datavorous/entrance-exam-dataset huggingface dataset:
An ideal gas is compressed isothermally until its pressure is doubled and then allowed to expand adiabatically to regain its original volume \(\left(\gamma=1.4\right.\) and \(\left.2^{-14}=0.38\right)\). The ratio of the final to initial pressure is

Options:
A \(0.76: 1\)
B \(1: 1\)
C \(0.66: 1\)
D \(0.86: 1\)

The correct answer is:
\(1: 1\)

Solution:
Hints:
\(\begin{aligned}
& \mathrm{P}_{\mathrm{f}} \mathrm{V}^\gamma=\left(2 \mathrm{P}_{\mathrm{i}}\right)\left(\frac{\mathrm{V}}{2}\right)^\gamma \\
& \frac{\mathrm{P}_{\mathrm{f}}}{\mathrm{P}_{\mathrm{i}}}=2\left(\frac{\gamma}{2 \gamma}\right)^\gamma=2(2)^{-\gamma} \\
& =2 \times 0.38=0.76 \\
&
\end{aligned}\)

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2011
An ideal gas of molar mass $\mathrm{M}$ is contained in a very tall vertical cylindrical column in the uniform gravitational field. Assuming the gas temperature to be $\mathrm{T}$, the height at which the centre of gravity of the gas is located is (R: universal gas constant)

Options:
A $\frac{\mathrm{RT}}{\mathrm{g}}$
B $\frac{\mathrm{RT}}{\mathrm{Mg}}$
C $\mathrm{MgR}$
D RTg

The correct answer is:
$\frac{\mathrm{RT}}{\mathrm{Mg}}$

Solution:
$\because$ field is uniform COM and COG coincides.
$$
\text { So } \mathrm{y}_{\mathrm{cem}}=\frac{\int_{0}^{\infty} \mathrm{ydm}}{\int_{0}^{\infty} \mathrm{dm}}=\frac{\int_{0}^{\infty} \mathrm{y} \mathrm{Ady} \rho}{\int_{0}^{\infty} \mathrm{Ady} \rho}=\frac{\int_{0}^{\infty} \mathrm{y} \rho \mathrm{dy}}{\int_{0}^{\infty} \rho \mathrm{dy}}
$$
From Barometric formula :- $\rho=\rho_{0} \mathrm{e}^{-\mathrm{Mgy} / \mathrm{RT}}$
So $y_{\text {cen }}=\frac{\int_{0}^{\infty} y e^{-M g y / R T}}{\int_{0}^{\infty} e^{-M g y / R T}}=\frac{R T}{M g}$

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2021
An ideal gas undergoes the cyclic process abca as shown in the given $\mathrm{P}-\mathrm{V}$ diagram. It rejects $50 \mathrm{~J}$ of heat during $a b$ and absorbs $80 \mathrm{~J}$ of heat during ca. During $b c$, there is no transfer of heat and $40 \mathrm{~J}$ of work is done by the gas. What should be the area of the closed curve abca?

Options:
A $30 \mathrm{~J}$
B $40 \mathrm{~J}$
C $10 \mathrm{~J}$
D $90 \mathrm{~J}$

The correct answer is:
$30 \mathrm{~J}$

Solution:
Hint:

Area of cycle $=\mathrm{W}_{\text {total }}$
$=\Delta Q_{\text {net }}=\Delta Q_{A B}+\Delta Q_{B C}+\Delta Q_{C A}=-50+0+80=30 \mathrm{~J}$
NOTE: In process $\mathrm{BC}, \mathrm{W}_{\mathrm{BC}}>0$ and $\Delta \mathrm{U}_{\mathrm{BC}}>0$
$\therefore \Delta \mathrm{Q}_{\mathrm{BC}}=\Delta \mathrm{U}_{\mathrm{BC}}+\mathrm{W}_{\mathrm{BC}}>0$
Thus $\Delta \mathrm{Q}_{\mathrm{eC}}=0$ as given in question isn't possible. But we have neglected this technical mistake while answering.

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2020
An ideal mono-atomic gas of given mass is heated at constant pressure. In this process, the fraction of supplied heat energy used for the increase of the internal energy of the gas is

Options:
A $3 / 8$
B $3 / 5$
C $3 / 4$
D $2 / 5$

The correct answer is:
$3 / 5$

Solution:
Fraction $=\frac{\Delta U}{\Delta Q}=\frac{C_{Y}}{C_{p}}$
For monoatomic gas $\frac{C_{p}}{C_{\gamma}}=\gamma=\frac{5}{3}$
So, $\quad \frac{C_{\gamma}}{C_{p}}=\frac{1}{\gamma}=\frac{3}{5}$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2013
An infinite sheet carrying a uniform surface charge density $\sigma$ lies on the $x y$ -plane. The work done to carry a charge $q$ from the point $\mathbf{A}=a(\mathbf{i}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$ to the point $\mathbf{B}=a(\mathbf{i}-2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})$
(where $a$ is a constant with the dimension of length and $\varepsilon_{0}$ is the permittivity of free space) is

Options:
A $\frac{3 \sigma a q}{2 \varepsilon_{0}}$
B $\frac{2 \sigma a q}{\varepsilon_{0}}$
C $\frac{5 \sigma a q}{2 \varepsilon_{0}}$
D $\frac{3 \sigma a q}{\varepsilon_{0}}$

The correct answer is:
$\frac{3 \sigma a q}{2 \varepsilon_{0}}$

Solution:
The given
$$
\begin{aligned}
\mathbf{A} &=a(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\
\mathbf{B} &=a(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \\
\mathbf{A B} &=\mathbf{O B}-\mathbf{O A} \\
&=a(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})-a(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\
\mathbf{A B} &=a(-4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})
\end{aligned}
$$
Work done $=q\left(\frac{\sigma}{2 \varepsilon_{0}}\right) \hat{\mathbf{k}} \cdot \mathbf{A B} \quad$ (along to $Z$ -axisi
$$
\begin{array}{r}
=q\left(\frac{\sigma}{2 \varepsilon_{0}}\right) \hat{\mathbf{k}} \cdot a(-4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})=\frac{3 q \sigma a}{2 \varepsilon_{0}} \\
(\because \mathbf{i} \cdot \mathbf{i}=\mathbf{j} \cdot \mathbf{j}=\mathbf{i} \cdot \mathbf{k}=1 \text { and } \mathbf{i} \cdot \mathbf{j}=\mathbf{j} \cdot \hat{\mathbf{k}}=\hat{\mathbf{k}} \cdot \hat{\mathbf{i}}=0)
\end{array}
$$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2014
An interference pattern is obtained with two coherent sources of intensity ratio \(\mathrm{n}: 1\). The ratio \(\frac{\mathrm{I}_{\text {Max }}-\mathrm{I}_{\text {Min }}}{\mathrm{I}_{\text {Max }}+\mathrm{I}_{\text {Min }}}\) will be maximum if

Options:
A \(n=1\)
B \(n=2\)
C \(n=3\)
D \(n=4\)

The correct answer is:
\(n=1\)

Solution:
Hint: \(\frac{I_1}{I_2}=n\)
\(\begin{aligned}
& \mathrm{I}_1=\mathrm{nI}_2 \\
& \mathrm{I}_{\max }=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2=\left(\sqrt{\mathrm{nI}} 2+\sqrt{\mathrm{I}_2}\right)^2=(\sqrt{\mathrm{n}}+1)^2\left(\sqrt{\mathrm{I}_2}\right)^2
\end{aligned}\)
\(\begin{aligned}
& I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2=\left(\sqrt{n I_2}-\sqrt{I_2}\right)^2=(\sqrt{n}-1)^2\left(\sqrt{I_2}\right)^2 \\
& \frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=\frac{(\sqrt{n}+1)^2\left(I_2\right)-(\sqrt{n}-1)^2 I_2}{(\sqrt{n}+1)^2 I_2+(\sqrt{n}-1)^2 I_2}=\frac{(\sqrt{n}+1)^2-(\sqrt{n}-1)^2}{(\sqrt{n}+1)^2+(\sqrt{n}-1)^2} \\
& =\frac{(n+1+2 \sqrt{n})-(n+1-2 \sqrt{n})}{n+1+2 \sqrt{n}+n+1-2 \sqrt{n}}=\frac{4 \sqrt{n}}{2(n+1)}=\frac{2 \sqrt{n}}{n+1}
\end{aligned}\)
\(\therefore\) decreases with increasing \(\mathrm{n}\).
\(\therefore\) It will be maximum if \(n=1\)

Topic: Physics, Wave Optics, WBJEE, WBJEE 2023
An object is located $4 \mathrm{m}$ from the first of two thin converging lenses of focal lengths $2 \mathrm{m}$ and $1 \mathrm{m},$ respectively. The lenses are separated by $3 \mathrm{m}$. The final image formed by the second lens is located from the source at a distance of

Options:
A $8.0 \mathrm{m}$
B $5.5 \mathrm{m}$
C $6.0 \mathrm{m}$
D $6.5 \mathrm{m}$

The correct answer is:
$5.5 \mathrm{m}$

Solution:
From the lens formula (for first lens)
$$
\begin{aligned}
\frac{1}{f_{1}} &=\frac{1}{v_{1}}-\frac{1}{u_{1}} \Rightarrow \frac{1}{2}=\frac{1}{v_{1}}-\frac{1}{(-4)} \\
\Rightarrow \quad \frac{1}{2}+\frac{1}{4} &=\frac{1}{v_{1}}=\frac{3}{4} \\
\Rightarrow \quad v_{1} &=\frac{4}{3}, u_{2}=3-\frac{4}{3}=5 / 3
\end{aligned}
$$
This image will be treated as the source for second lens, then again from lens formula, we have
$\frac{1}{f_{2}}=\frac{1}{v_{2}}-\frac{1}{u_{2}}4$
$\Rightarrow \quad \frac{1}{1}=\frac{1}{v_{2}}+5 / 3$
$\Rightarrow \quad 1-5 / 3=\frac{1}{v_{2}} \Rightarrow v_{2}=-3 / 2$
[By Eq. (i)]
This is the final image distance from 2 nd lens. So, the overall distance of image from the primary source(or object)
Let $d=4+3-1.5=5.5$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2015
An object is placed $30 \mathrm{cm}$ away from a convex lens of focal length $10 \mathrm{cm}$ and a sharp image is formed on a screen. Now a concave lens is placed in contact with the convex lens. The screen now has to be moved by $45 \mathrm{cm}$ to get $\underline{a}$ sharp image again. The magnitude of focal length of the concave lens is (in $\mathrm{cm}$ )

Options:
A 72
B 60
C 36
D 20

The correct answer is:
20

Solution:
For the first condition

$$
\frac{1}{f_{1}}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{10}=\frac{1}{v}+\frac{1}{30}
$$
$\frac{1}{v}=\frac{1}{10}-\frac{1}{30}$
$v=\frac{30}{2}=15 \mathrm{cm}$
$$
\begin{aligned}
\text { where } f_{1}=10 \mathrm{cm} & \begin{array}{l}
u=-30 \mathrm{cm} \\
v & =?
\end{array}
\end{aligned}
$$
For the first condition

For the second condition when concave lens is placed
$$
\begin{array}{l}
v^{\prime}=(15+45) \mathrm{cm}=60 \mathrm{cm} \\
\frac{1}{F}=\frac{1}{v^{\prime}}-\frac{1}{u}
\end{array}
$$
(where $F=$ focal length of combination)
$\therefore$
$$
\begin{array}{l}
\frac{1}{F}=\frac{1}{60}+\frac{1}{30} \\
F=\frac{60}{3} \mathrm{cm}=20 \mathrm{cm}
\end{array}
$$
The magnitude of focal length of concave lens $\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}} \Rightarrow \frac{1}{20}=\frac{1}{10}+\frac{1}{f_{2}} \Rightarrow f_{2}=-20 \mathrm{cm}$
(Negative sign for concave lens)

Topic: Physics, Ray Optics, WBJEE, WBJEE 2014
An object is placed $60 \mathrm{~cm}$ in front of a convex mirror of focal length $30 \mathrm{~cm}$. A plane mirror is now placed facing the object in between the object and the convex mirror such that it covers lower half of the convex mirror. What should be the distance of the plane mirror from the object so that there will be no parallax between the images formed by the two mirrors?

Options:
A $40 \mathrm{~cm}$
B $30 \mathrm{~cm}$
C $20 \mathrm{~cm}$
D $15 \mathrm{~cm}$

The correct answer is:
$40 \mathrm{~cm}$

Solution:
Hint:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}-\frac{1}{60}=\frac{1}{30}$
$\mathrm{v}=\frac{60}{3}=\pm 20 \mathrm{~cm}$
$\therefore$ Distance between object and image $=60+20=80 \mathrm{~cm}$ $\therefore x=40 \mathrm{~cm}$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2020
An object placed at a distance of $16 \mathrm{cm}$ from a $\begin{array}{llll}\text { convex lens } & \text { produces } & \text { an image } & \text { of }\end{array}$ magnification $m(m>1)$. If the object is moved towards the lens by $8 \mathrm{cm}$, then again an image of magnification $m$ is obtained. The numerical value of the focal length of the lens is

Options:
A $12 \mathrm{cm}$
B $14 \mathrm{cm}$
C $18 \mathrm{cm}$
D $20 \mathrm{cm}$

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

Solution:
Linear magnification, $m=\frac{f}{f+u}$
As given that magnification is same for both cases this is possible if the two different values of $u$ are of opposite sign.
$i, \theta,+\frac{f}{f-16}=\frac{-f}{f-8}$
$\Rightarrow \quad 16-f=f-8$
$\Rightarrow \quad 2 f=24$
or $\quad f=12 \mathrm{cm}$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2013
An object placed in front of a concave mirror at a distance of $x$ cm from the pole gives a 3 times magnified real image. If it is moved to a distance of $(x+5) \mathrm{cm},$ the magnification of the image becomes 2 . The focal length of the mirror is

Options:
A $15 \mathrm{cm}$
B $20 \mathrm{cm}$
C $25 \mathrm{cm}$
D $30 \mathrm{cm}$

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

Solution:
Given $y_{1}=3 x$ and $v_{2}=2(x+5)$ So here, Ist case
$$
\begin{array}{r}
\frac{1}{-x}+\frac{1}{-3 x}=\frac{1}{f} \\
\frac{1}{-(x+5)}+\frac{1}{-2(x+5)}=\frac{1}{f}
\end{array}
$$
On solving Eqs. (i) and (ii) we get $f=30 \mathrm{cm}$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2012
An object weighs \(\mathrm{m}_1\) in a liquid of density \(\mathrm{d}_1\) and that in liquid of density \(\mathrm{d}_2\) is \(\mathrm{m}_2\). The density \(d\) of the object is

Options:
A \(\mathrm{d}=\frac{\mathrm{m}_2 \mathrm{~d}_2-\mathrm{m}_1 \mathrm{~d}_1}{\mathrm{~m}_2-\mathrm{m}_1}\)
B \(d=\frac{\mathrm{m}_1 \mathrm{~d}_1-\mathrm{m}_2 \mathrm{~d}_2}{\mathrm{~m}_2-\mathrm{m}_1}\)
C \(\mathrm{d}=\frac{\mathrm{m}_2 \mathrm{~d}_1-\mathrm{m}_1 \mathrm{~d}_2}{\mathrm{~m}_1-\mathrm{m}_2}\)
D \(\mathrm{d}=\frac{\mathrm{m}_1 \mathrm{~d}_2-\mathrm{m}_2 \mathrm{~d}_1}{\mathrm{~m}_1-\mathrm{m}_2}\)

The correct answer is:
\(\mathrm{d}=\frac{\mathrm{m}_1 \mathrm{~d}_2-\mathrm{m}_2 \mathrm{~d}_1}{\mathrm{~m}_1-\mathrm{m}_2}\)

Solution:
Hints: \(V\left(d-d_1\right) g=m_1 g\)
\(\mathrm{V}\left(\mathrm{d}-\mathrm{d}_2\right) \mathrm{g}=\mathrm{m}_2 \mathrm{~g}\)
\(\frac{\mathrm{d}-\mathrm{d}_1}{\mathrm{~d}-\mathrm{d}_2}=\frac{\mathrm{m}_1}{\mathrm{~m}_2} \quad \therefore \mathrm{d}=\frac{\mathrm{m}_1 \mathrm{~d}_2-\mathrm{m}_2 \mathrm{~d}_1}{\mathrm{~m}_1-\mathrm{m}_2}\)

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2011
An objective type test paper has 5 questions. Out of these 5 questions, 3 questions have four options each $(a, b, c, d)$ with one option being the correct answer. The other 2 questions have two options each, namely true and false. A candidate randomly ticks the options. Then, the probability that he/she will tick the correct option in atleast four questions, is

Options:
A $\frac{5}{32}$
B $\frac{3}{128}$
C $\frac{3}{256}$
D $\frac{3}{64}$

The correct answer is:
$\frac{3}{64}$

Solution:
Total sample space, $n(S)=4^{3} \cdot 2^{2}$ and total number of favourable cases
$$
n(E)=\left({ }^{3} C_{1} \cdot 3+{ }^{2} C_{1} \cdot 1\right)+1
$$
$\therefore$ Required probability $=\frac{n(E)}{n(S)}$
$$
=\frac{{ }^{3} \mathrm{C} \cdot 3+{ }^{2} \mathrm{C}_{1} \cdot 1+1}{4^{3} \cdot 2^{2}}=\frac{3 \cdot 3+2+1}{4^{3} \cdot 4}
$$
$=\frac{12}{64 \cdot 4}=\frac{3}{64}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2013
An optically active compound having molecular formula $\mathrm{C}_{8} \mathrm{H}_{16}$ on ozonolysis gives acetone as one of the products. The structure of the compound is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2013
An organic compound made of $\mathrm{C}, \mathrm{H}$ and $\mathrm{N}$ contains $20 \%$ nitrogen. Its molecular weight is :

Options:
A $70$
B $140$
C $100$
D $65$

The correct answer is:
$70$

Solution:
Nitrogen at. wt. $=14$ in a molecule minimum one atom of $\mathrm{N}$ is present
i.e., $\begin{array}{rlr}20 \% & \equiv 14 & \text { Molecular weight }=70 \\ 100 \% & \equiv 14 \times 5=70 & \end{array}$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2009
An urn contains 8 red and 5 white balls. Three balls are drawn at random. Then, the probability that balls of both colours are drawn is

Options:
A $\frac{40}{143}$
B $\frac{70}{143}$
C $\frac{3}{13}$
D $\frac{10}{13}$

The correct answer is:
$\frac{10}{13}$

Solution:
Total number of selections of three balls in which balls of both colours are drawn
$=2$ red balls and 1 white ball +1 red ball and 2 white balls
$\begin{array}{l}
={ }^{8} C_{2} \times{ }^{5} C_{1}+{ }^{8} C_{1} \times{ }^{5} C_{2} \\
=140+80=220
\end{array}$
$\therefore$ Required probability
$=\frac{220} {{ }^{13} C_{3}}=\frac{220 \times 6}{13 \times 12 \times 11}=\frac{10}{13}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2012
Angle between an equipotential surface and electric lines of force is

Options:
A $0^{\circ}$
B $90^{\circ}$
C $180^{\circ}$
D $270^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
The electric lines of force are always perpendicular to the equpotential surtace
$$
\theta=90^{\circ}
$$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2016
Angle between the planes $x+y+2 z=6$ and $2 x-y+z=9$ is

Options:
A $\frac{\pi}{4}$
B $\frac{\pi}{6}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

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

Solution:
Equations of planes are
$x+y+2 z-6=0$
and $\quad 2 x-y+z-9=0$
$\cos \theta=\frac{a_{1} a_{2}+b_{1}b_{2}+c
_{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$=\frac{1 \times 2+1 \times(-1)+2 \times 1}{\sqrt{1+1+4} \sqrt{4+1+1}}$
$=\frac{2-1+2}{\sqrt{6} \times \sqrt{6}}=\frac{3}{6}=\frac{1}{2}$
$\Rightarrow \quad \cos \theta=\cos \left(\frac{\pi}{3}\right)$
$\therefore \quad \theta=\frac{\pi}{3}$

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2016
Angle between $\mathrm{y}^2=\mathrm{x}$ and $\mathrm{x}^2=\mathrm{y}$ at the origin is

Options:
A $2 \tan ^{-1}\left(\frac{3}{4}\right)$
B $\tan ^{-1}\left(\frac{4}{3}\right)$
C $\frac{\pi}{2}$
D $\frac{\pi}{4}$

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

Solution:
Hins : Angle between axes (since co-ordinate axes are the tangents for the given curve).

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2009
Anhydrous ferric chloride is prepared by

Options:
A Dissolving \(\mathrm{Fe}(\mathrm{OH})_3\) in concentrated \(\mathrm{HCl}\)
B Dissolving \(\mathrm{Fe}(\mathrm{OH})_3\) in dilute \(\mathrm{HCl}\)
C Passing dry \(\mathrm{HCl}\) over heated iron scrap
D Passing dry \(\mathrm{Cl}_2\) gas over heated iron scrap

The correct answer is:
Passing dry \(\mathrm{Cl}_2\) gas over heated iron scrap

Solution:
Hints: \(2 \mathrm{Fe}+3 \mathrm{Cl}_2 \stackrel{\Delta}{\longrightarrow} 2 \mathrm{FeCl}_3\)

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2011
Applying Lagrange's Mean Value Theorem for a suitable function $f(x)$ in $[0, h],$ we have $f(h)=f(0)+h f^{\prime}(\theta h), \quad 0 < \theta < 1 . \quad$ Then, for $f(x)=\cos x,$ the value of $\lim _{h \rightarrow 0^{*}} \theta$ is

Options:
A 1
B 0
C $1 / 2$
D $1 / 3$

The correct answer is:
$1 / 2$

Solution:
We know that in a Lagrange mean value theorem there exist $c \in(a, b)$ such that
$$
f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}
$$
$\therefore \quad f^{\prime}(\theta h)=\frac{f(h)-\cos 0}{h-0}$
$$
\begin{aligned}
\Rightarrow \quad-\sin (\theta h)=& \frac{\cos h-\cos 0}{h-0} \\
&[\because f(x)=\cos x] \\
=& \frac{\cos h-1}{h}
\end{aligned}
$$
$=\frac{\left(1-\frac{h^{2}}{2 !}\right)-1}{h}$
[neglecting higher power of $\Rightarrow \quad-\sin (\theta h)=\frac{-\frac{h^{2}}{2}}{h}$
$\Rightarrow \quad \sin (\theta h)=\frac{\frac{h}{2}}{1}$
$\Rightarrow \quad \theta h=\sin ^{-1}\left(\frac{h}{2}\right)$
$\Rightarrow \quad \theta=\frac{\sin ^{-1} \frac{h}{2} \times \frac{1}{2}}{h \times \frac{1}{2}}$
$\therefore \quad \lim _{h \rightarrow \theta^{+}} \theta=\frac{1}{2} \lim _{h \rightarrow 0^{+}} \frac{\sin ^{-1} \frac{h}{2}}{\frac{h}{2}}$
$=\frac{1}{2} \times 1=\frac{1}{2}$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2014
Aqueous solution of $\mathrm{HNO}_{3}, \mathrm{KOH}, \mathrm{CH}_{3} \mathrm{COOH}$ and $\mathrm{CH}_{3} \mathrm{COONa}$ of identical concentration are provided. The pair (s) of solutions which form a buffer upon mixing is (are)

Options:
A $\mathrm{HNO}_{3}$ and $\mathrm{CH}_{3} \mathrm{COOH}$
B $\mathrm{KOH}$ and $\mathrm{CH}_{3} \mathrm{COONa}$
C $\mathrm{HNO}_{3}$ and $\mathrm{CH}_{3} \mathrm{COONa}$
D $\mathrm{CH}_{3} \mathrm{COOH}$ and $\mathrm{CH}_{3} \mathrm{COONa}$

The correct answers are:
$\mathrm{HNO}_{3}$ and $\mathrm{CH}_{3} \mathrm{COONa}$, $\mathrm{CH}_{3} \mathrm{COOH}$ and $\mathrm{CH}_{3} \mathrm{COONa}$

Solution:
A Weak acid and its salt with a strong base forms a buffer.$\therefore$ (D)
$\mathrm{HNO}_{3}$ being a strong acid, displaces weaker acid $\mathrm{CH}_{3} \mathrm{COOH}$.
$\mathrm{HNO}_{3}+\mathrm{CH}_{3}-\mathrm{COO}^{-} \rightleftharpoons \mathrm{CH}_{3}-\mathrm{COOH}+\mathrm{NO}_{3}^{-}$
If $\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]>\left[\mathrm{HNO}_{3}\right]$, then solution contains weak acid $\left(\mathrm{CH}_{3} \mathrm{COOH}\right)$ and its conjugate base $\left(\mathrm{CH}_{3} \mathrm{COO}^{-}\right) . \quad \therefore(\mathrm{C})$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2021
Area in the first quadrant between the ellipses $x^{2}+2 y^{2}=a^{2}$ and $2 x^{2}+y^{2}=a^{2}$ is

Options:
A $\frac{a^{2}}{\sqrt{2}} \tan ^{-1} \frac{1}{\sqrt{2}}$
B $\frac{3 a^{2}}{4} \tan ^{-1} \frac{1}{2}$
C $\frac{5 a^{2}}{2} \sin ^{-1} \frac{1}{2}$
D $\frac{9 \pi a^{2}}{2}$

The correct answer is:
$\frac{a^{2}}{\sqrt{2}} \tan ^{-1} \frac{1}{\sqrt{2}}$

Solution:
Hint:
$A=\left(\frac{a^{2}}{6}+\int_{a / \sqrt{3}}^{a / \sqrt{2}} \sqrt{a^{2}-2 x^{2}} d x\right) \times 2=\frac{a^{2}}{\sqrt{2}} \tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2020
Area of the figure bounded by the parabola $y^2+8 x=16$ and $y^2-24 x=48$ is

Options:
A $\frac{11}{9}$ sq. unit
B $\frac{32}{3} \sqrt{6}$ sq. unit
C $\frac{16}{3}$ sq. unit
D $\frac{24}{5}$ sq. unit

The correct answer is:
$\frac{32}{3} \sqrt{6}$ sq. unit

Solution:
$$
\begin{gathered}
y^2=-8(x-2) \\
y^2=24(x+2) \cdots(2)
\end{gathered}
$$
Solving (1) and $(2) \Rightarrow x=-1$
Area $=2\left[\int_{-2}^{-1} 2 \sqrt{6} \sqrt{x+2} \mathrm{dx}+\int_{-1}^2 2 \sqrt{2} \sqrt{2-x} \mathrm{dx}\right]=\frac{32}{3} \sqrt{6}$ sq. unit

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2022
Area of the region bounded by $y=|x|$ and $y=-|x|+2$ is

Options:
A $4 \mathrm{sq}$ units
B $3 \mathrm{sq}$ units
C $2 \mathrm{sq}$ units
D 1 sq unit

The correct answer is:
$2 \mathrm{sq}$ units

Solution:
In figure, $B=(0,2) \Rightarrow O B=2$
and $O A B C$ is a square.

So, side of length $=\sqrt{2}$ units
$\therefore$ Area of $O A B C=2$ aq units

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2015
Arrange the following in order of increasing mass
I. 1 mole of \(\mathrm{N}_2\)
II. \(0.5 \mathrm{~mole}^2\) of \(\mathrm{O}_3\)
III. \(3.011 \times 10^{23} \mathrm{molecules}\) of \(\mathrm{O}_2\)
IV. 0.5 gram atom of \(\mathrm{O}_2\)

Options:
A IV \( < \) III \( < \) II \( < \) I
B IV \( < \) I \( < \) III \( < \) II
C III \( < \) II \( < \) IV \( < \) I
D I \( < \) III \( < \) II \( < \) IV

The correct answer is:
IV \( < \) III \( < \) II \( < \) I

Solution:
Hint : 1 mole \(\mathrm{N}_2=28 \mathrm{~g}\)
\(0.5 \mathrm{~mole} \mathrm{O}_3=24 \mathrm{~g}\)
\(3.011 \times 10^{23}\) molecules of \(\mathrm{O}_2=1 / 2\) mole \(\mathrm{O}_2=1\) mole \(\mathrm{O}=16 \mathrm{~g}\)
\(0.5 \mathrm{~g}\) atom \(\mathrm{O}_2=1 / 2\) mole of atoms of \(\mathrm{O}=8 \mathrm{~g}\)

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2023
As per de-Broglie's formula a macroscopic particle of mass $100 \mathrm{g}$ and moving at a velocity of $100 \mathrm{cm} \mathrm{s}^{-1}$ will have a wavelength of

Options:
A $6.6 \times 10^{-29} \mathrm{cm}$
B $6.6 \times 10^{-30} \mathrm{cm}$
C $6.6 \times 10^{-31} \mathrm{cm}$
D $6.6 \times 10^{-32} \mathrm{cm}$

The correct answer is:
$6.6 \times 10^{-31} \mathrm{cm}$

Solution:
According to de-Broglie. wavelength, $\lambda=\frac{h}{m v}$
Given, $\quad m=100 \mathrm{g}, v=100 \mathrm{cm} \mathrm{s}^{-1}$
and
$$
\begin{aligned}
h &=6.6 \times 10^{-34} \mathrm{J}-\mathrm{s} \\
&=6.6 \times 10^{-27} \mathrm{erg}-\mathrm{s}
\end{aligned}
$$
On substituting values, we get
$$
\lambda=\frac{6.6 \times 10^{-27}}{100 \times 100}=6.6 \times 10^{-31} \mathrm{cm}
$$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2014
As shown in figure, a rectangular loop of length ' a' and width ' $b$ ' and made of a conducting material of uniform crosssection is kept in a horizontal plane where a uniform magnetic field of intensity $B$ is acting vertically downward. Resistance per unit length of the loop is $\lambda \Omega / \mathrm{m}$. If the loop is pulled with uniform velocity ' $v$ ' in horizontal direction, which of the following statement is / are true?

Options:
A Current in the loop $\mathrm{I}=\frac{\mathrm{Bbv}}{\lambda(2 \mathrm{~b}+2 \mathrm{a})}$
B Current will be in clockwise direction, looking from the top.
C $V_P-V_S=V_Q-V_R$, where $V$ is the potential
D There cannot be any induction in part SR.

The correct answers are:
Current in the loop $\mathrm{I}=\frac{\mathrm{Bbv}}{\lambda(2 \mathrm{~b}+2 \mathrm{a})}$, There cannot be any induction in part SR.

Solution:
$\varepsilon=B \times b \times v$
$\mathrm{i}=\frac{\varepsilon}{\mathrm{R}_{\text {total }}}=\frac{\varepsilon}{\lambda(2 \mathrm{~b}+2 \mathrm{a})}$
Current is anticlockwise
$$
V_P-V_S \neq V_Q-V_R
$$
Also SR, $\vec{\ell} \| \vec{v} ;$ So $\varepsilon=0$

Topic: Physics, Magnetic Effects of Current, WBJEE, WBJEE 2022
As shown in figure below, a charge $+2 \mathrm{C}$ is situated at the origin $O$ and another charge $+5 \mathrm{C}$ is on the $x$ -axis at the point $A .$ The later charge from the point $A$ is then brought to a point $B$ on the $y$ -axis. The work done is $\left(\right.$ given $\left.\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{m} / \mathrm{F}\right)$

Options:
A $45 \times 10^{9} \mathrm{J}$
B $90 \times 10^{9} \mathrm{J}$
C zero
D $-45 \times 10^{9} \mathrm{J}$

The correct answer is:
zero

Solution:
Work done $=U_{f}-U_{i}$
$$
=\frac{1}{4 \pi \varepsilon_{0}} \times 2 \times 5 \times\left(\frac{1}{2}-\frac{1}{2}\right)=0
$$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2013