datavorous/entrance-exam-dataset huggingface dataset:
Out of 7 consonants and 4 vowels, words are formed each having 3 consonants and 2 vowels. The number of such words that can be formed is

Options:
A 210
B 25200
C 2520
D 302400

The correct answer is:
25200

Solution:
Out of 7 consonants, the number of ways of selecting 3 consonants $={ }^{7} C_{3}$
Similarly, number of ways of selecting
2 vowels out of 4 vowels $={ }^{4} C_{2}$
$\therefore$ Total number of words formed
$$
\begin{array}{l}
={ }^{7} C_{3} \times{ }^{4} C_{2} \times{ }^{5} P_{5} \\
=\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1} \times 51 \\
=7 \times 5 \times 2 \times 3 \times 120=25200
\end{array}
$$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2017
Out of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one?

Options:
A $(n-1) d^{8} n s^{2}$
B $(n-1) d^{5} n s^{2}$
C $(n-1) d^{3} n s^{2}$
D $(n-1) d^{5} n s^{1}$

The correct answer is:
$(n-1) d^{5} n s^{2}$

Solution:
Out of the given outer electronic configuration of atoms, the highest oxidation state is achieved by $(n-1) d^{5} n s^{2}$ i.e. 7
A large number of oxidation state is due to the fact the $(n-1) d$ -electrons may get involved along with $n s$ electrons in bonding as electrons in $(n-1) d-$ orbitals are in an energy state comparable to $n s$ -electrons. Oxidation state of other options are as follows:

Topic: Chemistry, Redox Reactions, WBJEE, WBJEE 2018
Oxidation of allyl alcohol with a peracid gives a compound of molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2}$, which contains an asymmetric carbon atom. The structure of the compound is

Options:
A
B
C
D

The correct answer is:

Solution:
Oxidation of allyl alcohol with a peracid gives a compound of molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2},$ which contains an asymmetric carbon atom. The structure of the compound is
So, the option (a) is correct answer.

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2019
Oxidation states of $\mathrm{Cr}$ in $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ and $\mathrm{CrO}_5$ are, respectively

Options:
A $+6,+5$
B $+6,+10$
C $+6,+6$
D None of these

The correct answer is:
$+6,+6$

Solution:
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7: 2(+1)+2 \mathrm{x}+7(-2)=0 \Rightarrow \mathrm{x}=+6$

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2022
Ozonolysis of an alkene produces only one dicarbonyl compound. The structure of the alkene is

Options:
A $\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}$
B
C
D $\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}$

The correct answer is:

Solution:
Only gives one compound on ozonolysis.

All other options give more than one aldehyde/ketone on ozonolysis.

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2016
$\mathrm{P}_4 \mathrm{O}_{10}$ is the anhydride of

Options:
A $\mathrm{H}_3 \mathrm{PO}_2$
B $\mathrm{H}_3 \mathrm{PO}_3$
C $\mathrm{H}_3 \mathrm{PO}_4$
D $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7$

The correct answer is:
$\mathrm{H}_3 \mathrm{PO}_4$

Solution:
Hints: $4 \mathrm{H}_3 \mathrm{PO}_4 \longrightarrow \mathrm{P}_4 \mathrm{O}_{10}+6 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2010
$[P] \stackrel{\text { Br }_{2}}{\longrightarrow} C_{2} \mathrm{H}_{4}$ Br $_{2} \frac{\text { NaNH }_{2}}{\mathrm{NH}_{3}}$ ? $[Q]$
$[Q] \frac{20 \% \mathrm{H}_{2} \mathrm{SO}_{4}}{\mathrm{Hg}^{2+} \cdot \Delta}[R] \stackrel{\mathrm{Zn}-\mathrm{Hg} / \mathrm{HC}}{\longrightarrow}[S]$
The species $P, Q, R$ and $S$ respectively are

Options:
A ethene, ethyne, ethanal, ethane
B ethane, ethyne, ethanal, ethene
C ethene, ethyne, ethanal, ethanol
D ethyne, ethane, ethene, ethanal

The correct answer is:
ethene, ethyne, ethanal, ethane

Solution:
Given reactions are
$[P] \stackrel{\mathrm{Br}_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Br}_{2} \frac{\mathrm{NaNH}_{2}}{\mathrm{NH}_{3}},[Q]$
$[Q] \frac{20 \% / \mathrm{H}_{2} \mathrm{SO}_{4}}{\mathrm{Hg}^{2+}, \Delta},[\mathrm{R}] \stackrel{\mathrm{Zn} \cdot \mathrm{Hg} / \mathrm{HG}}{\longrightarrow}[S]$
$[P]$ is an alkene with two carbon atoms i.e. ethene. When it reacts with Br $y$, addition reaction occurs and $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Br}_{2}$ is formed.

$\mathrm{C}_{\mathrm{H}}, \mathrm{Br}_{2}$(1, 2-dibromoethane) on reaction with $\operatorname{NaNH}_{2} / \mathrm{NH},$ gives ethyne $(\mathrm{C}, \mathrm{H}, \mathrm{y})$
Ethyne IQlin presence of $20 \% \mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{Hg}^{2+} \mathrm{at} 333 \mathrm{K}$
gives ethanal.

Ethanal $[R]$ undergoes reduction in presence of $$
\text { Zn-Hg /HCl to give ethane }[S]
$$
$\mathrm{CH}_{3}\mathrm{CHO} \xrightarrow[\mathrm{HCl}]{\mathrm{Zn-Hg}} \mathrm{CH}_{3}-\mathrm{CH}_{3}$

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2018
$P$ is the extremity of the latusrectum of ellipse $3 x^{2}+4 y^{2}=48$ in the first quadrant. The eccentric angle of $P$ is

Options:
A $\frac{\pi}{8}$
B $\frac{3 \pi}{4}$
C $\frac{\pi}{3}$
D $\frac{2 \pi}{3}$

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

Solution:
Given equation of ellipse is
$$
\begin{array}{l}
\quad 3 x^{2}+4 y^{2}=48 \\
\Rightarrow \quad \frac{3 x^{2}}{48}+\frac{4 y^{2}}{48}=1 \quad \Rightarrow \quad \frac{x^{2}}{16}+\frac{y^{2}}{12}=1 \\
\text { Here, } \quad a=4 \text { and } b=2 \sqrt{3}
\end{array}
$$
$\therefore$
$e=\sqrt{1-\frac{b^{2}}{a^{2}}}$
$=\sqrt{1-\frac{12}{16}}=\sqrt{\frac{4}{16}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$
$\therefore$ Coordinates of $P$ are $=(2,3)$ $\left.(4 \cos \theta, 2 \sqrt{3} \sin \theta)=(2,3) \quad [\because p=\left(a e, \frac{b^{2}}{a}\right)\right]$
By comparing, we get
$$
4 \cos \theta=2 \text { and } 2 \sqrt{3} \sin \theta=3
$$
$\Rightarrow \quad \cos \theta=\frac{1}{2}$ and $\sin \theta=\frac{\sqrt{3}}{2}$
$\Rightarrow \quad \cos \theta=\cos \frac{\pi}{3}$ and $\sin \theta=\sin \frac{\pi}{3}$
$\theta=\frac{\pi}{3}$ and $\theta=\frac{\pi}{3}$
$\therefore$ eccentric angle of $P$ is $\frac{\pi}{3}$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2019
Paracetamol is

Options:
A Methyl salicylate
B Phenyl salicylate
C N-acetyl p-amino phenol
D Acetyl salicylic acid

The correct answer is:
N-acetyl p-amino phenol

Solution:
Hints: Fact

Topic: Chemistry, Chemistry in Everyday Life, WBJEE, WBJEE 2011
Particle $A$ moves along $X$ -axis with a uniform velocity of magnitude $10 \mathrm{m} / \mathrm{s}$. Particle $B$ moves with uniform velocity $20 \mathrm{m} / \mathrm{s}$ along a direction making an angle of $60^{\circ}$ with the positive direction of $X$ -axis as shown in the figure. The relative velocity of $B$ with respect to that of $A$ is

Options:
A $10 \mathrm{m} / \mathrm{s}$ along $X$ -axis
B $10 \sqrt{3} \mathrm{m} / \mathrm{s}$ along $Y$ -axis (perpendicular to $X$ -axis)
C $10 \sqrt{5} \mathrm{m} / \mathrm{s}$ along the bisection of the velocities of A and $B$
D $30 \mathrm{m} / \mathrm{s}$ along negative $X$ -axis

The correct answer is:
$10 \sqrt{3} \mathrm{m} / \mathrm{s}$ along $Y$ -axis (perpendicular to $X$ -axis)

Solution:
$$
\begin{aligned}
&\text { The component of velocity of } B \text { along } x \text { -direction }\\
&\begin{aligned}
\mathbf{v}_{B x} &=20 \cos 60^{\circ}=20 \times \frac{1}{2}=10 \mathrm{m} / \mathrm{s} \\
\mathbf{v}_{A} &=10 \hat{\mathbf{i}} \\
\mathbf{v}_{B} &=10 \hat{\mathbf{i}}+10 \sqrt{3} \hat{\mathbf{j}} \\
\mathbf{v}_{B A} &=\mathbf{v}_{B}-\mathbf{v}_{A}=10 \hat{\mathbf{i}}+10 \sqrt{3} \hat{\mathbf{j}}-10 \hat{\mathbf{i}}=10 \sqrt{3} \hat{\mathbf{j}}
\end{aligned}
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2015
particle is moving in an elliptical orbit as shown in figure. If $\vec{p}, \vec{L}$ and $\vec{r}$ denote the linear momentum, angular momentum and position vector of the particle (from focus $\mathrm{O}$ ) respectively at a point $\mathrm{A}$, then the direction of $\vec{\alpha}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{L}}$ is along

Options:
A + ve $x$ axis
B - ve $x$ axis
C + ve y axis
D - ve y axis

The correct answer is:
+ ve $x$ axis

Solution:
$\vec{\alpha}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{L}}$
$\vec{L}$ is along Z-axis

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2022
$\mathrm{PbCl}_{2}$ is insoluble in cold water. Addition of $\mathrm{HCl}$ increases its solubility due to

Options:
A formation of soluble complex anions like $\left[\mathrm{PbCl}_{3}\right]^{-}$
B oxidation of Pb(II) to PB(iV)
C formation of $\left[\mathrm{Pb}(\mathrm{H}_{2}\mathrm{O})_{6}\right]^{2}$
D formation of polymeric lead complexes

The correct answer is:
formation of soluble complex anions like $\left[\mathrm{PbCl}_{3}\right]^{-}$

Solution:
$\mathrm{PbCl}_{2}$ react with HCl as follows $\mathrm{PbCl}_{2}(s)+\mathrm{Cl}^{-} \stackrel{\text { Cold }}{\longrightarrow}\left[\mathrm{PbCl}_{3}\right]^{-}(a q)$
$\mathrm{PbCl}_{2}(s)+2 \mathrm{Cl}^{-} \frac{\text { Excess }}{\text { of HCl }}\left[\mathrm{PbCl}_{4}\right]^{2-}(a q)$
Thus, addition of excess of $\mathrm{Cl}^{-}$ ions change the $\mathrm{PbCl}_{2}$ as soluble complex of $\left[\mathrm{PbCl}_{4}\right]^{-2}$. Hence, becomes soluble.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2017
Pick the correct statement.

Options:
A Relative lowering of vapour pressure is independent of $\mathrm{T}$.
B Osmotic pressure always depends on the nature of solute.
C Elevation of boiling point is independent of nature of the solvent.
D Lowering of freezing point is proportional to the molar concentration of solute.

The correct answer is:
Relative lowering of vapour pressure is independent of $\mathrm{T}$.

Solution:
Relative lowering of vapour pressure is $\frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}}$, which is equal to mole fraction of solute which is independent of temperature.

Topic: Chemistry, Solutions, WBJEE, WBJEE 2022
Platinum, Palladium and Iridium are called noble metals because

Options:
A Alfred Nobel discovered them
B They are shining lustrous and pleasing to look at
C They are found in native state
D They are inert towards many common reagents.

The correct answer is:
They are inert towards many common reagents.

Solution:
Hints: Fact

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2011
p-nitro-N, N - dimethylaniline cannot be represented by the resonating structures

Options:
A I and II
B II and IV
C I and III
D III and IV

The correct answer is:
II and IV

Solution:
Both the structures are impossible as $\mathrm{N}^{\circ}$ is pentavalent and we known that $\mathrm{N}$ can not expand its valency beyond 4 due to absence of vacant d-orbitals in its valence shell.

Topic: Chemistry, Amines, WBJEE, WBJEE 2021
$P Q$ is a double ordinate of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ such that $\triangle O P Q$ is an equilateral triangle, $O$ being the centre of the hyperbola. Then the eccentricity e of the hyperbola satisfies

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

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

Solution:
$\tan 30^{\circ}=\frac{b \tan \theta}{a \sec \theta}$
$\Rightarrow \frac{b}{a}=\frac{1}{\sin \theta \sqrt{3}}$
$e=\sqrt{1+\frac{1}{3 \sin ^2 \theta}} > \sqrt{1+\frac{1}{3}}$
$\Rightarrow e > \frac{2}{\sqrt{3}}\left(0 < \sin ^2 \theta < 1\right)$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2022
Pressure-volume $(p V)$ work done by an ideal gaseous system at constant volume is (where $E$ is internal energy of the system)

Options:
A $-\Delta p / p$
B zero
C $-V \Delta p$
D $-\Delta E$

The correct answer is:
zero

Solution:
From first law of thermodynamics
$$
\Delta E=q+W
$$
where, work do net $(W)=p \Delta V$
$$
\begin{array}{l}
=p \times \Delta V \\
=0
\end{array} \quad(\because \Delta V=0)
$$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2013
Product of any $r$ consecutive natural numbers is always divisible by

Options:
A $\mathrm{r}$ !
B $(r+4) !$
C $(r+1) !$
D $(r+2) !$

The correct answer is:
$\mathrm{r}$ !

Solution:
Hints : $(n+1)(n+2) \ldots \ldots .(n+r)$
$$
\begin{aligned}
& =\frac{(n+r) !}{n !} \\
& =\frac{(n+r) !}{n ! r !} r !=r !^{n+r} C_n
\end{aligned}
$$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2009
Radioactivity of a sample ( $\mathrm{z}=22$ ) decreases $90 \%$ after 10 years. What will be the half life of the sample?

Options:
A 5 years
B 2 years
C 3 years
D 10 years

The correct answer is:
3 years

Solution:
Hints: $t=10$ yrs $\quad t_{\frac{1}{2}}=?$
$$
\lambda=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{N}_0}{\mathrm{~N}_{\mathrm{t}}}
$$
Since radioactivity decreases $90 \%$ in 10 yrs. $\Rightarrow N_0=100 \& N_t=10$
Thus $\lambda=\frac{2.303}{10} \log \frac{100}{10} \Rightarrow \lambda=\frac{2.303}{10}$
since $\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{\lambda}=\frac{2.303 \times \log 2}{\lambda} \Rightarrow \mathrm{t}_{\frac{1}{2}}=\frac{2.303 \times \log 2}{2.303 / 10}$
$\Rightarrow \mathrm{t}_{\frac{1}{2}}=(\log 2) \times 10 \simeq 3$ years

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2010
Radon- 222 has a half-life of 3.8 days. If one starts with $0.064 \mathrm{kg}$ of radon-222, the quantity of radon- 222 left after 19 days will be

Options:
A $0.002 \mathrm{kg}$
B $0.062 \mathrm{kg}$
C $0.032 \mathrm{kg}$
D $0.024 \mathrm{kg}$

The correct answer is:
$0.002 \mathrm{kg}$

Solution:
Half life of radon $=3 \cdot 8$ days
Total half-lives in 19 days
$\begin{array}{l}
=\frac{19}{3 \cdot 8} \\
=\frac{190}{38}
\end{array}$
$=5$ half-lives
Initially mass of radon $=0.064 \mathrm{kg}$
$0 \cdot 064 \frac{T_{1}}{T_{3}} 0 \cdot 032 \stackrel{T_{2}}{\longrightarrow}$
$0 \cdot 016 \stackrel{T_{4}}{\longrightarrow} 0 \cdot 004 \stackrel{T_{5}}{\longrightarrow} 0 \cdot 002$
So in five half-life periods, the Radon sample will reduce to $0.002 \mathrm{kg}$
$\text {Or } \frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{5}$
$\Rightarrow \quad N=0 \cdot 064 \times \frac{1}{32}$
$=0 \cdot 002 \mathrm{kg}$

Topic: Physics, Nuclear Physics, WBJEE, WBJEE 2017
Ram is visiting a friend. Ram knows that his friend has 2 children and 1 of them is a boy. Assuming that a child is equally likely to be a boy or a girl, then the probability that the other child is a girl, is

Options:
A $1 / 2$.
B $1 / 3$
C $2 / 3$
D $7 / 10$

The correct answer is:
$2 / 3$

Solution:
Total sample space $=\left\{B_{1} B_{2}, B_{1} G_{1}, G_{1} B_{2}\right\}$
Favourable ways $=\left\{B_{1} G_{1}, G_{1} B_{2}\right\}$
$\therefore$ Required probability $=\frac{2}{3}$

Topic: Mathematics, Probability, WBJEE, WBJEE 2014
Reaction of benzene with $\mathrm{Me}_{3} \mathrm{CCOCl}$ in the presence of anhydrous AlCl ${ }_{3}$ gives

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2013
Reaction of formaldehyde and ammonia gives

Options:
A Hexamethylene tetramine
B Bakelite
C Urea
D Triethylene Tetramine

The correct answer is:
Hexamethylene tetramine

Solution:
Hints: \(6 \mathrm{HCHO}+4 \mathrm{NH}_3 \rightarrow\left(\mathrm{CH}_2\right)_6 \mathrm{~N}_4+6 \mathrm{H}_2 \mathrm{O}\)

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2011
Reaction of $\mathrm{R} \mathrm{OH}$ with $\mathrm{R}^{\prime} \mathrm{MgX}$ produces :

Options:
A $\mathrm{RH}$
B $\mathrm{R}^{\prime} \mathrm{H}$
C $\mathrm{R}-\mathrm{R}$
D $R^{\prime}-R^{\prime}$

The correct answer is:
$\mathrm{R}^{\prime} \mathrm{H}$

Solution:
Hints:

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2009
Reaction of silver nitrate solution with phosphorous acid produces:

Options:
A Silver phosphite
B Phosphoric acid
C Metallic silver
D Silver phosphate

The correct answers are:
Phosphoric acid, Metallic silver

Solution:
$2 \mathrm{AgNO}_{3}+\mathrm{H}_{3} \mathrm{PO}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}+2 \mathrm{Ag}+2 \mathrm{HNO}_{3}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2021
Reduction of the lactol S

OH with sodium borohydride gives

Options:
A
B
C
D

The correct answer is:

Solution:
The reaction occurs as follows:

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2017
Reflection of the line \(\bar{a} z+a \bar{z}=0\) in the real axis is given by

Options:
A \(\mathrm{az}+\overline{\mathrm{az}}=0\)
B \(\bar{a} z-a \bar{z}=0\)
C \(a z-\overline{a z}=0\)
D \(\frac{\mathrm{a}}{\mathrm{z}}+\frac{\overline{\mathrm{a}}}{\overline{\mathrm{z}}}=0\)

The correct answer is:
\(\mathrm{az}+\overline{\mathrm{az}}=0\)

Solution:
Hint:
Let, \(a=\alpha+i \beta\) and \(z=x+i y\)
\(\begin{aligned}
& \therefore \bar{a} z+a \bar{z}=0 \\
& \Rightarrow(\alpha-i \beta)(x+i y)+(\alpha+i \beta)(x-i y)=0 \\
& \Rightarrow 2(\alpha x+\beta y)=0 \Rightarrow \alpha x+\beta y=0 \text { which passes through origin } \therefore \text { slope }=-\frac{\alpha}{\beta}
\end{aligned}\)
\(\therefore\) Reflected line's slope \(=\frac{\alpha}{\beta}\)
\(\therefore\) Reflected line's equation : \(\alpha x -\beta y=0\)
\(\begin{aligned}
& \Rightarrow\left(\frac{a+\bar{a}}{2}\right) x-\left(\frac{a-\bar{a}}{2 i}\right) y=0 \\
& \Rightarrow\left(\frac{a+\bar{a}}{2}\right)\left(\frac{z+\bar{z}}{2}\right)-\left(\frac{a-\bar{a}}{2 i}\right)\left(\frac{z-\bar{z}}{2 i}\right)=0 \\
& \Rightarrow a z+\bar{a} \bar{z}=0 \\
& \Rightarrow a z+\bar{a} z=0
\end{aligned}\)

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2023
Ribose and 2 -deoxyribose can be differentiated by

Options:
A Fehling's reagent
B Tollen's reagent
C Barfoed's reagent
D Osazone formation

The correct answer is:
Osazone formation

Solution:
Ribose forms osazone with $\mathrm{Ph}-\mathrm{NH}-\mathrm{NH}_{2}$ whereas in deoxyribose, one - OH group is missing, due to which, it cannot form osazone.

Topic: Chemistry, Biomolecules, WBJEE, WBJEE 2013
Rolle's theorem is applicable in the interval [-2,2] for the function

Options:
A $f(x)=x^{3}$
B $f(x)=4 x^{4}$
C $f(x)=2 x^{3}+3$
D $f(x)=\pi|x|$

The correct answer is:
$f(x)=4 x^{4}$

Solution:
If we take $f(x)=4 x^{4}$, then
(i) $f(x)$ is continuous in (-2,2)
(ii) $f(x)$ is differentiable in (-2,2)
(iii) $f(-2)=f(2)$ So, $f(x)=4 x^{4}$ satisfies all the conditions of Rolle's theorem therefore $\exists$ a point $c$ such that
$f^{\prime}(c)=0$
$\Rightarrow \quad 16 c^{3}=0 \Rightarrow c=0 \in(-2,2)$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2012
$S_{1}$ and $S_{2}$ are the coherent point sources of light located in the $x y$ -plane at points (0,0) and $(0,3 \lambda)$ respectively. Here $\lambda$ is the wavelength of light. At which one of the following points (given as coordinates), the intensity of interference will be maximum?

Options:
A $(3 \lambda, 0)$
B $(4 \lambda, 0)$
C $(5 \lambda / 4,0)$
D $(2 \lambda / 3,0)$

The correct answer is:
$(4 \lambda, 0)$

Solution:
The given situation can be show as

The intensity will be maximum at those given points where the path difference between the two interfering waves is an integral multiple of wavelength $i . e ., \Delta x=n \lambda$
For the the given points, the intensily will be maximum for $(4 \lambda, 0)$

Topic: Physics, Wave Optics, WBJEE, WBJEE 2013
S and $T$ are the foci of an ellipse and B is end point of the minor axis. If STB is an equilateral triangle, the eccentricity of the ellipse is

Options:
A $\frac{1}{4}$
B $\frac{1}{3}$
C $\frac{1}{2}$
D $\frac{2}{3}$

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

Solution:
Hints : $\frac{\mathrm{b}}{\mathrm{ae}}=\sqrt{3} ; \quad \mathrm{b}=\sqrt{3} \mathrm{ae}$
$$
\mathrm{e}^2=\frac{\mathrm{a}^2-3 \mathrm{a}^2 \mathrm{e}^2}{\mathrm{a}^2}=1-3 \mathrm{e}^2 ; \quad 4 \mathrm{e}^2=1 \Rightarrow \mathrm{e}=\frac{1}{2}
$$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2010
$\int \frac{\mathrm{dx}}{\sin x+\sqrt{3} \cos x}$ equals
where $\mathrm{c}$ is an arbitrary constant

Options:
A $\frac{1}{2} \ln \left|\tan \left(\frac{x}{2}-\frac{\pi}{6}\right)\right|+c$
B $\frac{1}{2} \ln \left|\tan \left(\frac{\mathrm{x}}{4}-\frac{\pi}{6}\right)\right|+c$
C $\frac{1}{2} \ln \left|\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)\right|+c$
D $\frac{1}{2} \ln \left|\tan \left(\frac{\mathrm{x}}{4}+\frac{\pi}{3}\right)\right|+\mathrm{c}$

The correct answer is:
$\frac{1}{2} \ln \left|\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)\right|+c$

Solution:
Hints : $\int \frac{d x}{\sin x+\sqrt{3} \cos x}=\int \frac{d x}{2\left(\frac{1}{2} \sin x+\frac{\sqrt{3}}{2} \cos x\right)}=\frac{1}{2} \int \frac{d x}{\sin \left(x+\frac{\pi}{3}\right)}$
$$
\begin{aligned}
& =\frac{1}{2} \int \operatorname{cosec}\left(\mathrm{x}+\frac{\pi}{3}\right) \mathrm{dx}=\frac{1}{2} \log \left|\tan \left(\frac{\mathrm{x}}{2}+\frac{\pi}{6}\right)\right|+\mathrm{c} \\
& =\frac{1}{2} \ln \left|\tan \left(\frac{\mathrm{x}}{2}+\frac{\pi}{6}\right)\right|+\mathrm{c}
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2009
Same quantity of ice is filled in each of the two metal containers $P$ and $Q$ having the same size, shape and wall thickness but made of different materials. The containers are kept in identical surroundings. The ice in $P$ melts completely in time $t_{1}$ whereas in $Q$ takes a time $t_{2}$. The ratio of thermal conductivities of the materials of $P$ and $Q$ is

Options:
A $t_{2}: t_{1}$
B $t_{1}: t_{2}$
C $t_{1}^{2}: t_{2}^{2}$
D $t_{2}^{2}: t_{1}^{2}$

The correct answer is:
$t_{2}: t_{1}$

Solution:
We know that Relation between temperature gradient (TG) and thermal conductivity $(K)$
So,
$$
\frac{d \theta}{d t}=-K A \frac{d \theta}{d x}=-K A \times(\mathrm{TG})
$$
Le. $\mathrm{TG} \propto \frac{1}{K}\left(\frac{d \theta}{d t}=\right.$ constant $)$
or $K \times \frac{1}{T}$
$$
\begin{array}{l}
K_{1} \times \frac{1}{t_{1}} \\
K_{2} \times \frac{1}{t_{2}}
\end{array}
$$
From the Eqs. (i) and (ii), we get
$$
\frac{K_{1}}{K_{2}}=\frac{1 / t_{1}}{1 / t_{2}}
$$
$$
K_{1}: K_{2}=t_{2}: t_{1}
$$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2014
Sand $T$ are the foci of an ellipse and $B$ is the end point of the minor axis. If $S T B$ is equilateral triangle, the eccentricity of the ellipse is

Options:
A $\frac{1}{4}$
B $\frac{1}{3}$
C $\frac{1}{2}$
D $\frac{2}{3}$

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

Solution:
Equation of the ellipse is
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
$$

Foci are $S(a e, 0), T(-a e, 0)$
$B(0, b)$ is the end of the minor axis. STB is an equilateral triangle.
$$
\begin{aligned}
S B &=S T \\
S B^{2} &=S T^{2}
\end{aligned}
$$
$\Rightarrow \quad a^{2} e^{2}+b^{2}=4 a^{2} e^{2}$
$\Rightarrow \quad b^{2}=3 a^{2} e^{2}$
$\Rightarrow \quad a^{2}\left(1-e^{2}\right)=3 a^{2} e^{2}$
$\Rightarrow \quad 1-e^{2}=3 e^{2}$
$\Rightarrow \quad 4 e^{2}=1$
$\Rightarrow \quad e^{2}=\frac{1}{4}$
$\Rightarrow \quad e=\frac{1}{2}$
Eccentricity of the ellipse, $e=\frac{1}{2}$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2019
Select the molecule in which all the atoms may lie on a single plane is

Options:
A 4-Nitrobenzaldehyde
B 4-Methoxybenzaldehyde
C 4-Methylnitrobenzene
D 4-Nitroacetophenone

The correct answer is:
4-Nitrobenzaldehyde

Solution:
Hint :
All atoms other than Hydrogen are \(\mathrm{sp}^2\) hybridised.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2023
Silicon oil is obtained from the hydrolysis and polymerisation of

Options:
A trimetylchlorosilane and dimethyldichlorosilane
B trimethylchlorosilane and methyldichlorosilane
C methyltrichlorosilane and dimethyldichlorosilane
D triethylchlorosilane and dimethyidichlorosilane

The correct answer is:
trimetylchlorosilane and dimethyldichlorosilane

Solution:
Silicon oil in a polymer of trimethylchloro silane and dimethyldichloro silane. These are useful as broad spectrum antifoaming agents.

Topic: Chemistry, Polymers, WBJEE, WBJEE 2013