datavorous/entrance-exam-dataset huggingface dataset:
$\int \cos (\log x) d x=F(x)+C,$ where $C$ is an arbitrary constant. Here, $F(x)$ is equal to

Options:
A $x[\cos (\log x)+\sin (\log x)]$
B $x[\cos (\log x)-\sin (\log x)]$
C $\frac{x}{2}[\cos (\log x)+\sin (\log x)]$
D $\frac{x}{2}[\cos (\log x)-\sin (\log x)]$

The correct answer is:
$\frac{x}{2}[\cos (\log x)+\sin (\log x)]$

Solution:
Let $I=\int \cos (\log x) d x$
Put $\log x=t$
$\Rightarrow \quad x=e^{t}$
$\begin{aligned} \therefore \quad d x &=e^{t} d t \\ \therefore \quad I &=\int e^{t} \cos t d t \\ &=\frac{e^{t}}{1^{2}+1^{2}}[\cos t+\sin t]+C \end{aligned}$
$\left[\because \int e^{a x} \cos b x d x=\frac{e^{a x}}{a^{2}+b^{2}}[a \cos b x+b \sin b x]+c\right]$
$\Rightarrow \quad I=\frac{e^{t}}{2}[\cos t+\sin t]+C$
$\quad \quad=\frac{x}{2}[\cos (\log x)+\sin (\log x]+C$
$\therefore f(x)=\frac{x}{2}[\cos (\log x)+\sin (\log x)]$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2017
Cupric compounds are more stable than their cuprous counterparts in solid state. This is because

Options:
A the endothermic character of the 2nd IP of Cu is not so high
B size of $\mathrm{Cu}^{2+}$ is less than $\mathrm{Cu}^{+}$
C $\mathrm{Cu}^{2+}$ has stabler electronic contiguration as compared to $\mathrm{Cu}^{+}$
D the lattice energy released for cupric compounds is much higher than $\mathrm{Cu}^{+}$

The correct answers are:
the endothermic character of the 2nd IP of Cu is not so high, size of $\mathrm{Cu}^{2+}$ is less than $\mathrm{Cu}^{+}$, the lattice energy released for cupric compounds is much higher than $\mathrm{Cu}^{+}$

Solution:
Electronic configurations of cuprous $\left(\mathrm{Cu}^{*}\right)$ and cupric $\left(\mathrm{Cu}^{2}{ }^{+}\right)$ ions are as follows
$$
\begin{aligned}
\mathrm{Cu}^{+} &=[\mathrm{Ar}] 3 d^{10} 4 s^{\circ} \\
\mathrm{Cu}^{2+} &=[\mathrm{Ar}] 3 d^{9} 4 \mathrm{s}^{\circ}
\end{aligned}
$$
Thus, electronic configuration of $\mathrm{Cu}^{+}$ is more stable but it is less stable because in $\mathrm{Cu}^{2+}$ due to its small siag the nuclear charge is sufficient to hold 27 electrons but in $Cu^+$ such a condition is not true. Further, the IInd IP of Cu is not very high as compared to its Ist IP. Consequently a large amount of latsicy energy is released for cupric compounds as compared to $Cu^+$ compounds.

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2014
Current through \(\mathrm{ABC}\) and \(\mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C}^{\prime}\) is \(\mathrm{I}\). What is the magnetic field at \(\mathrm{P}\) ? \(\mathrm{BP}=\mathrm{PB}^{\prime}=\mathrm{r}\left(\right.\) Here \(\mathrm{C}^{\prime} \mathrm{B}^{\prime} \mathrm{PBC}\) are collinear)

Options:
A \(\mathrm{B}=\frac{1}{4 \pi} \frac{2 \mathrm{I}}{\mathrm{r}}\)
B \(\mathrm{B}=\frac{\mu_0}{4 \pi}\left(\frac{2 \mathrm{I}}{\mathrm{r}}\right)\)
C \(\mathrm{B}=\frac{\mu_{\mathrm{o}}}{4 \pi}\left(\frac{\mathrm{I}}{\mathrm{r}}\right)\)
D Zero

The correct answer is:
\(\mathrm{B}=\frac{\mu_0}{4 \pi}\left(\frac{2 \mathrm{I}}{\mathrm{r}}\right)\)

Solution:
Hints:
\(\mathrm{B}=2\left[\frac{\mu_0}{4 \pi} \frac{I}{\mathrm{r}}\right]\)

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2011
Cyclo [18] carbon is an allotrope of carbon with molecular formula $\mathrm{C}_{18}$. It is a ring of 18 carbon atoms, connected by single and triple bonds. The total number of triple bonds present in this cyclocarbon are

Options:
A 9
B 10
C 12
D 6

The correct answer is:
9

Solution:

Topic: Chemistry, p Block Elements (Group 13 & 14), WBJEE, WBJEE 2021
Cyclopentanol on reaction with NaH followed by $\mathrm{CS}_{2}$ and $\mathrm{CH}_{3} \mathrm{I}$ produces a/an

Options:
A ketone
B alkone
C ether
D xanthate

The correct answer is:
xanthate

Solution:
When cyclopentanol on reaction with NaH followed by $\mathrm{CS}_{2}$ and $\mathrm{CH}_{3} \mathrm{l}$ produces a xanthate.

So, the option (d) is correct answer.

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2019
$$
\text { Dipole moment of }
$$
is $1.5 \mathrm{D}$. The dipole moment of
is

Options:
A $1.5 \mathrm{D}$
B $2.25 \mathrm{D}$
C $1 \mathrm{D}$
D $3 \mathrm{D}$

The correct answer is:
$1.5 \mathrm{D}$

Solution:
$$
\text { Hints: Given for this molecule }
$$
$\mu_1=1.5 \mathrm{D}$
For
$\mu=0$ (as it is symmetrical)
Hence for
$\mu$ will be $1.5 \mathrm{D}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2010
Dissolving NaCN in de-ionised water will result in a solution having

Options:
A $\mathrm{pH} < 7$
B $\mathrm{pH}=7$
C $\mathrm{pOH}=7$
D $\mathrm{pH}>7$

The correct answer is:
$\mathrm{pH}>7$

Solution:
$\because$ On dissolving NaCN in de-ionised water, following reaction takes place.
$\mathrm{CN}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HCN}+\mathrm{OH}^{-}$
CN- ions come from the salt of strong base and weak acid.
$\therefore$ The solution becomes basic and $\mathrm{pH}>7$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2017
During a reversible adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio $\frac{C_{p}}{C_{V}}$ for the gas is

Options:
A $\frac{3}{2}$
B $\frac{7}{2}$
C $\frac{5}{3}$
D $\frac{9}{7}$

The correct answer is:
$\frac{3}{2}$

Solution:
For a reversible adiabatic process.
$$
p T^{\frac{\gamma}{1-\gamma}}=\text { constant }
$$
According to question, during a reversible adiabatic process the pressure of a gas is found to be proportional to the cube of its absolute temperature.
$$
\begin{array}{c}
p \propto T^{3} \\
p T^{-3}=\text {constant }
\end{array}
$$
Equating equation (i) and (ii), we get
$$
\frac{\gamma}{1-\gamma}=-3
$$
$$
\begin{array}{l}
\gamma=-3(1-\gamma) \\
\gamma=-3+3 \gamma \\
-2 \gamma=-3 \\
\gamma=\frac{3}{2}
\end{array}
$$
As we know, the ratio of molar heat capacities at constant pressure and constant volume is represented by $y$. So, the ratio of $\frac{C}{C_{V}}$ for the gas is $\frac{3}{2}$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2018
During electrolysis of molten $\mathrm{NaCl},$ some water was added. What will happen?

Options:
A Electrolysis will stop
B Hydrogen will be evolved
C Some amount of caustic soda will be formed
D A fire is likely

The correct answers are:
Hydrogen will be evolved, Some amount of caustic soda will be formed, A fire is likely

Solution:
During electroysis of molten $\mathrm{NaCl}$, when water is added, the following
statements are true
(a) false
(b) $\because \mathrm{Na}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{NaOH}+\frac{1}{2} \mathrm{H}_{2}$
and discharge potential of hydrogen is less than of sodium. So, $\mathrm{H}_{2}$ will evolve.
(c) Some amount of caustic soda (NaOH) will be formed [As shown in (b)]
(d) Fire can likely take place because relations given given in (b) is highly exothermic.
$\therefore(b),(c)$ and $(d)$ are correct.

Topic: Chemistry, s Block Elements, WBJEE, WBJEE 2017
During the preparation of $\mathrm{NH}_3$ in Haber's process, the promoter(s) used is / are -

Options:
A $\mathrm{PtO}_2$
B Mo
C Mix of $\mathrm{Al}_2 \mathrm{O}_3$ and $\mathrm{K}_2 \mathrm{O}$
D Fe and Mn

The correct answers are:
Mo, Mix of $\mathrm{Al}_2 \mathrm{O}_3$ and $\mathrm{K}_2 \mathrm{O}$

Solution:
Mo was used as promoter earlier. Now $\mathrm{K}_2 \mathrm{O}$ and $\mathrm{Al}_2 \mathrm{O}_3$ are used as promoter.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2022
$\beta-emission$ is always accompanied by

Options:
A formation of antineutrino and $\alpha$ -particle
B emission of $\alpha$ -particle and $\gamma$ -ray
C formation of antineutrino and $y$ -ray
D formation of antineutrino and positron

The correct answer is:
formation of antineutrino and $y$ -ray

Solution:
During the $\beta$ -emission, a neutron gets converted into a proton along with the formation of antineutrino and electron.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2014
$\int \mathrm{e}^{\mathrm{x}}\left(\frac{2}{\mathrm{x}}-\frac{2}{\mathrm{x}^2}\right) \mathrm{dx}$ is equal to

Options:
A $\frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{x}}+\mathrm{C}$
B $\frac{e^x}{2 x^2}+C$
C $\frac{2 \mathrm{e}^{\mathrm{x}}}{\mathrm{x}}+\mathrm{C}$
D $\frac{2 \mathrm{e}^{\mathrm{x}}}{\mathrm{x}^2}+\mathrm{C}$

The correct answer is:
$\frac{2 \mathrm{e}^{\mathrm{x}}}{\mathrm{x}}+\mathrm{C}$

Solution:
Hints: $\int \mathrm{e}^{\mathrm{x}}\left(\frac{2}{\mathrm{x}}-\frac{2}{\mathrm{x}^2}\right) \mathrm{dx}=2 \int \mathrm{e}^{\mathrm{x}}\left(\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}^2}\right) \mathrm{dx}=\frac{2 \mathrm{e}^{\mathrm{x}}}{\mathrm{x}}+\mathrm{c}$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2010
Each of $a$ and $b$ can take values 1 or 2 with equal probability. The probability that the equation $a x^{2}+b x+1=0$ has real roots, is equal to

Options:
A $\frac{1}{2}$
B $\frac{1}{4}$
C $\frac{1}{8}$
D $\frac{1}{16}$

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

Solution:
The given equation
$$
a x^{2}+b x+1=0
$$
has real roots.
$\therefore$ Discriminant $(D) \geq 0$
$$
\Rightarrow \quad b^{2}-4 a \geq 0
$$
From Eq. (ii), we observe that
a has to be 1 and $b$ has to be 2 .
So, the required probability $=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2013
Each of the two inputs $A$ and $B$ can assume values either 0 or 1. Then which of the following will be equal to $\bar{A} \cdot \bar{B} ?$

Options:
A $A+B$
B $\frac{}{A+B}$
C $\frac{}{A \cdot B}$
D $\bar{A}+\bar{B}$

The correct answer is:
$\frac{}{A+B}$

Solution:
By using the de-Morgan's law
$$
\overline{A+B}=\overline{A \cdot B}
$$
Hence, option (b) is the correct.

Topic: Physics, Semiconductors, WBJEE, WBJEE 2019
Electric field component of an EM radiation varies with time as $\mathrm{E}=\mathrm{a}\left(\cos \omega_{0} \mathrm{t}+\sin \omega \mathrm{t} \cos \omega_{0} \mathrm{t}\right)$, where ' $\mathrm{a}$ ' is a constant and $\omega=10^{15} \mathrm{sec}^{-1}, \omega_{0}=5 \times 10^{15} \mathrm{sec}^{-1}$. This radiation falls on a metal whose stopping potential is $-2 \mathrm{ev}$. Then which of the following statement $(\mathrm{s})$ is/are true $?\left(\mathrm{~h}=6.62 \times 10^{-34} \mathrm{~J}-\mathrm{S}\right)$

Options:
A For light of frequency $\omega$, photoelectric effect is not possible
B Stopping potential vs. frequency graph will be a straight line
C The work function of the metal is $-2 \mathrm{ev}$
D The maximum kinetic energy of the photo electrons is $1.95 \mathrm{ev}$.

The correct answers are:
For light of frequency $\omega$, photoelectric effect is not possible, Stopping potential vs. frequency graph will be a straight line

Solution:
Stopping potential $=-2 \mathrm{~V} \Rightarrow \mathrm{KE}_{\max }=2 \mathrm{ev}$ (Option D incorrect) $\mathrm{E}=\mathrm{a}\left(\cos \omega_{0} \mathrm{t}+\sin \omega \mathrm{t} \cos \omega_{0} \mathrm{t}\right)$ $=\mathrm{a} \cos \omega_{0} \mathrm{t}+\frac{\mathrm{a}}{2}\left\{\sin \left(\omega+\omega_{0}\right) \mathrm{t}+\sin \left(\omega-\omega_{0}\right) \mathrm{t}\right\}$
Corresponding frequencies are $\frac{\omega_{0}}{\alpha \pi}, \frac{\omega+\omega_{0}}{\alpha \pi} \& \frac{\omega-\omega_{0}}{2 \pi}$
$\left(\frac{\omega+\omega_{0}}{2 \pi}\right)$ determines maximum KE \& Stopping potential
Now Energy corresponding to $\frac{\omega+\omega_{0}}{2 \pi}$
Corresponding energy $=\mathrm{hf}$
Stopping potential $=V_{0}=2 V=3.95 \mathrm{ev}$
Now $\mathrm{KE}_{\max }=\mathrm{ev}_{0}=\mathrm{hf}-\phi \Rightarrow 2 \mathrm{ev}=3.95 \mathrm{ev}-\phi \Rightarrow \phi=1.95 \mathrm{ev}$
for $\mathrm{w}=10^{15} \mathrm{sec}^{-1}$
frequency $=\frac{10^{15}}{2 \pi}=\mathrm{f}_{1}$
When frequency is $\mathrm{w}, \mathrm{f}_{1}=\frac{\omega}{2 \pi}=\frac{10^{15}}{2 \pi} \quad \mathrm{E}=\frac{6.63 \times 10^{-34} \times 6 \times 10^{15}}{2 \pi \times 1.6 \times 10^{-19}} \mathrm{ev}=3.95 \mathrm{ev}$
Energy $=\mathrm{hf}_{1}=\frac{6.63 \times 10^{-34} \times 10^{15}}{2 \pi \times 1.6 \times 10^{-19}} \mathrm{ev}=0.66 \mathrm{ev} < \phi$ (work function)
Hence photoelectric effect not possible (option A correct)
Also ev $_{0}=h f-\phi$
(Straight line) (option B correct) $\quad \mathrm{KE}_{\max }=\mathrm{E}-\phi \Rightarrow \phi=\mathrm{E}-2 \mathrm{eV}=1.95 \mathrm{ev}$ (option $\mathrm{c}$ is incorrect)

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2021
"Electron" is an alloy of

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

The correct answer is:
$\mathrm{Mg}$ and $\mathrm{Zn}$

Solution:
Electron is an alloy of $\mathrm{Mg}(95 \%)+\mathrm{Zn}(4.5 \%)$ and $\mathrm{Cu}(0.5 \%)$

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2009
Electrons are emitted with kinetic energy $T$ from a metal plate by an irradiation of light of intensity $J$ and frequency v. Then, which of the following will be true?

Options:
A $T \propto J$
B $T$ linearly increasing with $v$
C $T \propto$ lime of irradiation
D Number of electrons emitted $\propto J$

The correct answers are:
$T$ linearly increasing with $v$, Number of electrons emitted $\propto J$

Solution:
In photoelectric effect, if the incident light had a frequency less than a minimum frequency $v_{0}$, then no electrons are ejected regardless of the light's amplitude. This minimum frequency is also called the threshold frequency, and the value of $v_{0}$ depends on the metal. For frequency greater than $v_{0},$ electrons would be ejected from the metal. Furthermore, the kinetic energy of the photoelectrons is proportional to the light frequency
The relationship between photoelectron kinetic energy $T$ and light frequency $v$ is shown in graph below

Graph shows that kinetic energy $T$ is linearly increasing with light frequency (v).
However, for a given photosensitive material and frequency of incident light, number of photoelectrons emitted per second is directly proportional to the intensity of incident light
i.e., Number of electrons emitted $\propto J$. Also, photoelectric emission is an instantaneous process.

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2019
Eleven apples are distributed among a girl and a boy, Then, which one of the following statements is true?

Options:
A atleast one of them will receive 7 apples
B the girl receives atleast 4 apples or the boy receives atleast 9 apples
C the girl receives atleast 5 apples or the boy receives atleast 8 apples
D the girl receives atleast 4 apples or the boy receives atleast 8 apples

The correct answer is:
the girl receives atleast 4 apples or the boy receives atleast 8 apples

Solution:
When eleven apples are distributed among a gil and a boy, then the girl receives atieast
14 apples or the boys receives atieast 8 apples. (by hypothesis)

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2013
Eleven equal point charges, all of them having a charge $+Q$, are placed at all the hour positions of a circular clock of radius $r$. except at the $10 \mathrm{h}$ position. What is the clectric field strength at the centre of the clock?

Options:
A $\frac{Q}{4 \pi \varepsilon_{0} r^{2}}$ from the centre towards the mark 10
B $\frac{Q}{4 \pi \varepsilon_{0} r^{2}}$ from the mark 10 towards the centre
C $\frac{Q}{4 \pi \varepsilon_{0} r^{2}}$ from the centre towards the mark 6
D Zero.

The correct answer is:
$\frac{Q}{4 \pi \varepsilon_{0} r^{2}}$ from the centre towards the mark 10

Solution:
The clock diagram is as shown below

In the above diagram, charge $+Q$ is not placed at
$10 \mathrm{h}$ position.
So, net electric ficld strength at centre $C$.
$$
\begin{array}{r}
\mathrm{E}_{\text {net }}=\mathrm{E}_{1}+\mathrm{E}_{2}+\mathrm{E}_{3}+\mathrm{E}_{4}+\mathrm{E}_{5}+\mathrm{E}_{6} \\
+\mathrm{E}_{7}+\mathrm{E}_{8}+\mathrm{E}_{9}+\mathrm{E}_{11}+\mathrm{E}_{12} \\
\Rightarrow \mathrm{E}_{\text {net }}=\mathrm{E}_{1}+\mathrm{E}_{2}+\mathrm{E}_{3}+\mathrm{E}_{4} \\
+\mathrm{E}_{5}+\mathrm{E}_{6}+\left(-\mathrm{E}_{1}-\mathrm{E}_{2}-\mathrm{E}_{3}-\mathrm{E}_{5}-\mathrm{E}_{6}\right)
\end{array}
$$
$\Rightarrow \mathrm{E}_{\mathrm{net}}=\mathrm{E}_{4}=\frac{Q}{4 \pi \varepsilon_{0} r^{2}},$ from centre towards the
mark $10 .$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2019
Equal masses of ethane and hydrogen are mixed in a empty container at $298 \mathrm{~K}$. The fraction of total pressure exerted by hydrogen is

Options:
A $15: 16$
B $1: 1$
C $1: 4$
D $1: 6$

The correct answer is:
$15: 16$

Solution:
Hint:
Let mass of ethane and $\mathrm{H}_{2}$ are $\mathrm{xg}$.
$\begin{array}{c}
n_{\mathrm{H}_{2}}=\frac{\mathrm{x}}{2} \text { and } \mathrm{n}_{\mathrm{C}_{2} \mathrm{H}_{\mathrm{z}}}=\frac{\mathrm{x}}{30} \\
\frac{\mathrm{P}_{\mathrm{H}_{2}}}{\mathrm{P}_{\mathrm{T}}}=\frac{\mathrm{n}_{\mathrm{H}_{2}}}{\mathrm{n}_{\mathrm{H}_{2}}+\mathrm{n}_{\mathrm{C}_{2} \mathrm{H}}}=\frac{\frac{\mathrm{x}}{2}}{\frac{\mathrm{X}}{2}+\frac{\mathrm{x}}{30}}=\frac{15}{16}
\end{array}$

Topic: Chemistry, Solutions, WBJEE, WBJEE 2020
Equal volumes of molar hydrochloric acid and sulphuric acid are neutralised by dilute $\mathrm{NaOH}$ solution and $\mathrm{x} \mathrm{kcal}$ and $\mathrm{y} \mathrm{kcal}$ of heat are liberated respectively. Which of the following is true?

Options:
A $x=y$
B $x=\frac{y}{2}$
C $x=2 y$
D none of the above

The correct answer is:
$x=\frac{y}{2}$

Solution:
Enthalpy of $1 \mathrm{~g}$ equivalent of strong acid and $1 \mathrm{~g}$ equivalent strong base $=13.7 \mathrm{kcal}$ Equal volume contains double eq. of $\mathrm{H}_2 \mathrm{SO}_4$ than $\mathrm{HCl}$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2009
Equal weights of $\mathrm{CH}_{4}$ and $\mathrm{H}_{2}$ are mixed in an empty container at $25^{\circ} \mathrm{C}$. the fraction of the total pressure exerted by $\mathrm{H}_{2}$ is

Options:
A $1 / 9$
B $1 / 2$
C $8 / 9$
D $16 / 17$

The correct answer is:
$8 / 9$

Solution:
Ratio of number of moles of $\mathrm{CH}_{4}$ and $\mathrm{H}_{2}$ are
$$
\begin{array}{r}
\mathrm{CH}_{4}: \mathrm{H}_{2}=\frac{x}{16} ; \frac{x}{2} \\
=1: 8
\end{array}
$$
Hence, partial pressure exerted by
$$
\mathrm{H}_{2}=\frac{8}{1+8} \times p_{\text {total }}=\frac{8}{9} \times p_{\text {lotal }}
$$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2012
Equal weights of ethane and hydrogen are mixed in an empty container at 25 C. The fraction of total pressure exerted by hydrogen is

Options:
A 1: 2
B 1: 1
C 01:16
D 15:16

The correct answer is:
15:16

Solution:
Given, equal weights of ethane and hydrogen are mixed in an empty container at $25^{\circ} \mathrm{C}$
Initial gram weight $=\begin{array}{ll}\mathrm{C}_{2} \mathrm{H}_{6} & \mathrm{H}_{2} \\ \mathrm{Wg} & \mathrm{Wg}\end{array}$
Number of moles $=\frac{w}{30} \quad \frac{w}{2}$ According to Henry's law,
$$
\begin{array}{l}\frac{p_{\mathrm{H}_{2}}}{p_{\text {total }}}=\chi_{\mathrm{H}_{2}}=\frac{n_{\mathrm{H}_{2}}}{n_{\mathrm{H}_{2}}+n_{\mathrm{C}_{2} \mathrm{H}_{6}}}=\frac{\frac{w}{2}}{\frac{w}{2}+\frac{w}{30}} \\ \frac{p_{\mathrm{H}_{2}}}{p_{\text {total }}}=\frac{\frac{w}{2}}{\frac{15 w+w}{30}}=\frac{w}{2} \times \frac{30}{16 w}=\frac{15}{16}\end{array}
$$
So, the fraction of total pressure exerted by
hydrogen is 15: 16

Topic: Chemistry, Solutions, WBJEE, WBJEE 2018
Equation of a tangent to the hyperbola $5 x^{2}-y^{2}=5$ and which passes through an external point (2,8) is

Options:
A $3 x-y+2=0$
B $3 x+y-14=0$
C $23 x-3 y-22=0$
D $3 x-23 y+178=0$

The correct answers are:
$3 x-y+2=0$, $23 x-3 y-22=0$

Solution:
Given equation of hyperbola can be write as
$\frac{x^{2}}{1}-\frac{y^{2}}{(\sqrt{5})^{2}}=1$
here, $a=1$ and $b=\sqrt{5}$
Since, $y=m x \pm \sqrt{a^{2} m^{2}-b^{2}}$
$y=m x \pm \sqrt{m^{2}-5}$
$\Rightarrow \quad(8-2 m)^{2}=\left(\sqrt{m^{2}-5}\right)^{2}$
$\Rightarrow \quad(8-2 m)^{2}=m^{2}-5$
$\Rightarrow 64+4 m^{2}-32 m=m^{2}-5$
$\Rightarrow 3 m^{2}-32 m+69=0$
$\Rightarrow \quad m=\frac{22}{3}, 3 \quad$ [by solving equation $]$
When $m=\frac{23}{3},$ then equation of tangent
$$
(y-8)=\frac{23}{3}(x-2)
$$
$\Rightarrow \quad 3 y-24=23 x-46$
$\Rightarrow \quad 23 x-3 y-46+24=0$
$\Rightarrow \quad 23 x-3 y-22=0$
When $m=3$, then equation of tangent
$$
\begin{aligned}
& &(y-8) &=3(x-2) \\
\Rightarrow & & y-8 &=3 x-6
\end{aligned}
$$
$\Rightarrow \quad 3 x-y-6+8=0$
$\Rightarrow \quad 3 x-y+2=0$
Hence, option (a) and (c) are correct.

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2019
Equilibrium constants for the following reactions at $1200 \mathrm{K}$ are given
$2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)$
$K_{1}=6.4 \times 10^{-8}$
$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$
$K_{2}=1.6 \times 10^{-6}$
The equilibrium constant for the reaction? $\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$
at $1200 \mathrm{K}$ will be

Options:
A 0.05
B 20
C 0.2
D 5

The correct answer is:
5

Solution:
For the given reactions, $2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 2 \mathrm{H}_{2}+\mathrm{O}_{2}, K_{1}=6.4 \times 10^{-8}$
or, $\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}, K_{1}^{\prime}=\sqrt{K_{1}}$
or $\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}, \longrightarrow \mathrm{H}_{2} \mathrm{O}, K_{1}^{\prime \prime \prime}=1 / \sqrt{K_{1}} \ldots$
$2 \mathrm{CO}_{2} \rightleftharpoons 2 \mathrm{CO}+\mathrm{O}_{2}, K_{2}=1.6 \times 10^{-6}$
or $\mathrm{CO}_{2} \rightleftharpoons \mathrm{CO}+\frac{1}{2} \mathrm{O}_{2}, K_{2}^{\prime}=\sqrt{K_{2}}$
From (i) and (ii) [on adding] $K=\frac{\sqrt{K_{2}}}{\sqrt{K_{1}}}=\frac{\sqrt{1.6 \times 10^{-6}}}{\sqrt{6.4 \times 10^{-8}}}$
$K=\sqrt{\frac{10^{2}}{4}}=\sqrt{25}=5$

Topic: Chemistry, Chemical Equilibrium, WBJEE, WBJEE 2017
Equivalent capacitance between $A$ and $B$ in the figure is

Options:
A $20 \mu \mathrm{F}$
B $8 \mu \mathrm{F}$
C $12 \mu \mathrm{F}$
D $16 \mu \mathrm{F}$

The correct answer is:
$8 \mu \mathrm{F}$

Solution:
$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}$
$C=2 \mu F$
$2 \mu \mathrm{F}$ will be parallel to $4 \mu \mathrm{F}$. so the resultant capacitance

$\begin{aligned} \frac{1}{C^{\prime \prime}} &=\frac{1}{4 \mu F}+\frac{1}{4 \mu F}=\frac{2}{4} \mu F \\ C^{-} &=2 \mu F \\ C_{A Q} &=2 \mu F+6 \mu F \\ &=8 \mu F \end{aligned}$

Topic: Physics, Capacitance, WBJEE, WBJEE 2016
Equivalent conductivity at infinite dilution for $\quad$ sodium-potassium oxalate $\left[\left(\mathrm{COO}^{-}\right)_{2} \mathrm{Na}^{+} \mathrm{K}^{+}\right]$ will be [given molar
conductivities of oxalate, $\mathrm{K}^{+}$ and $\mathrm{Na}^{+}$ ions at infinite dilution are $148.2, \quad 50.1,$ $73.5 \mathrm{S} \mathrm{cm}^{2} \mathrm{mol}^{-1}$ respectively $]$

Options:
A $271.8 \mathrm{S} \mathrm{cm}^{2} \mathrm{eq}^{-1}$
B $67.95 \mathrm{Scm}^{2} \mathrm{eq}^{-1}$
C $543.6 \mathrm{Scm}^{2} \mathrm{eq}^{-1}$
D $135.9 \mathrm{Scm}^{2} \mathrm{eq}^{-1}$

The correct answer is:
$135.9 \mathrm{Scm}^{2} \mathrm{eq}^{-1}$

Solution:
$\lambda_{m}^{-}=\lambda_{m}^{-}$ (oxalate) $+\lambda_{m}^{-}\left(\mathrm{Na}^{+}\right)+\lambda_{m}^{-}\left(\mathrm{K}^{+}\right)$
$$
\begin{array}{l}
=(148 \cdot 2+73 \cdot 5+50 \cdot 1) \\
=271.85 \mathrm{cm}^{2} \mathrm{mol}^{-1}
\end{array}
$$
$\therefore$
$$
\begin{aligned}
\lambda_{e q}^{-} &=\frac{\lambda_{m}^{-}}{n \cdot \text { factor }}=\frac{271.8}{2} \\
&=135.9 \mathrm{Scm}^{2} \mathrm{eq}^{-1}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2013
Ethelene can be separated from acetylene by passing the mixture through :

Options:
A fuming $\mathrm{H}_2 \mathrm{SO}_4$
B pyrogallol
C ammoniacal $\mathrm{Cu}_2 \mathrm{Cl}_2$
D Charcoal powder

The correct answer is:
ammoniacal $\mathrm{Cu}_2 \mathrm{Cl}_2$

Solution:
Hints : $\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}+\mathrm{Cu}_2 \mathrm{Cl}_2 \rightarrow \mathrm{Cu}^{+} \mathrm{C}^{-} \equiv \mathrm{C}^{-} \mathrm{Cu}^{+} \downarrow$
Red ppt.
$$
\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2+\mathrm{Cu}_2 \mathrm{Cl}_2 \rightarrow \mathrm{No} . \mathrm{ppt}
$$

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2009