datavorous/entrance-exam-dataset huggingface dataset:
As shown in the figure, a pump is designed as horizontal cylinder with a piston having area A and an outlet orifice having an area ' \(a\) '. The piston moves with a constant velocity under the action of force \(F\). If the density of the liquid is \(\rho\), then the speed of the liquid emerging from the orifice is, (assume \(A > > a\) )

Options:
A \(\sqrt{\frac{F}{\rho A}}\)
B \(\frac{a}{A} \sqrt{\frac{F}{\rho A}}\)
C \(\sqrt{\frac{2 F}{\rho A}}\)
D \(\frac{A}{a} \sqrt{\frac{2 F}{\rho A}}\)

The correct answer is:
\(\sqrt{\frac{2 F}{\rho A}}\)

Solution:
Hint: by principle of continuity
\(\mathrm{AV}=\mathrm{av}\)
by Bernoulis principle
\(\begin{aligned} & P+\frac{1}{2} \rho V^2=P_0+\frac{1}{2} \rho v^2 \\ & {\left[\frac{F}{A}+P_0\right]+\frac{1}{2} \rho V^2=P_0+\frac{1}{2} \rho v^2} \\ & \frac{F}{A}+\frac{1}{2} \rho\left[\frac{a v}{A}\right]^2=\frac{1}{2} \rho v^2 \\ & \frac{F}{A}=\frac{1}{2} \rho v^2\left[1-\frac{a^2}{A^2}\right]\end{aligned}\)
\(\begin{aligned} & v=\sqrt{\frac{2 F}{\rho A\left[1-\frac{a^2}{A^2}\right]}} \\ & a^2 \ll A^2 \\ & v=\sqrt{\frac{2 F}{\rho A}}\end{aligned}\)

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2023
As shown in the figure, a rectangular loop of a conducting wire is moving away with a constant velocity $v$ in a perpendiculat direction from a very long straight conductor carrying a steady current $l$. When the
breadth of the rectangular loop is very small compared to its distance from the straight conductor, how does the emf. $E$ induced in the loop vary with time $t ?$

Options:
A $E \propto \frac{1}{t^{2}}$
B $E \propto \frac{1}{t}$
C $E \alpha-\ln (t)$
D $E \propto \frac{1}{t^{3}}$

The correct answer is:
$E \propto \frac{1}{t^{2}}$

Solution:
$\therefore$ Motional emf in the loop.
$$
\begin{aligned}
E &=E_{1}-E_{2}=vl l\left(B_{1}-B_{2}\right) \\
&=v\left(\frac{\mu_{0}}{2 \pi} \cdot \frac{I}{y}-\frac{\mu_{0} I}{2 \pi(y+b)}\right] \\
&=\frac{\mu_{0}}{2 \pi} \cdot I \cdot v l\left[\frac{y+b-y}{y(y+b)}\right]=\frac{\mu_{0}}{2 \pi} \cdot \frac{v I l}{y(y+b)} b
\end{aligned}
$$
$\because \quad b< < y$
$\therefore \quad E=\frac{\mu_{0}}{2 \pi} \cdot \frac{v l I}{y^{2}} b$
$\because v=$ constant
$$
$$
Thus, $\quad E=\frac{\mu_{0}}{2 \pi} \cdot \frac{v l l}{v^{2} t^{2}} b$
Hence, $\quad E \propto \frac{1}{t^{2}}$

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2018
As shown in the figure, a single conducting wire is bent to form a loop in the form of a circle of radius 'r' concentrically inside a square of side ' $a$ ', where $a: r=8: \pi$. A battery B drives a current through the wire. If the battery B and the gap G are of negligible sizes, determine the strength of magnetic field at the common centre $O$.

Options:
A $\cdot \frac{\mu_{0} \mid}{2 \pi a} \sqrt{2}(\sqrt{2}-1)$
B $\frac{\mu_{0} I}{2 \pi a}(\sqrt{2}+1)$
C $\frac{\mu_{0} I}{\pi a} 2 \sqrt{2}(\sqrt{2}+1)$
D $\frac{\mu_{0} I}{\pi \mathrm{a}} 2 \sqrt{2}(\sqrt{2}-1)$

The correct answer is:
$\frac{\mu_{0} I}{\pi \mathrm{a}} 2 \sqrt{2}(\sqrt{2}-1)$

Solution:
Hint:
$\frac{a}{r}=\frac{8}{\pi}, r=\frac{\pi a}{8}$
$B=\frac{\mu_{0} \mid}{2 r}-\frac{4 \mu_{0} I}{4 \pi a / 2} \times \sqrt{2}=\frac{\mu_{0} I}{2 r}-\frac{2 \sqrt{2} \mu_{0} I}{\pi a}=\frac{\mu_{0} \mid \times 8}{2 \pi a}-\frac{2 \sqrt{2} \mu_{0} I}{\pi a}$
$=\frac{\mu_{0} \mid}{\pi a}(4-2 \sqrt{2})=\frac{\mu_{0} I}{\pi a} 2 \sqrt{2}(\sqrt{2}-1)$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2020
As shown in the figure, a wire is bent to form a D-shaped loop, carrying current I, where the curved part is semi-circle of radius $\mathrm{R}$. the loop is placed in a uniform magnetic field $\overrightarrow{\mathrm{B}}$, which is directed into the plane of the paper. The magnetic force left by the closed loop is

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

The correct answer is:
0

Solution:
Hint:
Zero (Net magnetic force on a closed current loop in a uniform $\overrightarrow{\mathrm{B}}$ is zero)

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2020
As shwon in the figure, a point charge $q_{1}=+1 \times 10^{-8} C$ is placed at the origin in $x$ -y plane and another point charge $\mathrm{q}_{2}=+3 \times 10^{-6} \mathrm{C}$ is placed at the co-ordinate $(10,0)$. In that case, which of the following graph(s) shows most correctly the electric field vector in $\mathrm{E}_{x}$ in $\mathrm{x}$ -direction?

Options:
A
B
C
D

The correct answer is:

Solution:
Hint:

Topic: Physics, Electrostatics, WBJEE, WBJEE 2020
Aspirin is

Options:
A Acetyl salicylic acid
B Benzoyl salicylic acid
C Chloro benzoic acid
D Anthranilic acid

The correct answer is:
Acetyl salicylic acid

Solution:
Hints:

Topic: Chemistry, Chemistry in Everyday Life, WBJEE, WBJEE 2010
Assume that each diode as shown in the tịgure has a forvard bias resistance of $50 \Omega$ and an infinite reverse bias resistance. The current through the resistance $150 \Omega$ is

Options:
A $0.66 \mathrm{A}$
B $0.05 \mathrm{A}$
C zero
D $0.04 \mathrm{A}$

The correct answer is:
$0.04 \mathrm{A}$

Solution:
The circuit is

As the lower diode attached to $100 \Omega$ resistance is in reversed biased so, $1 t$ is non-conducting
Now, the circuit can be redrawn as,

$\therefore$ Current through the circuit, $I=\frac{V}{R}$
$$
=\frac{10}{250}=0.04 \mathrm{A}
$$

Topic: Physics, Semiconductors, WBJEE, WBJEE 2015
Assume that the earth moves around the sun in a circular orbit of radius $R$ and there exists a planet which also move around the sun in a circular orbit with an angular speed twice as large as that of the earth. The radius of the orbit of the planet is

Options:
A $2^{-2 / 3} R$
B $2^{2 / 3} R$
C $ 2^{-1 / 3} R$
D $\frac{R}{\sqrt{2}}$

The correct answer is:
$2^{-2 / 3} R$

Solution:
According to the Kepler's third law
$$
T^{2} \propto r^{3}
$$
where, $T=$ time period of revolution
$r=$ radius
Now, $\left(\frac{T_{E}}{T_{P}}\right)^{2}=\left(\frac{r_{E}}{r_{P}}\right)^{3}=\frac{r_{E}}{r_{P}}=\left(\frac{T_{E}}{T_{P}}\right)^{2 / 3}$
$$
\frac{r_{E}}{r_{P}}=\left(\frac{2 \pi / \omega_{E}}{2 \pi / \omega_{P}}\right)^{2 / 3} \quad\left[\because T=\frac{2 \pi}{\omega}\right]
$$
$$
\frac{r_{E}}{r_{P}}=\left(\frac{\omega_{P}}{\omega_{E}}\right)^{2 / 3}
$$
According to the question, $\omega_{P}=2 \omega_{E}$ and $r_{E}=R$
$\Rightarrow \quad \frac{R}{r_{P}}=\left(\frac{2 \omega_{E}}{\omega_{E}}\right)^{2 / 3}$
$$
\begin{array}{l}
\frac{R}{r_{P}}=(2)^{2 / 3} \\
r_{P}=\frac{R}{(2)^{2 / 3}}=R(2)^{-2 / 3}
\end{array}
$$

Topic: Physics, Gravitation, WBJEE, WBJEE 2019
Assuming the compounds to be completely dissociated in aqueous solution, identify the pair of the solutions that can be expected to be isotonic at the same temperature.

Options:
A $0.01 \mathrm{M}$ urea and $0.01 \mathrm{M} \mathrm{NaCl}$
B $0.02 \mathrm{M} \mathrm{NaCl}$ and $0.01 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}$
C 0.03 M $\mathrm{NaCl}$ and $0.02 \mathrm{M} \mathrm{MgCl}_{2}$
D $0.01 \mathrm{M}$ sucrose and $0.02 \mathrm{M}$ glucose

The correct answer is:
0.03 M $\mathrm{NaCl}$ and $0.02 \mathrm{M} \mathrm{MgCl}_{2}$

Solution:
$\because$ For Isotonic solution
$$
\pi_{1}=\pi_{2} \text { i.e. } i_{1} C_{1} R T=i f_{2} R T
$$
For option $(c)$
$$
\begin{array}{l}
i_{1}=2, i_{2}=3 \\
C_{1}=0.03 \\
C_{2}=0.02
\end{array}
$$
Thus, $\quad i_{1} \times C_{1}=i_{2} \times C_{2}$
Hence, are isotonic, i.e.
$$
0.06=0.06
$$
In all other options, $i_{1} \times C_{1} \neq i_{2} \times C_{2}$ Thus, are not isotonic.

Topic: Chemistry, Solutions, WBJEE, WBJEE 2017
At $25^{\circ} \mathrm{C}, \mathrm{pH}$ of a $10^{-8} \mathrm{M}$ aqueous $\mathrm{KOH}$
solution will be

Options:
A 6
B 7.02
C 8.02
D 9.02

The correct answer is:
7.02

Solution:
$\mathrm{NaOH} \longrightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-} \quad\left[\mathrm{OH}^{-}\right]=10^{-8} \mathrm{M}$
$\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-}$ $\left[\mathrm{OH}^{-}\right]=10^{-7} \mathrm{M}$
$$
\begin{aligned}
\therefore\left[\mathrm{OH}^{-}\right]_{\text {total }} &=\left(10^{-8}+10^{-7}\right) \mathrm{M} \\
&=10^{-7}(1.1) \\
&=1.1 \times 10^{-7} \\
\therefore \quad \mathrm{pOH} &=\log 1.1 \times 10^{-7}=6.98 \\
\therefore \quad \mathrm{pH} &=14-6.98 \\
&=7.02
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2013
At $25^{\circ} \mathrm{C}$, the molar conductance of $0.007 \mathrm{M}$ hydrofluoric acid is 150 mho $\mathrm{cm}^{2} \mathrm{mol}^{-1}$ and its $\Lambda_{m}^{\circ}=500$mho$cm^2$$mol^{-1}$ The value of the dissociation constant of the acid at the given concentration at $25^{\circ} \mathrm{C}$ is

Options:
A $7 \times 10^{-4} \mathrm{M}$
B $7 \times 10^{-5} \mathrm{M}$
C $9 \times 10^{-3} \mathrm{M}$
D $9 \times 10^{-4} \mathrm{M}$

The correct answer is:
$9 \times 10^{-4} \mathrm{M}$

Solution:
Degree of dissociation, $\alpha=\frac{\lambda^{\circ} c}{\lambda^{\circ}m}=\frac{150}{500}=0.3$
Given, $C=0.007 \mathrm{M}$
Hydrofluoric acid dissociates in the following manner

Dissociation constant,
$$
K_{a}=\frac{\left[H^{+}\right]\left[F^{-}\right]}{[H F]}=\frac{C \alpha \cdot C \alpha}{C(1-\alpha)}=\frac{C \alpha^{2}}{(1-\alpha)}
$$
On substituting values, we get
$$
\begin{aligned}
K_{a}=\frac{0.007 \times(0.3)^{2}}{(1-0.3)} &=\frac{63 \times 10^{-3} \times 10^{-2}}{0.7} \\
&=9 \times 10^{-4} \mathrm{M}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2014
At $25^{\circ} \mathrm{C}$, the solubility produet of salt of $\mathrm{MX}_{2}$ type is $3.2 \times 10^{-8}$ in water. The solubility (in $\mathrm{mol} / \mathrm{L})$ of $M X_{2}$ in water at the sume temperature will be

Options:
A $1.2 \times 10^{-3}$
B $2 \times 10^{-3}$
C $3.2 \times 10^{-3}$
D $1.75 \times 10^{-3}$

The correct answer is:
$2 \times 10^{-3}$

Solution:
For $M X_{2}$ type salt,
Solubility product, $K_{\mathrm{sp}}=4 s^{3} ; 3.2 \times 10^{-8}=4 \mathrm{s}^{3}$
or $s=\sqrt[3]{\frac{3.2 \times 10^{-8}}{4}}=2 \times 10^{-3} \mathrm{mol} / \mathrm{L}$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2013
At $273 \mathrm{~K}$ temperature and $76 \mathrm{~cm~} \mathrm{Hg}$ pressure, the density of a gas is $1.964 \mathrm{~g} \mathrm{~L}^{-1} .$ The gas is

Options:
A $\mathrm{CH}_{4}$
B $\mathrm{CO}$
C $\mathrm{He}$
D $\mathrm{CO}_{2}$

The correct answer is:
$\mathrm{CO}_{2}$

Solution:
Hint:
Molar mass of gas $(\mathrm{M})=\frac{\mathrm{dRT}}{\mathrm{P}}=\frac{1.964 \times 0.0821 \times 273}{1}$
$=44 \mathrm{~g}$
Hence, the gas is $\mathrm{CO}_{2}$

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2020
At a certain place, the horizontal component of earth's magnetic field is $\sqrt{3}$ times the vertical component. The angle of dip at that place is

Options:
A $30^{\circ}$
B $60^{\circ}$
C $45^{\circ}$
D $90^{\circ}$

The correct answer is:
$30^{\circ}$

Solution:
$\tan \theta=\frac{V}{H}=\frac{1}{\sqrt{3}} \Rightarrow \theta=30^{\circ}$

Topic: Physics, Magnetic Properties of Matter, WBJEE, WBJEE 2009
At a certain temperature the time required for the complete diffusion of $200 \mathrm{mL}$ of $\mathrm{H}_{2}$ gas is 30 min. The time required for the complete diffusion of $50 \mathrm{mL}$ of $\mathrm{O}_{2}$ gas at the same temperature will be

Options:
A $60 \mathrm{min}$
B $30 \mathrm{min}$
C $45 \mathrm{min}$
D $15 \mathrm{min}$

The correct answer is:
$30 \mathrm{min}$

Solution:
According to Graham's law of diffusion.
Rate of diffusion, $r \propto \frac{1}{\sqrt{M}}=\frac{V}{t}$
Here, $M=$ molecular mass, $V=$ volume and $t=$ timel Thus, for $\mathrm{H}_{2}$ gas,
$$
\frac{200}{30}=\frac{1}{\sqrt{2}}
$$
for $O_{2}$ gas,
$$
\frac{50}{t}=\frac{1}{\sqrt{32}}
$$
Eqs. (i) and (ii), we get From
$$
\begin{aligned}
\frac{200 \times t}{30 \times 50} &=\frac{\sqrt{32}}{\sqrt{2}} \\
t &=\frac{\sqrt{16} \times 30 \times 50}{200} \\
&=\frac{4 \times 30}{4}=30 \mathrm{min}
\end{aligned}
$$

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2014
At a certain temperature, the value of the slope of the plot of osmotic pressure $(\pi)$ against concentration $\left(C\right.$ in $\mathrm{mol} \mathrm{L}^{-1}$ ) of a certain polymer solution is $291 \mathrm{R.}$ The temperature at which osmotic pressure is measured, is $(R$ is gas constant)

Options:
A $271 ^\circ \mathrm{C}$
B $18 ^\circ \mathrm{C}$
C $564 \mathrm{k}$
D $18 \mathrm{K}$

The correct answer is:
$18 ^\circ \mathrm{C}$

Solution:
$$
\begin{array}{l}
\text { Slope. } R T=291 R \text { or } T=291 \mathrm{K} \\
\therefore \text { Temperature }=291-273=18^{\circ} \mathrm{C}
\end{array}
$$

Topic: Chemistry, Solutions, WBJEE, WBJEE 2015
At a particular height, the velocity of an ascending body is $\mathbf{u}$. The velocity at the same height while the body falls freely is

Options:
A $2 \mathbf{u}$
B $-\mathbf{u}$
C $\mathbf{u}$
D $-2 \mathbf{u}$

The correct answer is:
$-\mathbf{u}$

Solution:
Given that velocity of ascending body = u

Let there is no air resistance when body falls freely in downward direction, the direction of falling body will opposite to the ascending body.
$\therefore$ The velocily at the same height while the body falls freely $=-u$

Topic: Physics, Motion In One Dimension, WBJEE, WBJEE 2016
At a particular temperature. the ratio of equivalent conductance to specific conductance of a 0.01 N NaCl solution is

Options:
A $10^{5} \mathrm{cm}^{3}$
B $10^{3} \mathrm{cm}^{3}$
C $10 \mathrm{cm}^{3}$
D $10^{5} \mathrm{cm}^{2}$

The correct answer is:
$10^{5} \mathrm{cm}^{3}$

Solution:
As we know that,
Equivalent conductance, ( $\lambda$ )
$=\frac{\text { Specific conductance }(K) \times 1000}{\text { Concentration }}$
or. $\quad \frac{\lambda}{K}=\frac{1000}{\text { conc. }}$
Or. $\frac{\lambda}{K}=\frac{1000 \Omega^{-1} \mathrm{cm}^{2} \mathrm{eq}^{-1}}{0.01 \Omega^{-1} \mathrm{cm}^{-1}}$
$\frac{\lambda}{K}=10^{5} \mathrm{cm}^{3} \mathrm{eq}^{-1}$
(Given, conc. $=0.01 \mathrm{N}$ )
Or,$\frac{\lambda}{K}=10^{5} \mathrm{cm}^{3} \mathrm{eq}^{-1}$

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2015
At constant pressure, the heat of formation of a compound is not dependent on temperature, when

Options:
A $\Delta C_{p}=0$
B $\Delta C_{v}=0$
C $\Delta C_{p}>0$
D $\Delta C_{p} < 0$

The correct answer is:
$\Delta C_{p}=0$

Solution:
Key Point According to Kirchhoff sequation
$$
\Delta H_{(f)}=\Delta H_{(i)}+C_{P}(\Delta T)
$$
At constant pressure
$$
\Delta C_{p}=O
$$
Therefore, heat of formation $\left(\Delta H_{f}\right)$ does not depend on temperature and value of $\Delta C_{p}$ i.e. when $\Delta C_{p}=0$
Hence, option (a) is the correct answer.

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2019
$$
\text { At identical temperature and pressure, the rate of diffusion of hydrogen gas is } 3 \sqrt{3} \text { times that }
$$
of a hydrocarbon having molecular formula $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}$. What is the value of ' $n$ '?

Options:
A 1
B 4
C 3
D 8

The correct answer is:
4

Solution:
$$
\begin{aligned}
& \text { Hints : } \frac{r_{H_2}}{r_{C_n H_{2 n-2}}}=\sqrt{\frac{M_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}}}{\mathbf{M}_{\mathrm{H}_2}}}=\sqrt{\frac{\mathbf{M}_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}}}{2}} \\
& \because \sqrt{\frac{\mathrm{M}_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}}}{2}}=3 \sqrt{3}=\sqrt{27} \\
& \Rightarrow \mathbf{M}_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}}=27 \times 2=54
\end{aligned}
$$
Hence, $12 n+(2 n-2) \times 1=54 \Rightarrow 14 n=56 \Rightarrow n=4$ Thus Hydrocarbon is $\mathrm{C}_4 \mathrm{H}_6$

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2010
At room temperature, the reaction between
water and fluorine produces

Options:
A $\mathrm{HF}$ and $\mathrm{H}_{2} \mathrm{O}_{2}$
B $\mathrm{HF}, \mathrm{O}_{2}$ and $\mathrm{F}_{2} \mathrm{O}_{2}$
C $F^{-}, O_{2}$ and $H^{+}$
D HOF and HF

The correct answer is:
$F^{-}, O_{2}$ and $H^{+}$

Solution:
At room temperature, the reaction between water and fluorine produces $\mathrm{F}^{-}, \mathrm{O}_{2}$ and $\mathrm{H}^{+}$ Fluorine, being non-polar molecule, readily dissolves with water and forms mixture of oxygen and ozone as shown below
$$
\begin{array}{l}
2 \mathrm{F}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{HF}+\mathrm{O}_{2} \\
3 \mathrm{F}_{2}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow 6 \mathrm{HF}+\mathrm{O}_{3}
\end{array}
$$
HF exist as $\mathrm{H}^{+}$ and $\mathrm{F}^{-}$ in a solution.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2018
At S.T.P. the volume of $7.5 \mathrm{g}$ of a gas is $5.6 \mathrm{L}$. The gas is

Options:
A NO
B $\mathrm{N}_{2} \mathrm{O}$
C CO
D $\mathrm{CO}_{2}$

The correct answer is:
NO

Solution:
Given, mass of gas $(W)=7.5 \mathrm{g}$
Volume of gas at $\operatorname{STP}(V)=56 \mathrm{L}$
$\because$ Moles of gas at $\mathrm{STP}=\frac{V(\mathrm{L})}{22.4(\mathrm{L})}=\frac{W}{M}$
Where $M=$ molar mass of gas.
$\therefore \quad M=\frac{W \times 22.4}{V}$
$$
=\frac{7.5 \times 22.4}{5.6}=30.00 \mathrm{g} \mathrm{mol}^{-1}
$$
Among the given options Molar mass $(M)$ of
$\mathrm{NO}=14+16=30.00 \mathrm{g} \mathrm{mol}^{-1}$
(b) $\mathrm{N}_{2} \mathrm{O}=28+16=44.00 \mathrm{g} \mathrm{mol}^{-1}$
(c) $\mathrm{CO}=12+16=28.00 \mathrm{g} \mathrm{mol}^{-1}$
(d) $\mathrm{CO}_{2}=12+32=44.00 \mathrm{g} \mathrm{mol}^{-1}$
$\because$ Molar mass of $\mathrm{NO}=30 \mathrm{g} \mathrm{mol}^{-1}$
Thus, the given gas is NO. Hence, is the correct option.

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2019
At temperature of $298 \mathrm{K}$, the emf of the following electrochemical cell. $\mathrm{Ag}(\mathrm{s})\left|\mathrm{Ag}^{+*}(0.1 \mathrm{M}) \| \mathrm{Zn}^{2 +}(0.1 \mathrm{M})\right| \mathrm{Zn}(\mathrm{s})$ will be (Given, $E^{\prime \prime}$ call $=-1.562 \mathrm{V}$ )

Options:
A $-1.532 \mathrm{V}$
B $-1.503 V$
C $1.532 \mathrm{V}$
D $-3.06 V$

The correct answer is:
$-1.532 \mathrm{V}$

Solution:
From the given cell, the cell reaction is
$2 \mathrm{Ag}(s)+Z \mathrm{n}^{2+}(01 \mathrm{M}) \longrightarrow$ $2 \mathrm{Ag}^{+}(0.1 \mathrm{M})+\mathrm{Zn}(s)$
The Nemst equation is,
$E_{\text {cel }}=E_{\text {cel }}^{\circ}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Ag}^{+}\right]^{2}}{\left.\mid \mathrm{Zn}^{2+}\right]}$
or, $E_{\text {cal }}=(-1.562)-\frac{(0.0591)}{2} \log \frac{(0.1)^{2}}{(0.1)}$
$\begin{aligned} \text { Or, } E_{\text {cel }} &=(-1.562)-\frac{0.0591}{2} \log 10^{-1} \\ &=-1.562+\frac{0.0591}{2} \\ &=-1.562+0.02955 \\ &=-1.532 \mathrm{V} \end{aligned}$

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2015
At two different places, the angles of dip are respectively $30^{\circ}$ and $45^{\circ} .$ At these two places the ratio of horizontal component of earth's magnetic field is

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

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

Solution:
The horizontal component of earth's magnetic field is given by, $H=R \cos \delta$
where, $\delta$ is the angle of dip so $H_{1}=R \cos 30^{\circ}$ and $\mathrm{H}_{2}=\mathrm{Rcos} 45^{\circ}$
$\therefore \quad \frac{H_{1}}{H_{2}}=\frac{R \cos 30^{\circ}}{R \cos 45^{\circ}}=\frac{\sqrt{3} / 2}{1 / \sqrt{2}}=\frac{\sqrt{3}}{\sqrt{2}}$

Topic: Physics, Magnetic Properties of Matter, WBJEE, WBJEE 2013
Atoms in a $\mathrm{P}_4$ molecule of white phosphorus are arranged regularly in the following way:

Options:
A at the corners of a cube
B at the corners of a octahedron
C at the corners of a tetrahedron
D at the centre and corners of a tetrahedron

The correct answer is:
at the corners of a tetrahedron

Solution:
Hints:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2009
Average distance of the Earth from the Sun is $L_{1}$ : If one year of the Earth $=D$ days, one year of another planet whose average distance from the Sun is $L_{2}$ will be

Options:
A $D\left(\frac{L_{2}}{L_{1}}\right)^{\frac{1}{2}}$ days
B $D\left(\frac{L_{2}}{L_{1}}\right)^{\frac{3}{2}}$ days
C $D\left(\frac{L_{2}}{L_{1}}\right)^{\frac{2}{3}}$ days
D $D\left(\frac{L_{2}}{L_{1}}\right)$ days

The correct answer is:
$D\left(\frac{L_{2}}{L_{1}}\right)^{\frac{3}{2}}$ days

Solution:
By Keplar's law,
$$
\begin{array}{l}
T^{2} \propto R^{3} \\
T=D\left(\frac{L_{2}}{L_{1}}\right)^{3 / 2}
\end{array}
$$

Topic: Physics, Gravitation, WBJEE, WBJEE 2012
Avogadro's law is valid for

Options:
A all gases
B ideal gas
C Van der Waals gas
D real gas

The correct answer is:
ideal gas

Solution:
Fact

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2022
$B$ is an extremity of the minor axis of an ellipse whose foci are $S$ and $S^{\prime} .$ If $\angle S B S^{\prime}$ is a right angle, then the eccentricity of the ellipse is

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

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

Solution:
Slope of $S B, m_{1}=\frac{b-0}{0-a e}=-\frac{b}{a e}$
and slope of $S^{\prime} B, m_{2}=\frac{b-0}{0-(-a e)}=\frac{b}{a e}$
since, $\angle S B S^{\prime}$ is a right angle.
$\therefore$ $$m_{1} m_{2}=-1$$
$\Rightarrow \quad \frac{-b}{a e} \times \frac{b}{a e}=-1$
$\Rightarrow \quad b^{2}=a^{2} e^{2}$
$\Rightarrow \quad \frac{b^{2}}{a^{2}}=e^{2}$
$\Rightarrow \quad 1-e^{2}=e^{2}$
$\Rightarrow \quad 2 e^{2}=1$
$\Rightarrow \quad e^{2}=\frac{1}{2}$
$\Rightarrow \quad e=\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \quad e=\frac{1}{\sqrt{2}}$ or $e=-\frac{1}{\sqrt{2}}$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2017
Baeyer's reagent is

Options:
A alkaline potassium permanganate
B acidified potassium permanganate
C neutral potassium permanganate
D alkaline potassium manganate

The correct answer is:
alkaline potassium permanganate

Solution:
Baeyer's reagent is $1 \%$ cold dilute alkaline potassium permanganate. It is used to identify unsaturation. All unsaturated compounds lose its purple colour.

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2013
Best reagent for nuclear iodination of aromatic compounds is

Options:
A $\mathrm{KI} / \mathrm{CH}_{3} \mathrm{COCH}_{3}$
B $I_{2} / C H_{3} C N$
C $\mathrm{KI} / \mathrm{CH}_{3} \mathrm{COOH}$
D $\mathrm{I}_{2} / \mathrm{HNO}_{3}$

The correct answer is:
$\mathrm{I}_{2} / \mathrm{HNO}_{3}$

Solution:
Best reagent for nuclear iodination of aromatic compounds is $\mathrm{I}_{2} / \mathrm{HNO}_{3}$

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2015
Blackened oil painting can be restored into original form by the action of :

Options:
A Chlorine
B $\mathrm{BaO}_2$
C $\mathrm{H}_2 \mathrm{O}_2$
D $\mathrm{MnO}_2$

The correct answer is:
$\mathrm{H}_2 \mathrm{O}_2$

Solution:
Blackening of oil painting is due to $\mathrm{PbS}$ which is oxidised by $\mathrm{H}_2 \mathrm{O}_2$ to form white $\mathrm{PbSO}_4$
$\underset{\text { (Black) }}{\mathrm{PbS}+\mathrm{H}_2 \mathrm{O}_2} \rightarrow \underset{\text { (white) }}{\mathrm{PbSO}_4}+\mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Hydrogen, WBJEE, WBJEE 2009
Block $B$ lying on a table weighs W. The coefficient of static friction between the block and the table is $\mu$. Assume that the cord between $B$ and the knot is horizontal. The maximum weight of the block $A$ for which the system will be stationary is

Options:
A $\frac{W \tan \theta}{\mu}$
B $\mu W tan \theta$
C $\mu W \sqrt{1+\tan ^{2} \theta}$
D $\mu W \sin \theta$

The correct answer is:
$\mu W tan \theta$

Solution:
Let weight of $A$ is $W^{\prime}$. From the free body diagram. For equilibrium of the system,

$$
\begin{aligned}
T \cos \theta &=\mu N=\mu W \\
T \sin \theta &=W^{\prime}
\end{aligned}
$$
where $T=$ tension in the thread lying between knot and the support.
On divindg, we get $\frac{T \sin \theta}{T \cos \theta}=\frac{W^{\prime}}{\mu W}$
$\Rightarrow \quad \tan \theta=\frac{W^{\prime}}{\mu W}$
$\Rightarrow W^{\prime}=\mu W \tan \theta$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2015
Boiling water reacts with $\mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2{ }^{+} \mathrm{Cl}^{-}$to give

Options:
A aniline
B benzylamine
C phenol
D benzaldehyde

The correct answer is:
phenol

Solution:
$$
\text { Hints : } \mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2^{+} \mathrm{Cl}^{-} \underset{(\text { Boil })}{\stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow}} \mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\left(\mathrm{S}_{\mathrm{N}} \mathrm{Ar}\right)
$$

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2010