datavorous/entrance-exam-dataset huggingface dataset:
The initial pressure and volume of a given mass of an ideal gas $\left(\right.$ with $\frac{C_{p}}{C_{V}}=\gamma),$ taken in a cylinder fitted with a piston, are $p_{0}$ and $V_{0}$ respectively. At this stage the gas has the same temperature as that of the surrounding medium which is $T_{0}$. It is adiabatically compressed to a volume equal to $\frac{V_{0}}{2}$ Subsequently the gas is allowed to come to thermal equilibrium with the surroundings. What is the heat released to the surrounding?

Options:
A 0
B $(2^{\gamma-1}-1) \frac{p_{0} V_{0}}{\gamma-1}$
C $\gamma {p_{0}} {V_{0}} ln2$
D $\frac{{p_{0}} V_{0}}{2(\gamma-1)}$

The correct answers are:
$(2^{\gamma-1}-1) \frac{p_{0} V_{0}}{\gamma-1}$

Solution:
The equation of state for an ideal gas undergoing adiabatic process is given as,
$$
T V^{\gamma-1}=\text { constant }
$$
Let the temperature after adiabatic compression as given in the question be $T$ then,
$$
T_{0} V^{\gamma-1}=T\left(\frac{V_{0}}{2}\right)^{\gamma-1}
$$
$\Rightarrow \quad T=T_{0} 2^{\gamma-1}$
Now, heat released at volume $\frac{V_{0}}{2}$ to achieve
temperature $T_{0}$ The net heat released can be determined by the
equation.
$$
\begin{aligned}
\Delta Q &=\mu C_{V} \Delta T \\
&=\mu \times \frac{R}{\gamma-1} \quad\left(T_{0} 2^{\gamma-1}-T_{0}\right) \\
&=\frac{\mu R T_{0}}{\gamma-1}\left(2^{\gamma-1}-1\right)\left(\because \mu T_{0} R=p_{0} V_{0}\right)
\end{aligned}
$$
$\therefore$ Heat released $=\frac{p_{0} V_{0}}{\gamma-1}\left(2^{\gamma-1}-1\right)$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2019
The inputs to the digital circuit are as shown below. The output $Y$ is

Options:
A $A+B+\bar{C}$
B $(A+B) \bar{C}$
C $\bar{A}+\bar{B}+\bar{C}$
D $\bar{A}+\bar{B}+C$

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

Solution:
$$
\begin{aligned}
&\text { The output of digital circuit can be given as, }\\
&Y=\overline{A B}+\bar{C}\\
&\text { It will be same as, } Y=\bar{A}+\bar{B}+\bar{C}
\end{aligned}
$$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2015
The insulated plates of a charged parallel plate capacitor (with small separation between the plates) are approaching each other due to electrostatic attraction. Assuming no other force to be operative and no radiation taking place, which of the following graphs approximately shows the variation with time $(t)$ of the potential difference $(V)$ between the plates?

Options:
A
B
C
D

The correct answer is:

Solution:
Force $=\frac{1}{2} \cdot \frac{\sigma}{\varepsilon_{0}} q$
$\therefore$ Relative acceleration of plates. $a_{\mathrm{rel}}=\frac{2 F}{M}$
Potential difference, $V=E d$ where, $d=d_{0}-\frac{1}{2} a_{\text {rel }} t^{2}$
(: plates are moving)
$\therefore$
$V=E\left[d_{0}-\frac{1}{2} a_{\mathrm{rel}} t^{2}\right]$
This is a equation of a parabola with downward concavity.

Topic: Physics, Capacitance, WBJEE, WBJEE 2018
The integrating factor of the differential equation $\left(1+x^{2}\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}$ is

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

The correct answer is:
$e^{\tan -1 x}$

Solution:
Given equation is
$$
\left(1+x^{2}\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}
$$
or it can be rewritten as
$$
\frac{d y}{d x}+\frac{1}{\left(1+x^{2}\right)} y=\frac{e^{\tan ^{-1} x}}{1+x^{2}}
$$
It is a linear differential equation of the form of
$$
\frac{d y}{d x}+P y=Q
$$
where, $P=\frac{1}{1+x^{2}}$ and $Q=\frac{e^{\tan ^{-1} x}}{1+x^{2}}$
$$
=e^{\int \frac{1}{1+x^{2}} d x}=e^{\tan ^{-1} x}
$$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2014
The integrating factor of the differential equation $3 x \log _{e} x \frac{d y}{d x}+y=2 \log _{e} x$ is given by

Options:
A $\left(\log _{e} x\right)^{3}$
B $\log _{e}\left(\log _{e} x\right)$
C $\log _{e} x$
D $\left(\log _{e} x\right)^{1 / 3}$

The correct answer is:
$\left(\log _{e} x\right)^{1 / 3}$

Solution:
Given, $3 x \log _{e} x \frac{d y}{d x}+y=2 \log _{e} x$
Dividing both sides by $3 x \log _{e} x,$ we get $\frac{d y}{d x}+\frac{1}{3 x \log _{e} x} y=\frac{2 \log _{e} x}{3 x \log _{e} x}$
$\Rightarrow \frac{d y}{d x}+\frac{1}{3 x \log _{e} x} y=\frac{2}{3 x}$
which is linear form $\frac{d y}{d x}+P y=Q,$ where $P$ and $Q$ are function of $x$ and the integrating factor is given by the following formula $e^{\int P d x}$.
$\therefore$
$\mathrm{IF}=e^{\int \frac{1}{3 x \log _{t} x}} d x$
$\begin{aligned} \text { Put } \log _{e} x &=t \Rightarrow \frac{1}{x} d x=d t \\ &=e^{1 / 3} \int \frac{d t}{t}=e^{1 / 3} \log t \\ &=e^{\log 1^{1 / 3}}=t^{1 / 3}=\left(\log _{e} x\right)^{-1 / 3} \end{aligned}$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2012
The integrating factor of the differential equation $\frac{d y}{d x}+\left(3 x^{2} \tan ^{-1} y -x^{3}\right)\left(1+y^{2}\right)=0$ is

Options:
A $e^{x^{2}}$
B $e^{x^{3}}$
C $e^{3 x^{2}} \quad$
D $e^{3 x^{3}}$

The correct answer is:
$e^{x^{3}}$

Solution:
Given, $\frac{d y}{d x}=-\left(3 x^{2} \tan ^{-1} y-x^{3}\right)\left(1+y^{2}\right)$
$\Rightarrow \quad \frac{d y}{d x}=x^{3}\left(1+y^{2}\right)-3 x^{2}\left(\tan ^{-1} y\right)\left(1+y^{2}\right)$
$\Rightarrow \frac{1}{\left(1+y^{2}\right)} \cdot \frac{d y}{d x}=x^{3}-3 x^{2} \tan ^{-1} y$
$\Rightarrow \frac{1}{1+y^{2}} \cdot \frac{d y}{d x}+3 x^{2} \tan ^{-1} y=x^{3}$
Put $\tan ^{-1} y=t \Rightarrow \frac{1}{1+y^{2}} \cdot \frac{d y}{d x}=\frac{d t}{d x}$
$\therefore \quad \frac{d t}{d x}+3 t x^{2}=x^{3}$
which is linear differential equation in $t$.
Now.
$\mathrm{I} \quad \mathrm{F}=e^{\int 3 \mathrm{x}^{2} d x}=e^{x^{3}}$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2015
The integrating factor of the differential equation $x \log x \frac{d y}{d x}+y=2 \log x$ is given by

Options:
A $\mathrm{e}^{\mathrm{x}}$
B $\log x$
C $\log (\log \mathrm{x})$
D $x$

The correct answer is:
$\log x$

Solution:
Hints : $\frac{d y}{d x}+\frac{1}{x \log x} \cdot y=\frac{2}{x}$
$$
\begin{aligned}
& \text { If }=e^{\int \frac{1}{x \log x} d x}=e^{\int \frac{1 / x}{\log x} d x} \\
& =e^{\log (\log x)}=\log x
\end{aligned}
$$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2009
The integrating factor of the first order differential equation $x^{2}\left(x^{2}-1\right) \frac{d y}{d x}+x\left(x^{2}+1\right) y=x^{2}-1$ is

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

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

Solution:
We have,
$x^{2}\left(x^{2}-1\right) \frac{d y}{d x}+x\left(x^{2}+1\right) y=x^{2}-1$
$\Rightarrow \quad \frac{d y}{d x}+\frac{x^{2}+1}{x\left(x^{2}-1\right)} y=\frac{1}{x^{2}}$
$\begin{aligned} I F=& e^{\int \frac{x^{2}+1}{x\left(x^{2}-1\right)} d x} \\ &=e^{\int \frac{x^{2}+1}{x(x-1)(x+1)}} d x \end{aligned}$
Let $\frac{x^{2}+1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$
$\Rightarrow x^{2}+1=A(x-1)(x+1)+B x(x+1)$ $\quad+\operatorname{Cx}(x-1)$
$\begin{array}{ll}\text { Put } x=0, & \\ \therefore \quad & 1=-A \\ \Rightarrow & A=-1 \\ \Rightarrow &
\end{array}$
Put $x=1
\therefore \quad 2=2 B$
$\begin{array}{ll}\therefore & B=1 \\ \Rightarrow & x=-1\end{array}$
$\therefore \quad 2=2 C$
$\Rightarrow C=1$
$\therefore \frac{x^{2}+1}{x(x-1)(x+1)}=\frac{-1}{x}+\frac{1}{x-1}+\frac{1}{x+1}$
$\therefore \quad I F=e^{\int\left(\frac{-1}{x}+\frac{1}{x-1}+\frac{1}{x+1}\right) d x}$
$=e^{[-\log x+\ln (x-1)+\log (x+1)]}$
$=e^{\ln \left(\frac{x^{2}-1}{x}\right)}=\frac{x^{2}-1}{x}$
$=x-\frac{1}{x}$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2017
The intensity of a sound appears to an observer to be periodic. Which of the following can be the cause of it?

Options:
A The intensity of the source is periodic
B The source is moving towards the observer
C The observer is moving away from the source
D The source is producing a sound composed of two nearby frequencies

The correct answers are:
The intensity of the source is periodic, The source is producing a sound composed of two nearby frequencies

Solution:
The intensity of a sound source appears be periodic due to
(i) source intensity is periodic.
(ii) source is producing a sound composed of two nearby frequencies.

Topic: Physics, Wave Optics, WBJEE, WBJEE 2018
The intensity of light emerging from one of the slits in a Young's double slit experiment is found to be $1.5$ limes the intensity of light emerging from the other slit. What will be the approximate ratio of intensity of an interference maximum to that of an interference minimum

Options:
A $2.25$
B 98
C 5
D $9.9$

The correct answer is:
98

Solution:
Hint:
$I_{1}=1.51_{2}$
$\frac{\left.\right|_{1}}{I_{2}}=\frac{3}{2} \cdot \quad \frac{l_{\max }}{I_{\min }}=\left(\frac{\sqrt{L_{1}}+\sqrt{L_{2}}}{\sqrt{I_{1}}-\sqrt{L_{2}}}\right)^{2}=\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)^{2}=98$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2020
The intensity of magnetization of a bar magnet is $5.0 \times 10^{4} \mathrm{Am}^{-1}$. The magnetic length and the area of cross-section of the magnet are $12 \mathrm{cm}$ and $1 \mathrm{cm}^{2}$ respectively. The magnitude of magnetic moment of this bar magnet is (in SI unit)

Options:
A 0.6
B 1.3
C 1.24
D 2.4

The correct answer is:
0.6

Solution:
We know that Intensity of magnetisation
$I=\frac{M}{V}$
[where $M=$ magnetic moment, $V=$ volume]
So
$\begin{aligned}
M=I V &=5.0 \times 10^{4} \times \frac{12}{100} \times \frac{1}{(100)^{2}} \\
&=60 \times 10^{4} \times 10^{-6}=0.6 \mathrm{Am}^{2}
\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, WBJEE, WBJEE 2014
The intercept on the line \(y=x\) by the circle \(x^2+y^2-2 x=0\) is \(A B\). Equation of the circle with \(A B\) as diameter is

Options:
A \(x^2+y^2=1\)
B \(\mathrm{x}(\mathrm{x}-1)+\mathrm{y}(\mathrm{y}-1)=0\)
C \(x^2+y^2=2\)
D \((x-1)(x-2)+(y-1)+(y-2)=0\)

The correct answer is:
\(\mathrm{x}(\mathrm{x}-1)+\mathrm{y}(\mathrm{y}-1)=0\)

Solution:
Hints: \(2 x^2-2 x=0 \quad x(x+1)=0 \quad x=0,1 ; y=0,1\)
\((0,0),(1,1)\) as diametric ends
\(\begin{aligned}
& (x-0)(x-1)+(y+0)(y-1)=0 \\
& x^2+y^2-x-y=0
\end{aligned}\)

Topic: Mathematics, Circle, WBJEE, WBJEE 2011
The intermediate image formed by the objective of a compound microscope is

Options:
A real, inverted and magnified
B real, erect and magnified
C virtual, erect and magnified
D virtual, inverted and magnified

The correct answer is:
real, inverted and magnified

Solution:
The intermediate image formed by the objective of a compound microscope is real, inverted and magnified.

Topic: Physics, Ray Optics, WBJEE, WBJEE 2014
The intermediate $J$ in the following Wittig reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2014
The ionization energy of hydrogen is $13.6 \mathrm{eV}$. The energy of the photon released when an electron jumps from the first excited state $(n=2)$ to the ground state of a hydrogen atom is

Options:
A $3.4 \mathrm{eV}$
B $4.53 \mathrm{eV}$
C $10.2 \mathrm{eV}$
D $13.6 \mathrm{eV}$

The correct answer is:
$10.2 \mathrm{eV}$

Solution:
Energy required to removed electron in th $(n=2)$ stage The energy of the photon
$\begin{array}{l}
=R\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right) \Rightarrow 13.6 \mathrm{eV}\left(\frac{1}{1}-\frac{1}{4}\right) \\
=13.6 \mathrm{eV}\left(\frac{4-1}{4}\right)=13.6 \mathrm{eV} \times \frac{3}{4}=\frac{40.8}{4} \mathrm{eV} \\
=10.2 \mathrm{eV}
\end{array}$

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2014
The ionization energy of the hydrogen atom is 13.6 eV. The potential energy of the electron in $n=2$ state of hydrogen atom is

Options:
A $+3.4 \mathrm{eV}$
B $-3.4 \mathrm{eV}$
C $+6.8 \mathrm{eV}$
D $-6.8 \mathrm{eV}$

The correct answer is:
$-6.8 \mathrm{eV}$

Solution:
Given, $E_{n}=13.6 \mathrm{eV}$
Energy of an electron in $n^{\text {th }}$ state
$E_{n}=\frac{-13.6 z^{2} \mathrm{eV}}{n^{2}}$
$\therefore$ Energy of an electron in $n=2$ state
$E_{2}=\frac{18-6 z^{2}}{(2)^{2}}=-34 e V$
So.
$\begin{aligned}
P E &=2 E_{n=2} \\
&=2 \times(-3.4) \\
&=-6.8 \mathrm{eV}
\end{aligned}$

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2013
The isomerisation of 1 -butyne to 2 -butyne can be achieved by treatment with

Options:
A hydrochloric acid
B ammoniacal silver nitrate
C ammoniacal cuprous chloride
D ethanolic potassium hydroxide

The correct answer is:
ethanolic potassium hydroxide

Solution:
Isomerisation of 1 -butyne to 2 -butyne can be achived by treatment with ethanolic KOH

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2017
The IUPAC name of
is :

Options:
A 3-Formyl-2-pentene
B 2-Ethylbut-2-enal
C 3-Ethylbut-3enal
D 2-Ethylcrotonaldehyde

The correct answer is:
2-Ethylbut-2-enal

Solution:
Hint :

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2023
The IUPAC name of the compound $X$ is

Options:
A 4 -cyano-4-methy 4 - oxopentane
B 2 -cyano-2-methyl-4.oxopentane
C 2,2 -dimethyl-4-oxopentanenitrile
D 4 - cyano-4-methyl-2-pentanone

The correct answer is:
2,2 -dimethyl-4-oxopentanenitrile

Solution:
2,2-dimethyl-4-oxopentanenitrile

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2013
The IUPAC name of the following molecule is

Options:
A 5,6 -dimethyihept-2-ene
B 2,3 -dimethyhept-5-ene
C 5,6 -dimethyhept-3-ene
D $5-$ iso-propylhex-2-ene

The correct answer is:
5,6 -dimethyihept-2-ene

Solution:
The longest chain contains 7 C-atom, so the root word is hept and suffix -ene is added for double bond along with its position. Since, two methyl goups are also present at 5 and 6 positions, so the correct IUPAC name of the compound is 5.6 -dimethylhept-2-ene.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2014
The kinetic energy $\left(\mathrm{E}_{\mathrm{k}}\right)$ of a particle moving along $X$-axis varies with its position $(X)$ as shown in the figure. The force acting on the particle at $X=10 \mathrm{~m}$ is

Options:
A $5 \hat{\mathrm{i} N}$
B $0 \mathrm{~N}$
C $97.5 \hat{\mathrm{i} N}$
D $-5 \hat{\mathrm{i} N}$

The correct answer is:
$-5 \hat{\mathrm{i} N}$

Solution:
$$
\frac{d k}{d x}=\frac{d\left(\frac{1}{2} m v^2\right)}{d x}=\frac{1}{2} m \not{2} v \frac{d v}{d x}=m a=F
$$
So, Force is slope at $x=10 m \quad F=-\frac{20}{4}=-5 \hat{i}$

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2022
The kinetic study of a reaction like $v A \rightarrow P$ at $300 \mathrm{K}$ provides the following curve, where concentration is taken in mol $\mathrm{dm}^{-3}$ and time in min.

Identify the correct order $(n)$ and rate constant $(k)$

Options:
A $n=0, k=4.0 \mathrm{mol} \mathrm{dm}^{-3} \mathrm{min}^{-1}$
B $n=1 / 2, k=2.0 \mathrm{mol}^{1 / 2} \mathrm{dm}^{-3 / 2} \mathrm{min}^{-1}$
C $n=1, k=80 \mathrm{min}^{-1}$
D $n=2, k=16.0 \mathrm{dm}^{3} \mathrm{mol}^{-1} \mathrm{min}^{-1}$

The correct answer is:
$n=2, k=16.0 \mathrm{dm}^{3} \mathrm{mol}^{-1} \mathrm{min}^{-1}$

Solution:
From the given graph Slope $=\frac{(\text { rate })^{1 / 2}}{[A]}=4$
$\because$ rate $(r)=k[A]^{n}$
Where, $k=$ rate constant $[A]=$ concentration at time $t$ (in min).
$n=$ order.
According to the graph $(n=2)$ $\therefore \quad k=\frac{\text { rate }}{[A]^{n}}=[4]^{2}=16$
Hence, $n=2$ and
$$
k=16.0 \mathrm{dm}^{3} \mathrm{mol}^{-1} \min^{-1}
$$
and option (d) is the correct answer.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2019
The law of motion of a body moving along a straight line is $x=\frac{1}{2}$ vt. $x$ being its distance from a fixed point on the line at time $t$ and $v$ is its velocity there, Then

Options:
A acceleration $f$ varies directly with $x$
B acceleration $f$ vares inversely with $x$
C acceleration $f$ is constant
D acceleration $f$ varies directly with t

The correct answer is:
acceleration $f$ is constant

Solution:
We have, $\quad x=\frac{1}{2} v t$ $\Rightarrow \quad x=\frac{1}{2} \frac{d x}{d t} t$
$\left[\because v=\frac{d x}{d t}\right]$
$\Rightarrow \quad \frac{2 d t}{t}=\frac{d x}{x}$
$\Rightarrow \quad 2 \cdot \int \frac{d t}{t}=\int \frac{d x}{x}$
$\Rightarrow \quad 2 \log |t|+\log |c|=\log |x|$
$\Rightarrow \quad \log \left(t^{2} \cdot c\right)=\log x$
$\Rightarrow \quad x=t^{2} c$
$\Rightarrow$
$\frac{d x}{d t}=2 t c$
$\Rightarrow$
$\frac{d^{2} x}{d t^{2}}=2$
$\Rightarrow$ acceleration $f$ is constant.

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2018
The least distance of vision of a long sighted person is $60 \mathrm{cm}$. By using a spectacle lens, this distance is reduced to $12 \mathrm{cm}$. The power of the lens is

Options:
A $+5.0 D$
B $(20 / 3) \mathrm{D}$
C $-(10 / 3)$ D
D $+2.0 \mathrm{D}$

The correct answer is:
$(20 / 3) \mathrm{D}$

Solution:
Here, $v=-60 \mathrm{cm}, u=-12 \mathrm{cm}$
$\therefore$ By using the relation, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
We have. $\frac{1}{f}=\frac{1}{-60}-\frac{1}{-12}$
$\Rightarrow \quad \frac{1}{f}=\frac{1}{15} \mathrm{cm}$ or $\frac{100}{15} \mathrm{m}$
So, the power of lens, $P=\frac{100}{15}=+\frac{20}{3} \mathrm{D}$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2013
The least positive integer $n$ such that $\left(\begin{array}{cc}\cos \frac{\pi}{4} & \sin \frac{\pi}{4} \\ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4}\end{array}\right)^{n}$ is an identity matrix of order 2 is

Options:
A 4
B 8
C 12
D 16

The correct answer is:
8

Solution:
We have, $\left(\begin{array}{cc}\cos \pi / 4 & \sin \pi / 4 \\ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4}\end{array}\right)^{n}$
Let $A=\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)$
$\Rightarrow \quad A=\left(\begin{array}{cc}k & k \\ -k & k\end{array}\right)$
$\Rightarrow A^{2}=\left(\begin{array}{cc}k & k \\ -k & k\end{array}\right)\left(\begin{array}{cc}k & k \\ -k & k\end{array}\right)=\left(\begin{array}{cc}0 & 2 k^{2} \\ -2 k^{2} & 0\end{array}\right)$
$A^{4}=\left(\begin{array}{cc}0 & 2 k^{2} \\ -2 k^{2} & 0\end{array}\right)\left(\begin{array}{cc}0 & 2 k^{2} \\ -2 k^{2} & 0\end{array}\right)$
$A^{4}=\left(\begin{array}{cc}-4 k^{4} & 0 \\ 0 & -4 k^{4}\end{array}\right)=\left(\begin{array}{cc}-4 \times \frac{1}{4} & 0 \\ 0 & -4 \times \frac{1}{4}\end{array}\right)$
$\Rightarrow A^{4}=\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)$
$\Rightarrow A^{B}=\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)\left(\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right)=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2018
The least positive value of $t$, so that the lines $x=t+\alpha, y+16=0 \quad$ and $\quad y=\alpha x$ are concurrent, is

Options:
A 2
B 4
C 16
D 8

The correct answer is:
8

Solution:
Consider the given equation of lines,
$$
\begin{array}{r}
x-(t+\alpha)=0 \\
y+16=0
\end{array}
$$
and
$$
-\alpha x+y=0
$$
Since, these lines are concurrent, therefore the system of equations is consistent.
Now,
$$
\left|\begin{array}{ccc}
1 & 0 & -(t+\alpha) \\
0 & 1 & 16 \\
-\alpha & 1 & 0
\end{array}\right|=0
$$
$$
\begin{array}{lr}
\Rightarrow & 1(0-16)-(t+\alpha)(0+\alpha)=0 \\
\Rightarrow & -16-\alpha(t+\alpha)=0 \\
\Rightarrow & \alpha(t+\alpha)+16=0 \\
\Rightarrow & \alpha^{2}+t \alpha+16=0
\end{array}
$$
Clearly, $\alpha$ should be real. $\therefore t^{2}-4 \times 16 \geq 0 \Rightarrow t^{2}-64 \geq 0$
Hence, least positive value of $t$ is 8.

Topic: Mathematics, Determinants, WBJEE, WBJEE 2015
The least value of $2 x^{2}+y^{2}+2 x y+2 x-3 y+8$ for real
numbers $x$ and $y,$ is

Options:
A 2
B 8
C 3
D $-1 / 2$

The correct answer is:
$-1 / 2$

Solution:
$2 x^{2}+y^{2}+2 x y+2 x-3 y+8$
$=\frac{1}{2}\left|4 x^{2}+2 y^{2}+4 x y+4 x-6 y+16\right|$
$=\frac{1}{2}\left[\left(y^{2}-8 y\right)+\left(4 x^{2}+y^{2}+4 x y+4 x+2 y\right)+16\right].$
$=\frac{1}{2}\left|(y-4)^{2}+(2 x+y+1)^{2}-1\right|$
So, least value will be $-\frac{1}{2}$ at $y=4$ and $x=\frac{-5}{2}$

Topic: Mathematics, Pair of Lines, WBJEE, WBJEE 2015
The length of a metal wire is $L_{1}$ when the tension is $T_{1}$ and $L_{2}$ when the tension is $T_{2}$ The unstretched length of wire is

Options:
A $\frac{L_{1}+L_{2}}{2}$
B $\sqrt{L_{1} L_{2}}$
C $\frac{T_{2} L_{1}-T_{1} L_{2}}{T_{2}-T_{1}}$
D $\frac{T_{2} L_{1}+T_{1} L_{2}}{T_{2}+T_{1}}$

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

Solution:
Let the initial length of the metal wire is $L$.
The strain at tension $T_{1}$ is $\Delta L_{1}=L_{1}-L$
The strain at tension $T_{2}$ is $\Delta L_{2}=L_{2}-L$
Suppose, the Young's modulus of the wire is $Y$, then
$\frac{\frac{T_{1}}{A}}{\frac{\Delta L_{1}}{L}}=\rac{\frac{T_{2}}{A}}{\frac{\Delta L}{L}}$
where, $A$ is an cross-section of the wire assumed to be same at all the situations. $\begin{aligned} \Rightarrow \frac{T_{1}}{A} \times \frac{L}{\Delta L_{1}} &=\frac{T_{2}}{A} \times \frac{L}{\Delta L_{2}} \\ \Rightarrow \quad \frac{T_{1}}{\left(L_{1}-L\right)} &=\frac{T_{2}}{\left(L_{2}-L\right)} \\ & T_{1}\left(L_{2}-L\right)=T_{2}\left(L_{1}-L\right) \cdot L=\frac{T_{2} L_{1}-T_{1} L_{2}}{T_{2}-T_{1}} \end{aligned}$

Topic: Physics, Mechanical Properties of Solids, WBJEE, WBJEE 2015
The length of conjugate axis of a hyperbola is greater than the length of transverse axis. Then, the eccentricity e is

Options:
A $=\sqrt{2}$
B $>\sqrt{2}$
C $ < \sqrt{2}$
D $ < \frac{1}{\sqrt{2}}$

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

Solution:
Let the length of conjugate axis $=b$
and the length of transvers axis $=a$ Given that, $b>a$
Now,$b^{2}>a^{2}$
$\therefore$ Eccentricity $(e)=\sqrt{1+\frac{b^{2}}{a^{2}}}$
$e^{2}=\left(\sqrt{1+\frac{b^{2}}{a^{2}}}\right)^{2}>2$
$e>\sqrt{2}$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2019
The length of the chord of the parabola $y^{2}=4 a x(a>0)$ which passes through the vertex and makes an acute angle $\alpha$ with the axis of the parabola is

Options:
A $\pm 4 a \cot \alpha \operatorname{cosec} \alpha$
B $4 a \cot \alpha \operatorname{cosec} \alpha$
C $-4 a \cot \alpha \operatorname{cosec} \alpha$
D $4 a \operatorname{cosec}^{2} \alpha$

The correct answer is:
$4 a \cot \alpha \operatorname{cosec} \alpha$

Solution:
Hint:
Equation of OP:-
$y=x \tan \alpha$
Solving with $y^{2}=4 a x$, we get :
$x^{2} \tan ^{2} \alpha=4 a x \quad \Rightarrow x=4 a \cot ^{2} \alpha$
Substituting, $y=4 a \cot \alpha$
$\therefore \mathrm{P} \equiv\left(4 \mathrm{a} \cot ^{2} \alpha, 4 \mathrm{a} \cot \alpha\right)$
So, $\overline{\mathrm{OP}}=\sqrt{16 \mathrm{a}^{2} \cot ^{4} \alpha+16 \mathrm{a}^{2} \cot ^{2} \alpha}=4 \mathrm{a} \cot \alpha \operatorname{cosec} \alpha\left(\right.$ as $0^{\circ} < \alpha < 90^{\circ}$, so $\left.\cot \alpha>0, \operatorname{cosec} \alpha>0\right)$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2020
The length of the latus rectum of the ellipse \(16 x^2+25 y^2=400\) is

Options:
A \(5 / 16\) unit
B \(32 / 5\) unit
C \(16 / 5\) unit
D \(5 / 32\) unit

The correct answer is:
\(32 / 5\) unit

Solution:
Hints : Length of latus rectum \(=2 \frac{b^2}{a}=\frac{2 \times 16}{5}=\frac{32}{5}\)
\(\begin{aligned}
& 16 x^2+25 y^2=400 \\
& \frac{x^2}{25}+\frac{y^2}{16}=1 \\
& a^2=25 ; b^2=16
\end{aligned}\)

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2011
The letters of the word COCHIN are permuted and all permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is

Options:
A 96
B 48
C 183
D 267

The correct answer is:
96

Solution:
In a dictionary, the words at each stage are arranged in aiphabetical order.
In the given problem, we must consider the words beginning with $\mathrm{C}, \mathrm{C}, \mathrm{H}, I, \mathrm{N}, \mathrm{O}$ in order.
So. Number of words starting with $C C=4 !$
Number of words starting with $\mathrm{CH}=4!$
Number of words starting with $\mathrm{Cl}=4!$
Number of words starting with $\mathrm{CN}=4!$
Next word staring with CO ie. COCHIN = 1
$\therefore$ Required number of words $=4 \times 41=96$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2016
The letters of the word 'COCHIN' are permuted and all the permutations are arranged in alphabetical order as in English dictionary. The number of words that appear before the word 'COCHIN', is

Options:
A 360
B 192
C 96
D 48

The correct answer is:
96

Solution:
The number of words starting with $\mathrm{CC}=4!$
The number of words starting with $\mathrm{CH}=4!$
The number of words starting with $\mathrm{Cl}=4!$
The number of words starting with $\mathrm{CN}=4!$ COCHIN is the first word in the list of words beginning with CO.
$\therefore$ Number of words that appear before the word COCHIN = 96

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2015