datavorous/entrance-exam-dataset huggingface dataset:
A current ' $I^{\prime}$ is flowing along an infinite, straight wire, in the positive $Z$ -direction and the same current is flowing along a similar parallel wire $5 \mathrm{m}$ apart, in the negative Z-direction. A point $P$ is at a perpendicular distance $3 \mathrm{m}$ from the first wire and $4 \mathrm{m}$ from the second. What will be magnitude of the magnetic field B of $P ?$

Options:
A $\frac{5}{12}\left(u_{0} I\right)$
B $\frac{7}{24}\left(\mu_{0} I\right)$
C $\frac{5}{24}\left(\mu_{0} I\right)$
D $\frac{25}{288}\left(\mu_{0} I\right)$

The correct answer is:
$\frac{5}{24}\left(\mu_{0} I\right)$

Solution:
According to the question.

Magnetic field due to first wire $(A)$ is $B_{1}=\frac{\mu_{0} \times I}{2 \pi \times 3}$
Magnetic field due to second wire $(B)$ is $B_{2}=\frac{\mu_{0} \times 1}{2 \pi \times 4}$
Net magnitude of magnetic field $B=\sqrt{B_{1}^{2}+B_{2}^{2}}$
$$
B=\frac{\mu_{0} I}{2 \pi} \sqrt{\frac{1}{9}+\frac{1}{16}}=\frac{\mu_{0} I \times 5}{2 \pi \times 3 \times 4}
$$
Magnetic field $B=\frac{5}{24} \times \frac{\mu_{0} I}{\pi}$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2019
A current of 1 A is flowing along positive $x$ -axis through a straight wire of length $0.5 \mathrm{m}$ placed in a region of a magnetic field given by $\mathbf{B}=(2 \mathbf{i}+4 \mathbf{j}) \mathrm{T} .$ The magnitude and the direction of the force experienced by the wire respectively are

Options:
A $\sqrt{18} \mathrm{N},$ along positive $z$ -axis
B $\sqrt{20} \mathrm{N}$, along positive $x$ -axis
C $2 \mathrm{N}$, along positive $z$ -axis
D $4 \mathrm{N},$ along positive $y$ -axis

The correct answer is:
$2 \mathrm{N}$, along positive $z$ -axis

Solution:
The force experienced by the wire placed in magnetic field, $F=B il$
$$
\begin{array}{l}
=(2 \hat{i}+4 \hat{j})\left(1 \hat{i} \times \frac{1}{2}\right) \\
=\hat{i} \times \hat{i}+4 \times \frac{1}{2}(\hat{i} \times \hat{j})=2 \hat{k} N
\end{array}
$$
Direction of this force can be find out by Fleming's left hand rule which is along pasitive z-axis

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2013
A curve passes through the point $(3,2)$ for which the segment of the tangent line contained between the co-ordinate axes is bisected at the point of contact. The equation of the curve is

Options:
A $y=x^2-7$
B $x=\frac{y^2}{2}+2$
C $x y=6$
D $x^2+y^2-5 x+7 y+11=0$

The correct answer is:
$x y=6$

Solution:
$\frac{x}{x_1}+\frac{y}{y_1}=2$
$\because\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=-\frac{y_1}{x_1}$
$\therefore \frac{d y}{d x}=-\frac{y}{x} \Rightarrow \frac{d y}{y}=-\frac{d x}{x} \Rightarrow \ln y=-\ln x+k \Rightarrow \ln x y=k \Rightarrow x y=e^k=C \Rightarrow x y=C$ passes through $(3,2)$
$\therefore C=3 \times 2=6 \therefore x y=6$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2022
A cyclic process is shown in \(\mathrm{p}-\mathrm{v}\) diagram and \(\mathrm{T}-\mathrm{S}\) diagram. Which of the following statement(s) is/are true?

Options:
A \(1 \rightarrow 2\) : Isobaric, \(2 \rightarrow 3\) : Isothermal
B \(3 \rightarrow 1\) : Isochoric, \(2 \rightarrow 3\) : adiabatic
C Work done by the system in the complete cyclic process in non-zero
D The heat absorbed by the system in the complete cyclic process in non-zero

The correct answers are:
\(3 \rightarrow 1\) : Isochoric, \(2 \rightarrow 3\) : adiabatic, Work done by the system in the complete cyclic process in non-zero, The heat absorbed by the system in the complete cyclic process in non-zero

Solution:
Hint :

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2023
A cylinder of height his filled with wator and is kept on a block of height $h / 2$ The level of water in the cylinder is kept constant. Four holes numbered 1,2,3 and 4 are at the side of the cylinder and at heights $0, h / 4, h / 2$ and $3 h / 4,$ respectively. When all four holes are openned together, the hole from which water will reach farthest distance on the plane $P Q$ is the hole number.

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Height of the cylinder from the level $P O$ is $H=h+\frac{h}{2}=\frac{3 h}{2}$
To cover maximum distance along the plane $P Q$, height of the hole is $=\frac{3 h}{2} \times \frac{1}{2}=\frac{3 h}{4}$
This height corresponds to hole (2).

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2015
A cylindrical block floats vertically in a liquid of density $\rho_{1}$ kept in a container such that the fraction of volume of the cylinder inside the liquid is $x_{1}$. Then some amount of another immiscible liquid of density $\rho_{2}\left(\rho_{2} < \rho_{1}\right)$ is added to the liquid in the container so that the cylinder now floats just fully immersed in the liquids with $x_{2}$, fraction of volume of the cylinder inside the liquid of density $\rho_{1}$. The ratio $\rho_{1} / \rho_{2}$ will be

Options:
A $\frac{1-x_{2}}{x_{1}-x_{2}}$
B $\frac{1-x_{1}}{x_{1}+x_{2}}$
C $\frac{x_{1}-x_{2}}{x_{1}+x_{2}}$
D $\frac{x_{2}}{x_{1}}-1$

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

Solution:
The given situation is shown as

As the Bouyant force in both the cases are same
$\therefore$
$$
\rho_{1} x_{1} g=\rho_{1} x_{2} g+\rho_{2}\left(1-x_{2}\right) g
$$
On solving $\quad \frac{\rho_{1}}{\rho_{2}}=\frac{\left(1-x_{2}\right)}{\left(x_{1}-x_{2}\right)}$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2013
A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the determinant chosen is non-zero is

Options:
A $\frac{3}{16}$
B $\frac{3}{8}$
C $\frac{1}{4}$
D $\frac{5}{8}$

The correct answer is:
$\frac{3}{8}$

Solution:
$\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|=a d-b c=0$
Case $1: a d=b c=0 \rightarrow 3 \times 3=9$ cases
Case $2:$ ad $=\mathrm{bc}=1 \rightarrow 1$ case only
$\therefore$ Probability $=1-\frac{9+1}{2^4}=\frac{3}{8}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2022
A diver at a depth of \(12 \mathrm{~m}\) in water \(\left(\mu=\frac{4}{3}\right)\) sees the sky in a cone of semivertical angle :

Options:
A \(\sin ^{-1}\left(\frac{4}{3}\right)\)
B \(\tan ^{-1}\left(\frac{4}{3}\right)\)
C \(\sin ^{-1}\left(\frac{3}{4}\right)\)
D \(90^{\circ}\)

The correct answer is:
\(\sin ^{-1}\left(\frac{3}{4}\right)\)

Solution:
Hints: \(c=\sin ^{-1}\left(\frac{1}{\mu}\right)\)
\(=\sin ^{-1}\left(\frac{3}{4}\right)\)

Topic: Physics, Ray Optics, WBJEE, WBJEE 2011
A double ordinate $\mathrm{PQ}$ of the hyperbola $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ is such that $\Delta \mathrm{OPQ}$ is equilateral, $\mathrm{O}$ being the centre of the hyperbola. Then the eccentricity e satisfies the relation

Options:
A $1 < \mathrm{e} < \frac{2}{\sqrt{3}}$
B $\mathrm{e}=\frac{2}{\sqrt{3}}$
C $\mathrm{e}=\frac{\sqrt{3}}{2}$
D $\mathrm{e}>\frac{2}{\sqrt{3}}$

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

Solution:
Hint:
$P \equiv(a \sec \theta, b \tan \theta) \angle \mathrm{POM}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{\mathrm{b}}{\mathrm{a}} \sin \theta$
$\Rightarrow \sin \theta=\frac{a}{b \sqrt{3}}$
$\Rightarrow \frac{b^{2}}{a^{2}}>\frac{1}{3} \quad \Rightarrow 1+\frac{b^{2}}{a^{2}}>\frac{4}{3} \quad \Rightarrow e>\frac{2}{\sqrt{3}}$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2020
A drop of some liquid of volume $0.04 \mathrm{cm}^{3}$ is placed on the surface of a glass slide. Then another glass slide is placed on it in such a way that the liquid forms a thin layer of area $20 \mathrm{cm}^{2}$ between the surfaces of the two slides. To separate the slides a force of $16 \times 10^{5}$ dyne has to be applied normal to the surfaces. The surface tension of the liquid is (in dyne-cm $^{-1}$ )

Options:
A 60
B 70
C 80
D 90

The correct answer is:
80

Solution:
Let, thickness of layer be $x$
So, volume $V=$ Area $\times x$
$$
\begin{array}{ll}
V=A \times x & (\because)=2 r) \\
x=V / A
\end{array}
$$
$2 r=\frac{V}{A} \Rightarrow r=\frac{V}{2 A}$
(i)
and $\quad \Delta P=\frac{T}{r}$
We know that $F=\Delta P \times A$
$$
=\frac{T}{r} \times A
$$
$$
\begin{array}{l}
F=\frac{T}{\left(\frac{V}{2 A}\right)} \times A \\
T=\frac{F \times V}{2 A^{2}}
\end{array}
$$
[from
where $F=16 \times 10^{5}$ dyne $, V=0.04 \mathrm{cm}^{3}$
$$
\begin{aligned}
A=20 \mathrm{cm}^{2}=\frac{16 \times 10^{5} \times 0.04}{2 \times 20^{2}} \\
=\frac{8 \times 10^{5} \times 4}{20^{2} \times 100}=\frac{8 \times 10^{5} \times 4}{400 \times 100} \\
=8 \times 10^{5} \times 10^{-4} &=80 \mathrm{dyne} / \mathrm{cm} \\
&=80 \text { dyne } \mathrm{cm}^{-1}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2014
A drop of water detaches itself from the exit of a tap when $(\sigma=$ surface tension of water, $\rho=$ density of water, $R=$ radius of the tap exit, $r=$ radius of the drop $)$

Options:
A $r>\left(\frac{2}{3} \frac{R \sigma}{\rho g}\right)^{1 / 3}$
B $r=\frac{2}{3} \frac{\sigma}{\rho g}$
C $\frac{2 \sigma}{r}>$ atmospheric pressure
D None of the above

The correct answers are:
None of the above

Solution:
Let mass of the drop $=m$
Weight of the drop = mg
It will act downward. The force due to surface tension on the drop $=\sigma2 \pi R$
Where, $R=$ radius of tap If will act in upward direction. The drop of water will detach when
$$
m g>\sigma 2 \pi R
$$
$$
\begin{array}{l}
\text { But, } \quad m=v \times \rho=\frac{4}{3} \pi^{3} p g \\
\therefore \quad \frac{4}{3} \pi^{3} \rho g>\sigma \times 2 \pi A
\end{array}
$$
$$
r>\left[\frac{3}{2} \frac{\sigma \cdot R}{\rho g}\right]^{1 / 3}
$$
Hence, none of the given options is correct.

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2016
A fair coin is tossed at a fixed number of times. If the probability of getting exactly 3 heads equals the probability of getting exactly 5 heads, then the probability of getting exactly one head is

Options:
A $1 / 64$
B $1 / 32$
C $1 / 16$
D $1 / 8$

The correct answer is:
$1 / 32$

Solution:
Let the coin be tossed $n$ times
Let getting head is consider to be success $\therefore \quad p=\frac{1}{2}, q=1-p=1-\frac{1}{2}=\frac{1}{2}$
It is given that,
$P(X=3)=P(X=5)$
$\Rightarrow \quad{ }^{n} C_{3}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{n-3}={ }^{n} C_{5}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{n-5}$
$\Rightarrow \quad{ }^{n} C_{3}={ }^{n} C_{5}$
$\Rightarrow \quad n=3+5 \quad\left[\because^{n} C_{x}={ }^{n} C_{y} \Rightarrow x+y=n \mid\right.$
$\Rightarrow \quad n=8$
Now, $P(X=1)={ }^{8} C_{1}\left(\frac{1}{2}\right)^{1}\left(\frac{1}{2}\right)^{8-1}$
$={ }^{8} C_{1} \times\left(\frac{1}{2}\right)^{8}=\frac{1}{32}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2015
A fair six-faced die is rolled 12 times. The probability that each face turns up twice is equal to

Options:
A $\frac{12 !}{6 ! 6 ! 6^{12}}$
B $\frac{2^{12}}{2^{6} 6^{12}}$
C $\frac{12 !}{2^{6} 6^{12}}$
D $\frac{12 !}{6^{2} 6^{12}}$

The correct answer is:
$\frac{12 !}{2^{6} 6^{12}}$

Solution:
Required probability
$$
\begin{aligned}={ }^{12} C_{2} \times{ }^{10} C_{2} \times{ }^{8} C_{2} \times{ }^{6} C_{2} \times{ }^{4} C_{2} \times{ }^{2} C_{2} \times\left(\frac{1}{6}\right)^{12} \\=\frac{12 !}{10 ! \times 2 !} \times \frac{10 !}{8 ! \times 2 !} \times \frac{8 !}{6 ! \times 2 !} \times \frac{6 !}{4 ! \times 2 !} \\ \times \frac{4 !}{2 ! \times 2 !} \times \frac{2 !}{2 ! \times 1 !} \times\left(\frac{1}{6}\right)^{12} \end{aligned}
$$
$=\frac{12 !}{2^{6} \times 6^{12}}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2014
A family of curves is such that the length intercepted on the $y$ -axis between the origin and the tangent at a point is three times the ordinate of the point of contact. The family of curves is

Options:
A $x=C$, $C$ is a constant
B $x y^{2}=C, C$ is a constant
C $x^{2} y=C, C$ is a constant
D $x^{2} y^{2}=C, C$ is a constant

The correct answer is:
$x^{2} y=C, C$ is a constant

Solution:
Let the general equation of tangent.
Which passes through the point $(x, y)$ is
$(Y-y)=\frac{d y}{d x}(x-x)$
$\Rightarrow \quad Y-y=X \frac{d y}{d x}-x \frac{d y}{d x}$
for length of $Y$ -intercept, put $X=0$ in Eq. $(1),$ we get $Y-y=-x \frac{d y}{d x}$
$\Rightarrow \quad Y=y-x \frac{d y}{d x}$
Now,according to the guestion Y-intercept $=3 \times$ ordinate of the point of contact $\Rightarrow \quad y-x \frac{d y}{d x}=3 y$
$\Rightarrow \quad-x \frac{d y}{d x}=2 y$
$\Rightarrow$
$\int \frac{d y}{y}=-\int \frac{2 d x}{x}$ (on integrating)
$\Rightarrow \quad \log y=-2 \log x+\log C$
$\Rightarrow \quad \log y+\log x^{2}=\log C$
$\Rightarrow y x^{2}=C$. where $C$ is a constant.

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2013
A fighter plane, flying horizontally with a speed $360 \mathrm{kmph}$ at an altitude of $500 \mathrm{~m}$ drops a bomb for a target straight ahead of it on the ground. The bomb .should be dropped at what approximate distance ahead of the target ? Assume that acceleration due to gravity $(\mathrm{g})$ is $10 \mathrm{~ms}^{-2}$. Also neglect air drag

Options:
A $1000 \mathrm{~m}$
B $50 \sqrt{5} \mathrm{~m}$
C $500 \sqrt{5} \mathrm{~m}$
D $866 \mathrm{~m}$

The correct answer is:
$1000 \mathrm{~m}$

Solution:
Hint:
$v=360 \mathrm{~km} / \mathrm{h}=100 \mathrm{~m} / \mathrm{s} .$
$\mathrm{h}=500 \mathrm{~m}$
$\mathrm{R}=\mathrm{u} \sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=100 \times \sqrt{\frac{2 \times 500}{10}}=1000 \mathrm{~m}$

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2020
A frictionless piston-cylinder based enclosure contains some amount of gas at a pressure of $400 \mathrm{kPa}$. Then heat is transferred fo the gas at constant pressure in a quasi-static process. The piston moves up slowly through a height of $10 \mathrm{cm} .$ If the piston has a cross-sectional area of $0.3 \mathrm{m}^{2},$ the work done by the gas in this process is

Options:
A $6 \mathrm{kJ}$
B $12 \mathrm{kJ}$
C $7.5 \mathrm{kJ}$
D $24 \mathrm{kJ}$

The correct answer is:
$12 \mathrm{kJ}$

Solution:
$W_{\text {constant pressure }}=p \times \Delta V$ $=400 \times 10^{3} \times 0.3 \times 10 \times 10^{-2}$
$=400 \times 10 \times 3$
$=12000=12 \mathrm{kJ}$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2013
A galvanometer can be converted to a voltmeter of full-scale deflection $\mathrm{V}_{0}$ by connecting a series resistance $\mathrm{R}_{1}$ and can be converted to an ammeter of full-scale deflection $I_{0}$ by connecting a shunt resistance $\mathrm{R}_{2}$. What is the current flowing through the galvanometer at its full-scale deflection?

Options:
A $\frac{\mathrm{V}_{0}-\mathrm{I}_{0} \mathrm{R}_{2}}{\mathrm{R}_{1}-\mathrm{R}_{2}}$
B $\frac{\mathrm{V}_{0}+\mathrm{I}_{0} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$
C $\frac{\mathrm{V}_{0}-\mathrm{I}_{0} \mathrm{R}_{2}}{\mathrm{R}_{2}-\mathrm{R}_{1}}$
D $\frac{V_{0}+l_{0} R_{1}}{R_{1}+R_{2}}$

The correct answer is:
$\frac{\mathrm{V}_{0}-\mathrm{I}_{0} \mathrm{R}_{2}}{\mathrm{R}_{1}-\mathrm{R}_{2}}$

Solution:
Hint:
$R_{1}=\frac{V_{0}}{I_{g}}-G_{\ldots} \ldots \ldots . .(1)$
$\mathrm{R}_{2}=\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{I}_{0}-\mathrm{I}_{\mathrm{g}}} \ldots \ldots \ldots . .(2)$
Eliminating G we get
$I_{g}=\frac{V_{0}-I_{0} R_{2}}{R_{1}-R_{2}}$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2020
A galvanometer having internal resistance $10 \Omega$ requires 0.01 A for a full scale deflection. To convert this galvanometer to a voltmeter of full-scale deflection at $120 \mathrm{V}$, we need to connect a resistance of

Options:
A $11990 \Omega$ in series
B $11990 \Omega$ in parallel
C $12010 \Omega$ in series
D $12010 \Omega$ in parallel

The correct answer is:
$11990 \Omega$ in series

Solution:
Resistance $R=\frac{V}{i_{g}}-G=\frac{120}{0.01}-10$
$$
\begin{aligned}
&=12000-10 \\
R=11990 \Omega
\end{aligned}
$$
To convert a galvanometer into a voltmeter, high resistance should connect in series.

Topic: Physics, Current Electricity, WBJEE, WBJEE 2014
A gas bubble of $2 \mathrm{cm}$ diameter rises througha liquid of density $1.75 \mathrm{g} \mathrm{cm}^{-3}$ with a fixed speed of $0.35 \mathrm{cms}^{-1}$. Neglect the density of the gas. The coefficient of viscosity of the liquid is

Options:
A 870 poise
B 1120 poise
C 982 poise
D 1089 poise

The correct answer is:
1089 poise

Solution:
The gas bubble is rising through the liquid of densily $1.75 \mathrm{g} / \mathrm{cm}^{3}$ with constant speed. The acting forces on bubble are

Buoyancy force of liquid $=F_{0}$ (in upward direction) Viscous force due to liquid = $F_{V}$, (in downward direction)
$$
\therefore \quad F_{V}=F_{b}
$$
For free falling in a viscous medium
$$
F_{V}=6 \pi \eta {V_{T}}
$$
$\eta$ = viscosity coefficient of the liquid
$r=$ radius of bubble
$$
\begin{aligned}
F_{D} &=m g=v \times \rho g \\
&=\frac{4}{3} \pi r^{3} \rho_{e} \cdot g
\end{aligned}
$$
$\begin{aligned} 6 \pi \eta r V_{t} &=\frac{4}{3} \pi r^{3} \rho_{e} \cdot g \\ \eta &=\frac{24 \pi \cdot r^{3} \rho_{e} \cdot g}{3 \times 6 \pi rV_{T}}=\frac{2}{9} \cdot \frac{r^{2} \rho_{e} \cdot g}{v_{T}} \end{aligned}$
$\begin{aligned} D &=2 \mathrm{cm}, r=1 \mathrm{cm}=1 \times 10^{-2} \mathrm{m} \\ \rho_{e} &=1.75 \mathrm{g} / \mathrm{cm}^{3}=1.75 \times 10^{* 3} \mathrm{kg} / \mathrm{m}^{3} \\ g &=10 \mathrm{m} / \mathrm{s}^{2} \\ \mathrm{v}_{T} &=\text { terminal velocity } \\ &=0.35 \times 10^{-2} \mathrm{m} / \mathrm{s} \\ \eta &=\frac{2}{9} \times \frac{\left(10^{-2}\right)^{2} \times 1.75 \times 10^{+3} \times 10}{0.35 \times 10^{-2}} \\ &=\frac{2}{9} \times \frac{10^{-4} \times 1.75 \times 10^{3} \times 10}{0.35 \times 10^{-2}} \\ &=\frac{2 \times 175 \times 10 \times 10}{9 \times 35} \times 10 \text { poise } \\ &=1089 \text { poise } \end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2016
A gas can be liquefied at temperature T and pressure $p$ provided

Options:
A $T=T_{c}$ and $p < p_{c}$
B $T < T_{c}$ and $p>p_{c}$
C $T>T_{c}$ and $p>p_{c}$
D $T>T_{c}$ and $p < p_{c}$

The correct answer is:
$T < T_{c}$ and $p>p_{c}$

Solution:
A gas can be liquefied at temperature $T$ when it is lower than critical temperature, $T_{c}$ and pressure $p$ is greater than cntical pressure.
Hence. $T < T_{c}$ and $p>p_{c}$

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2015
A given amount of $\mathrm{Fe}^{2+}$ is oxidized by $\mathrm{x}$ mol of $\mathrm{MnO}_{4}$ in acidic medium. The number of moles of $\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}$ required to oxidize the same amount of $\mathrm{Fe}^{2+}$ in acidic medium is

Options:
A $\mathrm{x}$
B $0.83 \mathrm{x}$
C $2.0 \mathrm{x}$
D $1.2 \mathrm{x}$

The correct answer is:
$0.83 \mathrm{x}$

Solution:
As per Law of Equivalence
no. of eq. of $\mathrm{Fe}^{+2}=$ no. of eq. of $\mathrm{KMnO}_{4}=$ no. of eq. of $\mathrm{Cr}_{2} \mathrm{O}_{7}^{-2}$
$\begin{array}{l}
\mathrm{MnO}_{4}^{-}(\mathrm{O} \cdot \text { Agent })+5 \mathrm{e}^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{+2}+4\mathrm{H}_{2} \mathrm{O} \\
(\mathrm{n} . \mathrm{f}=5) \\
\mathrm{Cr}_{2} \mathrm{O}_{7}^{-2}(\mathrm{O} \cdot \mathrm{A})+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{+3}+7 \mathrm{H}_{2} \mathrm{O} \\
(\mathrm{n} . \mathrm{f}=6)
\end{array}$
no. of eq. of $\mathrm{KMnO}_{4}=$ no. of moles $\times \mathrm{n}$. factor $=\mathrm{x} \times 5$
no. of eq. of $\mathrm{Cr}_{2} \mathrm{O}_{7}^{-2}=$ no. of moles $\times n . \mathrm{f}=\mathrm{y} \times 6$
From Law of equivalence
$\begin{array}{l}
6 y=5 x \\
\text { or } y=5 / 6 x=0.83 x
\end{array}$

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2021
A given quantity of gas is taken from \(A\) to \(C\) in two ways; \(a\) ) directly from \(A \rightarrow C\) along a straight line and \(b\) ) in two steps, from \(A \rightarrow B\) and then from \(B \rightarrow C\). Work done and heat absorbed along the direct path \(A \rightarrow C\) is \(200 \mathrm{~J}\) and \(280 \mathrm{~J}\) respectively

Options:
A \(80 \mathrm{~J}\)
B 0
C \(160 \mathrm{~J}\)
D \(120 \mathrm{~J}\)

The correct answer is:
\(160 \mathrm{~J}\)

Solution:
Hint : \(\Delta W=200 \mathrm{~J} \quad \Delta Q=280 \mathrm{~J} \quad\) for path \(A C\)
\(\therefore \quad \Delta U=\Delta Q-\Delta W=280-200=80 \mathrm{~J}\)
\((\Delta U)\) is same for both paths.
\(\therefore \Delta Q=\Delta W+\Delta U=80+80=160 \mathrm{~J}\)

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2023
A glass slab consists of thin uniform layers of progressively decreasing refractive indices RI (see figure) such that the RI of any layer is $\mu-m \Delta \mu$. Here, $\mu$ and $\Delta \mu$ denote the $\mathrm{RI}$ of 0 th layer and the difference in RI between any two consecutive layers, respectively. The integer $m=0,1,2,3, \ldots$ denotes the numbers of the successive layers. A ray of light from the oth layer enters the 1st layer at an angle of incidence of $30^{\circ}$. After undergoing the mth refraction, the ray emerges parallel to the interface. If $\mu=1.5$ and $\Delta \mu=0.015,$ the value of $m$ is

Options:
A 20
B 30
C 40
D 50

The correct answer is:
50

Solution:
By Snell's law. $\mu \sin i=$ constant
$$
\begin{aligned}
\mu \sin i &=(\mu-m \Delta \mu) \sin r \\
\text { where } \mu=1.5, i &=30^{\circ}, r=90^{\circ}, \Delta \mu=0.015 \\
1.5 \sin 30^{\circ} &=(1.5-m \times 0.15) \sin 90^{\circ} \\
\frac{3}{2} \times \frac{1}{2} &=(1.5-m \times 0.015) \times 1 \\
\frac{3}{4} &=\frac{3}{2}-\frac{15 m}{1000} \\
15 \mathrm{m} &=\left(\frac{3}{2}-\frac{3}{4}\right) \times 1000 \\
15 \mathrm{m} &=\frac{6-3}{4} \times 1000 \Rightarrow m=\frac{3}{4} \times \frac{1000}{15} \\
m &=\frac{3000}{60} \Rightarrow m=50
\end{aligned}
$$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2014
A golf ball of mass $50 \mathrm{gm}$ placed on a tee, is struck by a golf-club. The speed of the golf ball as it leaves the tee is $100 \mathrm{~m} / \mathrm{s}$, the time of contact on the ball is $0.02 \mathrm{~s}$. If the force decreases to zero linearly with time, then the force at the beginning of the contact is

Options:
A $100 \mathrm{~N}$
B $200 \mathrm{~N}$
C $250 \mathrm{~N}$
D $500 \mathrm{~N}$

The correct answer is:
$500 \mathrm{~N}$

Solution:
Impulse $=$ Area under F-t graph $=$ change in linear momentum
$$
=\frac{1}{2} \times F \times 0.02=\frac{50}{1000}(100-0) \Rightarrow F=500 \mathrm{~N}
$$

Topic: Physics, Laws of Motion, WBJEE, WBJEE 2022
A hemisphere of radius $\mathrm{R}$ is placed in a uniform electric field $\mathrm{E}$ so that its axis is parallel to the field. Which of the following statement(s) is / are true?

Options:
A Flux through the curved surface of hemisphere is $\pi R^2 E$.
B Flux through the circular surface of hemisphere is $\pi R^2 E$.
C Total flux enclosed is zero
D Work done in moving a point charge $q$ from $A$ to $B$ via the path $A C B$ depends upon $R$.

The correct answers are:
Flux through the curved surface of hemisphere is $\pi R^2 E$., Total flux enclosed is zero

Solution:
$$
\begin{aligned}
&\quad \text { as } \phi_{\text {in }}=0, \quad \phi_{\text {total }}=0 \\
&\phi_{\text {curved }}+\phi_{\text {flat }}=0 \\
&\phi_{\text {flat }}=-E \times \pi R^2 \\
&\therefore \phi_{\text {curved }}=E \times \pi R^2
\end{aligned}
$$
Also $\Delta \mathrm{V}=-\overrightarrow{\mathrm{E}} \cdot \Delta \overrightarrow{\mathrm{r}}$ for uniform eletrified.
$$
\begin{aligned}
&\text { as } \vec{E} \perp \Delta \vec{r} \\
&\Delta V=0 \\
&W=q \Delta V=0
\end{aligned}
$$
so work is independent of $R$ in moving charge $q$ from $A+B$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2022
A hollow metal sphere of radius $R$ is charged with a charge $Q .$ The electric potential and intensity inside the sphere are respectively

Options:
A $\frac{Q}{4 \pi \varepsilon_{0} R^{2}}$ and $\frac{Q}{4 \pi \varepsilon_{0} R}$
B $\frac{Q}{4 \pi \varepsilon_{0} R}$ and zero
C zero and zero
D $\frac{4 \pi \varepsilon_{0} Q}{R}$ and $\frac{Q}{4 \pi \varepsilon_{0} R^{2}}$

The correct answer is:
$\frac{Q}{4 \pi \varepsilon_{0} R}$ and zero

Solution:
Inside a hollow charged sphere, the electric field intensity is always zero.
$\therefore$ $E=0$ but potential is constant.

Potential on the surtace of the hollow sphere
$$
V=\frac{K Q}{R}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R}=\frac{Q}{4 \pi \varepsilon_{0} R}
$$
This is the potential $(\mathrm{V})$ inside the charged hollow meta sphere.

Topic: Physics, Electrostatics, WBJEE, WBJEE 2016