datavorous/entrance-exam-dataset huggingface dataset:
The area of the region, bounded by the curves $y=\sin ^{-1} x+x(1-x)$ and
$y=\sin ^{-1} x-x(1-x)$ in the first quadrant, is

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

The correct answer is:
$\frac{1}{3}$

Solution:
$\sin ^{-1} x$ is defined, if $-1 \leq x \leq 1$ In first quadrant $0 \leq x \leq 1$ and $x(1-x) \geq 0$
$\therefore$
$y=\sin ^{-1} x+x(1-x)$
Lies above $y=\sin ^{-1} x-x(1-x)$
On solving, we get $2 x(1-x)=0$
$\Rightarrow \quad \quad \quad \quad \quad \quad \quad \quad \quad=0,1$
$\therefore$ Required area $=\int_{0}^{1}\left(y_{1}-y_{2}\right) d x$
$=\int_{0}^{1}\left[\left\{\sin ^{-1} x+x(1-x)\right\}\right.$
$\left.-\left\{\sin ^{-1} x-x(1-x)\right\}\right] d x$
$=2 \int_{0}^{1}\left(x-x^{2}\right) d x$
$=2\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}=2\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{1}{3}$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2012
The area of the region bounded by the curves $y=x^{2}$ and $x=y^{2}$ is

Options:
A $1 / 3$
B $1 / 2$
C $1 / 4$
D 3

The correct answer is:
$1 / 3$

Solution:
Given curves are $y=x^{2}$ and $x=y^{2}$, which is the form of parabola.

The point of intersection, $x=\left(x^{2}\right)^{2}$$
\begin{array}{l}
\Rightarrow \quad x=x^{4} \Rightarrow x\left(1-x^{3}\right)=0 \\
\Rightarrow \quad x=0 \text { and } 1=x^{3} \\
\Rightarrow \quad x=0 \text { and } x=1 \\
\text { When } x=0, \text { then } y=0 \\
\text { When } x=1, \text { then } y=1^{2}=1
\end{array}
$
$\therefore$ The point of intersection is (0,0) and (1,1)
$\therefore$ Area of shaded region
$$
\begin{array}{l}
=\int_{0}^{1}\left(y_{2}-y_{1}\right) d x \\
=\int_{0}^{1}\left[\sqrt{x}-x^{2}\right] d x=\left[\frac{x^{3 / 2}}{3 / 2}-\frac{x^{3}}{3}\right]_{0}^{1} \\
=\frac{2}{3}(1)^{3 / 2}-\frac{(1)^{3}}{3}-0-0 \\
=\frac{2}{3}-\frac{1}{3}=\frac{1}{3} \mathrm{sq} \text { units }
\end{array}
$$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2014
The area of the region bounded by the curves $y=x^{3}, y=\frac{1}{x}, x=2$ is

Options:
A $4-\log _{e} 2$
B $\frac{1}{4}+\log _{e} 2$
C $3-\log _{e} 2$
D $\frac{15}{4}-\log _{e} 2$

The correct answer is:
$\frac{1}{4}+\log _{e} 2$

Solution:
First of all we draw the graph $y=x^{3}, y=\frac{1}{x}, x=2$

Required area i.e., (OMPNO) $=\int_{0}^{1} x^{3} d x+\int_{1}^{2} \frac{1}{x} d x$
$=\left[\frac{x^{4}}{4}\right]_{0}+\left[\log _{4} x\right]_{1}^{2}=\frac{1}{4}+\log _{c} 2$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2012
The area of the region bounded by the parabola $y=x^{2}-4 x+5$ and the straight line $y=x+1$ is

Options:
A $\frac{1}{2}$
B 2
C 3
D $\frac{9}{2}$

The correct answer is:
$\frac{9}{2}$

Solution:
Given equation of parabola is
$$
y=x^{2}-4 x+5
$$
$\Rightarrow \quad y=(x-2)^{2}+1$
$\Rightarrow \quad(x-2)^{2}=(y-1)$
and equation of line is
$$
\begin{array}{r}
y=x+1 \\
\Rightarrow \quad x-y=-1
\end{array}
$$

On putting the value of $(y-1)$ from Eqs. (ii) in (1). we get
$$
(x-2)^{2}=x \Rightarrow x^{2}+4-4 x=x
$$
$\Rightarrow \quad x^{2}-5 x+4=0$
$\Rightarrow x^{2}-4 x-x+4=0$
$\Rightarrow \quad x(x-4)-1(x-4)=0$
$\Rightarrow \quad(x-1)(x-4)=0$ or $x=1,4$
then from is (ii) $y=2,5$
$\therefore$ Required area $=\int_{1}^{4}\left\{(x+1)-\left(x^{2}-4 x+5\right)\right\} d x$
$=\int_{1}^{4}\left(-x^{2}+5 x-4\right) d x=\left[\frac{-x^{3}}{3}+\frac{5 x^{2}}{2}-4 x\right]_{1}^{4}$
$=\left[-\frac{64}{3}+40-16+\frac{1}{3}-\frac{5}{2}+4\right]$
$=\left(-21-\frac{3}{2}+28\right)=\frac{9}{2}$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2013
The area of the region bounded by $y^2=x$ and $y=|x|$ is

Options:
A $\frac{1}{3}$ sq.unit
B $\frac{1}{6}$ sq.unit
C $\frac{2}{3}$ sq. unit
D 1 sq.unit

The correct answer is:
$\frac{1}{6}$ sq.unit

Solution:
Hints: $\mathrm{y}^2=\mathrm{x}$
$$
\left.\int_0^1(\sqrt{x}-x) d x=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^2}{2}\right]_0^1=\frac{3}{2}-\frac{1}{2}=\frac{4-3}{6}=\frac{1}{6}
$$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2010
The area of the region enclosed between parabola $y^{2}=x$ and the line $y=m x$ is $\frac{1}{48}$
Then, the value of $m$ is

Options:
A -2
B -1
C 1
D 2

The correct answers are:
-2, 2

Solution:
Equation of parabola is $y^{2}=x$ and line $y=m x$
For intersection point of both curves put $x=y^{2}$,
we get
$$
y=m y^{2} \Rightarrow y(m y-1)=0
$$
$\Rightarrow \quad y=0$ or $y=\frac{1}{m}$
Then, $\quad x=0$ or $x=\frac{1}{m^{2}}$
$\therefore$ Intersection points are (0,0) and $P\left(\frac{1}{m^{2}}, \frac{1}{m}\right)$

$\therefore$ Required area $=\int_{0}^{1 / m}\left(\frac{y}{m}-y^{2}\right)|d y=|\left[\frac{y^{2}}{2 m}-\frac{y^{3}}{3}\right]_{0}^{1 / m} \mid$
$=\left|\frac{1}{2 m^{3}}-\frac{1}{3 m^{3}}\right|=\left|\frac{1}{6 m^{3}}\right|=\frac{1}{48}$
$\Rightarrow \quad \frac{1}{6 m^{3}}=\pm \frac{1}{48} \Rightarrow m^{3}=\pm 8$
(given)
Now, if $m^{3}=8$
$\Rightarrow \quad m^{3}=(2)^{3} \Rightarrow m=2$
If $\quad m^{3}=-8$
$\Rightarrow \quad m^{3}=(-2)^{3} \Rightarrow m=-2$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2013
The area of the region lying above $X$ -axis, and included between the circle $x^{2}+y^{2}=2 a x$ and the parabola $y^{2}=a x, a>0$ is

Options:
A $8 \pi a^{2}$
B $a^{2}\left(\frac{\pi}{4}-\frac{2}{3}\right)$
C $\frac{16 \pi a^{2}}{9}$
D $\pi\left(\frac{27}{8}+3 a^{2}\right)$

The correct answers are:
$a^{2}\left(\frac{\pi}{4}-\frac{2}{3}\right)$

Solution:
Given, equation of circle $x^{2}+y^{2}=2 a x$
$\Rightarrow$
$(x-a)^{2}+y^{2}=a^{2}$
and equation of parabola is $y^{2}=a x, a>0$
Intersection points of circle and parabola
$\begin{array}{lr}\Rightarrow & x^{2}+a x=2 a x \\ \Rightarrow & x^{2}=a x\end{array}$
$\Rightarrow \quad x^{2}-a x=0$
$\Rightarrow \quad x(x-a)=0$
$\Rightarrow \quad x=0, a$
Intersecting points are $(0,0)$ and $(a, a)$
$\therefore$ Required area $=\frac{\pi a^{2}}{4}-\int_{0}^{a} \sqrt{a x} d x$
$=\frac{\pi a^{2}}{4}-\sqrt{a}\left(\frac{x^{3 / 2}}{3 / 2}\right)_{0}^{a}$
$=\frac{\pi a^{2}}{4}-\frac{2 a^{2}}{3}=a^{2}\left(\frac{\pi}{4}-\frac{2}{3}\right)$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2018
The area of the region $\left\{(x, y): x^{2}+y^{2} \leq 1 \leq x+y\right\}$ is

Options:
A $\frac{\pi^{2}}{2}$
B $\frac{\pi}{4}$
C $\frac{\pi}{4}-\frac{1}{2}$
D $\frac{\pi^{2}}{3}$

The correct answer is:
$\frac{\pi}{4}-\frac{1}{2}$

Solution:
Hint:

Area of shaded region $=\frac{\pi}{4}-\frac{1}{2}$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2020
The area of the triangle formed by the intersection of a line parallel to $X$ -axis and passing through $P(h, k),$ with the lines $y=x$ and $x+y=2$ is $h^{2}$. The locus of the point $P$ is

Options:
A $x=y-1$
B $x=-(y-1)$
C $x=1+y$
D $x=-(1+y)$

The correct answers are:
$x=y-1$, $x=-(y-1)$

Solution:
Given, ar $(\Delta A B C)=h^{2}$
$\Rightarrow \quad \frac{1}{2}\left|\begin{array}{ccc}1 & 1 & 1 \\ k & k & 1 \\ 2-k & k & 1\end{array}\right|=\pm h^{2}$
$\Rightarrow 1(k-k)-1(k-2+k)+1\left(k^{2}-2 k+k^{2}\right)=\pm 2 h^{2}$
$\Rightarrow \quad-(2 k-2)+\left(2 k^{2}-2 k\right)=\pm 2 h^{2}$
$\Rightarrow \quad 2-2 k+2 k^{2}-2 k=\pm 2 h^{2}$
$\Rightarrow \quad 2 k^{2}-4 k+2=\pm 2 h^{2}$
$\Rightarrow$
$k^{2}-2 k+1=\pm h^{2}$
Hence, locus of a point is
$\Rightarrow$
$(k-1)^{2}=h^{2}$
$y-1=\pm x$
$\Rightarrow \quad x=y-1$ or $x=-(y-1)$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2018
The atomic masses of helium and neon are $4.0$ and $20.0$ amu respectively. The value of the de Broglie wavelength of helium gas at $-73^{\circ} \mathrm{C}$ is $\mathrm{M}$ times the de Broglie wavelength of neon at $727^{\circ} \mathrm{C}$. The value of $\mathrm{M}$ is

Options:
A 5
B 25
C $\frac{1}{5}$
D $\frac{1}{25}$

The correct answer is:
5

Solution:
$\lambda_{\mathrm{He}}(200 \mathrm{~K})=\mathrm{M} \times \lambda_{\mathrm{Ne}}(1000 \mathrm{~K})$ $\left(\lambda=\frac{\mathrm{h}}{\mathrm{m} \cdot \mathrm{v}}\right)$
$\frac{\mathrm{h}}{4 \mathrm{amu} \times \mathrm{C}_{\mathrm{mp}}(200)}=\mathrm{M} \times \frac{\mathrm{h}}{20 \mathrm{amu} \times \mathrm{C}_{\mathrm{mp}}(1000)} \quad\left(\mathrm{C}_{\mathrm{mp}}=\sqrt{\frac{2 \mathrm{KT}}{\mathrm{m}}}\right)$
$\Rightarrow \frac{1}{4 \times \sqrt{\frac{2 \mathrm{~K} \times 200}{4}}}=\frac{\mathrm{M}}{20 \times \sqrt{\frac{2 \mathrm{~K} \times 1000}{20}}} \Rightarrow \frac{1}{4 \times \sqrt{50}}=\frac{\mathrm{M}}{20 \times \sqrt{50}} \Rightarrow \mathrm{M}=5$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2021
The atomic number of cerium $(\mathrm{Ce})$ is 58 . The correct electronic confipuration of $\mathrm{Ce}^{3+}$ ion is

Options:
A $[\mathrm{Xe}] 4 f^{1}$
B $[\mathrm{Kr}] 4 f^{1}$
C $[\mathrm{Xe}] 4 f^{13}$
D $[\mathrm{Kr}] 4 d^{1}$

The correct answer is:
$[\mathrm{Xe}] 4 f^{1}$

Solution:
Atomic number of cerium $=58$
Electronic configuration of Ce
$$
=[\mathrm{Xe}] 4 f^{2}, 5 d^{0}, 6 s^{2}
$$
Electronic configuration of $\mathrm{Ce}^{3+}$
$$
=[\mathrm{Xe}] 4 f^{1}
$$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2015
The average length of all vertical chords of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, a \leq x \leq 2 a\), is

Options:
A \(\mathrm{b}\{2 \sqrt{3}+\ell n (2+\sqrt{3})\}\)
B \(b\{3 \sqrt{2}+\ell \ln (3+\sqrt{2})\}\)
C \(\mathrm{a}\{2 \sqrt{5}-\ell \mathrm{n}(2+\sqrt{5})\}\)
D None of these

The correct answer is:
None of these

Solution:
Hint : \(A_L=\frac{2 \int_a^{2 a} y d x}{\int_a^{2 a} d x}\)
\(=\frac{2 \int_a^{2 a} \frac{b}{a} \sqrt{x^2-a^2} d x}{\int_a^{2 a} d x}=b[2 \sqrt{3}-\ell \mathrm{n}(2+\sqrt{3})]\)

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2023
The average ordinate of \(y=\sin x\) over \([0, \pi]\) is

Options:
A \(\frac{2}{\pi}\)
B \(\frac{3}{\pi}\)
C \(\frac{4}{\pi}\)
D \(\pi\)

The correct answer is:
\(\frac{2}{\pi}\)

Solution:
Hint : As \(\sin x\) is a continuous function in \([0, \pi]\)
Average ordinate will be \(=\frac{1}{\pi-0} \int_0^\pi \sin x d x=\frac{[-\cos x]_0^\pi}{\pi}=\frac{2}{\pi}\)

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2023
The average speed of $\mathrm{H}_2$ at $T_1 \mathrm{~K}$ is equal to that of $\mathrm{O}_2$ at $T_2 \mathrm{~K}$. The ratio $T_1: T_2$ is

Options:
A $1: 6$
B $16: 1$
C $1: 4$
D None of these

The correct answer is:
None of these

Solution:
$\left(\mathrm{C}_{\mathrm{av}}\right)_{\mathrm{H}_2}=\left(\mathrm{C}_{\mathrm{av}}\right)_{\mathrm{O}_2}$
$\sqrt{\frac{8 \mathrm{RT}_1}{\pi \mathrm{M}_{\mathrm{H}_2}}}=\sqrt{\frac{8 \mathrm{RT}_2}{\pi \mathrm{M}_{\mathrm{O}_2}}}$
$\therefore \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{\mathrm{M}_{\mathrm{H}_2}}{\mathrm{~N}_{\mathrm{O}_2}}=\frac{2}{32}=\frac{1}{16}$

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2022
The axis of the parabola $x^{2}+2 x y+y^{2}-5 x+5 y-5=0$ is

Options:
A $x+y=0$
B $x+y-1=0$
C $x-y+1=0$
D $x-y=\frac{1}{\sqrt{2}}$

The correct answer is:
$x+y=0$

Solution:
Given, equation $x^{2}+2 x y+y^{2}-5 x+5 y-5=0$
$\Rightarrow(x+y)^{2}=5 x-5 y+5$
$\Rightarrow(x+y)^{2}=5(x-y+1)$
$\therefore$ Axis of the parabola is $x+y=0$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2017
The basicity of aniline is weaker in comparison to that of methyl amine due to

Options:
A hyperconjugative effect of Me-group in $\mathrm{MeNH}_{2}$
B resonance effect of phenyl group in aniline
C lower molecular weight of methyl amine as compared to that of aniline
D resonance effect of $-\mathrm{NH}_{2}$ group in $\mathrm{MeNH}_{2}$

The correct answer is:
resonance effect of phenyl group in aniline

Solution:
The lone pair of electrons, present in N-atom in aniline involve in ring resonance. So it is less basic than methyl amine.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2012
The best method for preparation of $Me_{3}\mathrm{CCN}$ is

Options:
A to react $Me_{3} \mathrm{COH}$ with HCN
B to react $Me_{3}$ CBr with NaCN
C to react $Me_{3} \mathrm{CMgBr}$ with CICN
D to react $Me_{3} \mathrm{Cl}$ with $\mathrm{NH}_{2} \mathrm{CN}$

The correct answer is:
to react $Me_{3} \mathrm{CMgBr}$ with CICN

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, WBJEE, WBJEE 2013
The best method for the preparation of 2, 2-dimethylbutane is via the reaction of

Options:
A $\mathrm{Me}_{3} \mathrm{CBr}$ and $\mathrm{MeCH}_{2}$ Br in Na/ether
B $\left(\mathrm{Me}_{3} \mathrm{C}\right)_{2} \mathrm{CuLi}$ and $\mathrm{MeCH}_{2} \mathrm{Br}$
C $\left(\mathrm{MeCH}_{2}\right)_{2} \mathrm{CuLi}$ and $\mathrm{Me}_{3} \mathrm{CBr}$
D $\mathrm{Me}_{3} \mathrm{CMgl}$ and $\mathrm{MeCH}_{2}l$

The correct answer is:
$\left(\mathrm{Me}_{3} \mathrm{C}\right)_{2} \mathrm{CuLi}$ and $\mathrm{MeCH}_{2} \mathrm{Br}$

Solution:
2,2 - dimethyl butane is prepared from Corey House alkane synthesis. $\left(\mathrm{Me}_{3} \mathrm{C}\right)_{2} \mathrm{Cul}_{2} \mathrm{Li}+\mathrm{MeCH}_{2} \mathrm{Br} \stackrel{\mathrm{S}_{\mathrm{N}}{2}}{\longrightarrow}$
$$
\mathrm{Me}_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{CH}_{3}+\mathrm{Me}_{3} \mathrm{C}CuLi
$$

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2013
The bob of a pendulum of mass $m$, suspended by an inextensible string of length $L$ as shown in the figure carries a small charge $a$. An infinite horizontal plane conductor with uniform surface charge density $\sigma$ is placed below it. What will be the time period of the pendulum for small amplitude oscillations?

Options:
A $2 \pi \sqrt{\frac{\mathrm{L}}{\left(\mathrm{g}-\frac{\mathrm{mq}}{\varepsilon_{0} \sigma}\right)}}$
B $\sqrt{\frac{\mathrm{L}}{\left(\mathrm{g}-\frac{\mathrm{mq} \sigma}{\varepsilon_{0}}\right)}}$
C $\frac{1}{2 \pi} \sqrt{\frac{\mathrm{L}}{\left(\mathrm{g}-\frac{\mathrm{q} \sigma}{\varepsilon_{0} \mathrm{~m}}\right)}}$
D $2 \pi \sqrt{\frac{\mathrm{L}}{\left(\mathrm{g}-\frac{\mathrm{q} \sigma}{\varepsilon_{0} \mathrm{~m}}\right)}}$

The correct answer is:
$2 \pi \sqrt{\frac{\mathrm{L}}{\left(\mathrm{g}-\frac{\mathrm{q} \sigma}{\varepsilon_{0} \mathrm{~m}}\right)}}$

Solution:
$\therefore$ Apparent weight of the bob, $w^{\prime}=m g-q E$
$$
m y^{\prime}=m g-q E
$$
$\therefore$
$g^{\prime}=g-\frac{q E}{m}$
Time period of a pendulum. $T=2 \pi \sqrt{\frac{L}{g_{e f f}}}$ $=2 \pi \sqrt{\frac{L}{g-\frac{q E}{m}}}=2 \pi \sqrt{\frac{L}{\left(g-\frac{q \sigma}{\varepsilon_{0} m}\right)}}$

Topic: Physics, Oscillations, WBJEE, WBJEE 2018
The bob of a swinging seconds pendulum (one whose time period is $2 \mathrm{~s}$ ) has a small speed $\mathrm{v}_{0}$ at its lowest point. Its height from this lowest point $2.25 \mathrm{~s}$ after passing through it is given by

Options:
A $\frac{\mathrm{v}_{0}^{2}}{2 \mathrm{~g}}$
B $\frac{\mathrm{v}_{0}^{2}}{\mathrm{~g}}$
C $\frac{v_{0}^{2}}{4 g}$
D $\frac{9 v_{0}^{2}}{4 g}$

The correct answer is:
$\frac{v_{0}^{2}}{4 g}$

Solution:
Hint:
$T=2$ sec.
at $\mathrm{t}=\mathrm{T}+\mathrm{T} / 8$
$\mathrm{t}=\mathrm{T}+\mathrm{T} / 8 \quad \mathrm{v}=\mathrm{A} \omega \cos \omega \mathrm{t}=\mathrm{v}_{0} / \sqrt{2}$
By Mechanical energy conservation:
$\left(\frac{v_{0}}{\sqrt{2}}\right)^{2}=v_{0}^{2}-2 g h \quad h=\frac{v_{0}^{2}}{4 g}$

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2020
The boiling point of the water is higher than liquid $\mathrm{HF}$. The reason is that

Options:
A Hydrogen bonds are stronger in water
B Hydrogen bonds are stronger in $\mathrm{HF}$.
C Hydrogen bonds are larger in number in $\mathrm{HF}$
D Hydrogen bonds are larger in number in water

The correct answer is:
Hydrogen bonds are larger in number in water

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2022
The boiling points of $\mathrm{HF}, \mathrm{HCl}, \mathrm{HBr}$ and $\mathrm{HI}$ follow the order

Options:
A HF $>\mathrm{HCl}>\mathrm{HBr}>\mathrm{Hl}$
B $\mathrm{HF}>\mathrm{Hl}>\mathrm{HBr}>\mathrm{HCl}$
C $\mathrm{Hl}>\mathrm{HBr}>\mathrm{HCl}>\mathrm{H}$
D $\mathrm{HCl}>\mathrm{HF}>\mathrm{HBr}>\mathrm{Hl}$

The correct answer is:
$\mathrm{HF}>\mathrm{Hl}>\mathrm{HBr}>\mathrm{HCl}$

Solution:
Due to hydrogen bonding, HF shows highest boiling point.
Thereafter van der Waals' force decide the boiling point.
Larger the size or molecules mass, greater is the van der Waals' forces. Hence, higher is the boiling point.
$\because$ van der Waals' forces is more for HI than HBr, which is in turn is more than HCl.
Hence, correct order is $\mathrm{HF}>\mathrm{Hl}>\mathrm{HBr}>\mathrm{HCl}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2016
The bond angle in $\mathrm{NF}_{3}\left(102.3^{\circ}\right)$ is smaller than $\mathrm{NH}_{3}\left(107.2^{\circ}\right) .$ This is because of

Options:
A large size of F compared to H
B large size of N compared to F
C opposite polarity of $\mathrm{N}$ in the two molecules
D small size of $\mathrm{H}$ compared to $\mathrm{N}$

The correct answer is:
opposite polarity of $\mathrm{N}$ in the two molecules

Solution:
In $\mathrm{NF}_{3}$, $\mathrm{N}$ is less electronegative as compared to F but in $\mathrm{NH}_{3}$, it is more electronegative than $\mathrm{H}$. And in case of same central atom. as the electronegativity of other atoms increases, bond angle decreases, Thus, bond angle in $\mathrm{NF}_{3}$ is smaller than that in $\mathrm{NH}_{3}$ because of the difference in the polarity of $\mathrm{N}$ in these molecules.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2014
The bond order of $\mathrm{CO}$ molecule is

Options:
A 2
B 2.5
C 3
D 3.5

The correct answer is:
3

Solution:
Hints : $\mathrm{CO} \rightarrow \sigma(1 \mathrm{~S})^2, \sigma^*(1 \mathrm{~S})^2, \sigma(2 \mathrm{~S})^2, \sigma\left(2 \mathrm{P}_z\right)^2, \pi\left(2 \mathrm{P}_{\mathrm{z}}\right)^2=\pi\left(2 \mathrm{P}_y\right)^2, \sigma^*(2 \mathrm{~S})^2$
$$
\text { B.O }=\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{o}}}{2}=\frac{10-4}{2}=3
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2010
The catalyst used for olefin polymerization is

Options:
A Ziegler-Natta Catalyst
B Wilkinson Catalyst
C Raney nickel catalyst
D Merrifield resin

The correct answer is:
Ziegler-Natta Catalyst

Solution:
$\mathrm{TiCl}_3+\left(\mathrm{C}_2 \mathrm{H}_5\right)_3 \mathrm{Al}$

Topic: Chemistry, Polymers, WBJEE, WBJEE 2009
The change in the gravitational potential energy when a body of mass $m$ is raised to a height $n \mathrm{R}$ above the surface of the Earth is (here $R$ is the radius of the Earth)

Options:
A $\left(\frac{n}{n+1}\right) m g \mathrm{R}$
B $\left(\frac{n}{n-1}\right) m g \mathrm{R}$
C $n m g R$
D $\frac{m g R}{n}$

The correct answer is:
$\left(\frac{n}{n+1}\right) m g \mathrm{R}$

Solution:
Hints: $\Delta \mathrm{U}=\frac{m g h}{1+\frac{h}{\mathrm{R}}}=\frac{m g \times n \mathrm{R}}{1+\frac{n \mathrm{R}}{\mathrm{R}}}=\frac{n m g \mathrm{R}}{n+1}$

Topic: Physics, Gravitation, WBJEE, WBJEE 2010
The change of entropy $(d S)$ is defined as

Options:
A $d S=\frac{\delta q}{T}$
B $d S=\frac{d H}{T}$
C $d S=\frac{\delta q_{eqv}}{T}$
D $d S=\frac{d H-d G}{T}$

The correct answer is:
$d S=\frac{\delta q_{eqv}}{T}$

Solution:
The entropy is the measure of disorder or randomness in a system. When a system changes from one state to another, the change of entropy, $d S$ is given by
$$
d S=\frac{\delta q_{\mathrm{rev}}}{T}
$$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2013
The charge carried by 1 millimole of $M^{n+}$ ions is 193 coulombs. The value of $n$ is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
One millimole $=0.001 \mathrm{mol}$
i.e. $1 \times 10^{-3} \mathrm{mol}$
$\because 96500 \mathrm{C}$ is required for $1 \mathrm{mol}$ of electrons.
$\therefore 193 \mathrm{C}$ give $\frac{193 \times 1}{96500}=0.002 \mathrm{mol},$ i.e. $2 \times 10^{-3} \mathrm{mol}$ of
electrons.
Thus, value of $n$ in $M^{n+}$ is $=\frac{0.002}{0.001}=2$
Hence, option (b) is the correct answer.

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2019
The charge on the capacitor of capacitance \(\mathrm{C}\) shown in the figure below will be

Options:
A \(\mathrm{CE}\)
B \(\frac{\mathrm{CE} \mathrm{R}}{\mathrm{R}_1+r}\)
C \(\frac{C E R_2}{R_2+r}\)
D \(\frac{C E R_1}{R_2+r}\)

The correct answer is:
\(\frac{C E R_2}{R_2+r}\)

Solution:
Hints: \(I=\frac{E}{R_2+r}\) (Since finally no current flows through capacitor)
\(\therefore\) Potential difference across \(\mathrm{R}_2, \mathrm{~V}=\mathrm{R}_2=\frac{\mathrm{ER}_2}{\mathrm{R}_2+\mathrm{r}}\)
\(\therefore\) Charge on the capacitor \(\mathrm{Q}=\mathrm{CV}=\frac{\mathrm{CER}_2}{\mathrm{R}_2+\mathrm{r}}\)

Topic: Physics, Capacitance, WBJEE, WBJEE 2011
The chord of the curve $y=x^{2}+2 a x+b$ joining the points where $x=\alpha$ and $x=\beta,$ is parallel to the tangent to the curve at abscissa $x$ is equal to

Options:
A $\frac{a+b}{2}$
B $\frac{2 a+b}{3}$
C $\frac{2 \alpha+\beta}{3}$
D $\frac{\alpha+\beta}{2}$

The correct answer is:
$\frac{\alpha+\beta}{2}$

Solution:
Given curve,
At $y=x^{2}+2 a x+b$
$x=a$
$y=\alpha^{2}+2 a \alpha+b$
and at $x=\beta$
$y=\beta^{2}+2 a \beta+b$
$\therefore$ Slope of line joining $\left(\alpha, \alpha^{2}+2 a \alpha+b\right)$ and
$\left(\beta, \beta^{2}+2 a \beta+b\right)$ is
$=\frac{\left(\alpha^{2}+2 a \alpha+b\right)-\left(\beta^{2}+2 a \beta+b\right)}{\alpha-\beta}$
$=\frac{\alpha^{2}+2 a \alpha+b-\beta^{2}-2 a \beta-b}{\alpha-\beta}$
$=\frac{\left(\alpha^{2}-\beta^{2}\right)+2 a(\alpha-\beta)}{\alpha-\beta}$
$\frac{=(\alpha-\beta)(\alpha+\beta)+2 a(\alpha-\beta)}{\alpha-\beta}$
$=\alpha+\beta+2 a$
Slope of given curve $=\frac{d y}{d x}$ $=2 x+2 a$
Now, according to question, tangent is parallel to the chord. Therefore,
$$
\begin{array}{r}
2 x+2 a=\alpha+\beta+2 a \\
\Rightarrow \quad 2 x=\alpha+\beta \Rightarrow x=\frac{\alpha+\beta}{2}
\end{array}
$$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2017
The circles $\mathrm{x}^2+\mathrm{y}^2-10 \mathrm{x}+16=0$ and $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2$ intersect at two distinct points if

Options:
A $\mathrm{a} < 2$
B $2 < $ a $ < 8$
C $a>8$
D $\mathbf{a}=2$

The correct answer is:
$2 < $ a $ < 8$

Solution:
Hints : $C_1(5,0) \quad r_1=\sqrt{25-16}=3$
$$
\begin{aligned}
& \mathrm{C}_2(0,0) \quad \mathrm{r}_2=\mathbf{a} \\
& \mathrm{r}_1 \& \mathrm{r}_2 < \mathrm{C}_1 \mathrm{C}_2 < \mathrm{r}_1+\mathrm{r}_2 \\
& |\mathbf{a}-3| < \sqrt{25} < \mathrm{a}+3 \\
& |\mathbf{a}-3| < 5 < \mathrm{a}+3 \\
& -5 < \mathrm{a}-3 < 5 \\
& -2 < \mathrm{a} < 8 \\
& 2 < \mathrm{a} < 8
\end{aligned}
$$

Topic: Mathematics, Circle, WBJEE, WBJEE 2009
The co-efficient of $a^{3} b^{4} c^{5}$ in the expansion of $(b c+c a+a b)^{6}$ is

Options:
A $\frac{12 !}{3 ! 4 ! 5 !}$
B $\frac{6 !}{3 !}$
C 33
D 3. $\left(\frac{6 !}{3 ! 3 !}\right)$

The correct answer is:
3. $\left(\frac{6 !}{3 ! 3 !}\right)$

Solution:
$\left\{\frac{6 !}{p ! q ! r !} a^{q+r} b^{p+r} c^{p+q}\right.$
$q+r=3, p+r=4, p+q=5, p+q+r=6$
$\Rightarrow p=3, q=2, r=1$
co-efficient $=\frac{6 !}{3 ! 2 !}$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2021
The coefficient of $x^{3}$ in the infinite series expansion of $\frac{2}{(1-x)(2-x)},$ for $|x| < 1,$ is

Options:
A $-\frac{1}{16}$
B $\frac{15}{8}$
C $-\frac{1}{8}$
D $\frac{15}{16}$

The correct answer is:
$\frac{15}{8}$

Solution:
Given expression can be rewritten as
$$
\begin{array}{l}
2(1-x)^{-1} \times 2^{-1}\left(1-\frac{x}{2}\right)^{-1} \\
\quad=\left[1+x+x^{2}+x^{3}+\ldots\right] \\
\quad\left[1+\left(\frac{x}{2}\right)+\left(\frac{x}{2}\right)^{2}+\left(\frac{x}{2}\right)^{3}+\ldots\right.
\end{array}
$$
$\therefore$ Coefficient of $x^{3}$ in the above expression is
$$
\begin{array}{l}
=\left[\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)+1\right] \\
=\left[\frac{1}{8}+\frac{1}{4}+\frac{1}{2}+1\right] \\
=\left[\frac{1+2+4+8}{8}\right]=\frac{15}{8}
\end{array}
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2014
The coefficient of $x^{10}$ in the expansion of $1+(1+x)+\dots+(1+x)^{20}$

Options:
A ${ }^{19} C_{9}$
B ${ }^{20} C_{10}$
C ${ }^{21} C_{11}$
D ${ }^{22} C_{12}$

The correct answer is:
${ }^{21} C_{11}$

Solution:
The given series is in GP. Hence, its sum $S=\frac{1\left\{(1+x)^{20+1}-1\right\}}{(1+x)-1}=\frac{(1+x)^{21}-1}{x}$
Therefore, the required coefficient of $x^{10}$ in
the expansion of $(1+x)^{21}-1$
$=$ Coefficient of $x^{11}$ in the expansion of
$$
={ }^{21} C_{11}
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2012