\frac{1}{M_{1}}\right)}=0.1$
$\Rightarrow \frac{\frac{1}{\mathrm{M}_{2}}}{\frac{\mathrm{M}_{1}+\mathrm{M}_{2}}{\mathrm{M}_{1} \mathrm{M}_{2}}}=0.1 \Rightarrow \frac{\mathrm{M}_{1}}{\mathrm{M}_{1}+\mathrm{M}_{2}}=\frac{1}{10} \Rightarrow \frac{\mathrm{M}_{1}}{\mathrm{M}_{1}+\mathrm{M}_{2}-\mathrm{M}_{1}}=\frac{1}{10-1} \Rightarrow \frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}=\frac{1}{9} \Rightarrow \frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}=9$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2021
The mole fraction of ethanol in water is $0.08$. Its molality is

Options:
A $6.32 \mathrm{~mol} \mathrm{~kg}^{-1}$
B $4.83 \mathrm{~mol} \mathrm{~kg}^{-1}$
C $3.82 \mathrm{~mol} \mathrm{~kg}^{-1}$
D $2.84 \mathrm{~mol} \mathrm{~kg}^{-1}$

The correct answer is:
$4.83 \mathrm{~mol} \mathrm{~kg}^{-1}$

Solution:
Hint:
$\begin{aligned} \text { Molality } &=\frac{\mathrm{x}_{\mathrm{EtOH}}}{\mathrm{x}_{\mathrm{H}_{2} \mathrm{O}}} \times \frac{1000}{\mathrm{MW}_{\text {solvent }}} \\ &=\frac{0.08}{0.92} \times \frac{1000}{18} \\ &=4.83 \mathrm{~mol} \mathrm{~kg}^{-1} \end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2020
The molecular shapes of \(\mathrm{SF}_4, \mathrm{CF}_4\) and \(\mathrm{XeF}_4\) are

Options:
A the same with 2, 0 and 1 lone pairs of electrons on the central atoms, respectively.
B the same with 1,1 and 1 lone pairs of electrons on the central atoms, respectively
C different with 0,1 and 2 lone pairs of electrons on the central atoms, respectively
D different with 1, 0 and 2 lone pairs of electrons on the central atoms, respectively

The correct answer is:
different with 1, 0 and 2 lone pairs of electrons on the central atoms, respectively

Solution:
Hint :

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2023
The molecule/molecules that has/have delocalised lone pair(s) of electrons is/are

Options:
A I, II and III
B I, II and IV
C I and III
D only III

The correct answer is:
only III

Solution:
The molecule in which a lone pair of electrons undergoes conjugation with an alternate double bond is said to be delocalised. The molecule that has delocalised lone-pair of electrons is,

So, option (d) is correct.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2019
The monomer used to produce orlon is

Options:
A $\mathrm{CH}_2=\mathrm{CHF}$
B $\mathrm{CH}_2=\mathrm{CCl}_2$
C $\mathrm{CH}_2=\mathrm{CHCl}$
D $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CN}$

The correct answer is:
$\mathrm{CH}_2=\mathrm{CH}-\mathrm{CN}$

Solution:
Orlon or PAN
Monomer $\Rightarrow \mathrm{CH}_2=\mathrm{CH}-\mathrm{CN}$

Topic: Chemistry, Polymers, WBJEE, WBJEE 2009