datavorous/entrance-exam-dataset huggingface dataset:
$\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x$ is equal to

Options:
A 1
B 0
C 2
D None of the above

The correct answer is:
0

Solution:
Let $\quad I=\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x$
$\therefore$
$=\int_{0}^{1} \log \left(\frac{1-x}{x}\right) d x$
$I=\int_{0}^{1} \log \left(\frac{x}{1-x}\right) d x=-1$
$\quad\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]$
$2 l=0 \Rightarrow I=0$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2016
$\int_{0}^{2}\left[\mathrm{x}^{2}\right] \mathrm{dx}$ is equal to where [] if GIF

Options:
A 1
B $5-\sqrt{2}-\sqrt{3}$
C $3-\sqrt{2}$
D $8 / 3$

The correct answer is:
$5-\sqrt{2}-\sqrt{3}$

Solution:
Hint:
$\int_{0}^{1} 0 \mathrm{dx}+\int_{1}^{\sqrt{2}} 1 \cdot \mathrm{dx}+\int_{\sqrt{2}}^{\sqrt{3}} 2 \cdot \mathrm{d} x+\int_{\sqrt{3}}^{2} 3 \mathrm{~d} x$
$=5-\sqrt{2}-\sqrt{3}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2020
$\int_{0}^{100} e^{x-[x]} d x$ is equal to

Options:
A $\frac{e^{100}-1}{100}$
B $\frac{e^{100}-1}{e-1}$
C $100(e-1)$
D $\frac{e-1}{100}$

The correct answer is:
$100(e-1)$

Solution:
Let $I=\int_{0}^{100} e^{x-[x]} d x$
$=100 \int_{0}^{1} e^{x-[x]} d x$
$[\because x-[x]$ is a periodic function of period 1 and $\int_{0}^{mT} f(x) d x=m \int_{0}^{T} f(x) d x,$ where $T$ is period of $f(x)]$ $=100 \int_{0}^{1} e^{x} d x[\because x-[x]=x$ for $0 < x < 1]$
$=100\left[e^{x}\right]_{0}^{1}$
$=100\left[e^{1}-e^{0}\right]$
$=100(e-1)$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2017
$\int_0^{1000} e^{x-[x]}$ is equal to

Options:
A $\frac{e^{1000}-1}{e-1}$
B $\frac{e^{1000}-1}{1000}$
C $\frac{e-1}{1000}$
D $1000(\mathrm{e}-1)$

The correct answer is:
$1000(\mathrm{e}-1)$

Solution:
Hins : $I=1000 \int_0^1 \mathrm{e}^{x-[x]}$
$=1000 \int_0^1 e^x d x=1000\left(e^x\right)_0^1=100(e-1)$
Period of function is 1

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2009
$1 \times 10^{-3}$ mole of $\mathrm{HCl}$ is added to a buffer solution made up of $0.01 \mathrm{M}$ acetic acid and 0.01 M sodium acetate. The final pH of the buffer will be (given, $\mathrm{p} K_{a}$ of acetic acid is 4.75 at $\left.25^{\circ} \mathrm{C}\right)$

Options:
A 4.6
B 4.66
C 4.75
D 4.8

The correct answer is:
4.66

Solution:
$\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}$
$$
\begin{aligned}
\mathrm{pH}=& \mathrm{pK}_{a}+\log \frac{(\text { salt })}{(\text { acid })}=4.75+\log \frac{0.009}{0.011} \\
&=4.66
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2013
1.

2.

3.

4.

The dipeptides which may be obtained from the aminoacids glycine and alanine are

Options:
A only 1
B only 2
C both 1 and 2
D all of them

The correct answer is:
all of them

Solution:

Topic: Chemistry, Biomolecules, WBJEE, WBJEE 2021
$\int_{1}^{3} \frac{|x-1|}{|x-2|+|x-3|} d x=$

Options:
A $1+\frac{4}{3} \log _{e} 3$
B $1+\frac{3}{4} \log _{e} 3$
C $1-\frac{4}{3} \log _{e} 3$
D $1-\frac{3}{4} \log _{e} 3$

The correct answer is:
$1+\frac{3}{4} \log _{e} 3$

Solution:
$\int_{1}^{3} \frac{|x-1|}{|x-2|+|x-3|} d x=\int_{1}^{2} \frac{x-1}{-x+2-x+3} d x+\int_{2}^{3} \frac{x-1}{x-2-x+3} d x=\int_{1}^{2} \frac{x-1}{-2 x+5}+\int_{2}^{3} \frac{x-1}{1} d x=1+\frac{3}{4} \ln 3$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2021
$\int \sqrt{1+\cos x} d x$ is equal to

Options:
A $2 \sqrt{2} \cos \frac{\mathrm{x}}{2}+\mathrm{C}$
B $2 \sqrt{2} \sin \frac{\mathrm{x}}{2}+\mathrm{C}$
C $\sqrt{2} \cos \frac{x}{2}+C$
D $\sqrt{2} \sin \frac{x}{2}+C$

The correct answer is:
$2 \sqrt{2} \sin \frac{\mathrm{x}}{2}+\mathrm{C}$

Solution:
Hints: $\int \sqrt{1+\cos x} d x=\sqrt{2} \int \cos \left(\frac{x}{2}\right) d x=2 \sqrt{2} \sin \left(\frac{x}{2}\right)+c$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2010
$0.1(\mathrm{M}) \mathrm{HCI}$ and $0.1(\mathrm{M}) \mathrm{H}_2 \mathrm{SO}_4$ each of volume $2 \mathrm{ml}$ are mixed and the volume is made up to $6 \mathrm{ml}$ by adding $2 \mathrm{ml}$ of $0.01(\mathrm{~N}) \mathrm{NaCl}$ solution. The $\mathrm{pH}$ of the resulting mixture is

Options:
A 1.17
B 1.0
C 0.3
D $\log 2-\log 3$

The correct answer is:
1.0

Solution:
Hints: Mili moles of $\mathrm{H}^{+}=0.1 \times 2+0.1 \times 2 \times 2=0.6$
Total volume in $\mathrm{ml}=6$
$$
\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]=-\log \left(\frac{0.6}{6}\right)=-\log 0.1=1
$$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2010
$1 \mathrm{~mL}$ of water has 25 drops. Let $\mathrm{N}_0$ be the Avogadro number. What is the number of molecules present in 1 drop of water ? (Density of water $=1 \mathrm{~g} / \mathrm{mL}$ )

Options:
A $\frac{0.02}{9} \mathrm{~N}_0$
B $\frac{18}{25} \mathrm{~N}_0$
C $\frac{25}{18} \mathrm{~N}_0$
D $\frac{0.04}{25} \mathrm{~N}_0$

The correct answer is:
$\frac{0.02}{9} \mathrm{~N}_0$

Solution:
Volume of one drop $=\left(\frac{1}{25}\right) \mathrm{mL}$
$\therefore$ Mass of $1 \mathrm{drop}=\mathrm{V} \times \mathrm{d}$
$=\left(\frac{1}{25} \mathrm{~mL}\right)(1 \mathrm{~g} / \mathrm{mL})$
$=\frac{1}{25} \mathrm{~g}$
Number of moles of $\mathrm{H}_2 \mathrm{O}=\frac{\text { Massof water inonedrop }}{\text { Molarmassof water }}=\frac{\frac{1}{25}}{18}=\frac{1}{25 \times 18}$
$\therefore$ Number of $\mathrm{H}_2 \mathrm{O}$ Molecule $=\frac{1}{25 \times 18} \mathrm{~N}_0=\frac{1}{50 \times 9} \mathrm{~N}_0=\frac{0.02}{9} \mathrm{~N}_0$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2022
1 mole of photon, each of frequency $2500 \mathrm{~S}^{-1}$, would have approximately a total energy of:

Options:
A $1 \mathrm{erg}$
B 1 Joule
C $1 \mathrm{eV}$
D $1 \mathrm{MeV}$

The correct answer is:
$1 \mathrm{erg}$

Solution:
Total Energy $=\mathrm{Nhv}=6.022 \times 10^{23} \times 6.626 \times 10^{-34} \mathrm{~J} . \mathrm{S} . \times 2500 \mathrm{~s}^{-1}=9.9 \mathrm{erg} \approx 10 \mathrm{erg}$ In (A) option, it should be 10 erg instead of $1 \mathrm{erg}$.

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2009
$1+{ }^{n} C_{1} \cos \theta+{ }^{n} C_{2} \cos 2 \theta+\ldots+{ }^{n} C_{n} \cos n \theta$ equals

Options:
A $\left(2 \cos \frac{\theta}{2}\right)^{n} \cos \frac{n \theta}{2}$
B $2 \cos ^{2} \frac{n \theta}{2}$
C $2 \cos ^{2 n} \frac{\theta}{2}$
D $\left(2 \cos ^{2} \frac{\theta}{2}\right)^{n}$

The correct answer is:
$\left(2 \cos \frac{\theta}{2}\right)^{n} \cos \frac{n \theta}{2}$

Solution:
Given, $1+{ }^{n} C_{1} \cos \theta+{ }^{n} C_{2} \cos 2 \theta+\ldots+{ }^{n} C_{n} \cos n \theta$
Which is real part of complex number. $\left({ }^{n} C_{0}+{ }^{n} C_{1} e^{i\theta}+\ldots\right)$
ie. $\operatorname{Re}\left({ }^{n} \mathrm{C}_{0}+{ }^{n} \mathrm{C}_{1} e^{i\theta}+\ldots\right)$
$=\operatorname{Re}\left(1+e^{\theta i}\right)^{n}=\operatorname{Ro}(1+\cos \theta+i \sin \theta)^{n}$
$=$ Re $\left(2 \cos ^{2} n \frac{\theta}{2}+i 2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}\right)$
$\left[\begin{array}{c}\because 1+\cos \theta=2 \cos ^{2} \frac{\theta}{2} \text { and } \\ \sin \theta=2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}\end{array}\right]$
$=\left(2 \cos \frac{\theta}{2}\right)^{n} \operatorname{Re}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)^{n}$
$=\left(2 \cos \frac{\theta}{2}\right)^{n} \operatorname{Re}\left(\cos \frac{n \theta}{2}+i \sin \frac{\theta}{2}\right)$
[by De moivre's theorem]
$=\left(2 \cos \frac{\theta}{2}\right)^{n} \cdot \cos \frac{n \theta}{2}$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2016
$\int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x$ is equal to
where $c$ is an arbitrary constant

Options:
A $\log \left(\sin ^{-1} \mathrm{x}\right)+\mathrm{c}$
B $\frac{1}{2}\left(\sin ^{-1} x\right)^2+c$
C $\log \left(\sqrt{1-x^2}\right)+c$
D $\sin \left(\cos ^{-1} \mathrm{x}\right)+\mathrm{c}$

The correct answer is:
$\frac{1}{2}\left(\sin ^{-1} x\right)^2+c$

Solution:
$$
\text { Hints : } \begin{array}{rlr}
& I=\int t d t & \sin ^{-1} x=t \\
& =\frac{1}{2} t^2+c & \frac{1}{\sqrt{1-x^2}} d x=d t \\
& =\frac{1}{2}\left(\sin ^{-1} x\right)^2+c &
\end{array}
$$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2009
$\left(2^{3 \mathrm{n}}-1\right)$ will be divisible by $(\forall \mathrm{n} \in \mathrm{N})$

Options:
A 25
B 8
C 7
D 3

The correct answer is:
7

Solution:
$$
\begin{aligned}
& \text { Hints: } 2^{3 \mathrm{n}}=(8)^{\mathrm{n}}=(1+7)^{\mathrm{n}}=={ }^{\mathrm{n}} \mathrm{C}_0+{ }^{\mathrm{n}} \mathrm{C}_1 7+{ }^{\mathrm{n}} \mathrm{C}_2 7^2+\ldots \ldots .+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} 7^{\mathrm{n}} \\
& \Rightarrow 2^{3 \mathrm{n}}-1=7\left[{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} \mathrm{C}_2 7+\ldots \ldots \ldots \ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} 7^{\mathrm{n}-1}\right] \\
& \therefore \text { divisible by } 7
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2010
2 gm of metal carbonate is neutralized completely by \(100 \mathrm{ml}\) of \(0.1(\mathrm{~N}) \mathrm{HCl}\). The equivalent weight of metal carbonate is

Options:
A 50
B 100
C 150
D 200

The correct answer is:
200

Solution:
Hints: Number of gram equivalents of \(\mathrm{HCl}\)
\(=\frac{100 \times 0.1}{1000}=0.01\)
Number of gram equivalents of metal carbonate required for neutralisation must also be 0.01 . Thus, mass of 1 gram eqivalent of carbonate salt \(\frac{2}{0.01}=200 \mathrm{~g}\)
\(\therefore\) Equivalent mass of carbonate salt \(=200\)

Topic: Chemistry, Redox Reactions, WBJEE, WBJEE 2011
2 moles of an ideal monoatomic gas is carried from a state $\left(p_{0}, V_{0}\right)$ to state $\left(2 p_{0}, 2 V_{0}\right)$ along a straight line path in a $p-V$ diagram. The amount of heat absorbed by the gas in the process is given by

Options:
A $3 p_{0} V_{0}$
B $\frac{9}{2} p_{0} V_{0}$
C $6 p_{0} V_{0}$
D $\frac{3}{2} p_{0} V_{0}$

The correct answer is:
$6 p_{0} V_{0}$

Solution:
The internal energy $\Delta U=n C_{r} \Delta T$
$C_{v}=$ Specific heat of gas at constant volume
$$
\begin{aligned}
\Rightarrow \quad \Delta U &=n \cdot \frac{3 R}{2}\left(\frac{4 p_{0} V_{0}}{n R}-\frac{p_{0} V_{0}}{n R}\right) \\
&=n \cdot \frac{3 R}{2}\left(\frac{4 p_{0} V_{0}-p_{0} V_{0}}{n R}\right) \\
&=n \cdot \frac{3 R}{2} \cdot \frac{3 p_{0} V_{0}}{n R} \\
&=\frac{9}{2} p_{0} V_{0}
\end{aligned}
$$
Work done by the gas $W=\left(2 p_{0}+p_{0}\right) \frac{V_{0}}{2}=\frac{3 p_{0} V_{0}}{2}$
From first law of thermodynamics,
$$
\begin{aligned}
\Delta Q &=d W+d U \\
&=\frac{3 p_{0} V_{0}}{2}+\frac{9}{2} p_{0} V_{0}
\end{aligned}
$$
[from Eqs. (i) and (ii)]
$$
\begin{array}{l}
=\frac{3 p_{0} V_{0}}{2}+\frac{9}{2} p_{0} V_{0} \\
=\frac{12 p_{0} V_{0}}{2}=6 p_{0} V_{0}
\end{array}
$$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2017
$\int 2^{x}\left[f^{\prime}(x)+f(x) \log 2\right] d x$ is equal to

Options:
A $2^{x} f^{\prime}(x)+C$
B $2^{x} \log 2+C$
C $2^{x} f(x)+C$
D $2^{x}+C$

The correct answer is:
$2^{x} f(x)+C$

Solution:
Let $I=\int 2^{n}\left[f^{\prime}(x)+f(x) \log 2\right] d x$
$\begin{array}{ll}\text { Consider } & g(x)=2^{x} f(x) \\ \Rightarrow & g^{\prime}(x)=2^{x} f^{\prime}(x)+2^{x} f(x) \log 2 \\ \Rightarrow & g^{\prime}(x)=2^{x}\left[f^{\prime}(x)+f(x) \log 2\right] \\ \therefore \quad & \quad I=\int g^{\prime}(x) d x=g(x)+C=2^{x} f(x)+C\end{array}$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2016
(+)-2-chloro-2-phenylethane in toluene racemises slowly in the presence of small amount of $\mathrm{SbCl}_{5},$ due to the formation of

Options:
A carbanion
B carbene
C free-radical
D carbocation

The correct answer is:
carbocation

Solution:
A planar carbocation is generated when $\mathrm{SbCl}_{5}$ remove $\mathrm{Cl}^{-}$ from the substrate. This carbocation is subsequently attacked by $\mathrm{Cl}^{-}$ (nucleophile) from both the sides (j.e. from top and bottom) to produce a racemic mixture.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2013
2-methylpropane on monochlorination under photochemical condition give

Options:
A 2 -chloro-2-methyipropane as major product
B $(1: 1)$ mixture of 1 -chioro- 2 -methylpropane and 2 -chloro-2-methylpropane
C 1-chioro-2-methylpropane as a major product
D $(1: 9)$ mixture of 1 -chioro- 2 -methylpropane and 2-chloro-2-methylpropane

The correct answer is:
1-chioro-2-methylpropane as a major product

Solution:
There are nine primary hydrogens and one tertiary hydrogen in 2 -methyl propane. Tertiary hydrogen atoms react with Cl about 5.5 times as fast as primary. Even though the primary hydrogens are less reactive, there are so many of them that the primary product is the major product.

Topic: Chemistry, Haloalkanes and Haloarenes, WBJEE, WBJEE 2013
3 moles of mono-atomic gas $(\gamma=5 / 3)$ is mixed with 1 mole of a diatomic gas $(\gamma=7 / 3) .$ The value of $\gamma$ for the mixture will be

Options:
A $\frac{9}{11}$
B $\frac{11}{7}$
C $\frac{12}{7}$
D $\frac{15}{7}$

The correct answer is:
$\frac{11}{7}$

Solution:
The number of degrees of ireedom for the mixture is
$\therefore$$\begin{aligned} f_{\text {mix }} &=\frac{n_{1} f_{1}+n_{2} f_{2}}{n_{1}+n_{2}} \\ &=\frac{3 \times 3+1 \times 5}{4}=\frac{7}{2} \\ \gamma &=1+\frac{2}{f} \end{aligned}$
$\Rightarrow \quad \gamma=1+\frac{4}{7}=\frac{11}{7}$

Topic: Physics, Kinetic Theory of Gases, WBJEE, WBJEE 2013
$\begin{array}{llll}\text { 3. The sum } & \text { of } & \text { the } & \text { series }\end{array}$ $1+\frac{1}{2}{ }^{n} C_{1}+\frac{1}{3}{ }^{n} C_{2}+\quad+\frac{1}{n+1}{ }^{n} C_{n}$ is equal to

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

The correct answer is:
$\frac{2^{n+1}-1}{n+1}$

Solution:
$1+\frac{1}{2}{ }^{n} C_{1}+\frac{1}{3}{ }^{n} C_{2}+\quad+\frac{1}{n+1}{ }^{n} C_{n}$
$=\frac{1}{n+1}\left[(n+1)+\frac{(n+1) n}{21}\right.$
$+\frac{(n+1) n(n-1)}{3 !}+(1]$
$=\frac{1}{n+1}\left[^{n+1} C_{1}+{ }^{n+1} C_{2}+\quad+{ }^{n+1} C_{n+1}\right]$
\(=\frac{2^{n+1}-1}{n+1}\)

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2012
$\int \frac{\log \sqrt{x}}{3 x} d x$ is equal to

Options:
A $\frac{1}{3}(\log \sqrt{x})^{2}+C$
B $\frac{2}{3}(\log \sqrt{x})^{2}+C$
C $\frac{2}{3}(\log x)^{2}+C$
D $\frac{1}{3}(\log x)^{2}+C$

The correct answer is:
$\frac{1}{3}(\log \sqrt{x})^{2}+C$

Solution:
Let $I=\int \frac{\log \sqrt{x}}{3 x} d x$
Aqain, let $\log \sqrt{x}=z \Rightarrow \frac{1}{2 x} d x=d z$
$I=\int \frac{2 z}{3} d z=\frac{2}{3} \int z d z$
$=\frac{2}{3} \cdot \frac{z^{2}}{2}+C=\frac{1}{3} \left(\log \sqrt{x}\right)^{2}+C$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2016
$\int \frac{\log \sqrt{\mathrm{x}}}{3 \mathrm{x}} \mathrm{dx}$ is equal to

Options:
A $\frac{1}{3}(\log \sqrt{\mathrm{x}})^2+\mathrm{C}$
B $\frac{2}{3}(\log \sqrt{\mathrm{x}})^2+\mathrm{C}$
C $\frac{2}{3}(\log x)^2+C$
D $\frac{1}{3}(\log x)^2+C$

The correct answer is:
$\frac{1}{3}(\log \sqrt{\mathrm{x}})^2+\mathrm{C}$

Solution:
Hints : $\mathrm{x}=\mathrm{t}^2 \Rightarrow \int \frac{\ell \mathrm{nt}}{3 \mathrm{t}^2}(2 \mathrm{tdt})=\frac{2}{3} \int \frac{\ell \mathrm{nt}}{\mathrm{t}} \mathrm{dt}=\frac{2}{3} \frac{(\ell \mathrm{nt})^2}{2}+\mathrm{c}=\frac{(\ln \sqrt{\mathrm{x}})^2}{3}+\mathrm{c}$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2010
4 boys and 2 girls occupy seats in a row at random. Then the probability that the two girls occupy seats side by side is

Options:
A \(\frac{1}{2}\)
B \(\frac{1}{4}\)
C \(\frac{1}{3}\)
D \(\frac{1}{6}\)

The correct answer is:
\(\frac{1}{3}\)

Solution:
\(\begin{gathered} \text {Hints: }
\mathrm{n}(\mathrm{e})=\lfloor 5 \cdot\lfloor 2 \\
\mathrm{n}(\mathrm{s})=\lfloor 6 \\
\mathrm{p}=\frac{\lfloor 5 \cdot \lfloor 2}{\lfloor 6}=\frac{2}{6}=\frac{1}{3}
\end{gathered}\)

Topic: Mathematics, Probability, WBJEE, WBJEE 2011
\(4 f^2\) electronic configuration is found in

Options:
A \(\mathrm{Pr}\)
B \(\mathrm{Pr}^{3+}\)
C \(\mathrm{Nd}^{3+}\)
D \(\mathrm{Pm}^{3+}\)

The correct answer is:
\(\mathrm{Pr}^{3+}\)

Solution:
Hint : \(\operatorname{Pr}(59)=[X e] 4 f^3 6 s^2\)
\(\therefore \mathrm{Pr}^{+3}=[\mathrm{Xe}] 4 \mathrm{f}^2\)

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2023
$5 \mathrm{~m}$ of $0.1 \mathrm{M} ~ \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ is mixed with $10 \mathrm{ml}$ of $0.02 \mathrm{M} ~ \mathrm{KI}$. The amount of $\mathrm{Pbl}_{2}$ precipitated will be about

Options:
A $10^{-2} \mathrm{~mol}$
B $10^{-4} \mathrm{~mol}$
C $2 \times 10^{-4} \mathrm{~mol}$
D $10^{-3} \mathrm{~mol}$

The correct answer is:
$10^{-4} \mathrm{~mol}$

Solution:
Hint:

mole of $\mathrm{Pbl}_{2} \quad \mathrm{PPt}=\frac{0.1}{1000}=1 \times 10^{-4}$

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2020
$7^{2 n}+16 n-1(n \in N)$ is divisible by

Options:
A 65
B 63
C 61
D 64

The correct answer is:
64

Solution:
We have,
$7^{2 n}+16 n-1$
Put $n=1,$ we get
$\begin{aligned} 7^{2}+16-1 & \\ =& 49+15 \\ =& 64 \end{aligned}$
which is divisible by 64.

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2019
${ }_{11} Na $^{24}$ is radioactive and it decays to

Options:
A ${ }_{9} \mathrm{F}^{20}$ and $\alpha$ -particles
B ${ }_{13} \mathrm{Al}^{24}$ and positron
C ${ }_{11} \mathrm{Na}^{23}$ and neutron
D ${ }_{12} \mathrm{Mg}^{24}$ and $\beta$ - particles

The correct answer is:
${ }_{12} \mathrm{Mg}^{24}$ and $\beta$ - particles

Solution:
$_{11} \mathrm{Na}^{24} \longrightarrow _{12} \mathrm{Mg}^{24}+_{-1}\beta^{0}$

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2012
\(12 \mathrm{mC}\) and \(6 \mathrm{mC}\) charges are given to the two conducting plates having same cross-sectional area and placed face to face close to each other as shown in the figure. The resulting charge distribution in \(\mathrm{mC}\) on surface \(A, B, C\) and \(D\) are respectively,

Options:
A \(9,3,-3,9\)
B \(3,9,-9,3\)
C \(6,6,-6,12\)
D \(6,6,3,3\)

The correct answer is:
\(9,3,-3,9\)

Solution:
Hint :

Topic: Physics, Capacitance, WBJEE, WBJEE 2023
1,4 -dimethylbenzene on heating with anhydrous $\mathrm{AlCl}_{3}$ and HCl produces

Options:
A 1,2 -dimethylbenzene
B 1,3 -dimethylbenzene
C $1,2,$ 3-trimethylbenzene
D ethylbenzene

The correct answer is:
1,3 -dimethylbenzene

Solution:

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2015
\({ }^{15} \mathrm{C}_3+{ }^{15} \mathrm{C}_5+\ldots \ldots+{ }^{15} \mathrm{C}_{15}=\)

Options:
A \(2^{14}\)
B \(2^{14}-15\)
C \(2^{14}+15\)
D \(2^{14}-1\)

The correct answer is:
\(2^{14}-15\)

Solution:
Hints: \({ }^{15} \mathrm{C}_3+{ }^{15} \mathrm{C}_5+ \ldots\ldots+{ }^{15} \mathrm{C}_{15}=2^{14}-{ }^{15} \mathrm{C}_1=2^{14}-15\)

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2011
$19 \mathrm{~g}$ of water at $30^{\circ} \mathrm{C}$ and $5 \mathrm{~g}$ of ice at $-20^{\circ} \mathrm{C}$ are mixed together in a calorimeter. What is the final temperature of the mixture? Given specific heat of ice $=0.5 \mathrm{cal} \mathrm{g}^{-1}\left({ }^{\circ} \mathrm{C}\right)^{-1}$ and latent heat of fusion of ice $=80 \mathrm{cal} \mathrm{g}^{-1}$

Options:
A $0^{\circ} \mathrm{C}$
B $-5^{\circ} \mathrm{C}$
C $5^{\circ} \mathrm{C}$
D $10^{\circ} \mathrm{C}$

The correct answer is:
$5^{\circ} \mathrm{C}$

Solution:
$5 \times .5 \times 20+5 \times 80+5 t=19 \times 1 \times(30-t)$
$\mathrm{t}=5^{\circ} \mathrm{C}$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2009
$20 \mathrm{mL~} 0 .1$ (N) acetic acid is mixed with $10 \mathrm{mL}$ $0.1(\mathrm{N})$ solution of NaOH. The pH of the resulting solution is $\left(p K_{a}\right.$ of acetic acid is $\left.4.74\right)$

Options:
A 3.74
B 4.74
C 5.74
D 6.74

The correct answer is:
4.74

Solution:
From Henderson's equation,
$$
\begin{aligned}
p H &=p K_{a}+\log \frac{\left.\mid C H_{j} C O O N a\right]}{\left|C H_{3} C O O H\right|} \\
&=4.74+\log \frac{1}{1}=4.74
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2012
$2.5 \mathrm{ml}$ of $0.4(\mathrm{M})$ weak monoacidic base $\left(\mathrm{k}_{\mathrm{b}}=1 \times 10^{-12}\right.$ at $\left.25^{\circ} \mathrm{C}\right)$ is titrated with $\frac{2}{15}(\mathrm{M}) \mathrm{HCl}$ in water at $25^{\circ} \mathrm{C}$. The concentration of $\mathrm{H}^{+}$at equivalence point is $\left(\mathrm{k}_{\mathrm{w}}=1 \times 10^{-14}\right.$, at $\left.25^{\circ} \mathrm{C}\right)$,

Options:
A $3.7 \times 10^{-13}$ (M)
B $3.2 \times 10^{-7}(\mathrm{M})$
C $3.2 \times 10^{-2}(\mathrm{M})$
D $2.7 \times 10^{-2}(\mathrm{M})$

The correct answer is:
$3.2 \times 10^{-2}(\mathrm{M})$

Solution:
$\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2}$ (At equivalence point)
$0.4 \times 2.5=\frac{2}{15} \times \mathrm{V}_{2}$
$\mathrm{BOH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{B} \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(I)$
$\therefore \mathrm{V}_{2}=\frac{0.4 \times 2.5}{\frac{2}{15}}=\frac{0.4 \times 2.5}{2} \times 15=7.5 \mathrm{ml}$
$\therefore$ Concentration of salt $=\frac{0.4 \times 2.5}{2.5+7.5}$
or, Salt of weak base and strong acid = $0.1$
$\therefore\left[\mathrm{H}^{+}\right]=\mathrm{Ch}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \mathrm{C}}{\mathrm{K}_{\mathrm{b}}}}=\sqrt{\frac{10^{-14} \times 0.1}{10^{-12}}}=\sqrt{10^{-3}}=3.2 \times 10^{-2} \mathrm{M}$
as $C h^{2}=\frac{K_{w}}{K_{b}} \quad \therefore h=\sqrt{\frac{K_{w}}{K_{b} C}}$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2021
27 drops of mercury coalesce to form a bigger drop. What is the relative increase in surface energy?

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

The correct answer is:
$-\frac{2}{3}$

Solution:
Let the initial radius $r$ and finial radius $R . \frac{4}{3} \pi R^3=27 \times \frac{4}{3} \pi r^3 \Rightarrow R=3 r$
$$
U_i=\left(S 4 \pi r^2\right) \times 27 \quad U_f=S 4 \pi 9 r^2
$$
Relative increacase $=\left(\frac{U_f-U_i}{U_i}\right)=\frac{9-27}{27}=-\frac{18}{27}=-\frac{2}{3}$

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2022
$\left({ }_{32} \mathrm{Ge}^{76},{ }_{34} \mathrm{Se}^{76}\right) \text { and } \left.{(}_{14} \mathrm{Si}^{30},{ }_{16} \mathrm{S}^{32}\right) \text { are examples of }$

Options:
A isotopes and isobars
B isobars and isotones
C isotones and isotopes
D isobars and isotopes

The correct answer is:
isobars and isotones

Solution:
${ }_{32} \mathrm{Ge}^{76}$ and $\mathrm{sq} \mathrm{Se}^{76}$ have the same mass number but different atomic number, so they are isobars. Number of neutrons,
in $_{14} e^{\mathrm{Si}^{30}}=30-14=16$
in $_{16} e^{\mathrm{Si}^{32}}=32-16=16$
Because of the presence of same number of neutrons, $_{14} \mathrm{Si}^{30}$ and ${ }_{16} \mathrm{S}^{32}$ are called isotones.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2014
\(63 \mathrm{~g}\) of a compound (Mol. Wt. \(=126)\) was dissolved in \(500 \mathrm{~g}\) distilled water. The density of the resultant solution as \(1.126 \mathrm{~g} / \mathrm{ml}\). The molarity of the solution is

Options:
A \(1.25 \mathrm{~M}\)
B \(1.0 \mathrm{~M}\)
C \(0.75 \mathrm{~M}\)
D \(1.1 \mathrm{~M}\)

The correct answer is:
\(1.0 \mathrm{~M}\)

Solution:
Hint : Mass of compound (solute) \(=63 \mathrm{~g}\)
Mole of compound \(=\frac{63}{126}=\frac{1}{2}\) mole
\(\begin{aligned}
\text{Mass of solution}& =\text { Mass of solute }+ \text { Mass of solvent } \\
& =63+500 \\
& =563 \mathrm{~g}
\end{aligned}\)
Volume of solution \(=\frac{\text { Mass }}{\text { Density }}=\frac{563}{1.126} \mathrm{ml}\)
Molarity \(=\frac{\text { mole of compound }}{\text { volume of solution }(\text { inL })}\)
\(\begin{aligned} & =\frac{1 / 2 \times 1000}{563 / 1.126} \\ & =\frac{1.126 \times 1000}{2 \times 563}=1\end{aligned}\)

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2023