datavorous/entrance-exam-dataset huggingface dataset:
The determinant $\left|\begin{array}{ccc}a^{2}+10 & a b & a c \\ a b & b^{2}+10 & b c \\ a c & b c & c^{2}+10\end{array}\right|$ is

Options:
A divisible by 10 but not by 100
B divisible by 100
C not divisible by 100
D not divisible by 10

The correct answer is:
divisible by 100

Solution:
$=\frac{1}{a b c}\left|\begin{array}{ccc}a\left(a^{2}+10\right) & a^{2} b & a^{2} c \\ a b^{2} & b\left(b^{2}+10\right) & b^{2} c \\ a c^{2} & b c^{2} & c\left(c^{2}+10\right)\end{array}\right|=\frac{a b c}{a b c}\left|\begin{array}{ccc}a^{2}+10 & a^{2} & a^{2} \\ b^{2} & b^{2}+10 & b^{2} \\ c^{2} & c^{2} & c^{2}+10\end{array}\right|$
$\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$
$=\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}+10\right)\left|\begin{array}{ccc}1 & 1 & 1 \\ \mathrm{~b}^{2} & \mathrm{~b}^{2}+10 & \mathrm{~b}^{2} \\ \mathrm{c}^{2} & \mathrm{c}^{2} & \mathrm{c}^{2}+10\end{array}\right|$
$=\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}+10\right)\left|\begin{array}{ccc}1 & 0 & 0 \\ \mathrm{~b}^{2} & 10 & 0 \\ \mathrm{c}^{2} & 0 & 10\end{array}\right|\left[\begin{array}{c}\mathrm{C}_{2}-\mathrm{C}_{1} \\ \mathrm{C}_{3}-\mathrm{C} 1\end{array}\right]$
$=\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}+10\right) 100$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2021
The difference between orbital angular momentum of an electron in a 4 f orbital and another electron in a 4 s orbital is

Options:
A $2 \sqrt{3}$ $\times \frac{h}{(2 \times 3.14)}$
B $3 \sqrt{2}$ $\times \frac{h}{(2 \times 3.14)}$
C $\sqrt{3}$ $\times \frac{h}{(2 \times 3.14)}$
D 2 $\times \frac{h}{(2 \times 3.14)}$

The correct answer is:
$2 \sqrt{3}$ $\times \frac{h}{(2 \times 3.14)}$

Solution:
Hint:
Orbital angular momentum $=\sqrt{\ell(\ell+1)} \frac{\mathrm{h}}{2 \pi}$
For 4 f orbital; $\ell=3 \therefore$ Orbital angular momentum $=\sqrt{3(3+1)} \frac{\mathrm{h}}{2 \pi}=2 \sqrt{3} \frac{\mathrm{h}}{2 \pi}$
For 4 s orbital; $\ell=0 \therefore$ Orbital angular momentum $=0$
$\therefore$ Difference $=(2 \sqrt{3}) \frac{\mathrm{h}}{2 \pi}$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2020
The different colours of litmus in acidic, neutral and basic solutions are, respectively.

Options:
A red, orange and blue
B blue, violet and red
C red, colouriess and blue
D red, violet and blue

The correct answer is:
red, violet and blue

Solution:
\begin{array}{c|c}
\hline Medium & Colour of Litmus Paper \\
\hline Acidic & Red \\
Neutral & Violet \\
Basic & Blue \\
\hline
\end{array}

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2013
The differential equation of all the ellipses centred at the origin and have axes as the co-ordinate axes is

Options:
A $y^{2}+x y^{\prime 2}-y y^{\prime}=0$
B $x y y^{\prime \prime}+x y^{\prime 2}-y y^{\prime}=0$
C $y y^{\prime \prime}+x y^{\prime 2}-x y^{\prime}=0$
D $x^{2} y^{\prime}+x y^{\prime \prime}-3 y=0$
Where $y^{\prime} \equiv \frac{d y}{d x}, y^{\prime \prime} \equiv \frac{d^{2} y}{d x^{2}}$

The correct answer is:
$x y y^{\prime \prime}+x y^{\prime 2}-y y^{\prime}=0$

Solution:
equation of ellipse centred at origin and have axes as the coordinate axes is
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
$\Rightarrow \frac{x}{a^{2}}+\frac{y y^{\prime}}{b^{2}}=0$
$\Rightarrow \quad \frac{-b^{2}}{a^{2}}=\frac{y y^{\prime}}{x}$
$\Rightarrow \quad 0=\frac{x\left(y^{\prime \prime} y^{\prime}+y^{\prime 2}\right)-y y^{\prime}}{x^{2}}$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2021
The differential equation of \(\mathrm{y}=\mathrm{ae}^{\mathrm{bx}}\) (a & \(\mathrm{b}\) are parameters) is

Options:
A \(\mathrm{yy}_1=\mathrm{y}_2^2\)
B \(\mathrm{yy}_2=\mathrm{y}_1^2\)
C \(\mathrm{yy}_1^2=\mathrm{y}_2\)
D \(\mathrm{yy}_2^2=\mathrm{y}_1\)

The correct answer is:
\(\mathrm{yy}_2=\mathrm{y}_1^2\)

Solution:
\(\begin{aligned}
& \text { Hints: } y=a \cdot e^{b x} \ldots \ldots \ldots \ldots \text { (i) } \\
& y_1=a b e^{b x} \\
& y_1=b y \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \text { (ii) } \\
& y_2=b_1 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \text { (iii) } \\
& \text { Dividing (ii) & (iii) } \frac{y_1}{\mathrm{y}_2}=\frac{\mathrm{y}}{\mathrm{y}_1} \Rightarrow \mathrm{y}_1{ }^2=\mathrm{yy}_2
\end{aligned}\)

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2011
The differential equation of the family of curves $y=e^{x}(A \cos x+B \sin x)$ where $A, B$ are arbitrary constants is

Options:
A $\frac{d^{2} y}{d x^{2}}-9 x=13$
B $\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$
C $\frac{d^{2} y}{d x^{2}}+3 y=4$
D $\left(\frac{d y}{d x}\right)^{2}+\frac{d y}{d x}-x y=0$

The correct answer is:
$\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0$

Solution:
Hint:
$y=e^{x}(A \cos x+B \sin x)$
Differentiating w.r.t. x:-
$y^{\prime}=y+e^{x}(-A \sin x+B \cos x)$
Differentiating w.r.t. $x$ once again:-
$\begin{array}{c}
y^{\prime \prime}=y^{\prime}+\left(y^{\prime}-y\right)+e^{x}(-A \cos x-B \sin x) \\
=2 y^{\prime}-y-y \quad \Rightarrow y^{\prime \prime}-2 y^{\prime}+2 y=0
\end{array}$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2020
The differential equation representing the family of curves $y^{2}=2 d(x+\sqrt{d})$ where $d$ is a parameter, is of

Options:
A order 2
B degree 2
C degree 3
D degree 4

The correct answer is:
degree 3

Solution:
Given, $\quad y^{2}=2 d(x+\sqrt{d})$
$\Rightarrow$
2y $y_{1}=2 d \Rightarrow d=y y_{1}$
From Eq. (i),
$$
y^{2}=2 y y_{1}\left(x+\sqrt{y y_{1}}\right)
$$
$\Rightarrow \quad y^{2}-2 y y_{1} x=\sqrt{y y_{1}} \cdot 2 y y_{1}$
$\Rightarrow\left(y^{2}-2 y y_{1} x\right)^{2}=4\left(y y_{1}\right)^{3}$
So, degree of above equation is 3.

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2018
The differential of $f(x)=\log _{e}\left(1+e^{10 x}\right)-\tan ^{-1}\left(e^{5 x}\right)$ at $x=0$ and for $d x=0.2$ is

Options:
A $0.5$
B $0.3$
C $-0.2$
D $-0.5$

The correct answer is:
$0.5$

Solution:
$\begin{array}{l}
\frac{f(x+d x)-f(x)}{d x}=f^{\prime}(x) \\
f(x+d x)-f(x)=f^{\prime}(x) d x
\end{array}$
Now,
$\begin{array}{l}
f^{\prime}(x)=\frac{10 e^{10 x}}{1+e^{10 x}}-\frac{5 \cdot e^{5 x}}{1+e^{10 x}}=\frac{10 e^{10 x}-5 e^{5 x}}{1+e^{10 x}} \\
\left(f^{\prime}(x)\right)_{x-0}=\frac{5}{2} \\
\Delta f(x)=0.2 \times \frac{5}{2}=0.5
\end{array}$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2021
The digit in the unit's place of the number $1 !+2 !+3 !+\ldots \ldots \ldots+99 !$ is

Options:
A 3
B 0
C 1
D 7

The correct answer is:
3

Solution:
$1+2+6+24$ $\quad\quad\quad$unit digit is zero
$=33$
$\therefore$ Unit digit = 3

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2021
The dimension of angular momentum is

Options:
A $\left[\mathrm{M}^{0} \mathrm{L}^{1} \mathrm{T}^{-1}\right]$
B $\left[\mathrm{M}^{1} \mathrm{L}^{2} \mathrm{T}^{-2}\right]$
C $\left[\mathrm{M}^{1} \mathrm{L}^{2} \mathrm{T}^{1}\right]$
D $\left[\mathrm{M}^{2} \mathrm{L}^{1} \mathrm{T}^{-2}\right]$

The correct answer is:
$\left[\mathrm{M}^{1} \mathrm{L}^{2} \mathrm{T}^{1}\right]$

Solution:
Dimension of angular momentum
$\mathbf{L}=\mathbf{r} \times \mathbf{p}$
$=[\mathrm{L}]\left[\mathrm{MLT}^{-1}\right]=\left[\mathrm{ML}^{2} \mathrm{T}^{-1}\right]$

Topic: Physics, Units and Dimensions, WBJEE, WBJEE 2012
The dimension of the universal constant of gravitation, $G$ is

Options:
A $\left[\mathrm{ML}^{2} \mathrm{T}^{-1}\right]$
B $\left[\mathrm{M}^{-1} \mathrm{L}^{3} \mathrm{T}^{-2}\right]$
C $\left[\mathrm{M}^{-1} \mathrm{L}^{2} \mathrm{T}^{-2}\right]$
D $\left[\mathrm{ML}^{3} \mathrm{T}^{-2}\right]$

The correct answer is:
$\left[\mathrm{M}^{-1} \mathrm{L}^{2} \mathrm{T}^{-2}\right]$

Solution:
The dimensions of universal constant $G$, is obtained by Newton's law of gravitation
$F=\frac{G m_{1} m_{2}}{r^{2}}$
$\begin{array}{l}
G=\frac{F \times r^{2}}{m_{1} m_{2}} \\
F=\text { Force }
\end{array}$
$r=$ Distance $m_{1}$ and $m_{2}$ are masses
$\begin{aligned}
G &=\frac{\left[M^{1} L^{1} T^{-2}\right]\left[L^{2}\right]}{[M][M]} \\
&=\frac{\left[M^{\prime} L^{2} T^{-2}\right]}{\left[M^{2}\right]}=\left[M^{-1} L^{2} T^{-2}\right]
\end{aligned}$

Topic: Physics, Units and Dimensions, WBJEE, WBJEE 2017
The direction ratios of the normal to the plane passing through the points (1,2,-3) ,(-1,-2,1) and parallel to $\frac{x-2}{2}=\frac{y+1}{3}=\frac{z}{4}$ is

Options:
A (2, 3, 4)
B $(14,-8,-1)$
C $(-2,0,-3)$
D $(1,-2,-3)$

The correct answer is:
$(14,-8,-1)$

Solution:
Direction ratios of line joining the points (1,2,-3) and (-1,-2,1) is $(-1-1,-2-2,1+3)$
i.e. (-2,-4,4)
Let the direction ratios of the normal to the plane is $(a, b, c),$ then
$$
\begin{array}{r}
2 a+3 b+4 c=0 ....(i)\\
-2 a-4 b+4 c=0 ....(ii)
\end{array}
$$
and
By Eqs.
(i) and (ii), we get
$\begin{aligned} \frac{a}{3(4)-(4)(-4)} &=\frac{b}{(-2)(4)-(2)(4)} \\ &=\frac{c}{(2)(-4)-(-2)(3)} \\ \Rightarrow \quad \frac{a}{12+16} &=\frac{b}{-8-8}=\frac{c}{-8+6} \\ \Rightarrow \quad \frac{a}{28} &=\frac{b}{-16}=\frac{c}{-2} \\ \Rightarrow \quad \frac{a}{14} &=\frac{b}{-8}=\frac{c}{-1} \end{aligned}$
Hence, direction ratios are (14,-8,-1)

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2019
The dispersed phase and dispersion medium of fog respectively are

Options:
A solid, liquid
B liquid, liquid
C liquid, gas
D gas, liquid

The correct answer is:
liquid, gas

Solution:
The dispersed phase is liquid and dispersion medium is gas in the fog

Topic: Chemistry, Surface Chemistry, WBJEE, WBJEE 2015
The displacement of a particle at time $t$ is $x$, where $x=t^4-k t^3$. If the velocity of the particle at time $t=2$ is minimum, then

Options:
A $\mathrm{k}=4$
B $\mathrm{k}=-4$
C $\mathrm{k}=8$
D $\mathrm{k}=-8$

The correct answer is:
$\mathrm{k}=4$

Solution:
Hints : $\frac{d x}{d t}=4 t^3-3 k t^2$
$\frac{d v}{d t}=12 t^2-6 \mathrm{kt}$ at $\mathrm{t}=2$
$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=0,48-12 \mathrm{k}=0 \quad ; \mathrm{k}=4$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2010
The displacement of a particle in a periodic motion is given by $y=4 \cos ^{2}\left(\frac{t}{2}\right) \sin (1000 t)$. This displacement may be considered as the result of superposition of $n$ independent harmonic oscillations. Here $n$ is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Given, $Y=4 \cos ^{2}\left(\frac{t}{2}\right) \cdot \sin (1000 t)$
$=2 \times 2 \cos ^{2}\left(\frac{t}{2}\right) \cdot \sin (1000 t)$
$\quad\left[\because 2 \cos ^{2} \frac{t}{2}=(1+\cos t)\right]$
$=2(1+\cos t) \sin (1000 t)$
$=2 \sin (1000 t)+2 \cos t \cdot \sin (1000 t)$
$=2 \sin (1000 t)+2 \sin (1000 t) \cdot \cos t$
$=2 \sin (1000 t)+\sin (1000 t+t)$
$$
\begin{aligned}[\because-2 \sin A \cdot \cos B=\sin (A+B)+\sin (A-B)] \\=2 \sin (1000 t)+\sin (1001 t)+\sin (999 t) \end{aligned}
$$
So, $n=3$

Topic: Physics, Oscillations, WBJEE, WBJEE 2014
The displacement of a particle in S.H.M. varies according to the relation \(x=4(\cos \pi t+\sin \pi \mathrm{t})\). The amplitude of the particle is

Options:
A -4
B 4
C \(4 \sqrt{2}\)
D 8

The correct answer is:
\(4 \sqrt{2}\)

Solution:
Hints: \(\mathrm{R} \sin \delta=4\)
\(\mathrm{R} \cos \delta=4\)
\(R=4 \sqrt{2}\)

Topic: Physics, Oscillations, WBJEE, WBJEE 2011
The displacement of a plane progressive wave in a medium, travelling towards positive \(x\)-axis with velocity \(4 \mathrm{~m} / \mathrm{s}\) at \(t=0\) is given by \(y=3 \sin 2 \pi\left(-\frac{x}{3}\right)\). Then the expression for the displacement at a later time \(t=4\) sec will be

Options:
A \(y=3 \sin 2 \pi\left(-\frac{x-16}{3}\right)\)
B \(y=3 \sin 2 \pi\left(\frac{-x-16}{3}\right)\)
C \(y=3 \sin 2 \pi\left(\frac{-x-1}{3}\right)\)
D \(y=3 \sin 2 \pi\left(\frac{-x-1}{3}\right)\)

The correct answer is:
\(y=3 \sin 2 \pi\left(-\frac{x-16}{3}\right)\)

Solution:
Hint : Let \(y=3 \sin [\omega t-\mathrm{kx}]\)
\(\frac{\text { at } t=0}{y=3 \sin (-k x)}\)
\(k=\frac{2 \pi}{3}\)
\(v=\frac{\omega}{k}\)
\(4=\frac{\omega}{2 \pi / 3} \quad \Rightarrow \quad \omega=\frac{8 \pi}{3}\)
\(\begin{gathered}\therefore \quad y=3 \sin \left[\frac{8 \pi}{3} t-\frac{2 \pi}{3} x\right] \\ \therefore t=4 \\ y=3 \sin \left[\frac{8 \pi}{3} \times 4-\frac{2 \pi}{3} x\right] \\ =3 \sin 2 \pi\left[\frac{-x+16}{3}\right] \\ y=3 \sin \left[2 \pi\left(-\frac{x-16}{3}\right)\right]\end{gathered}\)

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2023
The displacement $\mathrm{x}$ of a particle at time $\mathrm{t}$ is given by $\mathrm{x}=\mathrm{At}^2+\mathrm{Bt}+\mathrm{C}$ where $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are constants and $\mathrm{v}$ is velocity of a particle, then the value of $4 \mathrm{Ax}-\mathrm{v}^2$ is

Options:
A $4 \mathrm{AC}+\mathrm{B}^2$
B $4 \mathrm{AC}-\mathrm{B}^2$
C $2 \mathrm{AC}-\mathrm{B}^2$
D $2 \mathrm{AC}+\mathrm{B}^2$

The correct answer is:
$4 \mathrm{AC}-\mathrm{B}^2$

Solution:
$$
\begin{aligned}
& \text { Hints: } \mathrm{x}=\mathrm{At}^2+\mathrm{Bt}+\mathrm{c} \\
& \mathrm{v}=2 \mathrm{At}+\mathrm{B} \Rightarrow \mathrm{v}^2=4 \mathrm{~A}^2 \mathrm{t}^2+4 \mathrm{AB} \mathrm{t}+\mathrm{B}^2 \\
& 4 \mathrm{Ax}=4 \mathrm{~A}^2 \mathrm{t}^2+4 \mathrm{AB} \mathrm{t}+4 \mathrm{AC} \\
& \Rightarrow \mathrm{v}^2-4 \mathrm{ax}=\mathrm{B}^2-4 \mathrm{AC} \\
& \Rightarrow 4 A \mathrm{Ax}-\mathrm{v}^2=4 A C-\mathrm{B}^2
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2010
The distance between an object and a divergent lens is $\mathrm{m}$ times the focal length of the lens. The linear magnification produced by the lens is

Options:
A $\mathrm{m}$
B $\frac{1}{\mathrm{~m}}$
C $m+1$
D $\frac{1}{m+1}$

The correct answer is:
$\frac{1}{m+1}$

Solution:
$\mathrm{u}=-\mathrm{mf}$
$\frac{1}{\mathrm{v}}-\frac{1}{(-m f)}=-\frac{1}{f} \Rightarrow \frac{1}{\mathrm{v}}=-\frac{1}{f}\left(1+\frac{1}{m}\right) \Rightarrow-\frac{\mathrm{v}}{\mathrm{u}}=\left(\frac{1}{1+m}\right)$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2009
The distance covered by a particle in $t$ seconds is given by $x=3+8 t-4 t^2$. After 1 second velocity will be

Options:
A 0 unit/second
B 3 units/second
C 4 units/second
D 7 units/second

The correct answer is:
0 unit/second

Solution:
$$
\begin{array}{r}
\text { Hints : } v=\frac{d x}{d t}=8-8 t \\
t=1, v=8-8=0
\end{array}
$$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2009
The ease of hydrolysis in the compounds $\mathrm{CH}_{3} \mathrm{COCl}(\mathrm{I}), \mathrm{CH}_{3}-\mathrm{CO}-\mathrm{O}-\mathrm{COCH}_{3}(\mathrm{II})$
$\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(\mathrm{III})$ and $\mathrm{CH}_{3} \mathrm{CONH}_{2}$ (IV) is of
the order

Options:
A $I>II>III>IV$
B $IV>III>II>I$
C $I>II>IV>III$
D $II>I>IV>III$

The correct answer is:
$I>II>III>IV$

Solution:
The ease of hydrolysis of carbonyl compounds depend on the group attached to the carbonyl carbon. Among the given options, chlorine $(-$ Cl) group is clectron withdrawing group which makes the carbonyl group electrophilic. Hence, hydrolysis can occur most easily. The order for electron withdrawing tendency among given options is as follows:
$$
\begin{array}{l}
-\mathrm{Cl}>-\mathrm{OCOCH}_{3}>-\mathrm{OC}_{2} \mathrm{H}_{5}>-\mathrm{NH}_{2} \\\end{array}$$
So, the correct order for the ease of hydrolysis in the given compounds is as follows:
$\mathrm{CH}_{3} \mathrm{COCl}>\mathrm{CH}_{3} \mathrm{COOCOCH}_{3}$
$>\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}>\mathrm{CH}_{3}{\mathrm{IV}} \mathrm{CONH}_{2}$

Topic: Chemistry, Carboxylic Acid Derivatives, WBJEE, WBJEE 2018
The ease of Nitration of the following three hydrocarbons follows the order

Options:
A II \(=\) III \(\approx\) I
B II \( > \) III \( > \) I
C III \( > \) II \( > \) I
D I \(=\) II \( > \) II

The correct answer is:
I \(=\) II \( > \) II

Solution:
Hints: Stability order of Arenium ion
\(\text { II } > \text { III } > \text { I }\)

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2011
The eccentric angle in the first quadrant of a point on the ellipse $\frac{x^{2}}{10}+\frac{y^{2}}{8}=1$ at a distance 3 units from the centre of the ellipse is

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

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

Solution:
Let $P(\sqrt{10} \cos \theta, \sqrt{8} \sin \theta)$ be the required point
on $\frac{x^{2}}{10}+\frac{y^{2}}{8}=1$
Whose distance from centre (0,0) is 3 units. $\therefore 10 \cos ^{2} \theta+8 \sin ^{2} \theta=9=9\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$
$$
\Rightarrow \quad \tan ^{2} \theta=1 \Rightarrow \theta=\frac{\pi}{4}
$$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2012
The eccentricity of the hyperbola \(4 x^2-9 y^2=36\) is

Options:
A \(\frac{\sqrt{11}}{3}\)
B \(\frac{\sqrt{15}}{3}\)
C \(\frac{\sqrt{13}}{3}\)
D \(\frac{\sqrt{14}}{3}\)

The correct answer is:
\(\frac{\sqrt{13}}{3}\)

Solution:
Hints: \(\frac{x^2}{9}-\frac{y^2}{4}=1\)
\(\begin{aligned}
& a=3, b=2 \\
& \therefore e=\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{13}{9}}=\frac{\sqrt{13}}{3}
\end{aligned}\)

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2011
The electric field of a plane electromagnetic wave of wave number \(k\) and angular frequency \(\omega\) is given by \(\vec{E}=E_0(\hat{i}+\hat{j}) \sin (k z-\omega t)\). Which of the following gives the direction of the associated magnetic field \(\vec{B}\) ?

Options:
A \(\hat{k}\)
B \(-\hat{i}+\hat{j}\)
C \(-\hat{i}-\hat{j}\)
D \(\hat{\mathrm{i}}-\hat{\mathrm{k}}\)

The correct answer is:
\(-\hat{i}+\hat{j}\)

Solution:
Hint: \(\vec{E} \cdot \vec{B}=0\) and in the plane of \(X Y\)

Topic: Physics, Electromagnetic Waves, WBJEE, WBJEE 2023
The electric potential for an electric field directed parallel to $\mathrm{X}$-axis is shown in the figure. Choose the correct plot of electric field strength.

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Physics, Electrostatics, WBJEE, WBJEE 2022
The electronic configuration of $\mathrm{Cu}$ is

Options:
A [Ne] $3 s^{2}, 3 p^{6}, 3 d^{9}, 4 s^{2}$
B [Ne] $3 s^{2}, 3 p^{6}, 3 d^{10}, 4 s^{1}$
C [Ne] $3 s^{2}, 3 p^{6}, 3 d^{3}, 4 s^{2}, 4 p^{6}$
D $[\mathrm{Ne}] 3 \mathrm{s}^{2}, 3 p^{6}, 3 d^{5} \cdot 4 s^{2} \cdot 4 p^{4}$

The correct answer is:
[Ne] $3 s^{2}, 3 p^{6}, 3 d^{10}, 4 s^{1}$

Solution:
Atomic number of Cu is 29 .
Thus, its electronic configuration can be written as
$$
\mathrm{Cu}_{29}=1s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}, 3 d^{10}, 4 s^{1}
$$
(Since, fully filled orbitals are more stable, so one electron from 4 s orbital transfers into $3 d$ orbital.
The electronic configuration of neon is
$$
\mathrm{Ne}_{10}=1 \mathrm{s}^{2}, 2 s^{2}, 2 p^{6}
$$
Thus, using inert gas, the configuration of Cu can also be represented as
$$
\mathrm{Cu}_{29}=[\mathrm{Ne}] 3 \mathrm{s}^{2}, 3 p^{6}, 3 d^{10}, 4 \mathrm{s}^{1}
$$

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2014
The electronic transitions from \(\mathrm{n}=2\) to \(\mathrm{n}=1\) will produce shortest wavelength in (where \(\mathrm{n}=\) principal quantum state)

Options:
A \(\mathrm{Li}^{+2}\)
B \(\mathrm{He}^{+}\)
C \(\mathrm{H}\)
D \(\mathrm{H}^{+}\)

The correct answer is:
\(\mathrm{Li}^{+2}\)

Solution:
Hints: \(\frac{1}{\lambda}=\mathrm{z}^2 \cdot \mathrm{R}_{\mathrm{H}}\left[1 / \mathrm{n}_1^2-1 / \mathrm{n}_2^2\right]\)
\(\begin{aligned}
& \Rightarrow \frac{1}{\lambda}=(\mathrm{z})^2 \cdot \mathrm{R}_{\mathrm{H}}\left\{\frac{1}{1}-\frac{1}{4}\right\}=\frac{3}{4} \mathrm{R}_{\mathrm{H}} \mathrm{z}^2 \\
& \therefore \lambda \propto 1 / \mathrm{z}^2
\end{aligned}\)
Hence, for shortest \(\lambda, \mathrm{z}\) must be maximum, which is for \(\mathrm{Li}^{+2}\).

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2011
The emission spectrum of hydrogen discovered first and the region of the electromagnetic spectrum in which it belongs, respectively are

Options:
A Lyman, ultraviolet
B Lyman, visible
C Balmer, ultraviolet
D Balmer, visible

The correct answer is:
Balmer, visible

Solution:
In 1885 , Balmer for the first time showed that the wave numbers of spectral lines present in the visible region in hydrogen spectrum are given by
$$
\bar{v}\left(\mathrm{cm}^{-1}\right)=109677\left(\frac{1}{(2)^{2}}-\frac{1}{n^{2}}\right)
$$
Here, $n=3,4,5, \ldots$
Thus, Balmer spectrum of hydrogen was discovered first and it lies in the visible region.

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2014
The energy of an electron in first Bohr orbit of \(\mathrm{H}\)-atom is \(-13.6 \mathrm{eV}\). The possible energy value of electron in the excited state of \(\mathrm{Li}^{2+}\) is

Options:
A \(-122.4 \mathrm{eV}\)
B \(-30.6 \mathrm{eV}\)
C \(-30.6 \mathrm{eV}\)
D \(13.6 \mathrm{eV}\)

The correct answer is:
\(-30.6 \mathrm{eV}\)

Solution:
Hints: \(E_n=\frac{E_1}{n^2} \times z^2\)
\(=\frac{-13.6}{4} \times 9=-30.6 \mathrm{eV}\)
For the excited state, \(\mathrm{n}=2\) and for \(\mathrm{Li}^{++}\)ion, \(\mathrm{z}=3\)

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2011
The energy of gamma $(\gamma)$ ray photon is $E$, and that of an X-ray photon is $E_{X}$. If the visible light photon has an energy of $E_{v}$, then we can say that

Options:
A $E_{x}>E_{y}>E_{v}$
B $E_{\gamma}>E_{y}>E_{x}$
C $E_{\gamma}>E_{x}>E_{v}$
D $E_{x}>E_{\gamma}>E_{y}$

The correct answer is:
$E_{\gamma}>E_{x}>E_{v}$

Solution:
$E_{y} \geq 100 \mathrm{keV}$
$\begin{aligned}
E_{X} &=100 \mathrm{eV} \text { to } 100 \mathrm{keV} \\
E_{v} &=2.48 \mathrm{eV}
\end{aligned}$
So, we can say that
$E_{y}>E_{X}>E_{v}$

Topic: Physics, Electromagnetic Waves, WBJEE, WBJEE 2014
The energy released by the fission of one uranium atom is $200 \mathrm{MeV}$. The number of fissions per second required to produce $3.2 \mathrm{~W}$ of power is (Take $1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$ )

Options:
A $10^7$
B $10^{10}$
C $10^{15}$
D $10^{11}$

The correct answer is:
$10^{11}$

Solution:
Hints: $u=200 \mathrm{MeV}=200 \times 10^6 \mathrm{eV}=200 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J}$
$\mathrm{E}=3.2 \mathrm{~J}$
No of fissions $=\frac{3.2}{2 \times 1.6 \times 10^{-11}}=10^{11}$

Topic: Physics, Nuclear Physics, WBJEE, WBJEE 2010
The energy required to break one mole of hydrogen-hydrogen bonds in $\mathrm{H}_{2}$ is $436 \mathrm{kJ}$. What is the longest wavelength of light required to break a single hydrogenhydrogen bond?

Options:
A $68.5 \mathrm{nm}$
B $137 \mathrm{nm}$
C $274 \mathrm{nm}$
D $548 \mathrm{nm}$

The correct answer is:
$274 \mathrm{nm}$

Solution:
Given, energy of one mole of $\mathrm{H}_{2}$ bonds $=436 \mathrm{k} \mathrm{J}$
$$
\begin{array}{l}
\because E=\frac{n \cdot h c}{\lambda}\\
\therefore \quad \lambda=\frac{n.h c}{E} \\
\lambda=\frac{6.634 \times 10^{-} 34 \mathrm{s} \times 3 \times 10^{3} \mathrm{ms}^{-1} \times 6.023 \times 10^{23}}{436 \times 10^{3} \mathrm{J}} \mathrm{m} \\
=0.2736 \times 10^{-34+8+23-3} \\
=0.2736 \times 10^{-6} \mathrm{m} \\
=273.6 \mathrm{nm}=274 \mathrm{nm}
\end{array}
$$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2016
The enol form in which ethyle-3-oxobutanoate exists is

Options:
A
B
C
D

The correct answer is:

Solution:
Stablised by H-bonding and Ketonic group enolises relatively more than ester group.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2022
The enthalpy of vaporisation of a certain liquid at its boiling point of $35^{\circ} \mathrm{C}$ is $24.64 \mathrm{kJ} \mathrm{mol}^{-1}$. The value of change in entropy for the process is

Options:
A $704 \mathrm{K}\mathrm{mol}^{-1} \mathrm{mol}^{-1}$
B $80 \mathrm{K}\mathrm{mol}y^{-1} \mathrm{mol}^{-1}$
C $24.64 \mathrm{K}\mathrm{mol}^{-1} \mathrm{mol}^{-1}$
D $7.04 \mathrm{K}\mathrm{mol}^{-1} \mathrm{mol}^{-1}$

The correct answer is:
$80 \mathrm{K}\mathrm{mol}y^{-1} \mathrm{mol}^{-1}$

Solution:
Entropy of vaporisation ( $\Delta$ S)
$$
=\frac{\text { enthalpy of vaporisation }(\Delta H)}{\text { boiling point (in K)} }
$$
Given, $\Delta H=24.64 \mathrm{kJ} \mathrm{mol}^{-1}$ and
$\text { boiling point }=35+273=308 \mathrm{k}$
$\therefore$
$$
\begin{aligned}
\Delta S &=\frac{24.64 \times 10^{3} \mathrm{J} \mathrm{mol}^{-1}}{308 \mathrm{K}} \\
&=80 \mathrm{JK}^{-1} \mathrm{mol}^{-1}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2014