datavorous/entrance-exam-dataset huggingface dataset:
The position vectors of the points $A, B, C$ and $D$ are $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \mathbf{k}, \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $4 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\lambda \hat{\mathbf{k}},$ respectively. If the points $A, B, C$ and $D$ lie on a plane, the value of $\lambda$ is

Options:
A 0
B 1
C 2
D $-4$

The correct answer is:
$-4$

Solution:
The position vectors of the points $A, B, C$ and $D$ are $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, 5 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and
$4 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\lambda \hat{\mathbf{k}}$ respectively
$\begin{aligned}
\mathbf{B A} &=(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\mathbf{k})-(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \mathbf{k})=\hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \mathbf{k} \\
\mathbf{C A} &=(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})-(5 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \mathbf{k})=-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\
\mathbf{D A} &=(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})-(4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}) \\
&=-\hat{\mathbf{i}}-\hat{\mathbf{j}}-(1+\lambda) \mathbf{k}
\end{aligned}$
These are coplanar, so
$\left|\begin{array}{ccc}1 & 1 & -3 \\ -2 & -1 & -3 \\ -1 & -1 & -(1+\lambda)\end{array}\right|=0$
$=\left|\begin{array}{ccc}-1 & -1 & 3 \\ 2 & 1 & 3 \\ 1 & 1 & 1+\lambda\end{array}\right|=0$
$-1[1+\lambda-3]+1[2+2 \lambda-3]+3[2-1]=0$
$\Rightarrow \quad-1[\lambda-2]+1[2 \lambda-1]+3=0$
$\Rightarrow -\lambda+2+2 \lambda-1+3=0$
$\Rightarrow \lambda+4=0$
$\Rightarrow \lambda=-4$

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2019
The possible product(s) to be obtained from the reaction of cyclobutyl amine with HNO $_{2}$ is/are

Options:
A
B
C
D

The correct answers are:
,

Solution:
The possible product $(s)$ to be obtained from the reaction of cyclobutyl amine with $\mathrm{HNO}_{2}$, is/are

Topic: Chemistry, Amines, WBJEE, WBJEE 2018
The possible values of $x$, which satisfy the trigonometric equation
$\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$ are

Options:
A $\pm \frac{1}{\sqrt{2}}$
B $\pm \sqrt{2}$
C $\pm \frac{1}{2}$
D $\pm 2$

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

Solution:
We have, $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$
$\Rightarrow \quad \tan ^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x-1}{x-2} \cdot \frac{x+1}{x+2}}\right]=\frac{\pi}{4}$
$\Rightarrow \frac{(x-1)(x+2)+(x-2)(x+1)}{(x-2)(x+2)-(x-1)(x+1)}=\tan \frac{\pi}{4}$
$\Rightarrow \quad \frac{x^{2}+x-2+x^{2}-x-2}{x^{2}-4-x^{2}+1}=1$
$\Rightarrow$ $\frac{2 x^{2}-4}{-3}=1$
$\Rightarrow$ $2 x^{2}-4=3$
$\Rightarrow$ $2 x^{2}=1$
$x^{2}=\frac{1}{2}$
$x=\pm \frac{1}{\sqrt{2}}$

Topic: Mathematics, Inverse Trigonometric Functions, WBJEE, WBJEE 2017
The potential difference $V$ required for accelerating an electron to have the de-Broglie wavelength of $1 Å$ is

Options:
A $100 \mathrm{V}$
B $125 \mathrm{V}$
C $150 \mathrm{V}$
D $200 \mathrm{V}$

The correct answer is:
$150 \mathrm{V}$

Solution:
The de-Broglie wavelength associated with electron is given by.
$$
\lambda=\frac{h}{\sqrt{2 m e V}}
$$
where, $h=$ Planck's constant
$$
\begin{aligned}
m &=\text { mass of electron } \\
\mathrm{e} &=\text { charge on electron } \\
\mathrm{V} &=\text { potential difference } \\
\lambda &=1 \dot{\mathrm{A}}=10^{-10} \mathrm{m} \\
\lambda^{2} &=\frac{h^{2}}{2 \mathrm{meV}} \Rightarrow \mathrm{V}=\frac{h^{2}}{2 \mathrm{me} \lambda^{2}} \\
&=\frac{\left(6.6 \times 10^{-34}\right)^{2}}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times\left(1 \times 10^{-10}\right)^{2}} \\
&=\frac{6.6 \times 10^{-34} \times 6.6 \times 10^{-34}}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 10^{-20}} \\
&=\frac{6.6 \times 6.6 \times 10^{2}}{2 \times 9.1 \times 1.6}=\frac{4356 \times 10^{2}}{2912}=150 \mathrm{V}
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2016
The potential energy of a particle of mass $0.02 \mathrm{~kg}$ moving along $x$-axis is given by $V=A x(x-4) J$ where $x$ is in metres and $\mathrm{A}$ is a constant. Which of the following is/are correct statement(s)?

Options:
A The particle is acted upon by a constant force
B The particle executes simple harmonic motion
C The speed of the particle is maximum at $x=2 m$
D The period of oscillation of the particle is $\frac{\pi}{5} \mathrm{sec}$

The correct answers are:
The particle executes simple harmonic motion, The speed of the particle is maximum at $x=2 m$

Solution:
$V=A x(x-4) J m=0.02=A x^{2}-4 A x$
$$
\begin{array}{l}
\mathrm{F}=-\frac{\mathrm{d} \mathrm{V}}{\mathrm{dx}}=-2 \mathrm{Ax}+4 \mathrm{~A} \text { (not constant) } \\
\mathrm{F}=-2 \mathrm{~A}(\mathrm{x}-2)
\end{array}
$$
Hence position execute S.H.M mean position is when force $=0$ i.e. $x=2$ and speed is maximum at mean position
$$
\omega^{2}=\frac{2 \mathrm{~A}}{\mathrm{~m}} \mathrm{~T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{~A}}}=2 \pi \sqrt{\frac{1}{100 \mathrm{~A}}}=\frac{\pi}{5} \sqrt{\frac{1}{\mathrm{~A}}}
$$

Topic: Physics, Oscillations, WBJEE, WBJEE 2021
The potential of a hydrogen electrode at $\mathrm{pH}=10$ is

Options:
A $0.59 \mathrm{~V}$
B $0.00 \mathrm{~V}$
C $-0.59 \mathrm{~V}$
D -0.059

The correct answer is:
$-0.59 \mathrm{~V}$

Solution:
Hints : $\mathrm{H}^{+}(\mathrm{pH}=10)\left|\mathrm{H}_2(1 \mathrm{~atm})\right| \operatorname{Pt}(\mathrm{s})$
Reaction: $2 \mathrm{H}^{+}\left(\mathrm{p}^{\mathrm{H}}=10\right)+2 \mathrm{e} \rightarrow \mathrm{H}_2(1 \mathrm{~atm})$
$$
\begin{aligned}
& \mathrm{E}=\mathrm{E}^0-\frac{0.0591}{2} \log \left(\frac{\mathrm{P}_{\mathrm{H}_2}}{\left[\mathrm{H}^{+}\right]^2}\right) \\
& =0-\frac{0.0591}{2} \log \frac{1}{\left(10^{-10}\right)^2}=-\frac{0.0591}{2} \times 2 \log \frac{1}{10^{-10}}=-0.0591 \times 10=-0.591 \\
& \text { i.e. } E=-0.591 \mathrm{~V} \\
&
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2010
The probability that at least one of $A$ and $B$ occurs is 0.6 . If A and $B$ occur simultaneously with probability 0.3 , then $P^{\prime}\left(A^{\prime}\right)+P_{(}\left(B^{\prime}\right)$ is

Options:
A 0.9
B 0.15
C 1.1
D 1.2

The correct answer is:
1.1

Solution:
$$
\begin{array}{cl}
\text { Hints: } \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.6 & \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=\mathrm{P}(\mathrm{A} \cup \mathrm{B})+\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.9 \\
\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.3 & \mathrm{P}\left(\mathrm{A}^{\prime}\right)+\mathrm{P}\left(\mathrm{B}^{\prime}\right)=2-0.9=1.1
\end{array}
$$

Topic: Mathematics, Probability, WBJEE, WBJEE 2010
The product of the above reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
It is an example of intramolecular Cannizzaro reaction.

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2015
The product of the following hydrogenation reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:
The double bonds will continue to get reduced in one molecule one after another. This process will continue till all the hydrogens are exhausted.

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2022
The product of the following reaction is :

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2022
The product of the reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
It is a substitution reaction, so the product
formed is

Topic: Chemistry, Haloalkanes and Haloarenes, WBJEE, WBJEE 2015
The product (s) from the above reaction will be

Options:
A
B
C
D

The correct answers are:
,

Solution:
Hint:

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2020
The product (s) in the following sequence of reactions will be
$\mathrm{Me}-\mathrm{C} \equiv \mathrm{C}-\mathrm{Me} \frac{\begin{array}{l}\text { 1. } \mathrm{Na} / \mathrm{NH}_{3} \text { (liq.) } \\ \text { ethanol, }-33^{\circ} \mathrm{C}\end{array}}{\text { 2. dil. alkaline } \mathrm{KMnO}_{4}}$ Product (s)

Options:
A
B
C
D

The correct answers are:
,

Solution:

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2021
The products $\mathrm{X}$ and $\mathrm{Y}$ which are formed in the following sequence of reactions are respectively $\mathrm{Phenol} \stackrel{\text { dil } \mathrm{HNO}_{3}}{\longrightarrow} \mathrm{X} \frac{1 . \mathrm{Zn} / \mathrm{HCl}, \Delta}{2 \cdot\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}(1 \text { equiv })} \longrightarrow \mathrm{Y}$

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2021
The quadratic equation
$2 x^{2}-\left(a^{3}+8 a-1\right) x+a^{2}-4 a=0$ has roots of opposite sign. Then,

Options:
A $a \leq 0$
B $0 < a < 4$
C $4 \leq a < 8$
D $a \geq 8$

The correct answer is:
$0 < a < 4$

Solution:
Since, roots are opposite sign.
Product of the roots $\leq 0$ $\begin{array}{ll}\Rightarrow & a^{2}-4 a < 0 \\ \Rightarrow & a(a-4) < 0 \\ \Rightarrow & 0 < a < 4\end{array}$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2012
The quadratic equation whose roots are three times the roots of $3 a x^2+3 b x+c=0$ is

Options:
A $a x^2+3 b x+3 c=0$
B $\mathrm{ax}^2+3 \mathrm{bx}+\mathrm{c}=0$
C $9 a x^2+9 b x+c=0$
D $a x^2+b x+3 c=0$

The correct answer is:
$a x^2+3 b x+3 c=0$

Solution:
Hints: $3 a \alpha^2+3 b \alpha+c=0$
$$
x=3 \alpha \Rightarrow \alpha=\frac{x}{3}
$$
$$
\begin{aligned}
& 3 a \frac{x^2}{9}+3 b \cdot \frac{x}{3}+c=0 \\
& a x^2+3 b x+3 c=0
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2009
The quadratic equation $x^2+15|x|+14=0$ has

Options:
A only positive solutions
B only negative solutions
C no solution
D both positive and negative solution

The correct answer is:
no solution

Solution:
Hints: $x^2+15|x|+14>0 \forall x$
Hence no solution

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2010
The quadratic expression $(2 x+1)^{2}-p x+q \neq 0$ for any real $x,$ if

Options:
A $p^{2}-16 p-8 q < 0$
B $p^{2}-8 p+16 q < 0$
C $p^{2}-8 p-16 q < 0$
D $p^{2}-16 p+8 q < 0$

The correct answer is:
$p^{2}-8 p-16 q < 0$

Solution:
Given equation is $(2 x+1)^{2}-p x+q \neq 0$
$$
\begin{array}{lr}
\Rightarrow & 4 x^{2}+4 x+1-p x+q \neq 0 \\
\Rightarrow & 4 x^{2}+(4-p) x+(1+q) \neq 0
\end{array}
$$
Now. D $ < $ 0
$\Rightarrow \quad(4-p)^{2}-4(4)(1+q) < 0$
$\Rightarrow \quad 16-8 p+p^{2}-16-16 q < 0$
$$
\Rightarrow \quad p^{2}-8 p-16 q < 0
$$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2015
The quantity $h v / K_{\mathrm{B}}$ corresponds to

Options:
A wavelength
B velocity
C temperature
D angular momentum

The correct answer is:
temperature

Solution:
The quantity $h v / K_{\mathrm{B}}$ corresponds to temperature. $h v=3 / 2 K_{\mathrm{B}} T$
where, $T=$ temperature $K_{\mathrm{B}}=$ Boltzmann constant $$
\begin{array}{l}
\frac{h v}{K_{B}}=\frac{3}{2} T
\end{array}
$$
since, $\frac{3}{2}$ is a constant, the value of $\frac{h v}{K_{\mathrm{B}}}$ corresponds
to temperature. Therefore option
(c) is correct.

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2019
The quantity of electricity needed to separately electrolyze 1 M solution of $\mathrm{ZnSO}_{4}$, $\mathrm{AlCl}_{3}$ and $\mathrm{AgNO}_{3}$ completely is in the ratio of

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

The correct answer is:
$2: 3: 1$

Solution:
From the Faradiay's IInd law, we know that
$$
W \propto \text { Zit, } \quad W \propto \frac{E}{96500} i t
$$
or $\quad$ it $\propto \frac{1}{E}$ or electricity $\propto \frac{1}{E}$
Thus, the amount of electricity required for $\mathrm{ZnSO}_{4}, \mathrm{AlCl}_{3}$ and $\mathrm{AgNO}_{3}$ is in the ratio of
2: 3: 1 . (Here, 2,3 and 1 are the valencies of metal atom in the given compounds).
Alternatively
In $Z n S O_{4}$,
$Z n^{2+}+2 e^{-}$ $\rightarrow Z \mathrm{n}$
$1mol$ $\text{2 mol or 2 Faraday}$
In $\mathrm{AlCl}_{3}$,
$\mathrm{Al}^{3+}+3 e^{-} \longrightarrow \mathrm{Al}$
$1mol$ $\text{3F}$
In $\mathrm{AgNO}_{3}$,
$\mathrm{Ag}^{+}+e^{-} \longrightarrow \mathrm{Ag}$
$1mol$ $\text{1F}$
Thus, the ratio of amounts of electricity required is 2: 3: 1

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2014
The radius of the first Bohr orbit of a hydrogen atom is $0.53 \times 10^{-8} \mathrm{~cm}$. The velocity of the electron in the first Bohr orbit is

Options:
A $2.188 \times 10^{9} \mathrm{~cm} \mathrm{~s}^{-1}$
B $4.376 \times 10^{8} \mathrm{~cm} \mathrm{~s}^{-1}$
C $1.094 \times 10^{8} \mathrm{~cm} \mathrm{~s}^{-1}$
D $2.188 \times 10^{9} \mathrm{~cm} \mathrm{~s}^{-1}$

The correct answer is:
$2.188 \times 10^{9} \mathrm{~cm} \mathrm{~s}^{-1}$

Solution:
Hint:
$r=0.53 \times 10^{-9} \mathrm{~cm}=0.53 \times 10^{-9} n^{2}$
$\therefore \mathrm{n}=1$
$\mathrm{~V}=2.18 \times 10^{9} \mathrm{cm} / \mathrm{s}$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2020
The radius of the light circle observed by a fish at a depth of 12 meter is (refractive index of water $=4 / 3$ )

Options:
A $36 \sqrt{7}$
B $\frac{36}{\sqrt{7}}$
C $36 \sqrt{5}$
D $4 \sqrt{5}$

The correct answer is:
$\frac{36}{\sqrt{7}}$

Solution:
$$
\text { Hints : } r=\frac{h}{\sqrt{\mu^2-1}}=\frac{12}{\sqrt{\frac{16}{9}-1}}=\frac{12 \times 3}{\sqrt{7}}=\frac{36}{\sqrt{7}}
$$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2010
The rate of a certain reaction is given by, rate $=k\left[\mathrm{H}^{+}\right]^{n}$. The rate increases 100 times when the pH changes from 3 to 1 . The order $(n)$ of the reaction is

Options:
A 2
B 0
C 1
D 1.5

The correct answer is:
1

Solution:
Given, rate $=k\left[\mathrm{H}^{*}\right]^{n}$
Initial $\mathrm{pH}=3 \mathrm{so}\left[\mathrm{H}^{+}\right]=1 \times 10^{-3}$, initial rate $=\mathrm{r}_{\mathrm{1}}$
Final $\mathrm{pH}=1$ so $\left[\mathrm{H}^{+}\right]=1 \times 10^{-1}$
Final rate $=r_{2}=100 r_{1}$
On substituting values, we get
$$
\begin{array}{l}
r_{1}=k\left[1 \times 10^{-3}\right]^{n} \\
r_{2}=100 r_{1}=k\left[10^{-1}\right]^{n}
\end{array}
$$
Dividing eqs (i) by (ii).
$$
\frac{r_{1}}{100 r_{1}}=\left[\frac{1 \times 10^{-3}}{1 \times 10^{-1}}\right]^{n}
$$
$\therefore$
$$
\begin{array}{c}
\frac{1}{100}=\left[\frac{10^{-2}}{1}\right]^{n} \Rightarrow\left[\frac{1}{100}\right]^{1}=\left[\frac{1}{100}\right]^{n} \\
n=1
\end{array}
$$
Thus, the reaction is of first order.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2014
The ratio of accelerations due to gravity $g_{1}:g_{2}$ on the surfaces of two planets is 5: 2 and the ratio of their respective average densities $\rho_{1}: \rho_{2}$ is 2: 1 . What is the ratio of respective escape velocities $v_{1}: v_{2}$ from the surface of the planets?

Options:
A 5: 2
B $\sqrt{5}: \sqrt{2}$
C $5: 2 \sqrt{2}$
D 25.4

The correct answer is:
$5: 2 \sqrt{2}$

Solution:
Escape velocity from a planet. $v_{0}=\sqrt{\frac{2 G M}{R}}=\sqrt{\frac{2 G M}{R^{2}}} R$
$$
=\sqrt{2 g R}
$$
According due to gravity $g=\frac{G M}{R^{2}}=\frac{G\frac{4}{3}\pi R^{3} \cdot \rho}{R^{2}}=\frac{4}{3} G\pi R_{p}$
$\therefore \quad$ Radius $R=\frac{3 g}{4 \pi G \rho}$
From Eqs. (i) and (ii), we get $v_{e}=\sqrt{2 g \cdot \frac{3 g}{4 \pi G \rho}}=\sqrt{\frac{3}{2} \frac{g^{2}}{\pi G \rho}}$
Thus, $\quad v_{e} \propto \frac{g}{\sqrt{\rho}}$
$\therefore \quad \frac{v_{n}}{v_{r_{2}}}=\frac{g_{1}}{\sqrt{\rho_{1}}} \times \frac{\sqrt{\rho_{2}}}{g_{2}}=\frac{5}{2} \times \frac{1}{\sqrt{2}}$
(given, $\frac{g_{1}}{g_{2}}=\frac{5}{2}$ and $\frac{Q}{\rho_{2}}=2: 1$ )
$=\frac{5}{2 \sqrt{2}}$

Topic: Physics, Gravitation, WBJEE, WBJEE 2018
The ratio of magnetic field and magnetic moment at the centre of a current carrying circular loop is $x$. When both the current and radius is doubled the ratio will be

Options:
A $x / 8$
B $x / 4$
C $x / 2$
D $2 x$

The correct answer is:
$x / 8$

Solution:
$$
\begin{aligned}
& \text { Hints: } \mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 a} \quad \quad \mathrm{M}=\mathrm{I}\left(\pi a^2\right) \\
& \frac{\mathrm{B}}{\mathrm{M}}=\frac{\mu_0 \mathrm{I}}{2 a} \times \frac{1}{\mathrm{I} \pi a^2}=\frac{\mu_0}{2 \pi a^3}=x
\end{aligned}
$$
Again, Ratio $=\frac{\mu_0}{2 \pi(2 a)^3}=\frac{1}{8}\left(\frac{\mu_0}{2 \pi a^3}\right)=\frac{x}{8}$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2010
The ratio of the diameter of the sun to the distance between the earth and the sun is approximately $0.009 .$ The approximate diameter of the image of the sun formed by a concave spherical mirror of radius of curvature $0.4 \mathrm{m}$ is

Options:
A $4.5 \times 10^{-6} \mathrm{m}$
B $4.0 \times 10^{-6} \mathrm{m}$
C $3.6 \times 10^{-3} \mathrm{m}$
D $1.8 \times 10^{-3} \mathrm{m}$

The correct answer is:
$1.8 \times 10^{-3} \mathrm{m}$

Solution:
The angle subtended by sun on mirror
;
$$
\begin{array}{l}
Q=\frac{D}{d_{S E}}=\frac{d}{f} \\
d=\frac{D}{d_{s E}} \times f
\end{array}
$$
According to the question,
$$
\begin{aligned}
\frac{D}{d_{s k}} &=0 \cdot 009 \\
f &=\frac{r}{2}=\frac{0.4}{2}=0 \cdot 2 \mathrm{m}
\end{aligned}
$$
$\therefore \quad d=0 \cdot 009 \times 0 \cdot 2 \mathrm{m}$
$$
\begin{array}{l}
=9 \times 2 \times 10^{-4} \mathrm{m} \\
=18 \times 10^{-4} \mathrm{m} \\
=1.8 \times 10^{-3} \mathrm{m}
\end{array}
$$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2017
The ratio of volumes of $\mathrm{CH}_{3} \mathrm{COOH } \text{ 0.1}$ (N) to CH COONa 0.1 (N) required to prepare a buffer solution of $\mathrm{pH} 5.74$ is (Given, $p K_{a}$ of $C H_{3} C O O H$ is $\left.4.74\right)$

Options:
A $10 : 1$
B $5 : 1$
C $1 : 5$
D $1 : 10$

The correct answer is:
$1 : 10$

Solution:
Given, that $\mathrm{pH}=5.74, \mathrm{pK}_{a}=4.74$
Suppose that volume of acid solution $=x \mathrm{L}$ Volume of salt solution $=y \mathrm{L}$
From Henderson equation,
$$
\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}
$$
or, $\mathrm{pH}-\mathrm{p} K_{\mathrm{a}}=\log \frac{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}$
or, $5.74-4.74=1=\log \frac{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}$
or $\frac{\left.\mathrm{CH}_{3} \mathrm{COONa}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}=10$
or $\frac{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{COONa}\right]}=\frac{1}{10}=\frac{\frac{x+y}{0.1 \mathrm{y}}}{\mathrm{x}+\mathrm{y}}$
Thus, $\frac{x}{y}=\frac{1}{10}$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2015
The reaction for obtaining the metal (M) from its oxide $\left(\mathrm{M}_{2} \mathrm{O}_{3}\right)$ ore is given by
$\mathrm{M}_{2} \mathrm{O}_{3}(\mathrm{~s})+2 \mathrm{Al}(\ell) \stackrel{\text { Heat }}{\rightarrow} \mathrm{Al}_{2} \mathrm{O}_{3}(\ell)+2 \mathrm{M}(\mathrm{s}),(\mathrm{s}=$ solid, $\ell=$ liquid $)$
in that case, $M$ is

Options:
A Copper
B Calcium
C Iron
D Zinc

The correct answer is:
Iron

Solution:
Hint:
Thermite Process

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2020
The reaction of aniline with chloroform under alkaline conditions leads to the formation of

Options:
A phenylcyanide
B phenylisonitrile
C phenylcyanate
D phenylisocyanate

The correct answer is:
phenylisonitrile

Solution:
Aniline is a primary amine (contains - $\mathrm{NH}_{2}$ group). When it is treated with chloroform under alkaline conditions, it results in the formation of a bad smelling compound, known by the name isonitrile or carbylamine. This reaction is known as carbylamine reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, WBJEE, WBJEE 2014
The reaction of methyltrichloroacetate $\left(\mathrm{Cl}_{3} \mathrm{CCO}_{2} \mathrm{Me}\right)$ with sodium methoxide $\mathrm({NaOMe})$ generates

Options:
A carbocation
B carbene
C carbanion
D carbon radical

The correct answer is:
carbene

Solution:

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2015
The reaction of nitroprusside anion with sulphide ion gives purple colouration due to the formation of

Options:
A the tetranionic complex of iron (II) coordinating to one $NOS^{-}$ ion
B the dianoinic complex of iron (II) coordinating to one $NCS^{-}$ ion
C the trianoionic complex of iron (III) coordinating to one $NOS^{-}$ ion
D the tetranionic complex of iron (III) coordinating to one $NCS^{-}$ ion

The correct answer is:
the tetranionic complex of iron (II) coordinating to one $NOS^{-}$ ion

Solution:

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2013
The reaction sequence given below gives product $R$
The structure of the product $R$ is

Options:
A
B
C
D

The correct answer is:

Solution:
The reaction occurs as follows :
The above reaction is an example of Borodine Hunsdiecker reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, WBJEE, WBJEE 2017
The reaction that takes place when $\mathrm{Cl}_2$ gas is passed through conc. $\mathrm{NaOH}$ solution is :

Options:
A Oxidation
B Reduction
C Displacement
D Disproportionation

The correct answer is:
Disproportionation

Solution:
Hence the reaction is disproportionation

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2009
The reactive species in chlorine bleach is

Options:
A $\mathrm{Cl}_{2} \mathrm{O}$
B $\mathrm{OCl}^{-}$
C $\mathrm{Cl} \mathrm{O}_{2}$
D $\mathrm{HCl}$

The correct answer is:
$\mathrm{OCl}^{-}$

Solution:
Chlorine bleach is $\mathrm{NaOCl}$. The hypochlorite ion in chlorine bleach dissociates to give nascent oxygen as shown below.
$$
\mathrm{OCl}^{-} \longrightarrow[\mathrm{O}]+\mathrm{Cl}^{-}
$$
Therefore, the reactive species in chlorine bleach is $\mathrm{OCl}^{-}$
So, the option (b) is correct.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2019
$$
\text { The reading of the ammeter in the following figure will be }
$$

Options:
A $0.8 \mathrm{~A}$
B $0.6 \mathrm{~A}$
C $0.4 \mathrm{~A}$
D $0.2 \mathrm{~A}$

The correct answer is:
$0.4 \mathrm{~A}$

Solution:
Hints: $\frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1 \Omega$
$$
\begin{aligned}
& \mathrm{R}_{\text {eq }}=1+4=5 \Omega \\
& \mathrm{I}=\frac{2}{5}=0.4 \mathrm{~A}
\end{aligned}
$$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2010
The reagent with which the following reaction is best accomplished is

Options:
A $\mathrm{H}_{3} \mathrm{PO}_{2}$
B $\mathrm{H}_{3} \mathrm{PO}_{3}$
C $\mathrm{H}_{3} \mathrm{PO}_{4}$
D $NaHSO_{3}$

The correct answer is:
$\mathrm{H}_{3} \mathrm{PO}_{2}$

Solution:
$\mathrm{H}_{3} \mathrm{PO}_{2}$ is the reagent that converts $-\mathrm{N}_{2}^{+} \mathrm{Cl}^{-}$ group into H. During this process $\mathrm{N}_{2}$ and $\mathrm{HCl}$ gases are released.

Topic: Chemistry, Amines, WBJEE, WBJEE 2014
The reagents to carry out the following conversion are

Options:
A $\mathrm{HgSO}_{4} / \mathrm{dil} \cdot \mathrm{H}_{2} \mathrm{SO}_{4}$
B $\mathrm{BH}_{3} ; \mathrm{H}_{2} \mathrm{O}_{2} / \mathrm{NaOH}$
C $0 s O_{4} ; H I_{4}$
D $\mathrm{NaNH}_{2} / \mathrm{CH}_{3} \mathrm{I} ; \mathrm{HgSO}_{4} / \mathrm{dil} \cdot \mathrm{H}_{2} \mathrm{SO}_{4}$

The correct answer is:
$\mathrm{NaNH}_{2} / \mathrm{CH}_{3} \mathrm{I} ; \mathrm{HgSO}_{4} / \mathrm{dil} \cdot \mathrm{H}_{2} \mathrm{SO}_{4}$

Solution:
(Here, Me means methyl group) Thus, the correct set of reagents is $\mathrm{NaNH}_{2} / \mathrm{CH}_{3} \mathrm{I} ; \mathrm{HgSO}_{4} / \mathrm{dil} \mathrm{H}_{2} \mathrm{SO}_{4}$

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2014
The reduction of benzenediazonium chloride to phenyl hydrazine can be accomplished by

Options:
A $\mathrm{SnCl}_{2}, \mathrm{HCl}$
B $\mathrm{Na}_{2} \mathrm{SO}_{3}$
C $\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}$
D $\mathrm{H}_{3} \mathrm{PO}_{2}$

The correct answers are:
$\mathrm{SnCl}_{2}, \mathrm{HCl}$, $\mathrm{Na}_{2} \mathrm{SO}_{3}$

Solution:
Reduction of benzene diazoniums chloride occurs as follows:

Hence, (a) and (b) are the correct answers.

Topic: Chemistry, Amines, WBJEE, WBJEE 2017