datavorous/entrance-exam-dataset huggingface dataset:
Experimental investigations show that the intensity of solar radiation is maximum for a wavelength $480 \mathrm{~nm}$ in the visible region. Estimate the surface temperature of sun. Given Wein's constant $b=2.88 \times 10^{-3} \mathrm{mK}$.

Options:
A $4000 \mathrm{~K}$
B $6000 \mathrm{~K}$
C $8000 \mathrm{~K}$
D $10^6 \mathrm{~K}$

The correct answer is:
$6000 \mathrm{~K}$

Solution:
Hints: $\lambda_{\mathrm{m}} \times \mathrm{T}=\mathrm{b}$
$\lambda_{\mathrm{m}}=480 \mathrm{~nm}$
$\mathrm{T}=\frac{b}{\lambda_m}=\frac{2.88 \times 10^{-3}}{480 \times 10^{-9}}=6000 \mathrm{~K}$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2010
Extraction of a metal ( $M$ ) from its sulfide ore $\left(M_{2} S\right)$ involves the following chemical reactions:
$\begin{array}{l}
2 \mathrm{M}_{2} \mathrm{~S}+3 \mathrm{O}_{2} \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{M}_{2} \mathrm{O}+2 \mathrm{SO}_{2} \uparrow \\
\mathrm{M}_{2} \mathrm{~S}+2 \mathrm{M}_{2} \mathrm{O} \stackrel{\text { heat }}{\longrightarrow} 6 \mathrm{M}+\mathrm{SO}_{2} \uparrow
\end{array}$
The metal (M) may be

Options:
A $\mathrm{Zn}$
B $\mathrm{Cu}$
C $\mathrm{Fe}$
D $\mathrm{Ca}$

The correct answer is:
$\mathrm{Cu}$

Solution:
$\begin{array}{l}
2 \mathrm{Cu}_{2} \mathrm{~S}+3 \mathrm{O}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{Cu}_{2} \mathrm{O}+\mathrm{SO}_{2} \\
2 \mathrm{Cu}_{2} \mathrm{O}+\mathrm{Cu}_{2} \mathrm{~S} \longrightarrow 6 \mathrm{Cu}+\mathrm{SO}_{2}
\end{array}$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, WBJEE, WBJEE 2021
"Extraction of gold $(\mathrm{Au})$ involves the formation of complex ions $X$ and $Y$

$X$ and $Y$ respectively are

Options:
A $Au(CN) ^{-}_{2}$ and $\mathrm{Zn} (\mathrm{CN)}_{4}^{2-}$
B $Au(CN) ^{3-}_{4}$ and $\mathrm{Zn} (\mathrm{CN)}_{4}^{2-}$
C $Au(CN) ^{-}_{3}$ and $\mathrm{Zn} (\mathrm{CN)}_{6}^{4-}$
D $Au(CN) ^{-}_{4}$ and $\mathrm{Zn} (\mathrm{CN)}_{3}^{-}$

The correct answer is:
$Au(CN) ^{-}_{2}$ and $\mathrm{Zn} (\mathrm{CN)}_{4}^{2-}$

Solution:
Extraction of gold (Au) involves the following sequence of reaction,
$\begin{array}{l}
\text { Gold ore } \frac{\text { Roasting }}{\mathrm{CN}^{-}, \mathrm{H}_{2} \mathrm{O}, \mathrm{O}_{2}} \quad \mathrm{HO}^{-}+\mathrm{Au}(\mathrm{CN})_{2}^{-}
\quad \stackrel{\mathrm{Zn}}{\longrightarrow} \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+\mathrm{Au} \\
Y
\end{array}$
Hence, $X=\mathrm{Au}(\mathrm{CN})_{2}^{-}, Y=\mathrm{Zn}(\mathrm{CN})_{4}^{2-}$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, WBJEE, WBJEE 2015
$f(x)$ is real valued function such that $2 f(x)+3 f(-x)=15-4 x$ for all $x \in R$. Then $f(2)=$

Options:
A $-15$
B 22
C 11
D 0

The correct answer is:
11

Solution:
$2 f(x)+3 f(-x)=15-4 x$...(1)
Replace $x$ by $-x$
$2 f(-x)+3 f(x)=15+4 x$ ...(2)
After solving (1) and (2), we get
$\begin{array}{l}
f(x)=4 x+3 \\
\therefore f(2)=8+3=11
\end{array}$

Topic: Mathematics, Functions, WBJEE, WBJEE 2021
$f: X \rightarrow \mathbb{R}, X=\{x \mid 0 < x < 1\}$ is defined as $f(x)=\frac{2 x-1}{1-|2 x-1|}$. Then

Options:
A $\mathrm{f}$ is only injective
B $f$ is only surjective
C $\mathrm{f}$ is bijective
D f is neither injective nor surjective

The correct answer is:
$\mathrm{f}$ is bijective

Solution:
Put $2 \mathrm{x}-1=\mathrm{t}$, then the function becomes
$\begin{aligned}
&f(t)=\frac{t}{1-|t|},-1 < t < 1 \\
&\therefore f(t)=\left\{\begin{array}{l}
\frac{t}{1+t},-1 < t \leq 0 \\
\frac{t}{1-t}, 0 < t < 1
\end{array}\right.
\end{aligned}$
$\because$ It is continuous and $f\left(-1^{+}\right)=-\infty, f\left(1^{-}\right)=+\infty$
$\therefore$ Range $=(-\infty,-\infty)=\mathbb{R}$
It is differentiable also,
$f^{\prime}(t)=\left\{\begin{array}{l}
\frac{1}{(1+t)^2},-1 < t < 0 \\
\frac{1}{(1-t)^2}, 0 < t < 1
\end{array}\right.$
$\because f^{\prime}(t) > 0 \forall-1 < t < 1$
$\therefore$ f is injective

Topic: Mathematics, Functions, WBJEE, WBJEE 2022
$\int \frac{\mathrm{f}(\mathrm{x}) \varphi^{\prime}(\mathrm{x})+\varphi(\mathrm{x}) \mathrm{f}^{\prime}(\mathrm{x})}{(\mathrm{f}(\mathrm{x}) \varphi(\mathrm{x})+1) \sqrt{\mathrm{f}(\mathrm{x}) \varphi(\mathrm{x})-1}} \mathrm{dx}=$

Options:
A $\sin ^{-1} \sqrt{\frac{\mathrm{f}(\mathrm{x})}{\varphi(\mathrm{x})}}+\mathrm{c}$
B $\cos ^{-1} \sqrt{(f(x))^{2}-(\varphi(x))^{2}}+c$
C $\sqrt{2} \tan ^{-1} \sqrt{\frac{f(x) \varphi(x)-1}{2}}+c$
D $\sqrt{2} \tan ^{-1} \sqrt{\frac{f(x) \varphi(x)+1}{2}}+c$

The correct answer is:
$\sqrt{2} \tan ^{-1} \sqrt{\frac{f(x) \varphi(x)-1}{2}}+c$

Solution:
Hint:
Let $f(x) \phi(x)=t$
$\int \frac{d t}{(t+1) \sqrt{t-1}}$
Let $t-1=p^{2}, d t=2 p d p$
$\Rightarrow \int \frac{2 d p}{p^{2}+2}=\sqrt{2} \tan ^{-1}\left|\sqrt{\frac{f(x) \phi(x)-1}{2}}\right|+c$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2020
$f(x)=x+|x|$ is continuous for

Options:
A $\quad x \in(-\infty, \infty)$
B $x \in(-\infty, \infty)-\{0\}$
C only $x>0$
D no value of $x$

The correct answer is:
$\quad x \in(-\infty, \infty)$

Solution:
$$
\text { Hints: } f(x)=\left\{\begin{array}{lc}
2 x ; & x \geq 0 \\
0 ; & x < 0
\end{array}\right.
$$

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2009
Ferric ion forms a prussian blue precipitate due to the formation of

Options:
A $\mathrm{K}_{4}\left[\mathrm{Fe}\left(\mathrm{CN})_{6}\right]\right.$
B $\mathrm{K}_{3}\left[\mathrm{Fe}\left(\mathrm{CN})_{6}\right)\right.$
C $Fe(CNS)_{3}$
D $Fe_{4}[Fe(CN)_{6}]_{3}$

The correct answer is:
$Fe_{4}[Fe(CN)_{6}]_{3}$

Solution:
Ferric ion forms a prussian blue precipitate due to the formation of $\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$. This complex is formed during the determine action of presence of nitrogen in the given sample. In this method, to portion of sodium fusion extract. freshly prepared ferrous sulphate, FeSO $_{4}$ solution is added and warmed. Then about 2 to 3 drops of $\mathrm{FeCl}_{3}$ solution are added and acidified with conc. HCl. The appearance of a prussian blue colour indicate the presence of nitrogen.
$$
\mathrm{FeSO}_{4}+2 \mathrm{NaOH} \longrightarrow \mathrm{Fe}(\mathrm{OH})_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4}
$$
$6 \mathrm{NaCN}+\mathrm{Fe}(\mathrm{OH})_{2} \longrightarrow \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+2 \mathrm{NaOH}$
$$
\begin{array}{r}
3 \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+4 \mathrm{FeCl}_{3} \longrightarrow \\
\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}+12 \mathrm{NaCl}
\end{array}
$$

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2018
Figure below shows the distance-time graph of the motion of a car. If follows from the graph that the car is

Options:
A at rest
B in uniform motion
C in non-uniform acceleration
D uniformly accelerated

The correct answer is:
uniformly accelerated

Solution:
Slope is increasing with constant rate. i.e motion is uniformaly accelerated
$\mathrm{x}=1.2 \mathrm{t}^2 \Rightarrow \mathrm{v}=2.4 \mathrm{t} \Rightarrow \mathrm{a}=2.4 \mathrm{~m} / \mathrm{s}^2$

Topic: Physics, Motion In One Dimension, WBJEE, WBJEE 2009
Find the correct statement(s) about photoelectric effect.

Options:
A There is no significant Ime delay between the absorption of a suitable tadiation and the emission of electrons
B Einstein analysis gives a threshold frequency above which no electron can be emitted
C The maximum kinetic energy of the emitted photoelectrons is proportional to the frequency of incident radiation
D The maximum kinetic energy of electrons does not depend on the intensity of radiation

The correct answers are:
There is no significant Ime delay between the absorption of a suitable tadiation and the emission of electrons, The maximum kinetic energy of electrons does not depend on the intensity of radiation

Solution:
There is no significant time delay between the absorption of a suitable radiation and the emission of electrons and the maximum kinetic energy of electrons does not depend on the intensity of radiation.
So, (a) and (d) are correct options.

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2014
Find the equivalent capacitance between $A$ and $B$ of the following arrangement :

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

The correct answer is:
3C

Solution:
Cnet. $=C+C+C=3 C$

Topic: Physics, Capacitance, WBJEE, WBJEE 2022
Find the maximum value of $|z|$ when $\left|z-\frac{3}{z}\right|=2,$ where $z$ being a complex number.

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

The correct answer is:
3

Solution:
We have, $|z|=\left|z-\frac{3}{z}+\frac{3}{z}\right|$
$\Rightarrow \quad|z| \leq\left|z-\frac{3}{z}\right|+\left|\frac{3}{z}\right|$
[using triangle inequality] $\Rightarrow \quad|z| \leq 2+\left|\frac{3}{z}\right|$
$\left[\because\left|z-\frac{3}{z}\right|=2\right]$
$\Rightarrow \quad|z| \leq 2+\frac{3}{|z|}$
$\begin{array}{lr}\Rightarrow & |z|^{2}-2|z|-3 \leq 0 \\ \Rightarrow & (|z|-3)(|z|+1) \leq 0\end{array}$
$\Rightarrow \quad-1 \leq|z| \leq 3$
Thus, the maximum value of $|z|$ is 3 .

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2015
Find the right condition(s) for Fraunhoffer difiraction due to a single slit.

Options:
A Source is at infinite distance and the incident beam has converged at the slit
B Source is near to the slit and the incidont beam is parallel
C Source is at infinity and the incident beam is parallel
D Source is near to the slit and the incident beam has converged at the slit

The correct answers are:
Source is near to the slit and the incidont beam is parallel, Source is at infinity and the incident beam is parallel

Solution:
For Fraunhofer diffraction, incident beam should
be parallel. This can be achieved when
(a) Source is placed at focus of converging lens.
(b) Source is at infinite distance.

Topic: Physics, Wave Optics, WBJEE, WBJEE 2015
Five equal resistance, each of resistance \(\mathrm{R}\), are connected as shown in figure below. A battery of \(\mathrm{V}\) volt is connected between \(A\) and \(B\). The current flowing in \(\mathrm{FC}\) will be

Options:
A \(\frac{3 \mathrm{~V}}{\mathrm{R}}\)
B \(\frac{V}{R}\)
C \(\frac{\mathrm{V}}{2 \mathrm{R}}\)
D \(\frac{2 \mathrm{~V}}{\mathrm{R}}\)

The correct answer is:
\(\frac{\mathrm{V}}{2 \mathrm{R}}\)

Solution:
Hints:
\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}} \therefore \text { Current in } \mathrm{FC}=\frac{\mathrm{I}}{2}=\frac{\mathrm{V}}{2 \mathrm{R}}\)

Topic: Physics, Current Electricity, WBJEE, WBJEE 2011
Five identical capacitors, of capacitance $20 \mu \mathrm{F}$ each, are connected to a battery of $150 \mathrm{V}$, in a combination as shown in the diagram. What is the total amount of charge stored?

Options:
A $15 \times 10^{-3} \mathrm{C}$
B $12 \times 10^{-3} \mathrm{C}$
C $10 \times 10^{-3} \mathrm{C}$
D $3 \times 10^{-3} \mathrm{C}$

The correct answer is:
$3 \times 10^{-3} \mathrm{C}$

Solution:
The given circuit shows a balanced Wheatstone bridge. Now, the circuit becomes

In the branch $A B,$ both capacitor are arranged in the series combination. Hence, its cquivalent capacitance is given by
$$
\begin{array}{l}
\frac{1}{C_{A B}}=\frac{1}{20}+\frac{1}{20}=\frac{2}{20}=\frac{1}{10} \\
C_{A B}=10 \mu \mathrm{F}
\end{array}
$$
Similarly, in branch $\mathrm{CD}$,
$$
C_{C D}=10 \mu \mathrm{F}
$$
Now, $C_{A B}$ and $C_{C D}$ are connected in parallel combination. Hence, the equivalent capacitance of the circuit,
$$
C_{\mathrm{eq}}=10+10=20 \mu \mathrm{F}
$$
We know that, charge $Q=C_{\mathrm{eq}} V$
$$
Q=20 \times 10^{-6} \times 150 \mathrm{V}
$$
Amount of charge stored $Q=3 \times 10^{-3} C$

Topic: Physics, Capacitance, WBJEE, WBJEE 2019
Five letter words, having distinct letters, are to be constructed using the letters of the word 'EQUATION' so that each word contains exactly three vowels and two consonants. How many of them have all the vowels together?

Options:
A 3600
B 1800
C 1080
D 900

The correct answer is:
1080

Solution:
Vowels = E, U, I, O, A
Consonants = Q, T, N

$=10 \times 3 \times 6 \times 6=1080$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2021
Five numbers are in AP with common difference $\neq 0 .$ If the 1 st, 3 rd and 4 th terms are in GP, then

Options:
A the 5th term is always $0 .$
B the 1 st term is always $0 .$
C the middle term is always $0 .$
D the middle term is always -2.

The correct answer is:
the 5th term is always $0 .$

Solution:
Let the tine number is in AP are
$$
(a-2 d),(a-d), a,(a+d),(a+2 d)
$$
where, $d \neq 0$
Given, 1 st, $3 r d$ and 4 th terms are in GP.
$$
\begin{array}{lr}
\Rightarrow \quad a^{2}=(a-2 d)(a+d) \\
\Rightarrow \quad a^{2}=a^{2}-2 a d+a d-2 d^{2} \\
\Rightarrow \quad 2 d^{2}+a d=0 \Rightarrow d(2 d+a)=0 \\
\because \quad d \neq 0
\end{array}
$$
$\therefore \quad a+2 d=0 \Rightarrow a=-2 d$
Hence, terms are
$$
-4 d,-3 d,-2 d,-d, 0
$$
$\therefore$ The fifth term is always $0 .$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2013
Five numbers are in HP. The middle term is 1 and the ratio of the second and the fourth terms is 2: 1 . Then, the sum of the first three terms is

Options:
A $\frac{11}{2}$
B 5
C 2
D $\frac{14}{3}$

The correct answer is:
$\frac{11}{2}$

Solution:
Let $a-2 d, a-d, a, a+d, a+2 d$ are in $A P$. then $\frac{1}{a-2 d} \cdot \frac{1}{a-d} \cdot \frac{1}{a} \cdot \frac{1}{a+d} \cdot \frac{1}{a+2 d}$ are in $\mathrm{HP}$.
Given, middle term $=1$ $\Rightarrow \quad \frac{1}{a}=1 \Rightarrow a=1$
and $\quad \frac{\text { Second term }}{\text { Fourth term }}=\frac{2}{1}$
$\Rightarrow \quad \frac{1}{a-d} \times a+d=\frac{2}{1}$
$\Rightarrow \quad a=3 d \Rightarrow d=\frac{1}{3} \quad(\because a=1)$
$\therefore$ Sum of first three terms $=\frac{1}{1-\frac{2}{3}}+\frac{1}{1-\frac{1}{3}}+1=3+\frac{3}{2}+1=\frac{11}{2}$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2013
For $0 \leq p \leq 1$ and for any positive $a, b$, let $I(p)=(a+b)^{p}, J(p)=a^{P}+b^{P},$ then

Options:
A $l(p)>J(p)$
B $l(p) \leq J(p)$
C $l(p) < J{(} p)$ in $\left[0 ,\frac{p}{2}\right]$ and $l(p)>J(p)$ in $\left[\frac{p}{2}, \infty\right)$
D $l(p) < J(p)$ in $\left[\frac{p}{2}, \infty\right)$ and $J(p) < l(p)$ in $\left[0, \frac{p}{2}\right]$

The correct answer is:
$l(p) \leq J(p)$

Solution:
Here, let $p=\frac{1}{m}$
then $\left(a^{p}+b^{p}\right)^{1 / p}=\left(a^{1 / m}+b^{1 / m}\right)^{m}$
$=a+b+k, k \geq 0$
$\therefore \quad a^{p}+b^{p} \geq(a+b)^{p} \geq(a+b)$
$\Rightarrow \quad J(p) \geq I(p)$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2018
For $0 \leq P, Q \leq \frac{\pi}{2},$ if $\sin P+\cos Q=2$, then the
value of $\tan \left(\frac{\vec{P}+Q}{2}\right)$ is equal to

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

The correct answer is:
1

Solution:
Given, $0 \leq P, Q \leq \frac{\pi}{2}$
and $\sin P+\cos Q=2$
This equation hold only when, $\quad P=\frac{\pi}{2}$
and $\quad Q=0$
$\mathrm{LHS}=\sin P+\cos Q=\sin \frac{\pi}{2}+\cos 0$
$$
=1+1=2=\mathrm{RHS}
$$
$\therefore \quad \tan \left(\frac{P+Q}{2}\right)=\tan \left(\frac{\frac{\pi}{2}+0}{2}\right)=\tan \frac{\pi}{4}=1$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2013
For $-\frac{\pi}{2} < x < \frac{3 \pi}{2}, \quad$ the $\quad$ value $\quad$ of $\frac{d}{d x}\left\{\tan ^{-1} \frac{\cos x}{1+\sin x}\right\}$ is equal to

Options:
A $\frac{1}{2}$
B $-\frac{1}{2}$
C 1
D $\frac{\sin x}{(1+\sin x)^{2}}$

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

Solution:
$\frac{d}{d x}\left\{\tan ^{-1} \frac{\cos x}{1+\sin x}\right\}$
$=\frac{1}{1+\left(\frac{\cos x}{1+\sin x}\right)^{2}} \frac{d}{d x} \frac{\cos x}{1+\sin x}$
$=\frac{(1+\sin x)^{2}}{1+\sin ^{2} x+2 \sin x+\cos ^{2} x}$
$=\frac{(1+\sin x)(-\sin x)-\cos (\cos x)}{(1+\sin x)^{2}}$
$=\frac{1}{2+2 \sin x} \times\left[-\left(\sin x+\sin ^{2} x+\cos ^{2} x\right)\right]$
$=\frac{1}{2(1+\sin x)} \times-(1+\sin x)=-\frac{1}{2}$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2012
For a chemical reaction at $27^{\circ} \mathrm{C},$ the activation energy is $600 \mathrm{R}$. The ratio of the rate constants at $327^{\circ} \mathrm{C}$ to that of at $27^{\circ} \mathrm{C}$
will be

Options:
A 2
B 40
C $e$
D $e^{2}$

The correct answer is:
$e$

Solution:
From Arrhenius equation
$$
\begin{array}{l}
\operatorname{In} \frac{k_{2}}{k_{1}}=\frac{E_{a}}{R}\left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right] \\
\operatorname{In} \frac{k_{2}}{k_{1}}=\frac{600 \mathrm{R}}{R}\left(\frac{1}{300}-\frac{1}{600}\right)
\end{array}
$$
or $\ln \frac{k_{2}}{k_{1}}=600\left(\frac{2-1}{600}\right)$
of $\ln \frac{k_{2}}{k_{1}}=1$ or $\frac{k_{2}}{k_{1}}=e$

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2013
For a plane eletromagnetic wave, the electric field is given by $\overrightarrow{\mathrm{E}}=90 \sin \left(0.5 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \mathrm{k} . \mathrm{v} / \mathrm{m} .$ The corresponding magnetic field $\vec{B}$ will be

Options:
A $\vec{B}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \hat{i} \mathrm{~T}$
B $\vec{B}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \hat{j} \mathrm{~T}$
C $\vec{B}=27 \times 10^{9} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \hat{j} \mathrm{~T}$
D $\overrightarrow{\mathrm{B}}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \hat{\mathrm{k}} \mathrm{T}$

The correct answer is:
$\vec{B}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \hat{j} \mathrm{~T}$

Solution:
$E=90 \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} t\right) \hat{k} v / m$
Direction of propagation is $-\hat{i}$
$\& E_{0}=90$
$\frac{\mathrm{E}_{0}}{\mathrm{~B}_{0}}=\mathrm{C} \Rightarrow \mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{C}}=\frac{90}{3 \times 10^{8}} \mathrm{~T}=3 \times 10^{-7} \mathrm{~T}$
$\overrightarrow{\mathrm{E}}$ is along $\hat{\mathrm{k}} \& \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$ should be along $-\hat{\mathrm{i}} \Rightarrow \overrightarrow{\mathrm{B}}$ is should be along $\hat{\mathrm{j}}$ Hence equation of magnetic field is
$\mathrm{B}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \hat{\mathrm{j}} \mathrm{T}$

Topic: Physics, Electromagnetic Waves, WBJEE, WBJEE 2021
For a reaction $2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{P}$, when concentration of $\mathrm{B}$ alone is doubled, rate does not change and when concentrations of both $\mathrm{A}$ and $\mathrm{B}$ is doubled, rate increases by a factor of 4 . The unit of rate constant is,

Options:
A $\mathrm{s}^{-1}$
B $\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}$
C $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$
D $\mathrm{L}^{2} \mathrm{~mol}^{-2} \mathrm{~s}^{-1}$

The correct answer is:
$\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}$

Solution:
Hint :
Rate law: $\mathrm{r}=\mathrm{k}[\mathrm{A}]^{\alpha}[\mathrm{B}]^{\beta} \quad \mathrm{B}$ follows first order kinetics. $\therefore \beta=1$
$\frac{r_{2}}{r_{1}}=\frac{[2 A]^{\alpha}[2 B]^{p}}{[A]^{\alpha}[B]^{\beta}}$
or, $4=2^{a+\beta}$
or, $\alpha+\beta=2$
or, $\alpha=1$
$\therefore$ overall order of reaction $=2$
Hence unit of rate constant $=\mathrm{L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{S}^{-1}$

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2020
For a reversible chemical reaction where the forward process is exothermic, which of the following statements is correct?

Options:
A The backward reaction has higher activation energy than the forward reaction
B The backward and the forward processes have the same activation energy
C The backward reaction has lower activation energy
D No activation anergy is required at all since energy is liberated in the process.

The correct answer is:
The backward reaction has higher activation energy than the forward reaction

Solution:
Hints:

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2010
For a spontaneous process, the correct statement(s) is (are)

Options:
A $\left(\Delta G_{\text {system }}\right)_{T, p}>0$
B $\left(\Delta S_{\text {system }}\right)+\left(\Delta S_{\text {surroundings }}\right)>0$
C $\left(\Delta G_{\text {system }}\right)_{T, p} < 0$
D $\left(\Delta U_{\text {system }}\right)_{T, v}>0$

The correct answers are:
$\left(\Delta S_{\text {system }}\right)+\left(\Delta S_{\text {surroundings }}\right)>0$, $\left(\Delta G_{\text {system }}\right)_{T, p} < 0$

Solution:
For a process to be spontaneous.
$\begin{aligned} \Delta S_{\text {system }}+\Delta S_{\text {surrounding }} &>0 \\ &\left(\Delta G_{\text {system }}\right)_{T, p} < 0 \\ &\left(\Delta U_{\text {system }}\right)_{T, V} < 0 \end{aligned}$.

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2014
For a spontaneous reaction at all temperatures which of the following is correct?

Options:
A Both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are positive
B $\Delta \mathrm{H}$ is positive and $\Delta \mathrm{S}$ is negative
C $\Delta \mathrm{H}$ is negative and $\Delta \mathrm{S}$ is positive
D Both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ are negative

The correct answer is:
$\Delta \mathrm{H}$ is negative and $\Delta \mathrm{S}$ is positive

Solution:
For spontancity $\Delta G$ must be negative $\because \Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$
To be negative at all Temperature, $\Delta \mathrm{H} < 0, \Delta \mathrm{S}>0$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2021
For a van der Waals' gas, the term $\left(\frac{a b}{V^{2}}\right)$ represents some

Options:
A pressure
B energy
C critical density
D molar mass

The correct answer is:
energy

Solution:
Dimensions of $\left(\frac{a b}{V^{2}}\right)$ are shown below.
$$
\begin{aligned}
\frac{a}{V^{2}} \times b &=\text { pressure } \times \text { volume } \\
&=\frac{\text { force }}{\text { area }} \times \text { volume } \\
&=\frac{\mathrm{MLT}^{-2}}{\mathrm{L}^{2}} \times \mathrm{L}^{3}=\mathrm{ML}^{2} \mathrm{T}^{-2}
\end{aligned}
$$
The units of energy (joule) is,
$$
1 \mathrm{J}=1 \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-2}=\mathrm{ML}^{2} \mathrm{T}^{-2}
$$
Therefore, the dimensions of $\left(\frac{a b}{V^{2}}\right)$ is same as that of energy. So, the option (b) is correct.

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2019
For air at room temperature, the atmospheric pressure is $1.0 \times 10^{5} \mathrm{Nm}^{-2}$ and density of ait is $1.2 \mathrm{kgm}^{-3}$. For a tube of length $1.0 \mathrm{m}$, closed at one end, the lowest frequency $\begin{array}{lllllll}\text { generated is } & 84 & \text { Hz. } & \text { The } & \text { value } & \text { of } \boldsymbol{\gamma}\end{array}$ (ratio of two specific heats) for air is

Options:
A 2.1
B 1.5
C 1.8
D 1.4

The correct answer is:
1.4

Solution:
The tube is closed at one end. So, it will act like a closed organ pipe. The lowest frequency produced in the tube

$$
\begin{array}{l}
l=\frac{\lambda}{4}, \lambda=4 \\
l=1 \mathrm{m} \\
f=84 \mathrm{Hz} \\
V=n \lambda=84 \times 4
\end{array}
$$
When wave propagate in air, the velocity
Here,
$$
\begin{array}{l}
v=\sqrt{\frac{\gamma p}{\rho}} \\
p=1.0 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}
\end{array}
$$
$$
\begin{aligned}
\rho &=1.2 \mathrm{kg} / \mathrm{m}^{3} \\
V^{2} &=\frac{\gamma P}{\rho} \\
\gamma &=\frac{\mathrm{v}^{2} \cdot \rho}{\mathrm{P}} \\
&=\frac{(84 \times 4)^{2} \times 1.2}{1.0 \times 10^{5}} \\
&=\frac{135472.2}{10^{5}}=1.354 \\
&=1.4
\end{aligned}
$$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2016
For all real values of $a_{0}, a_{1}, a_{2}, a_{3}$ satisfying $a_{0}+\frac{a_{1}}{2}+\frac{a_{z}}{3}+\frac{a_{3}}{4}=0,$ the equation $a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}=0$ has a real root in the interval

Options:
A [0,1]
B [-1,0]
C [1,2]
D [-2,-1]

The correct answer is:
[0,1]

Solution:
Let $f(x)=\frac{a_{3} x^{4}}{4}+\frac{a_{2} x^{3}}{3}+\frac{a_{1} x^{2}}{2}+a_{0} x$
$f(0)=0, f(1)=\frac{a_{3}}{4}+\frac{a_{2}}{3}+\frac{a_{1}}{2}+a_{0}=0$
$\Rightarrow f(0)=f(1)$
$\Rightarrow f^{\prime}(x)=0$ has atleast one real root in |0,1| $\therefore f^{\prime}(x)=a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$
Hence, $a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$ must has a real root in the interval $[0,1 \mid$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2015