datavorous/entrance-exam-dataset huggingface dataset:
Nitrogen dioxide is not produced on heating

Options:
A $\mathrm{KNO}_{3}$
B $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$
C $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$
D $\mathrm{AgNO}_{3}$

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

Solution:
Only KNO $_{3}$ on heating do not give $\mathrm{NO}_{2}$, but all other give
(a) $2 \mathrm{KNO}_{3} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{KNO}_{2}+\mathrm{O}_{2}$
(b) $2 \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} \stackrel{\mathrm{A}}{\longrightarrow} 2 \mathrm{P} \mathrm{bO}+4 \mathrm{NO}_{2}+\mathrm{O}_{2}$
(c) $2 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} \stackrel{\mathrm{A}}{\longrightarrow} 2 \mathrm{CuO}+4 \mathrm{NO}_{2}+\mathrm{O}_{2}$
(d) $2 \mathrm{AgNO}_{3} \frac{\Delta}{\text { Aed het }} 2 \mathrm{Ag}+\mathrm{O}_{2}+2 \mathrm{NO}_{2}$
Heavy metal nitrates liberate nitrogen dioxide on heating.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2016
Number of common tangents of $y=x^{2}$ and $y=-x^{2}+4 x-4$ is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
We have, equation of parabola $y=x^{2}$ Let $P\left(\alpha, \alpha^{2}\right)$ is a point on the parabola,
$\therefore$
$y-\alpha^{2}=2 \alpha(x-\alpha)$
$\because$
$\left[\because \frac{d y}{d x}=2 x \Rightarrow \frac{d y}{d x\left(\alpha, a^{2}\right)}=2 \alpha\right]$
$\Rightarrow \quad y=2 \alpha x-\alpha^{2}$
Also, given $y=-x^{2}+4 x-4$
$\therefore-x^{2}+4 x-4=2 a x-a^{2}$
$\Rightarrow \quad x^{2}+2 x(\alpha-2)+\left(4-a^{2}\right)=0$
Discriminant $=0$
$4(\alpha-2)^{2}-4\left(4-a^{2}\right)=0$
$\Rightarrow$
$(\alpha-2)^{2}-\left(4-\alpha^{2}\right)=0$
$\Rightarrow \quad \alpha^{2}-4 \alpha+4-4+\alpha^{2}=0$
$\alpha^{2}-2 \alpha=0$
$\alpha=0, \alpha=2$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2018
Number of hydrogen ions present in 10 milionth part of $1.33 \mathrm{cm}^{3}$ of pure water at $25^{\circ} \mathrm{C}$ is

Options:
A 6.023 million
B 60 million
C 8.01 million
D 80.23 million

The correct answer is:
8.01 million

Solution:
10 million $=10^{-7}$
$\therefore$ 10 $million^{th}$ part of $1.33 \mathrm{cm}^{3}=1.33 \times 10^{-7} \mathrm{cm}^{3}$ $=1.33 \times 10^{-7} \mathrm{mL}$
For pure water,
$$
\begin{aligned}
\left[\mathrm{H}^{+}\right] &=10^{-7} \mathrm{mol} / \mathrm{L}
\end{aligned}
$$
$\Rightarrow 1$ L water contains $\left[\mathrm{H}^{+}\right]=10^{-7} \mathrm{mol}$
or $1 \mathrm{mL}$ water contains $\left[\mathrm{H}^{+}\right]=\frac{10^{-7}}{1000}=10^{-10} \mathrm{mol}$
or 10 million $^{\text {th }}$ part of $1.33 \mathrm{cm}^{3}$ water contains $\begin{aligned}\left(\mathrm{H}^{+}\right] &=1.33 \times 10^{-7} \times 10^{-10} \mathrm{mol} \\ &=1.33 \times 10^{-17} \mathrm{mol} \end{aligned}$
$\therefore$ Number of $\mathrm{H}^{+}$ ions $=1.33 \times 10^{-17} \times N_{A}$
$=1.33 \times 10^{-17} \times 6.022 \times 10^{23}$
$=8.009 \times 10^{6}=8.01$ million

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2013
Number of intersecting points of the conics $4 x^{2}+9 y^{2}=1$ and $4 x^{2}+y^{2}=4$ is

Options:
A 1
B 2
C 3
D 0

The correct answer is:
0

Solution:
For curve I.
$4 x^{2}+9 y^{2}=1 \Rightarrow \frac{x^{2}}{\left(\frac{1}{2}\right)^{2}}+\frac{y^{2}}{\left(\frac{1}{3}\right)^{2}}=1$
which is an equation of ellipee with $a=\frac{1}{2}$ and $b=\frac{1}{3}$
For curve $I I, 4 x^{2}+y^{2}=4$ $\Rightarrow \quad \frac{x^{2}}{1}+\frac{y^{2}}{4}=1$
$\Rightarrow \quad \frac{x^{2}}{(1)^{2}}+\frac{y^{2}}{(2)^{2}}=1$
which is also an equation of ellipse with $a=1$ and $b=2$
Hence, there is no intersecting point.

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2015
Number of points having distance $\sqrt{5}$ fron the straight line $x-2 y+1=0$ and a distance $\sqrt{13}$ from the line $2 x+3 y-1=0,$ is

Options:
A 1
B 2
C 4
D 5

The correct answer is:
4

Solution:
Let the point be $(h, k)$
Then, $\left|\frac{h-2 k+1}{\sqrt{1^{2}+(-2)^{2}}}\right|=\sqrt{5}$
$h-2 k+1=\pm 5$
Also, $\left|\frac{2 h+3 k-1}{\sqrt{2^{2}+3^{2}}}\right|=$
$\Rightarrow \quad 2 h+3 k-1=\pm 13$
On solving Eqs
(i) and (ii), we get four points So. there are much 4 points.

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2015
Number of solutions of the equation $\tan x+\sec x=2 \cos x, x \in[0, \pi]$ is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
$\tan x+\sec x=2 \cos x$
$\Rightarrow \quad \frac{\sin x}{\cos x}+\frac{1}{\cos x}=2 \cos x$
$\Rightarrow \quad \sin x+1=2 \cos ^{2} x=2\left(1-\sin ^{2} x\right)$
$=2-2 \sin ^{2} x$
$\Rightarrow \quad 2 \sin ^{2} x+\sin x-1=0$
$\Rightarrow \quad 2 \sin ^{2} x+2 \sin x-\sin x-1=0$
$\Rightarrow \quad 2 \sin x(\sin x+2)-1(\sin x+1)=0$
$\Rightarrow \quad(\sin x+2)(2 \sin x-1)=0$
$\Rightarrow \quad \sin x=-2,$ which is not possible. or $\sin x=\frac{1}{2} \Rightarrow x=\frac{\pi}{6}, \frac{5 \pi}{6}$
$x \in[0, \pi]$
$\therefore$ number of solution $=2$.

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2012
Of the following acids the one which has the capability to form complex compound and also possesses oxidizing and reducing properties is :

Options:
A $\mathrm{HNO}_3$
B $\mathrm{HNO}_2$
C $\mathrm{HCOOH}$
D HCN

The correct answer is:
$\mathrm{HNO}_2$

Solution:
$\mathrm{HNO}_2$
Hints : Here oxidation state of $\mathrm{N}$ lies between -3 to +5

Topic: Chemistry, Redox Reactions, WBJEE, WBJEE 2009
Of the following compounds the most acidic is

Options:
A $\mathrm{As}_2 \mathrm{O}_3$
B $\mathrm{P}_2 \mathrm{O}_5$
C $\mathrm{Sb}_2 \mathrm{O}_3$
D $\mathrm{Bi}_2 \mathrm{O}_3$

The correct answer is:
$\mathrm{P}_2 \mathrm{O}_5$

Solution:
In a group as we go downwards, the oxide basic character increases hence maximum acidic oxide is $\mathrm{P}_2 \mathrm{O}_5$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2009
Of the following compounds, which one is the strongest Bronsted acid in a aqueous solution?

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

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

Solution:
$\because$ Higher be the oxidation number of central atom in oxo-acids, more strongly it behave as Bronsted-acid. Oxidation number for $\mathrm{Cl}$ in $\mathrm{HClO}_{3} \rightarrow+5$
Oxidation number for $\mathrm{Cl}$ in $\mathrm{HClO}_{2} \rightarrow+3$
Oxidation number for $\mathrm{Cl}$ in $\mathrm{HClO} \rightarrow+1$
Oxidation number for $\mathrm{Cl}$ in $\mathrm{HBrO} \rightarrow+1$
Hence, $\mathrm{HClO}_{3}$, is the strongest Bronsted acid in aqueous solution.

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2017
Of the following molecules, which have shape similar to $\mathrm{CO}_{2} ?$

Options:
A $\mathrm{HgCl}_{2}$
B $\mathrm{SnCl}_{2}$
C $\mathrm{C}_{2} \mathrm{H}_{2}$
D $\mathrm{NO}_{2}$

The correct answers are:
$\mathrm{HgCl}_{2}$, $\mathrm{C}_{2} \mathrm{H}_{2}$

Solution:
$\mathrm{HgCl}_{2}$ and $\mathrm{C}_{2} \mathrm{H}_{2}$ both show linear shape as of $\mathrm{CO}_{2}$.
($\because$ both have zero lone pair of electrons)

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2016
On heating, chloric acid decomposes to

Options:
A $\mathrm{HClO}_{4} . \mathrm{Cl}_{2}, \mathrm{O}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$
B $\mathrm{HClO}_{2}$, $\mathrm{Cl}_{2}, \mathrm{O}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$
C $\mathrm{HClO}, \mathrm{Cl}_{2} \mathrm{O}$ and $\mathrm{H}_{2} \mathrm{O}_{2}$
D $\mathrm{HCl}, \mathrm{HClO}, \mathrm{Cl}_{2} \mathrm{O},$ and $\mathrm{H}_{2} \mathrm{O}$

The correct answer is:
$\mathrm{HClO}_{4} . \mathrm{Cl}_{2}, \mathrm{O}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$

Solution:
$3 \mathrm{HClO}_{3} \stackrel{\mathrm{\Delta}}{\longrightarrow} \mathrm{HClO}_{4}+\mathrm{Cl}_{2}+2 \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2013
On mixing an alkane with chlorine and irradiating with ultra-violet light, it forms only one mono-chloro-alkane. The alkane is

Options:
A Propane
B Pentane
C Isopentane
D Neopentane

The correct answer is:
Neopentane

Solution:
Hints : Neopentane
contains all hydrogen atom equivalent

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2010
On passing 'C ampere of current for time 't' sec through 1 L of 2 (M) $\mathrm{CuSO}_{4}$ solution (atomic weight of $\mathrm{Cu}=63.5$ ), the amount ${ }^{\prime} \mathrm{m}^{\prime}$ of Cu (in gram) deposited on cathode will be

Options:
A $m=C t /(63.5 \times 96500)$
B $m=C t /(31.25 \times 96500)$
C $m=(C \times 96500) /(31.25 \times t)$
D $m=(31.25 \times C \times t) / 96500$

The correct answer is:
$m=(31.25 \times C \times t) / 96500$

Solution:
$m=\frac{E G}{F}$ $=\frac{\frac{63.5}{2} \times C \times t}{96500}=\frac{31.75 \times C \times t}{96500}$

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2012
On $R$, a relation $\rho$ is defined by xpy if and only if $x-y$ is zero or irrational. Then.

Options:
A $\rho$ is equivalence relation
B $\rho$ is reflexive but neither symmetric nor transitive
C $\rho$ is reflexive and symmetric but not transitive
D $\rho$ is symmetric and transitive but not reflexive

The correct answer is:
$\rho$ is reflexive and symmetric but not transitive

Solution:
On the set $R$ $x p y \Rightarrow x-y$ is zero or irrational number.
Now, $x p x$
$\Rightarrow x-x=0$
$\Rightarrow \rho$ is reflexive.
If $x \rho y \Rightarrow x-y$ is zero or irrational.
$=-(y-x)$ is zero or irrational.
$\Rightarrow$ yox is zero or irrational.
$\Rightarrow \rho$ is symmetric.
And if $x \rho y \Rightarrow x-y$ is 0 or irrational.
$y \rho z \Rightarrow y-z$ is 0 or irrational.
Then, $(x-y)+(y-z)=x-z$ may be 0 or rational.
$\Rightarrow \rho$ is not transitive.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2018
On $R$, the relation $\rho$ be defined by 'xpy holds if and only if $x-y$ is zero or irrational'. Then,

Options:
A $\rho$ is reflexive and transitive but not symmetric
B $\rho$ is reflexive and symmetric but not transitive
C $\rho$ is symmetric and transitive but not reflexive
D $\rho$ is equivalence relation

The correct answer is:
$\rho$ is reflexive and symmetric but not transitive

Solution:
We have, $x p y \Rightarrow x-y$ is zero or irrational. Now, $x-x=0$
$\Rightarrow(x, x) \in \rho$
$\therefore \rho$ is reflexive.
Again, if $x-y$ is either zero or irrational, then $y-x$ will also be either zero or inational.
$(x, y) \in \rho$
$\Leftrightarrow \quad(y, x) \in \rho$
$\therefore \rho$ is symmetric.
Again, $(2, \sqrt{3}) \in \rho$ and $(\sqrt{3}, 4) \in \rho$
$But (2,4) \notin \rho$
$\therefore p$ is not transitive.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2017
On $R$, the set of real numbers, a relation $\rho$ is defined as $a \rho b$ if and only if $1+a b>0$. Then,

Options:
A $\rho$ is an equivalence relation
B $\rho$ is reflexive and transitive but not symmetric
C $\rho$ is reflexive and symmetric but not transitive
D $\rho$ is only symmetric

The correct answers are:
$\rho$ is reflexive and symmetric but not transitive

Solution:
We know that,
$1+a^{2}>0, a \in R$
$\Rightarrow \quad(a, a) \in \rho$
$\therefore \rho$ is reflexive
Again, $\quad$ let $(a, b) \in \rho$
$\Rightarrow \quad 1+a b>0$
$\Rightarrow \quad 1+b a>0$
$\Rightarrow \quad(b, a) \in \rho$
$\therefore \rho$ is symmetric Now, $\quad(1,-0 \cdot 1) \in \rho$
and $(-0 \cdot 1,-9) \in \rho$
but $\quad(1,-9) \notin \rho$
$\therefore \rho$ is not transitive.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2017
On set $A=\{1,2,3\},$ relations $R$ and $S$ are given by $R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$
$S=\{(1,1),(2,2),(3,3),(1,3),(3,1)\}$
Then,

Options:
A $R \cup S$ is an equivalence relation
B $R \cup S$ is reflexive and transitive but not symmetric
C $R \cup S$ is reflexive and symmetric but not transitive
D $R \cup S$ is symmetric and transitive but not reflexive

The correct answer is:
$R \cup S$ is reflexive and symmetric but not transitive

Solution:
We have, $R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$
$S=\{(1,1),(2,2),(3,3),(1,3),(3,1)\}$
$\therefore R \cup S=\{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)\}$
Since, $(2,1) \in R \cup S,(1,3) \in R \cup S$
but $(2,3) \in R \cup S$
$\therefore \mathrm{R} \cup \mathrm{S}$ is reflexive and symmetric but not transitive.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2017
On the ellipse $4 x^{2}+9 y^{2}=1$, the points at which the tangents are parallel to the line $8 x=9 y,$ are

Options:
A $\left(\frac{2}{5}, \frac{1}{5}\right)$
B $\left(-\frac{2}{5}, \frac{1}{5}\right)$
C $\left(-\frac{2}{5},-\frac{1}{5}\right)$
D $\left(\frac{2}{5},-\frac{1}{5}\right)$

The correct answers are:
$\left(-\frac{2}{5}, \frac{1}{5}\right)$, $\left(\frac{2}{5},-\frac{1}{5}\right)$

Solution:
Given equation of ellipse is
$$
4 x^{2}+9 y^{2}=1
$$
On differentiating. we get $\Rightarrow \quad 8 x+18 yy'=0$
$\Rightarrow \quad y'=\frac{-8 x}{18 y}=m$
Also, $8 x=9 y \quad$ [equation of line]
On differentiating. we got $\Rightarrow \quad 8=9 y^{\prime}$
$\Rightarrow \quad y^{\prime}=\frac{8}{9}=m$
$[\because \text{tangents are parallel to this line}]$
So, by Eqs, we get $\frac{-8 x}{18 y}=\frac{8}{9}$
$\Rightarrow \quad-x=2 y$
$\Rightarrow \quad x=-2 y$
On substituting $x=-2 y$, we get
$$
\begin{array}{r}
4(-2 y)^{2}+9 y^{2}=1 \\
\Rightarrow \quad 16 y^{2}+9 y^{2}=1 \Rightarrow 25 y^{2}=1 \\
\Rightarrow \quad y=\pm \frac{1}{5} \quad \therefore \quad x=\mp \frac{2}{5}
\end{array}
$$
So, required points are $\left(\frac{2}{5},-\frac{1}{5}\right)$ and $\left(-\frac{2}{5}, \frac{1}{5}\right)$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2016
On the occasion of Dipawali festival each student of a class sends greeting cards to others. If there are 20 students in the class. the number of cards sends by students is

Options:
A ${ }^{20} \mathrm{C}_{2}$
B ${ }^{20} \mathrm{P}_{2}$
C $2 \times{ }^{20} \mathrm{C}_{2}$
D $2 \times{ }^{20} \rho_{2}$

The correct answers are:
${ }^{20} \mathrm{P}_{2}$, $2 \times{ }^{20} \mathrm{C}_{2}$

Solution:
Required number of ways $={}^{20} C_{2} \times 2!$ $= {}^{20}P_{2}$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2018
On the set $R$ of real numbers, the relation $\rho$ is defined by $x \rho y,(x, y) \in R$

Options:
A If $|x-y| < 2$, then $\rho$ is reflexive but neither symmetric nor transitive.
B II $x-y < 2$, then $\rho$ is reflexeve and symmetric bu not transitive.
C $If(x \mid \geq y,$ then $\rho$ is reflexive and transitive but not symmetric.
D If $x>|y|$, then $p$ is transitive but neither reflexve nor symmetric.

The correct answer is:
If $x>|y|$, then $p$ is transitive but neither reflexve nor symmetric.

Solution:
On the set $R$ of real numbers
For reflexive,
$x \rho x \Rightarrow(x, x) \in R$
$\Rightarrow x > |x|$ which is not true. $\Rightarrow \rho$ is not reflexive.
For symmetric,
$(x, y) \in R \Rightarrow x > |y|$
and $(y, x) \in R \Rightarrow y > |x|$
So, $x > |y| \neq y > |x|$ $\Rightarrow \rho$ is not symmetric.
For transitive, $(x, y) \in R \Rightarrow x > |y|(y, z) \in R \Rightarrow y > |z|$
$\Rightarrow x > |z| \Rightarrow(x, z) \in R$
$\Rightarrow \rho$ is transitive.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2018
On the set $R$ of real numbers we define $x P y$ if and only if $x y \geq 0$. Then, the relation $P$ is

Options:
A reflexive but not symmetric
B symmetric but not reflexive
C transitive but not reflexive
D reflexive and symmetric but not transitive

The correct answer is:
reflexive and symmetric but not transitive

Solution:
For every real number $x, x^{2} \geq 0$
$\therefore$ $(x, x) \in P$
Hence, $P$ is reflexive.
Now let $(x,y) \in P \\
\Rightarrow xy \geq 0 \\
\Rightarrow yx \geq 0$
$(y, x) \in P$
Hence, $P$ is symmetric.
Again (-1,0)$\in P$ and (0,2)$\in P$
But (-1,2)$\notin P$
$as(-1)(2)=-2 < 0$
Pis not transitive.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2017
One $\mathrm{Kg}$ of copper is drawn into a wire of $1 \mathrm{~mm}$ diameter and a wire of $2 \mathrm{~mm}$ diameter. The resistance of the two wires will be in the ratio

Options:
A $2: 1$
B $1: 2$
C $16: 1$
D $4: 1$

The correct answer is:
$16: 1$

Solution:
Mass $=\left(\pi \mathrm{r}_1^2 \ell_1\right) \sigma$ (Ist wire)
$\begin{aligned} & \text { Mass }=\left(\pi \mathrm{r}_1^2 \ell_2\right) \sigma(2 \text { nd wire) } \\ & \left(\pi \mathrm{r}_1^2 \ell_1\right) \sigma=\left(\pi \mathrm{r}_2^2 \ell_2\right) \sigma\end{aligned}$
$\begin{aligned} & \left(\pi \mathrm{r}_1^2 \ell_1\right) \sigma=\left(\pi \mathrm{r}_2^2 \ell_2\right) \sigma \\ & \frac{\ell_1}{\ell_2}=\left(\frac{r_2}{r_1}\right)^2\end{aligned}$
$\frac{R_1}{R_2}=\frac{\rho \frac{\ell_1}{A_1}}{\rho \frac{\ell_2}{A_2}}=\frac{\ell_1}{\ell_2} \times \frac{A_2}{A_1}=\frac{\ell_1}{\ell_2} \times\left(\frac{r_2}{r_1}\right)^2$
$\begin{aligned} & =\left(\frac{r_2}{r_1}\right)^4 \\ & \Rightarrow 16: 1\end{aligned}$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2009
One mole of a diatomic ideal gas undergoes a process shown in $\mathrm{P}-\mathrm{V}$ diagram. The total heat given to the gas $(\ln 2=0.7)$ is

Options:
A $2.5 \mathrm{P}_0 \mathrm{~V}_0$
B $3.9 \mathrm{P}_0 \mathrm{~V}_0$
C $1.1 \mathrm{P}_0 \mathrm{~V}_0$
D $1.4 \mathrm{P}_0 \mathrm{~V}_0$

The correct answer is:
$3.9 \mathrm{P}_0 \mathrm{~V}_0$

Solution:
For a Diatomic gas the total heat given is $\Delta \mathrm{Q}_{\mathrm{AB}}+\Delta \mathrm{Q}_{\mathrm{BC}}$
$$
\begin{aligned}
&\Delta \mathrm{Q}_{\mathrm{AB}}(\text { Isochoric Process })=\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T} \\
&=\frac{\mathrm{P}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}} \mathrm{V}_{\mathrm{i}}}{\gamma-1}=\frac{2 \mathrm{P}_0 \mathrm{~V}_0-\mathrm{P}_0 \mathrm{~V}_0}{\frac{7}{5}-1}=\frac{\mathrm{P}_0 \mathrm{~V}_0}{\left(\frac{2}{5}\right)}=\frac{5}{2} \mathrm{P}_0 \mathrm{~V}_0=2.5 \mathrm{P}_0 \mathrm{~V}_0
\end{aligned}
$$
$\Delta \mathrm{Q}_{\mathrm{BC}}$ (Isothermal Process)
$$
Q_{B C}=n R T \ell n \frac{V_f}{V_i} \quad=P V \ell n \frac{V_f}{V_i}
$$
$$
=2 \mathrm{P}_0 \times \mathrm{V}_0 \ln \left(\frac{2 \mathrm{~V}_0}{\mathrm{~V}_0}\right)=2 \mathrm{P}_0 \mathrm{~V}_0 \ell n 2=1.4 \mathrm{P}_0 \mathrm{~V}_0
$$
$$
\Rightarrow \Delta \mathrm{Q}_{\text {total }}=3.9 \mathrm{P}_0 \mathrm{~V}_0
$$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2022
One mole of a monoatomic ideal gas undergoes a quasistatic process, which is depicted by a straight line joining points $\left(V_{0}, T_{0}\right)$ and $\left(2 V_{0}, 3 T_{0}\right)$ in a ${V} \cdot T$ diagram. What is the value of the heat capacity of the gas at the point $\left(V_{0}, T_{0}\right) ?$

Options:
A $R$
B $\frac{3}{2} R$
C $2 R$
D 0

The correct answer is:
$2 R$

Solution:
Heat capacity of an ideal gas in a thermodynamic process,
$\because$ Ideal gas is monoatomic,
$$
\begin{aligned}
C_{V} &=\frac{f R}{2}=\frac{3 R}{2} \\
C_{\text {process }} &=C_{V}+\frac{p}{n} \cdot \frac{d V}{d T} \\
&=\frac{3}{2} R+\frac{p}{n} \cdot \frac{V_{0}}{2 T} \\
\therefore C_{\text {process }}=\frac{3}{2} R+\frac{n R T}{n 2 T} &[\because p V=n R T] \\
=& \frac{3}{2} R+\frac{R}{2}=2 R
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, WBJEE, WBJEE 2018
One mole of a van der Waals' gas obeying the equation $\left(p+\frac{a}{V^{2}}\right)(V-b)=R T$ undergoes the quasi-static cyclic process which is shown in the $p \cdot V$ diagram. The net heat absorbed by the gas in this process is

Options:
A $\frac{1}{2}\left(p_{1}-\rho_{2}\right)\left(V_{1}-V_{2}\right)$
B $\frac{1}{2}\left(p_{1}+p_{2}\right)\left(V_{1}-V_{2}\right)$
C $\frac{1}{2}\left(A_{1}+\frac{a}{V_{1}^{2}}-p_{2}-\frac{a}{V_{2}^{2}}\right)\left(V_{1}-V_{2}\right)$
D $\frac{1}{2}\left(\rho_{1}+\frac{a}{V_{1}^{2}}+p_{2}+\frac{a}{V_{2}^{2}}\right)\left(V_{1}-V_{2}\right)$

The correct answer is:
$\frac{1}{2}\left(p_{1}-\rho_{2}\right)\left(V_{1}-V_{2}\right)$

Solution:
For the cyclic process

Heat absorbed = Work done
$$
\begin{array}{l}
=\text { Area }=\frac{1}{2}(\Delta p) \times \Delta V \\
=\frac{1}{2}\left(p_{1}-p_{2}\right) \times\left(V_{1}-V_{2}\right) \\
=\frac{1}{2}\left(p_{1}-p_{2}\right)\left(V_{1}-V_{2}\right)
\end{array}
$$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2014
One mole of an ideal monoatomic gas expands along the polytrope $PV^3=$ constant from $V_1$ to $V_2$ at a constant pressure $P_1$. The temperature during the process is such that molar specific heat $C_V=\frac{3 R}{2}$. The total heat absorbed during the process can be expressed as

Options:
A $\quad P_1 V_1\left(\frac{V_1^2}{V_2^2}+1\right)$
B $\mathrm{P}_1 \mathrm{~V}_1\left(\frac{\mathrm{V}_1^2}{\mathrm{~V}_2^2}-1\right)$
C $\mathrm{P}_1 \mathrm{~V}_1\left(\frac{\mathrm{V}_1^3}{\mathrm{~V}_2^2}-1\right)$
D $\quad P_1 V_1\left(\frac{V_1}{V_2^2}-1\right)$

The correct answer is:
$\mathrm{P}_1 \mathrm{~V}_1\left(\frac{\mathrm{V}_1^2}{\mathrm{~V}_2^2}-1\right)$

Solution:
Assuming $P_1$ as initial pressure.
$\mathrm{PV}^3=$ constant $n=3$
$$
\mathrm{C}=\mathrm{C}_{\mathrm{V}}+\frac{\mathrm{R}}{1-\mathrm{n}}=\frac{3 \mathrm{R}}{2}+\frac{\mathrm{R}}{1-3}=\mathrm{R}
$$
$$
\begin{aligned}
&T_i=\frac{P_1 V_1}{R} \\
&P_2 V_2{ }^3=P_1 V_1^3 \\
&P_2=P_1 \times\left(\frac{V_1}{V_2}\right)^3
\end{aligned}
$$
$$
T_2=\frac{P_2 V_2}{R}=\frac{P_1}{R} \times\left(\frac{V_1}{V_2}\right)^3 \times V_2=\frac{P_1 V_1^3}{R V_2^2}
$$
$$
\mathrm{Q}=\mathrm{nC} \Delta \mathrm{T}
$$
$$
=1 \times R \times\left(\frac{P_1 V_1^3}{R V_2^2}-\frac{P_1 V_1}{R}\right)
$$
$$
\mathrm{Q}=\mathrm{P}_1 \mathrm{~V}_1\left(\frac{\mathrm{V}_1^2}{\mathrm{~V}_2^2}-1\right)
$$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2022
One mole of an ideal monoatomic gas is heated at a constant pressure from $0^{\circ} \mathrm{C}$ to $100^\circ C$. Then the change in the internal enery of the gas is (Given, $R=8.32 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ )

Options:
A $0.83 \times 10^{3} \mathrm{J}$
B $46 \times 10^{3} \mathrm{J}$
C $2.08 \times 10^{3} \mathrm{J}$
D $1.25 \times 10^{3} \mathrm{J}$

The correct answer is:
$1.25 \times 10^{3} \mathrm{J}$

Solution:
We know that $\Delta U=n C_{v} \Delta T$
$$
\begin{array}{l}
\text { where } \begin{aligned}
\Delta T &=\left(T_{2}-T_{1}\right) \\
T_{1} &=0^{\circ} \mathrm{C}=273 \mathrm{K} \\
T_{2} &=100^{\circ} \mathrm{C}=373 \mathrm{K} \\
n &=1 \text { (monoatomic gas) }
\end{aligned} \\
\begin{aligned}
R=8.32 \mathrm{J} \mathrm{mol}^{-} \mathrm{k}^{-1}=1 \times \frac{3}{2} R \times(373-273) \\
=1 \times \frac{3}{2} \times 8.32 \times 100=3 \times 8.32 \times 50
\end{aligned}
\end{array}
$$
$\Rightarrow 1248 \mathrm{kJ}$
of
$$
=1.25 \times 10^{3} \mathrm{J}
$$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2014
One of the products of the following reaction is $P$.

Structure of $P$ is

Options:
A
B
C
D

The correct answer is:

Solution:
2,2,2 -trichloroacetophenone reacts with aqueous KOH to form potassium Salt of benzoic acid, which undergoes hydrolysis to form benzo acid as the final product.

The product of the above reaction $P$ is benzoic acid. Thus, option (c) is correct answer.

Topic: Chemistry, Carboxylic Acid Derivatives, WBJEE, WBJEE 2019
Ortho-and para-hydrogens have

Options:
A Identical chemical properties but different physical properties
B Identical physical and chemical properties
C Identical physical properties but different chemical properties
D Different physical and chemical properties

The correct answer is:
Identical chemical properties but different physical properties

Solution:
No Solution Available

Topic: Chemistry, Hydrogen, WBJEE, WBJEE 2010
Out of 7 consonants and 4 vowels, the number of words (not necessarily meaningful) that can be made, each consisting of 3 consonants and 2 vowels, is

Options:
A 24800
B 25100
C 25200
D 25400

The correct answer is:
25200

Solution:
3 consonants can be selected from 7 consonante
$$
={ }^{7} C_{3} \text { ways }
$$
2 vowels can be selected from 4 vowels
$$
={ }^{4} C_{2} \text { ways }
$$
$\therefore$ Required number of words
$$
={ }^{7} C_{3} \times{ }^{4} C_{2} \times 5 !
$$
[selected 5 letters can be arrange in $51,$ so got, a different words]
$$
=35 \times 6 \times 120=25200
$$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2014