datavorous/entrance-exam-dataset huggingface dataset:
The reduction potential of hydrogen half-cell will be negative if

Options:
A $\mathrm{p}\left(\mathrm{H}_{2}\right)=1 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}$
B $\mathrm{p}\left(\mathrm{H}_{2}\right)=1 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}$
C $\mathrm{p}\left(\mathrm{H}_{2}\right)=2 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}$
D $\mathrm{p}\left(\mathrm{H}_{2}\right)=2 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}$

The correct answer is:
$\mathrm{p}\left(\mathrm{H}_{2}\right)=2 \mathrm{~atm}$ and $\left[\mathrm{H}^{+}\right]=1.0 \mathrm{M}$

Solution:
Reaction $2 \mathrm{H}^{+}+2 \mathrm{e} \rightarrow \mathrm{H}_{2}$
$\therefore \mathrm{n}=2, \quad \mathrm{Q}=\frac{\mathrm{P}_{\mathrm{H}_{2}}}{\left[\mathrm{H}^{+}\right]^{2}}$
As $\mathrm{E}_{\mathrm{H}^{+} / \mathrm{H}_{2}}=\mathrm{E}_{\mathrm{H}^{+} / \mathrm{H}_{2}}^{0} \frac{-0.059}{\mathrm{n}} \log \mathrm{Q}=-\frac{0.059}{2} \log \mathrm{Q}$
$\therefore$ if $\mathrm{Q}>1 \quad \mathrm{E}_{\mathrm{H}^{+} / \mathrm{H}_{2}} < 0$
(A) $Q=\frac{1}{1}=1 \quad X$
(B) $Q=\frac{1}{2^{2}} < 1 \quad X$
(C) $Q=\frac{2}{12}>1$

(D) $Q=\frac{2}{2^{2}} < 1 \quad X$

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2021
The reduction product of ethyl 3-oxobutanoate by $\mathrm{NaBH}_{4}$ in methanol is

Options:
A
B
C
D

The correct answer is:

Solution:
Hint:
$\mathrm{NaBH}_{4}$ reduces Ketone but not ester

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2020
The relationship between the pair of compounds shown above are respectively

Options:
A Homomer (identical), enantiomer and constitutional isomer
B Enantiomer, enantiomer and diastereomer
C Homomer (identical), homomer (identical) and constitutional isomer
D Enantiomer, homomer (identical) and geometrical isomer

The correct answer is:
Homomer (identical), homomer (identical) and constitutional isomer

Solution:

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2021
The representation of the ground state electronic configuration of He by box - diagram as \(\uparrow \uparrow\) is wrong because it violates

Options:
A Hysenberg's Uncertainty Principle
B Bohr's Quantization Theory of Angular Momenta
C Pauli Exclusion Principle
D Hund's Rule

The correct answer is:
Hysenberg's Uncertainty Principle

Solution:
Hints : According to Pauli Exclusion Principle, In any orbital, maximum two electrons can exist, having opposite spin.

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2011
The resistance across \(A\) and \(B\) in the figure below will be

Options:
A \(3 \mathrm{R}\)
B \(\mathrm{R}\)
C \(\frac{R}{3}\)
D None of these

The correct answer is:
\(\frac{R}{3}\)

Solution:
Hints: Resistance are in parallel \(\therefore \operatorname{Req}=\frac{\mathrm{R}}{3}\)

Topic: Physics, Current Electricity, WBJEE, WBJEE 2011
The rms speed of oxygen is $v$ at a particular temperature. If the temperature is doubled and oxygen molecules dissociate into oxygen atoms, the rms speed becomes

Options:
A v
B $\sqrt{2} v$
C $2v$
D $4 v$

The correct answer is:
$2v$

Solution:
As the rms speed is given by
$$
\begin{aligned}
& v_{\text {rms }}=\sqrt{\frac{3 R T}{M}}{V_{\text {ms }}} \propto \sqrt{\frac{T}{M}}
\end{aligned}
$$
When temperature is doubled and molecules dissociates into atoms, then
$$
\begin{aligned}
\frac{v_{\mathrm{rms}_{1}}}{v_{\mathrm{rms}_{2}}} &=\frac{\sqrt{\frac{T}{M}}}{\sqrt{\frac{2 T}{M / 2}}}=\frac{\sqrt{\frac{T}{M}}}{\sqrt{\frac{4 T}{M}}} \\
&=\frac{\sqrt{T}}{\sqrt{4 T}}=\frac{1}{2}
\end{aligned}
$$
If $v_{\mathrm{rms}},$ is $v,$ then $v_{\mathrm{rms}_{2}}$ will be $2 v$.

Topic: Physics, Kinetic Theory of Gases, WBJEE, WBJEE 2015
The r.m.s speed of the molecules of a gas at $100^{\circ} \mathrm{C}$ is $v .$ The temperature at which the r.m.s. speed will be $\sqrt{3} v$ is

Options:
A $546^{\circ} \mathrm{C}$
B $646^{\circ} \mathrm{C}$
C $746^{\circ} \mathrm{C}$
D $846^{\circ} \mathrm{C}$

The correct answer is:
$846^{\circ} \mathrm{C}$

Solution:
$v_{\text {tms }}=\sqrt{\frac{3 R T}{M}}$
Given, $\quad T=100^{\circ} \mathrm{C}=373 \mathrm{K}$
$\therefore \quad v=\sqrt{\frac{3 R \times 373}{M}}$
For $\quad \sqrt{3} v=\sqrt{\frac{3 R T}{M}}$
$\therefore \quad \frac{\sqrt{3} v}{v}=\frac{\sqrt{3 R T / M}}{\sqrt{3 R \times 373 / M}}$
$\Rightarrow \quad \sqrt{3}=\sqrt{\frac{T}{373}}$
$\Rightarrow \quad 3=\frac{T}{373}$
$\Rightarrow \quad T=1119 \mathrm{K}$
or $\quad T=846^{\circ} \mathrm{C}$

Topic: Physics, Kinetic Theory of Gases, WBJEE, WBJEE 2013
The rms value of potential difference $v$ shown in the figure is

Options:
A $\frac{\mathrm{v}_{0}}{2}$
B $\mathrm{v}_{0}$
C $\frac{v_{0}}{\sqrt{3}}$
D $\frac{v_{0}}{\sqrt{2}}$

The correct answer is:
$\frac{v_{0}}{\sqrt{2}}$

Solution:
$\mathrm{V}_{\mathrm{rms}}^{2} \times \mathrm{T}=\mathrm{V}_{0}^{2} \times \frac{\mathrm{T}}{2}+0$
$\mathrm{~V}_{\mathrm{rms}}^{2}=\frac{\mathrm{V}_{0}^{2}}{2} \quad \mathrm{~V}_{\mathrm{rms}}=\frac{\mathrm{V}_{0}}{\sqrt{2}}$

Topic: Physics, Alternating Current, WBJEE, WBJEE 2021
The RMS velocity of CO gas molecules at $27^{\circ} \mathrm{C}$ is approximately $1000 \mathrm{m} / \mathrm{s}$. For molecules at $600 \mathrm{K}$, the RMS velocity is approximately

Options:
A $2000 \mathrm{m} / \mathrm{s}$
B $1414 \mathrm{m} / \mathrm{s}$
C $1000 \mathrm{m} / \mathrm{s}$
D $1500 \mathrm{m} / \mathrm{s}$

The correct answer is:
$1414 \mathrm{m} / \mathrm{s}$

Solution:
As trom the formula, $v_{\text {ms }}=\sqrt{\frac{3 R T}{M}}$
Given that $\left(v_{\text {rms }}\right)_{CO}=1000 \mathrm{m} / \mathrm{s}$
$(Temp )_{CO}=27^{\circ} \mathrm{C}=300 \mathrm{K}$
$(Temp.) _{\mathrm{N}_{\mathrm{2}}}=600 \mathrm{K}$
Now putting the values, we get
$\frac{\left(V_{\operatorname{mot}}\right)_{CO}}{\left(v_{m}\right)_{N_{2}}}=\sqrt{\frac{3 R}{3 R} \times \frac{T_{CO}}{M_{CO}} \times \frac{M_{N_{2}}}{T_{N_{2}}}}$
$\frac{1000}{\left(v_{\text {rad }}\right)_{N_{2}}}=\sqrt{\frac{300}{28} \times \frac{28}{600}}=\frac{1}{\sqrt{2}}$
or. $\left(v_{\text {rms }}\right)_{N_{2}}=1000 \times 1414$ $=1414 \mathrm{m} / \mathrm{s}$

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2015
The role of fluorspar, which is added in small quantities in the electrolytic reduction of alumina dissolved in fused cryolite is

Options:
A as a catalyst
B to make fused mixture conducting
C to lower the melting temperature of the mixture
D to decrease the rate of oxidation of carbon at anode

The correct answers are:
to make fused mixture conducting, to lower the melting temperature of the mixture

Solution:
Among the given statements (b) and (c) are correct these are the facts.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, WBJEE, WBJEE 2017
The Rolle's theorem is applicable in the interval $-1 \leq \mathrm{x} \leq 1$ for the function

Options:
A $f(x)=x$
B $f(x)=x^2$
C $\mathrm{f}(\mathrm{x})=2 \mathrm{x}^3+3$
D $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$

The correct answer is:
$f(x)=x^2$

Solution:
Hints : $f(x)=x^2 \mid$ and $\mathrm{f}(1)=\mathrm{f}(-1)$ for $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$ but at $\mathrm{x}=0, \mathrm{f}(\mathrm{x})=|\mathrm{x}|$ is not differentiable hence (2) is the correct option.
$$
f(1)=1=f(-1)
$$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2009
The root mean square (rms) speed of \(X_2\) gas is \(\mathrm{x} \mathrm{m/s}\) at a given temperature. When the temperature is doubled, the \(\mathrm{X}_2\) molecules dissociated completely into atoms. The root mean square speed of the sample of gas then becomes (in \(\mathrm{m} / \mathrm{s}\) )

Options:
A \(x / 2\)
B \(x\)
C \(2 x\)
D \(4 x\)

The correct answer is:
\(2 x\)

Solution:
Hint : \(C_{m m s}=\sqrt{\frac{3 R T}{M}}\)
\(\begin{array}{ll}
\mathrm{T}_1=\mathrm{T} & \mathrm{T}_2=2 \mathrm{~T} \\
\mathrm{M}_1=\mathrm{M} & \mathrm{M}_2=\mathrm{M} / 2 \\
\mathrm{C}_1=\mathrm{X} & \mathrm{C}_2=?
\end{array}\)
\(\frac{C_1}{C_2}=\sqrt{\frac{T_1}{M_1} \times \frac{\mathrm{M}_2}{\mathrm{~T}_2}}=\sqrt{\frac{\mathrm{T}}{\mathrm{M}} \times \frac{\mathrm{M} / 2}{2 T}}=\frac{1}{2}\)
\(\therefore \frac{\mathrm{x}}{\mathrm{C}_2}=\frac{1}{2}\), Hence \(\mathrm{C}_2=2 \mathrm{x~} \mathrm{m} / \mathrm{s}\)

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2023
The roots of the quadratic equation $x^2-2 \sqrt{3} x-22=0$ are :

Options:
A imaginry
B real, rational and equal
C real, irrational and unequal
D real, rational and unequal

The correct answer is:
real, irrational and unequal

Solution:
$$
\text { Hints: } \begin{aligned}
& x^2-2 \sqrt{3}-22=0 \\
& D=12+(4 \times 22)>0
\end{aligned}
$$
$\because$ coeffs are irrational,
$$
x=\frac{2 \sqrt{3} \pm \sqrt{12+88}}{2}
$$
$\therefore$ Roots are irrational, real, unequal.

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2010
The second ionisation energy of the following elements follows the order

Options:
A $\mathrm{Zn}>\mathrm{Cd} < \mathrm{Hg}$
B $Z n>C d>H g$
C $\mathrm{Cd}>\mathrm{Hg} < \mathrm{Zn}$
D $\mathrm{Zn} < \mathrm{Cd} < \mathrm{Hg}$

The correct answer is:
$\mathrm{Zn}>\mathrm{Cd} < \mathrm{Hg}$

Solution:
Key Point For similar electronic configuration of outer most shell, size will decide the value of ionisation energy. More the size, lesser is the value of ionisation energy.
Electronic configuration - IInd I.E
$\mathrm{Zn}^{+}$ ion $(Z=30)=[\mathrm{Ar}] 3 \mathrm{d}^{10} 4 \mathrm{s}^{1}, - \quad(1734 \mathrm{kJ} / \mathrm{mol})$
$\mathrm{Cd}^{+}$ ion $(Z=48)=[\mathrm{Kr}] 4 d^{10} 5 \mathrm{s}^{1} - (1631 \mathrm{kJ} / \mathrm{mol})$
$\mathrm{Hg}^{+}$ ion $(Z=80)=[\mathrm{Xe}] 4 f^{14} \cdot 5 d^{10} 6 \mathrm{s}^{1} - (1809 \mathrm{kJ} / \mathrm{mol})$
$\because$ Size of $\mathrm{Cd}^{+}>\mathrm{Zn}^{+}$, thus it has lower IInd ionisation energy while due to lanthanoid effect (i.e., poor screening by $4 f$ and $5 d$ electrons, $\mathrm{Hg}^{+}$ has higher IInd ionisation-energy. Hence, correct order is,
$\mathrm{Zn} > \mathrm{Cd} < \mathrm{Hg}$ and option (a) is the correct answer.

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2019
The second law of thermodynamics says that in a cyclic process :

Options:
A work cannot be converted into heat
B heat cannot be converted into work
C work cannot be completely converted into heat
D heat cannot be completely converted into work

The correct answer is:
heat cannot be completely converted into work

Solution:
Because $0 \mathrm{~K}$ temperature is unattainable.

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2009
The second order derivative of a $\sin ^3 t$ with respect to a $\cos ^3 t$ at $t=\frac{\pi}{4}$ is

Options:
A 2
B $\frac{1}{12 a}$
C $\frac{4 \sqrt{2}}{3 a}$
D $\frac{3 a}{4 \sqrt{2}}$

The correct answer is:
$\frac{4 \sqrt{2}}{3 a}$

Solution:
Hints : $y=a \sin ^3 \mathrm{t} ; \mathrm{x}=\mathrm{a} \cos ^3 \mathrm{t}$
$$
\begin{aligned}
& \frac{\mathrm{dy}}{\mathrm{dt}}=3 \sin ^2 \mathrm{t} \cos \mathrm{t} ; \frac{\mathrm{dx}}{\mathrm{dt}}=-3 \mathrm{a} \cos ^2 \mathrm{t} \sin \mathrm{t} \\
& \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{3 a \sin ^2 t \cos t}{-3 a \cos ^2 t \operatorname{sint}}=-\frac{\operatorname{sint}}{\cos t}=-\operatorname{tant} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx} \mathrm{x}^2}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}(-\tan t)=\frac{\mathrm{d}}{\mathrm{dt}}(-\operatorname{tant}) \cdot \frac{\mathrm{dt}}{\mathrm{dx}} \\
& =\left(-\sec ^2 \mathrm{t}\right) \frac{1}{-3 \cos ^2 \mathrm{t} \sin t}=\frac{1}{+3 \cos ^4 \mathrm{t} \sin t} \\
& \left.\frac{d^2 y}{d x^2}\right|_{t=\pi / 4}=\frac{1}{3 a\left(\frac{1}{\sqrt{2}}\right)^4 \cdot\left(\frac{1}{\sqrt{2}}\right)}=\frac{(\sqrt{2})^5}{3 a}=\frac{4 \sqrt{2}}{3 a}
\end{aligned}
$$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2009
The sequence \(\log \mathrm{a}, \log \frac{\mathrm{a}^2}{\mathrm{~b}}, \log \frac{\mathrm{a}^3}{\mathrm{~b}^2}, \ldots \ldots \ldots\). is

Options:
A a GP.
B an A.P.
C a H.P.
D both a G.P. and a H.P

The correct answer is:
an A.P.

Solution:
\(\begin{aligned}\text {Hints: }
& \log \mathrm{a} \cdot(2 \log \mathrm{a}-\log \mathrm{b})(3 \log \mathrm{a}-2 \log \mathrm{b}) \\
& =\mathrm{T}_2-\mathrm{T}_1=\log \mathrm{a}-\log \mathrm{b} \\
= & \mathrm{T}_3-\mathrm{T}_2=\log \mathrm{a}-\log \mathrm{b}
\end{aligned}\)

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2011
The shape of $XeF_{5}^{-}$ will be

Options:
A square pyramid
B trigonal bipyramidal
C planar
D pentagonal bipyramid

The correct answer is:
planar

Solution:
Shape of XeF $\rightarrow$

The Xe show $s p^{3} d^{3}$ -hybridisation in $XeF^{-}_{5}$, Thus, its geometry is Pentagonal-bipyramidal. But, number of electron pair $=\frac{8+5+1}{2}=7$ Number of bond pair $=5^{2},$ number of lone pair $=2$ A lot of electrons are present at axil position and all bonds are in same plane. Hence, the shape is planar.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2017
The side $A B$ of $\triangle A B C$ is fixed and is of length $2 a$ unit. The vertex moves in the plane such that the vertical angle is always constant and is $\alpha$. Let $x$-axis be along $A B$ and the origin be at $A$. Then the locus of the vertex is

Options:
A $x^2+y^2+2 a x \sin \alpha+a^2 \cos \alpha=0$
B $x^2+y^2-2 a x-2 a y \cot \alpha=0$
C $x^2+y^2-2 a x \cos \alpha-a^2=0$
D $x^2+y^2-a x \sin \alpha-$ ay $\cos \alpha=0$

The correct answer is:
$x^2+y^2-2 a x-2 a y \cot \alpha=0$

Solution:
Let, $c=z=x+i y$
$\begin{aligned}
&\arg \left(\frac{2 \mathrm{a}-\mathrm{z}}{0-\mathrm{z}}\right)=\alpha \\
&\Rightarrow \arg \left(\frac{(\mathrm{x}-2 \mathrm{a})+\mathrm{iy}}{\mathrm{x}+\mathrm{iy}}\right)=\alpha \\
&\Rightarrow \mathrm{x}^2+\mathrm{y}^2-2 \mathrm{ax}-2 \mathrm{ay} \cot \alpha=0
\end{aligned}$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2022
The sine of the angle between the straight line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ and the plane $2 x-2 y+z=5$ is

Options:
A $\frac{2 \sqrt{3}}{5}$
B $\frac{\sqrt{2}}{10}$
C $\frac{4}{5 \sqrt{2}}$
D $\frac{\sqrt{5}}{6}$

The correct answer is:
$\frac{\sqrt{2}}{10}$

Solution:
Hint:
Let the angle be $\theta$. Then, $\cos \left(90^{\circ}-\theta\right)=\frac{(3 \hat{\mathbf{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})}{|3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}| \times|2 \hat{\mathbf{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}|} \Rightarrow \sin \theta=\frac{3}{\sqrt{50} \sqrt{9}}=\frac{\sqrt{2}}{10}$

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2020
The slope at any point of a curve $y=f(x)$ is given by $\frac{d y}{d x}=3 x^2$ and it passes through $(-1,1)$. The equation of the curve is

Options:
A $y=x^3+2$
B $y=-x^3-2$
C $y=3 x^3+4$
D $y=-x^3+2$

The correct answer is:
$y=x^3+2$

Solution:
Hints: $\frac{d y}{d x}=3 x^2 \Rightarrow \int d y=\int 3 x^2 d x \Rightarrow y=x^3+\mathrm{C}$
Curve passes through $(-1,1)$. Hence $1=-1+C \Rightarrow C=2$
$$
\therefore y=x^3+2
$$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2009
The smallest positive root of the equation $\tan x-x=0$ lies in

Options:
A $\left(0, \frac{\pi}{2}\right)$
B $\left(\frac{\pi}{2}, \pi\right)$
C $\left(\pi, \frac{3 \pi}{2}\right)$
D $\left(\frac{3 \pi}{2}, 2 \pi\right)$

The correct answer is:
$\left(\pi, \frac{3 \pi}{2}\right)$

Solution:
Given equation is
$\tan x-x=0$
$\tan x=x$
Solutions are abscissae of points of intersection of the curves $y=\tan x$ and $y=x$. From the figure, it is clearty visible that solution lies in $\left(\pi, \frac{3 \pi}{2}\right)$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2016
The smallest value of $5 \cos \theta+12$ is

Options:
A 5
B 12
C 7
D 17

The correct answer is:
7

Solution:
$$
\begin{aligned}
& \text { Hints : } 5 \cos \theta+12,-1 \leq \cos \theta \leq 1 \\
& \Rightarrow-5 \leq 5 \cos \theta \leq 5 \\
& \therefore 5 \cos \theta+12 \geq-5+12 \Rightarrow 5 \cos \theta+12 \geq 7
\end{aligned}
$$

Topic: Mathematics, Functions, WBJEE, WBJEE 2009
The solubility of \(\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2\) in water is y moles / litre. Its solubility product is

Options:
A \(6 y^4\)
B \(36 y^4\)
C \(64 y^5\)
D \(108 \mathrm{y}^5\)

The correct answer is:
\(108 \mathrm{y}^5\)

Solution:
\(\begin{aligned}
& \text {Hints: } \mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2(\mathrm{~s}) \rightleftharpoons 3 \underset{3 \mathrm{~s}}{\mathrm{Ca}^{2+}}(\mathrm{aq})+\underset{2 \mathrm{~s}}{2 \mathrm{PO}_4}{ }^{3-}(\mathrm{aq}) \\
& \mathrm{~K}_{\mathrm{Sp}}=\left[\mathrm{Ca}^{2+}\right]^3 \cdot\left[\mathrm{PO}_4^{3-}\right]^2 \\
& =(3 \mathrm{~s})^3 \cdot(2 \mathrm{~s})^2 \\
& =27 \mathrm{~s}^3 \times 4 \mathrm{~s}^2 \\
& =108 \mathrm{~s}^5 \\
&
\end{aligned}\)

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2011
The solution of $25 \frac{d^{2} y}{d x^{2}}-10 \frac{d y}{d x}+y=0$, $y(0)=1, y(1)=2 e^{1 / 5}$ is

Options:
A $y=e^{5 x}+e^{-5 x}$
B $y=(1+x) e^{5 x}$
C $y=(1+x) e^{x / 5}$
D $y=(1+x) e^{-x / 5}$

The correct answer is:
$y=(1+x) e^{x / 5}$

Solution:
Let $y=e^{\pi x}$ be the solution of given differential equation,
$\Rightarrow \quad \frac{d y}{d x}=m e^{n x} \Rightarrow \frac{d^{2} y}{d x^{2}}=m^{2} e^{m x}$
$\therefore \quad 25 \frac{d^{2} y}{d x^{2}}-10 \frac{d y}{d x}+y=0$
$\Rightarrow \quad 25 \mathrm{m}^{2} e^{m x}-10 m e^{m x}+e^{m x}=0$
$\Rightarrow \quad e^{\operatorname{mux}}\left(25 m^{2}-10 m+1\right)=0$
$\Rightarrow$ Auxiliary equation
$\Rightarrow \quad(5 m-1)^{2}=0$
$\Rightarrow \quad m=\frac{1}{5}, \frac{1}{5}$
since, roots are real and euqal.
$\therefore$ General solution is $y=\left(c_{1}+c_{2} x\right) e^{x / 5} \ldots .$
$$
\begin{array}{l}
y(0)=1 \Rightarrow c_{1}=1 \\
y(1)=2 e^{1 / 5} \Rightarrow 2 c^{1 / 5}=\left(c_{1}+c_{2}\right) e^{1 / 5}
\end{array}
$$
$\Rightarrow \quad c_{1}+c_{2}=2 \Rightarrow c_{1}=1$
Putting the value of $c_{1}$ and $c_{2}$ in Eq. (i), we get particular solution
$$
y=(1+x) e^{x / 5}
$$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2012
The solution of $\cos y \frac{d y}{d x}=e^{x+\sin y}+x^2 e^{\sin y}$ is $f(x)+e^{-\sin y}=C$ (C is arbitrary real constant) where $f(x)$ is equal to

Options:
A $e^x+\frac{1}{2} x^3$
B $e^{-x}+\frac{1}{3} x^3$
C $e^{-x}+\frac{1}{2} x^3$
D $e^x+\frac{1}{3} x^3$

The correct answer is:
$e^x+\frac{1}{3} x^3$

Solution:
$-e^{-\sin y} \cos y \frac{d y}{d x}=-\left[e^x+x^2\right] \Rightarrow d\left(e^{-\sin y}\right)+\left(e^x+x^2\right) d x=0 \Rightarrow e^{-\sin y}+e^x+\frac{x^3}{3}=C$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2022
The solution of $\operatorname{det}\left(A-\lambda I_2\right)=0$ be 4 and 8 and $A=\left(\begin{array}{ll}2 & 3 \\ x & y\end{array}\right)$. Then

Options:
A $x=4, y=10$
B $x=5, y=8$
C $x=3, y=9$
D $x=-4, y=10$

The correct answer is:
$x=-4, y=10$

Solution:
$\operatorname{det}\left(A-\lambda I_2\right)=0 \Rightarrow\left|\begin{array}{cc}2-\lambda & 3 \\ x & y-\lambda\end{array}\right|=0$
$\begin{array}{ll}
\Rightarrow & (2-\lambda)(y-\lambda)-3 x=0 \\
\Rightarrow & 2 y-\lambda y-2 \lambda+\lambda^2-3 x=0 \\
\Rightarrow & \lambda^2-(y+2) \lambda+2 y-3 x=0 \\
& y+2=4+8 \Rightarrow y=10 \\
& \text { and } 2 y-3 x=4 \times 8 \Rightarrow x=-4
\end{array}$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2022
The solution of \(\frac{d y}{d x}=\frac{y}{x}+\tan \frac{y}{x}\) is

Options:
A \(x=c \sin (y / x)\)
B \(x=c \sin (x y)\)
C \(y=c \sin (y / x)\)
D \(x y=c \sin (x / y)\)

The correct answer is:
\(x=c \sin (y / x)\)

Solution:
Hints: \(\frac{d y}{d x}=\frac{y}{x}+\tan \frac{y}{x}\)
\(\begin{aligned}
& \text { Put } \frac{y}{x}=\theta, y=\theta x \\
& \frac{d y}{d x}=\theta+\frac{x d \theta}{d x} \\
& \theta+x \cdot \frac{d \theta}{d x}=\theta+\tan \theta, \frac{d \theta}{\tan \theta}=\frac{d y}{x}
\end{aligned}\)
\(\begin{aligned}
& \int \cot \theta d \theta=\int \frac{d x}{x} \\
& \log \sin \theta=\log x+\log c \\
& \sin \theta=x . c ., \sin \frac{y}{x}=x . c \\
& x=c \cdot \sin \frac{y}{x}
\end{aligned}\)

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2011
The solution of the differential equation $\frac{d y}{d x}+\frac{y}{x \log _{e} x}=\frac{1}{x}$ under the condition $y=1$ when $x=e$ is

Options:
A $2 y=\log _{e} x+\frac{1}{\log _{e} x}$
B $y=\log _{e} x+\frac{2}{\log _{e} x}$
C $y \log _{e} x=\log _{2} x+1$
D $y=\log _{e} x+e$

The correct answer is:
$2 y=\log _{e} x+\frac{1}{\log _{e} x}$

Solution:
Given, differential equation is $\frac{d y}{d x}+\frac{y}{x \log _{e} x}=\frac{1}{x}$
It is a linear equation of the form $\frac{d y}{d x}+P y=Q$
where $\quad P=\frac{1}{x \log _{e} x}$ and $Q=\frac{1}{x}$
$\therefore$ Integrating factor, $\mathrm{IF}=e$
$$
\begin{array}{l}
=e^{\int \frac{1}{x \log _{e} x} d x} \\
=e^{\left.\log \log _{e} x\right)} \\
=\log _{e} x
\end{array}
$$
$\therefore$ Solution of differential equation is $y \times \log _{e} x=\int \frac{1}{x} \log _{e} x d x$
$\Rightarrow y \times \log _{e} x=\frac{\left(\log _{e} x\right)^{2}}{2}+C$
When $y=1$ and $x=e,$ then
$$
\begin{aligned}
& 1 \times \log _{e} e=\frac{\left(\log _{e} e\right)^{2}}{2}+C \\
\Rightarrow \quad 1 &=\frac{1}{2}+C \\
\Rightarrow \quad C=\frac{1}{2}
\end{aligned}
$$
On putting $C=\frac{1}{2}$ in Eq. (i), we get
$$
y \times \log _{e} x=\frac{\left(\log _{e} x\right)^{2}}{2}+\frac{1}{2}
$$
$\Rightarrow \quad 2 y=\log _{e} x+\frac{1}{\log _{e} x}$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2014
The solution of the differential equation $\left(y^{2}+2 x\right) \frac{d y}{d x}=y$ satisfies $x=1, y=1$. Then the solution is

Options:
A $x=y^{2}\left(1+\log _{e} y\right)$
B $y=x^{2}\left(1+\log _{e} x\right)$
C $x=y^{2}\left(1-\log _{e} y\right)$
D $y=x^{2}\left(1-\log _{e} x\right)$

The correct answer is:
$x=y^{2}\left(1+\log _{e} y\right)$

Solution:
Given differential equation is
$\left(y^{2}+2 x\right) \frac{d y}{d x}=y \Rightarrow \frac{d x}{d y}=\left(\frac{y^{2}+2 x}{y}\right)$
$\Rightarrow \quad \frac{d x}{d y}=y+\frac{2 x}{y}$
$\Rightarrow \quad \frac{d x}{d y}-\frac{2}{y} \cdot x=y$
$\mathrm{IF}=e^{\int \frac{2}{y} d y}=e^{-2 \mathrm{bg} \gamma}=\mathrm{y}^{-2}=\frac{1}{\mathrm{y}^{2}}$
$\therefore$ Complete solution is $x \cdot \frac{1}{y^{2}}=\int y \cdot \frac{1}{y^{2}} d y+C$
$\Rightarrow \quad \frac{x}{y^{2}}=\int \frac{d y}{y}+C=\log _{e} y+c$
$\Rightarrow \quad x=y^{2} \log _{e} y+C y^{2}$
$A t+x=1 y=1$ then from Eq $(i),$ we get
$$
1=0+C \Rightarrow C=1
$$
$\therefore$ From Eq. (i), we get
$$
x=y^{2}\left(\log _{e} y+1\right)
$$
Which is the required solution.

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2013
The solution of the differential equation $y \frac{d y}{d x}=x\left[\frac{y^{2}}{x^{2}}+\frac{\phi\left(\frac{y^{2}}{x^{2}}\right)}{\phi^{\prime}\left(\frac{y^{2}}{x^{2}}\right)}\right]$ is (where, $c$ is a constant)

Options:
A $\phi\left(\frac{y^{2}}{x^{2}}\right)=c x$
B $x \phi\left(\frac{y^{2}}{x^{2}}\right)=c$
C $\phi\left(\frac{y^{2}}{x^{2}}\right)=c x^{2}$
D $x^{2} \phi\left(\frac{y^{2}}{x^{2}}\right)=c$

The correct answer is:
$\phi\left(\frac{y^{2}}{x^{2}}\right)=c x^{2}$

Solution:
Given differential equation can be rewritten as
$$
\frac{d y}{d x}=\frac{y}{x}+\frac{x \phi\left(\frac{y^{2}}{x^{2}}\right)}{y \phi^{\prime}\left(\frac{y^{2}}{x^{2}}\right)}
$$
Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore$ Given equation becomes.
$$
v+x \frac{d v}{d x}=\frac{v x}{x}+\frac{x \phi\left(\frac{v^{2} x^{2}}{x^{2}}\right)}{v x \phi^{\prime}\left(\frac{v^{2} x^{2}}{x^{2}}\right)}
$$
$\Rightarrow \quad x \frac{d v}{d x}=\frac{\phi\left(v^{2}\right)}{v \phi^{\prime}\left(v^{2}\right)}$
$\Rightarrow \quad \frac{v \psi^{\prime}\left(v^{2}\right)}{\phi\left(v^{2}\right)} d v=\frac{d x}{x}$
On integrating both sides, we get
$$
\frac{1}{2} \log \phi\left(v^{2}\right)=\log x+\log c_{1}
$$
$\Rightarrow \quad \log \phi\left(v^{2}\right)=2 \log x c_{1}$
$\Rightarrow \quad \phi\left(v^{2}\right)=\left(x c_{1}\right)^{2} \Rightarrow \phi\left(\frac{y^{2}}{x^{2}}\right)=x^{2} c_{1}^{2}$
$\Rightarrow \quad \phi\left(\frac{y^{2}}{x^{2}}\right)=x^{2} c \quad$ (put $\left.c_{1}^{2}=c\right]$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2014