datavorous/entrance-exam-dataset huggingface dataset:
Let $z=x+i y,$ where $x$ and $y$ are real. The points $(x, y)$ in the $X-Y$ plane for which $\frac{z+i}{z-i}$
is purely imaginary, lie on

Options:
A a straight line
B an ellipse
C a hyperbola
D a circle

The correct answer is:
a circle

Solution:
$\frac{z+i}{z-i}=\frac{(x+j)+i}{(x+i y)-i}$
$=\frac{x+(1+y) i}{x+(y-1) i}$
$=\frac{x+(y+1) i}{x+(y-1) i} \times \frac{x-(y-1) i}{x-(y-1) i}$
$=\frac{x^{2}-x(y-1) i+x(y+1) i-(y+1)(y-1) i^{2}}{x^{2}-(y-1)^{2} i^{2}}$
$=\frac{x^{2}+i[-x y+x+x y+x]+\left(y^{2}-1\right)}{x^{2}+(y-1)^{2}}$
$=\frac{x^{2}+y^{2}-1}{x^{2}+(y-1)^{2}}+\frac{2 x}{x^{2}+(y-1)^{2}} i$
Now, $\frac{z+i}{z-i}$ is purely imaginary.
$\therefore$
$\operatorname{Re}\left(\frac{z+i}{z-i}\right)=0$
$\begin{array}{lr}\Rightarrow & \frac{x^{2}+y^{2}-1}{x^{2}+(y-1)^{2}}=0 \\ \Rightarrow & x^{2}+y^{2}=1\end{array}$
$\therefore(x, y)$ lies on a circle.

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2017
Letf $: \mathbb{R} \rightarrow \mathbb{R}$, be twice continuously differentiable (or $f$ " exists and is continuous) such that $f(0)=f(1)=f^{\prime}(0)=0$. Then

Options:
A $\mathrm{f}^{\prime \prime}(\mathrm{c})=0$ for some $\mathrm{c} \in \mathbb{R}$
B there is no point for which $f^{\prime \prime}(x)=0$
C at all points $f^{\prime \prime}(x)>0$
D at all points $f^{\prime \prime}(x) < 0$

The correct answer is:
$\mathrm{f}^{\prime \prime}(\mathrm{c})=0$ for some $\mathrm{c} \in \mathbb{R}$

Solution:
Hint:
$f(0)=f(1)=0$
By Rolle's theorem $\mathrm{f}^{\prime}(\mathrm{c})=0$ for some $\mathrm{c} \in(0,1)$
Now, $\mathrm{f}^{\prime}(0)=\mathrm{f}^{\prime}(\mathrm{c})=0$. Again by Rolle's theorem $\mathrm{f}^{\prime \prime}\left(\mathrm{c}_{1}\right)=0 \quad \mathrm{c}_{1} \in(0, \mathrm{c})$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2020
Light travels through a glass plate of thickness $t$ and having refractive index $\mu$. If $c$ be the velocity of light in vacuum, time taken by the light to travel through this thickness of glass is

Options:
A $\frac{t}{\mu C}$
B $\frac{t c}{\mu}$
C $\frac{\mu t}{c}$
D $\mu \mathrm{tc}$

The correct answer is:
$\frac{\mu t}{c}$

Solution:
Thickness of glass plate $=t$
Refractive index $=\mu$
Velocity of light = c
$$
\begin{array}{r}
\text { Speed of light }=\frac{\text { distance }}{\text { time }}=\frac{t}{T} \\
\qquad T=\frac{t}{\text { speed of light }}
\end{array}
$$
Speed of light in glass plate
$$
\begin{array}{l}
v=\frac{c}{\mu} \\
T=\frac{t}{c / \mu}=\frac{\mu t}{c}
\end{array}
$$
$$
\left(\because \mu=\frac{c}{v}\right)
$$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2016
$\lim _{n \rightarrow \infty} \frac{3}{n}\left\{1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+\sqrt{\frac{n}{n+9}}\right.$$+\ldots+\sqrt{\frac{1}{n+3}(n-1)}$

Options:
A does not exist
B is 1
C is 2
D is 3

The correct answer is:
is 2

Solution:
$\lim _{n \rightarrow \infty} \frac{3}{n}\left[1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+\sqrt{\frac{n}{n+9}}+\ldots\right.+\sqrt{\frac{n}{n+3(n-1)}}$
$=\lim _{n \rightarrow \infty} \frac{3}{n}\left[\sqrt{\frac{n}{n+3 \times 0}}+\sqrt{\frac{n}{n+3 \times 1}}+\sqrt{\frac{n}{n+3 \times 2}}\right.$
$+\sqrt{\frac{n}{n+3 \times 3}}+\ldots+\sqrt{\frac{n}{n+3(n-1)}}$
$=\lim _{n \rightarrow \infty} \frac{3}{n}\left[\sum_{r=0}^{n-1} \sqrt{\frac{n}{n+3 r}}\right]$
$=\lim _{n \rightarrow \infty} \frac{3}{n}\left[\sum_{r=0}^{n-1}\left[\frac{1}{1+3\left(\frac{r}{n}\right)}\right]\right.$
$=3 \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} \left(\sqrt{1+3\left(\frac{r}{n}\right)}\right)^{-\frac{1}{2}}$
$=3 \int_{0}^{1} \sqrt{\frac{1}{1+3 x}} d x$
$=3 \int_{0}^{1}(1+3 x)^{-1 / 2} d x$
$=3\left[\frac{2 \sqrt{1+3 x}}{3}\right]_{0}^{1}$
$=[4-2]$
$=2$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2019
$\lim _{n \rightarrow \infty} \frac{\sqrt{n}}{\sqrt{\left(n^{3}\right)}}+\frac{\sqrt{n}}{\sqrt{(n+4)^{3}}}+\frac{\sqrt{n}}{\sqrt{(n+8)^{3}}}+\cdots \cdots \cdots+\frac{\sqrt{n}}{\sqrt{[n+4(n-1)]^{3}}}$ is

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

The correct answers are:
$\frac{5-\sqrt{5}}{10}$

Solution:
$\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{\sqrt{n}}{\sqrt{(n+4 r)^{3}}}$
$=\sum_{r=0}^{n-1} \frac{1}{n}\left(\frac{n \sqrt{n}}{\sqrt{(n+4 r)^{3}}}\right)$
$=\sum_{r=0}^{n-1} \frac{1}{n}\left(\frac{1}{\left(1+\frac{4 r}{n}\right)^{3 / 2}}\right)$
$=\int_{0}^{1} \frac{d x}{(1+4 x)^{3 / 2}}$
$=\frac{1}{4} \int_{1}^{5} \frac{d z}{z^{3 / 2}}=\left(1 / 4\left(\frac{-2}{\sqrt{z}}\right)\right)_{1}^{5}=\frac{5-\sqrt{5}}{10}$

Topic: Mathematics, Limits, WBJEE, WBJEE 2021
$\lim _{n \rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n-1}}{n \sqrt{n}}$ is equal to

Options:
A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{2}{3}$
D 0

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

Solution:
$\lim _{n \rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n-1}}{n \sqrt{n}}$
$=\lim _{n \rightarrow \infty}\left(\frac{\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n-1}+\sqrt{n}}{n \sqrt{n}}-\frac{n}{n \times n}\right)$
$=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n}\left(\frac{r}{n}\right)^{1 / 2}-\lim _{n \rightarrow \infty} \frac{1}{n} \times \frac{n}{n}=\int_{0}^{1} \sqrt{x} d x+0=\frac{2}{3}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2015
$\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}$

Options:
A does not exist
B equals $\log _{e}\left(\pi^{2}\right)$
C equals 1
D lies between 10 and 11

The correct answer is:
equals $\log _{e}\left(\pi^{2}\right)$

Solution:
$\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}$
$\frac{0}{0}$ form
$=\lim _{x \rightarrow 0} \frac{\pi^{x} \log _{e} \pi}{\frac{1}{2 \sqrt{1+x}}}$
$=\lim _{x \rightarrow 0} 2 \sqrt{1+x}\left(\pi^{x} \log _{e} \pi\right)$
$=2 \sqrt{1}\left(\pi^{0} \log _{e} \pi\right)=2 \log _{e} \pi=\log _{e} \pi^{2}$

Topic: Mathematics, Limits, WBJEE, WBJEE 2012
$\lim _{x \rightarrow 0^{+}}\left(e^{x}+x\right)^{1 / x}$

Options:
A Does not exist finitely
B is 1
C is $e^{2}$
D is 2

The correct answer is:
is $e^{2}$

Solution:
$$
\begin{aligned}
&\text { Let } L=\lim _{x \rightarrow 0^{+}}\left(e^{x}+x\right)^{1 / x}\\
&\Rightarrow \log L=\lim _{x \rightarrow 0^{+}} \frac{\log \left(e^{x}+x\right)}{x}\\
&\Rightarrow \log L=\lim _{x \rightarrow 0^{+}} \frac{\frac{1}{\left(e^{x}+x\right)} \cdot e^{x}+1}{1}\\
&\text { [using L' Hospital rule] }\\
&\Rightarrow \log L=\lim _{x \rightarrow 0^{+}} \frac{e^{x}+1}{e^{x}+x}\\
&\Rightarrow \log L=2\\
&\Rightarrow \quad L=e^{2}
\end{aligned}
$$

Topic: Mathematics, Limits, WBJEE, WBJEE 2019
$\lim _{x \rightarrow 0}(\sin x)^{2 \tan x}$ is equal to

Options:
A 2
B 1
C 0
D does not exist

The correct answer is:
1

Solution:
Let $y=\lim _{x \rightarrow 0}(\sin x)^{2 \tan x}$
$\Rightarrow \log y=\lim _{x \rightarrow 0} \log (\sin x)^{2 \tan x}$
$=2 \lim _{x \rightarrow 0} \tan x \log \sin x$
$=2 \lim _{x \rightarrow 0} \frac{\log \sin x}{\cot x}$
$=2 \lim _{x \rightarrow 0} \frac{\frac{1}{\sin x} \cdot \cos}{-cosec^{2}x} x$
$=2 \lim _{x \rightarrow 0}(-\sin x \cos x)$
$=2 \times 0=0$
$\therefore \quad \log y=0 \Rightarrow y=e^{0}=1$

Topic: Mathematics, Limits, WBJEE, WBJEE 2017
$\lim _{x \rightarrow 0}\left(\frac{1}{x} \ln \sqrt{\frac{1+x}{1-x}}\right)$ is

Options:
A $1 / 2$
B 0
C 1
D does not exist

The correct answer is:
1

Solution:
$\lim _{x \rightarrow 0} \frac{1 / 2(\ln (1+x)-\ln (1-x))}{x}$
$=\frac{1}{2} \lim _{x \rightarrow 0}\left(\frac{1}{1+x}+\frac{1}{1-x}\right)=1$

Topic: Mathematics, Limits, WBJEE, WBJEE 2022
$\lim _{x \rightarrow 0^{+}}\left(x^{n} \ln x\right), n>0$

Options:
A does not exist
B exists and is zero
C exists and is 1
D exists and is $e^{-1}$

The correct answer is:
exists and is zero

Solution:
We have,
$\lim _{x \rightarrow 0^{+}} x^{n} \ln x$
$=\lim _{x \rightarrow 0^{+}} \frac{\ln x}{x^{-n}}$
$=\lim _{x \rightarrow 0^{+}} \frac{x}{-n x^{-n-1}} \quad$ [using L'Hospital's rule $]$
$=\lim _{x \rightarrow 0^{+}} \frac{-1}{n x^{-n}}=0 \quad\left[\because \frac{1}{x-n}=0,\right.$ when $\left.x=0\right]$

Topic: Mathematics, Limits, WBJEE, WBJEE 2019
$\lim _{x \rightarrow 0} \frac{\sin |x|}{x}$ is equal to

Options:
A 1
B 0
C positive infinity
D does not exist

The correct answer is:
does not exist

Solution:
Hints: $\operatorname{Lim}_{x \rightarrow 0} \frac{\sin |x|}{x}$
LHL $=-1 \quad$ RHL $=1$
Limit does not exist

Topic: Mathematics, Limits, WBJEE, WBJEE 2010
$\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{(1-\sqrt{x})}{(1-x)}}$ is equal to

Options:
A 1
B does not exist
C $\sqrt{\frac{2}{3}}$
D ln 2

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

Solution:
We have, $\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1-\sqrt{x}}{1-x}}$
$=\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1-\sqrt{x}}{(1+\sqrt x)(1- \sqrt x)}}$
$=\lim _{x \rightarrow 1}\left(\frac{1+x}{2+x}\right)^{\frac{1}{1+\sqrt{x}}}$
$=\left(\frac{1+1}{2+1}\right)^{\frac{1}{1+1}}=\left(\frac{2}{3}\right)^{\frac{1}{2}}=\sqrt{\frac{2}{3}}$

Topic: Mathematics, Limits, WBJEE, WBJEE 2016
$\lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{1}{(x-1)}\right)$

Options:
A Does not exist
B 1
C $\frac{1}{2}$
D 0

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

Solution:
Hint:
$\operatorname{lt}_{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right)=\operatorname{lt}_{x \rightarrow 1} \frac{(x-1)-\ln x}{(x-1) \ln x}=\frac{1}{2}$
Using L.H. rule twice

Topic: Mathematics, Limits, WBJEE, WBJEE 2020
\(\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{x^2}=\)

Options:
A \(\pi^2\)
B \(3 \pi\)
C \(2 \pi\)
D \(\pi\)

The correct answer is:
\(\pi\)

Solution:
\(\begin{aligned}
& \text {Hints : }=\operatorname{Lt}_{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{x^2}=\pi_{x \rightarrow 0} \frac{\sin ^2 x}{x^2}=\pi \\
& =\pi
\end{aligned}\)

Topic: Mathematics, Limits, WBJEE, WBJEE 2011
\(\lim _{x \rightarrow \infty}\left\{x-\sqrt[n]{\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)}\right\}\) where \(a_1, a_2, \ldots, a_n\) are positive rational numbers. The limit

Options:
A does not exist
B is \(\frac{a_1+a_2+\ldots a_n}{n}\)
C is \(\sqrt[n]{a_1 a_2 \ldots a_n}\)
D is \(\frac{n}{a_1+a_2+\ldots+a_n}\)

The correct answer is:
is \(\frac{a_1+a_2+\ldots a_n}{n}\)

Solution:
Hint: \(\operatorname{lt}_{x \rightarrow \infty}\left\{x-\sqrt[x]{\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)}\right\}\)
\(\begin{aligned} & \operatorname{lt}_{x \rightarrow \infty}\left\{x-x^n \sqrt{\left(1-\frac{a_1}{x}\right)\left(1-\frac{a_2}{x}\right) \ldots\left(1-\frac{a_n}{x}\right)}\right\} \\ & \Rightarrow \operatorname{lt}_{x \rightarrow \infty} x\left\{1-\left(1-\frac{a_1}{x}\right)^{1 / n}\left(1-\frac{a_2}{x}\right)^{1 / n} \ldots\left(1-\frac{a_n}{x}\right)^{1 / n}\right\} \\ & \Rightarrow \operatorname{It}_{x \rightarrow \infty} x\left[1-\left(1-\frac{a_1}{n x}\right)\left(1-\frac{a_2}{n x}\right) \ldots\left(1-\frac{a_n}{x n}\right)\right] \\ & \Rightarrow \operatorname{lt}_{x \rightarrow \infty} x\left[\frac{a_1+a_2+\ldots+a_n}{n x}\right]=\frac{a_1+a_2+\ldots+a_n}{n} \\ & \end{aligned}\)

Topic: Mathematics, Limits, WBJEE, WBJEE 2023
$\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x+1}-a x-b\right),(a, b \in R)=0$. Then

Options:
A $a=0, b=1$
B $a=1, b=-1$
C $a=-1, b-1$
D $a=0, b=0$

The correct answer is:
$a=1, b=-1$

Solution:
$\lim _{x \rightarrow \infty} \frac{x^2+1-a x(x+1)-b(x+1)}{x+1}=0$
$=\lim _{x \rightarrow \infty} \frac{(1-a) x^2-(a+b) x+1-b}{x+1}=0$
for the limit to exist,
$\begin{aligned}
1-\mathrm{a} &=0 \text { and }-(\mathrm{a}+\mathrm{b})=0 \\
\therefore \mathrm{a}=1, \mathrm{~b} &=-1
\end{aligned}$

Topic: Mathematics, Limits, WBJEE, WBJEE 2022
Lines $x+y=1$ and $3 y=x+3$ intersect the ellipse $x^{2}+9 y^{2}=9$ at the points $P, Q$ and $R$. The area of the $\triangle P Q R$ is

Options:
A $\frac{36}{5}$
B $\frac{18}{5}$
C $\frac{9}{5}$
D $\frac{1}{5}$

The correct answer is:
$\frac{18}{5}$

Solution:
Given equation of ellipse
$$
x^{2}+9 y^{2}=9
$$
$\Rightarrow \quad \frac{x^{2}}{9}+\frac{y^{2}}{1}=1$
and equation of lines
$$
x+y=1 \text { and } 3 y=x+3
$$
or $\quad \frac{x}{(-3)}+\frac{y}{1}=1$

$\therefore$ Area of $\Delta P Q R=\left|\begin{array}{c|ccc}0 & 1 & 1 \\ -3 & 0 & 1 \\ 9 / 5 & -4 / 5 & 1\end{array}\right| \mid$
$=\frac{1}{2}\left[\frac{24}{5}+\frac{12}{5}\right]$
$=\frac{1}{2} \times \frac{36}{5}=\frac{18}{5}$

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2013
ln a$\Delta A B C,$ if $\angle C=90^{\circ}, r$ and $R$ are the inradius and circumrodius of the ABC respectively, then $2(r+R)$ is equal to

Options:
A $b+c$
B $c+a$
C $a+b$
D $a+b+c$

The correct answer is:
$a+b$

Solution:
Tangents drawn from external points are of equal length.
$\Rightarrow \quad B D=B E=a-r$
$[\because C D=r]$
and $\quad A F=A E=b-r$
$[\because C F=r]$
$A E+B E=A B$
$\Rightarrow \quad b-r+a-r=2 R$
$1 \because A B$ is a diameter of circumcircle]
$\Leftrightarrow \quad b+a=2 R+2 r$
$\Rightarrow \quad 2(r+R)=a+b$

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2015
Magnetic field intensity $\mathrm{H}$ at the centre of a circular loop of radius $\mathrm{r}$ carrying current $\mathrm{I}$ e m.u is

Options:
A r/I oersted
B $2 \pi \mathrm{I} / \mathrm{r}$ oersted
C $\mathrm{I} / 2 \pi \mathrm{r}$ oersted
D $2 \pi \mathrm{r} / \mathrm{I}$ oersted

The correct answer is:
$2 \pi \mathrm{I} / \mathrm{r}$ oersted

Solution:
$H=\frac{\mu_0 I}{2 r}=\frac{\mu_0}{4 \pi} \times \frac{2 \pi I}{r}$
In e m.u system $\frac{\mu_0}{4 \pi}=1$. So $H=\frac{2 \pi I}{r}$

Topic: Physics, Magnetic Properties of Matter, WBJEE, WBJEE 2009
Match the flame colours of the alkaline earth metal salts in the Bunsen burner
$\begin{array}{l|l}
\hline \text{A. Caicium} & \text{p. Brick red} \\
\hline \text{B. Strontium} & \text{q. Apple green} \\
\hline \text{C. Barium} & \text{r. Crimson} \\
\hline
\end{array}$

Options:
A p, r, q
B r, p, q
C q, r, p
D p, q, r

The correct answer is:
p, r, q

Solution:
Flame colours are produced from the movement of the electrons in the metal ions present in the compounds.
Calcium gives brick red colour. Strontium gives crimson colour. Barium gives apple green colour.

Topic: Chemistry, Practical Chemistry, WBJEE, WBJEE 2015
\(\mathrm{BrF}_3\) self ionises as following

Options:
A \(2 \mathrm{BrF}_3 \rightleftharpoons \mathrm{BrF}^{+}+\mathrm{BrF}_5^{-}\)
B \(2 \mathrm{BrF}_3 \rightleftharpoons \mathrm{BrF}_2^{+}+\mathrm{BrF}_4^{-}\)
C \(2 \mathrm{BrF}_3 \rightleftharpoons \mathrm{BrF}_4^{+}+\mathrm{BrF}_2^{-}\)
D \(2 \mathrm{BrF}_3 \rightleftharpoons \mathrm{BrF}_3^{+}+\mathrm{BrF}_3^{-}\)

The correct answer is:
\(2 \mathrm{BrF}_3 \rightleftharpoons \mathrm{BrF}_2^{+}+\mathrm{BrF}_4^{-}\)

Solution:
Hint : \(2 \mathrm{BrF}_3 \rightleftharpoons \mathrm{BrF}_2^{+}+\mathrm{BrF}_4^{-}\)(Relatively more stable structures.)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2023
\(\mathrm{n}\) objects are distributed at random among \(\mathrm{n}\) persons. The number of ways in which this can be done so that at least one of them will not get any object is

Options:
A \(n !-n\)
B \(n^n-n\)
C \(n^n-n^2\)
D \(n^n-n\) !

The correct answer is:
\(n^n-n\) !

Solution:
Hint : Total number of ways of distributing \(n\) objects randomly among \(n\) persons \(=\mathrm{n}^n\)
Number of ways in which each person gets exactly one object \(={ }^n P_n=n\) !
\(\therefore\) Required number of ways \(=n^n-n\) !

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2023
\(\mathrm{Na}_2 \mathrm{CO}_3\) is prepared by Solvay process but \(\mathrm{K}_2 \mathrm{CO}_3\) cannot be prepared by the same because

Options:
A \(\mathrm{K}_2 \mathrm{CO}_3\) is highly soluble in \(\mathrm{H}_2 \mathrm{O}\)
B \(\mathrm{KHCO}_3\) is sparingly soluble
C \(\mathrm{KHCO}_3\) is appreciably soluble
D \(\mathrm{KHCO}_3\) decomposes

The correct answer is:
\(\mathrm{KHCO}_3\) is appreciably soluble

Solution:
Hint: \(\left(\mathrm{NH}_4\right) \mathrm{HCO}_3+\mathrm{KCl} \longrightarrow \mathrm{KHCO}_3(\mathrm{aq})+\mathrm{NH}_4 \mathrm{Cl}(\mathrm{aq})\)
\(\mathrm{KHCO}_3\) being appreciably soluble cant be isolated from reaction medium easily.

Topic: Chemistry, s Block Elements, WBJEE, WBJEE 2023
\(\mathrm{NO}_2\) is not obtained on heating

Options:
A \(\mathrm{AgNO}_3\)
B \(\mathrm{KNO}_3\)
C \(\mathrm{Cu}\left(\mathrm{NO}_3\right)_2\)
D \(\mathrm{Pb}\left(\mathrm{NO}_3\right)_2\)

The correct answer is:
\(\mathrm{KNO}_3\)

Solution:
Hints: \(\mathrm{KNO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{KNO}_2+\frac{1}{2} \mathrm{O}_2\)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2011
Maximum value of the function $f(x)=\frac{x}{8}+\frac{2}{x}$ on the interval [1,6] is

Options:
A 1
B $\frac{9}{8}$
C $\frac{13}{12}$
D $\frac{17}{8}$

The correct answer is:
$\frac{17}{8}$

Solution:
$f(x)=\frac{x}{8}+\frac{2}{x}$
$\therefore \quad f^{\prime}(x)=\frac{1}{8}-\frac{2}{x^{2}}=\frac{x^{2}-16}{8 x^{2}}$
For maximum or minimum $f^{\prime}(x)$ must be vanish.
$$
\begin{array}{ll}
\therefore & f^{\prime}(x)=0 \\
\Rightarrow & \frac{x^{2}-16}{8 x^{2}}=0 \Rightarrow x=4,-4 \\
& x \in[1,6] \\
\therefore & x \neq-4
\end{array}
$$
Also, in $[1,4], f^{\prime}(x) < 0 \Rightarrow f(x)$ is decreasing. $\ln [4,6], f^{\prime}(x)>0 \Rightarrow f(x)$ is increasing.
$f(1)=\frac{1}{8}+\frac{2}{1}=\frac{17}{8}$
$f(8)=\frac{6}{8}+\frac{2}{6}=\frac{3}{4}+\frac{1}{3}=\frac{13}{12}$
Hence, maximum value of $f(x)$ in [1,6] is $\frac{17}{8}$.

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2012
Mean of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ is $\bar{x}$. If an observation $x_{q}$, is replaced by $x_{q}^{\prime}$ then the new mean is

Options:
A $\bar{x}-x_{q}+x_{q}^{\prime}$
B $\frac{(n-1) \bar{x}+x_{q}^{\prime}}{n}$
C $\frac{(n-1) \bar{x}-x_{q}^{\prime}}{n}$
D $\frac{n x-x_{q}+x_{q}^{\prime}}{n}$

The correct answer is:
$\frac{n x-x_{q}+x_{q}^{\prime}}{n}$

Solution:
We have,
$$
\bar{x}=\frac{x_{1}+x_{2}+\ldots+x_{q}+\ldots+x_{2}}{n}
$$
$\Rightarrow \quad \Sigma x=n \bar{x}$
If $x_{q}$, is replaced by $x_{q}^{\prime}$, then new total will be $\Sigma x'=\Sigma x-x_{4}-x_{4}^{\prime}$
New mean will be
$$
\begin{array}{l}
\bar{x}^{\prime}=\frac{\Sigma x^{\prime}}{n} \\
\bar{x}^{\prime}=\frac{\sum x-x_{q}+x_{i}^{\prime}}{n} \\
\bar{x}^{\prime}=\frac{n \bar{x}-x_{q}+x_{i}^{\prime}}{n}
\end{array}
$$

Topic: Mathematics, Statistics, WBJEE, WBJEE 2017
Metal ion responsible for the Minimata disease is

Options:
A $\mathrm{Co}^{2+}$
B $\mathrm{Hg}^{2+}$
C $\mathrm{Cu}^{2}$
D $Z n^{2+}$

The correct answer is:
$\mathrm{Hg}^{2+}$

Solution:
Eating animal poisoned with mercury $\left(\mathrm{Hg}^{2+}\right)$ Causes deformity. known as minamata (minimata) disease which is characterised by diarrhoea, hemolysis, impairment of various senses, meningitis and death.

Topic: Chemistry, Chemistry in Everyday Life, WBJEE, WBJEE 2014
Methoxybenzene on treatment with $\mathrm{HI}$ produces

Options:
A iodobenzene and methanol
B phenol and methyl iodide
C iodobenzene and methyl iodide
D phenol and methanol

The correct answer is:
phenol and methyl iodide

Solution:
Methoxybenzene on treatment with HI produces phenol and methyl iodide. This is due the more stable aryl oxygen bond. Here. methyl phenyl oxonium ion is formed by the protonation of ether. The bond between $\mathrm{O}-\mathrm{CH}_{3}$ is weaker than the bond between $\mathrm{O}-\mathrm{C}_{6} \mathrm{H}_{5}$ because the carbon of phenyl group is $s p^{2}$ -hybridised and there is a partial double bond character. Thus, the reaction yields phenol and alkyl halide.

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2018
Mixing of two different ideal gases under isothermal reversible condition will lead to

Options:
A increase of Gibbs free energy of the system
B no change of entropy of the system
C increase of entropy of the system
D increase of enthalpy of the system

The correct answer is:
increase of entropy of the system

Solution:
On mixing of two different ideal gases under isothermal reversible conditions, $\Delta S_{\operatorname{mix}}$ is always positive $i . e .$, increasing.

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2013
Molecular velocities of two gases at the same temperature $(\mathrm{T})$ are $\mathrm{u}_{1}$ and $\mathrm{u}_{2} .$ Their masses are $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$ respectively. Which of the following expressions is correct at temperature $\mathrm{T} ?$

Options:
A $\quad \frac{\mathrm{m}_{1}}{\mathrm{u}_{1}^{2}}=\frac{\mathrm{m}_{2}}{\mathrm{u}_{2}^{2}}$
B $\mathrm{m}_{1} \mathrm{u}_{1}=\mathrm{m}_{2} \mathrm{u}_{2}$
C $\frac{\mathrm{m}_{1}}{\mathrm{u}_{1}}=\frac{\mathrm{m}_{2}}{\mathrm{u}_{2}}$
D $\mathrm{m}_{1} \mathrm{u}_{1}{ }^{2}=\mathrm{m}_{2} \mathrm{u}_{2}{ }^{2}$

The correct answer is:
$\mathrm{m}_{1} \mathrm{u}_{1}{ }^{2}=\mathrm{m}_{2} \mathrm{u}_{2}{ }^{2}$

Solution:
We know $u \times \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$ but as $\mathrm{T}$ is same for both gases
Hence u $\propto \frac{1}{\sqrt{\mathrm{m}}}$
$\therefore \frac{\mathrm{u}_{1}}{\mathrm{u}_{2}}=\sqrt{\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}}$
or $\frac{\mathrm{u}_{1}^{2}}{\mathrm{u}_{2}^{2}}=\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}}$
or $\mathrm{m}_{1} \mathrm{u}_{1}^{2}=\mathrm{m}_{2} \mathrm{u}_{2}^{2}$

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2021
$\mathrm{N}_{2} \mathrm{H}_{4}$ and $\mathrm{H}_{2} \mathrm{O}_{2}$ show similarity in

Options:
A Density
B Reducing nature
C Oxidising nature
D Hybridisation of central atoms

The correct answers are:
Reducing nature, Oxidising nature, Hybridisation of central atoms

Solution:

Topic: Chemistry, Hydrogen, WBJEE, WBJEE 2021
$n$ identical droplets are charged to $v$ volt each. If they coalesce to form a single drop, then its potential will be

Options:
A $n^{2 / 3} v$
B $n^{1 / 3} v$
C $n v$
D $v / n$

The correct answer is:
$n^{2 / 3} v$

Solution:
$$
\begin{aligned}
& \text { Hints: } n \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi \mathrm{R}^3 \\
& \Rightarrow \mathrm{R}=r n^{1 / 3} \\
& \mathrm{C}_0=4 \pi \varepsilon_0 r \\
& q_0=\mathrm{C}_0 \mathrm{~V}=\left(4 \pi \varepsilon_0 r\right) \mathrm{V}
\end{aligned}
$$
Capacitance of Bigger drop,
$$
\mathrm{C}=4 \pi \varepsilon_0 \mathrm{R}
$$
So, $\mathrm{V}=\frac{n q_0}{\mathrm{C}}=\frac{n\left(4 \pi \varepsilon_0 r \mathrm{~V}\right)}{4 \pi \varepsilon_0 \mathrm{R}}=n\left(\frac{r}{\mathrm{R}}\right) \mathrm{V}=n\left(\frac{1}{n^{1 / 3}}\right) \mathrm{V}=n^{2 / 3} \mathrm{~V}$

Topic: Physics, Capacitance, WBJEE, WBJEE 2010
Nickel combines with a uninegative monodentate ligand \(\left(\mathrm{X}^{-}\right)\)to form a paramagnetic complex \(\left[\mathrm{NiX}_4\right]^{2-}\). The hybridisation involved and number of unpaired electrons present in the complex are respectively.

Options:
A \(\mathrm{sp}^3\) two
B \(\mathrm{dsp}^2\), zero
C \(\mathrm{dsp}^2\), one
D \(\mathrm{sp}^3\), one

The correct answer is:
\(\mathrm{sp}^3\) two

Solution:
Hint : \(\mathrm{Ni}^{2+}\left(\mathrm{d}^8\right)\)
it should be paramagnetic with 2 unpaired electrons

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2023
Nitric acid can be obtained from ammonia via the formations of the intermediate compounds

Options:
A nitric oxide and nitrogen dioxide
B nitrogen and nitric oxide
C nitric oxide and dinitrogen pentoxide
D nitrogen and nitrous oxide

The correct answer is:
nitric oxide and nitrogen dioxide

Solution:
Ostwald process of for manufacture of nitric acid, $4 \mathrm{NH}_{3}+\mathrm{SO}_{2} \frac{\mathrm{Pt} \text { gauge }}{800-900^{\circ} \mathrm{C}} \underset{\text { nitric oxide }}{4 \mathrm{NO}}+6 \mathrm{H}_{2} \mathrm{O}$
$$
\Delta H=-21.5 \mathrm{kcal}
$$
$2 \mathrm{NO}+\mathrm{O}_{2} \stackrel{50^{\circ} \mathrm{C}}{\longrightarrow} \quad 2 \mathrm{NO}_{2}$
$4 \mathrm{NO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} \longrightarrow \underset{\text { nitric acid }}{4 \mathrm{HNO}_{3}}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2013