datavorous/entrance-exam-dataset huggingface dataset:
\( 1 \mathrm{Mole} \) of \( \mathrm{CO}_{2} \) gas at \( 300 \mathrm{~K} \) expanded under the reversible adiabatic condition such that its volume becomes \( 27 \) times. The magnitude of work done (in \( \mathrm{kJ} / \mathrm{mol} \) ) is: (Given \( \gamma=1.33 \) and \( \mathrm{C}_{\mathrm{v}}=25.10 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1} \) for \( \mathrm{CO}_{2} \) ) report your answer by rounding it up to nearest whole number

Options:

The correct answer is:
-5

Solution:
Number of moles =1
T1=300 K
V2=27V1
TV1γ-1=T2V2γ-1
T1T2=V2V1γ–1
T2=30012713
T2=100 K
Adiabatic condition, q=0;  ΔE=w=nCvT2–T1
w=1×25.10100–300=-5020 J/mole
w=-5.02  kJ/mole

Topic: Chemistry, Thermodynamics (C), JEE Main
1-methyl ethylene oxide when treated with an excess of HBr produces

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2020 (07 Jan Shift 1)
1 mol of an octahedral metal complex with formula MCl3·2 L on reaction with excess of AgNO3 gives 1 mol of AgCl. The denticity of Ligand L is___________ (Integer answer)

Options:

The correct answer is:
2

Solution:
As 1 mole complex give 1 mole mole AgCl precipitate. So only one Cl ion is in ionisation sphere.

As complex is octahedral so denticity of Ligand L is =2.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (27 Aug Shift 1)
1 molal aqueous solution of an electrolyte A2 B3 is 60% ionised. The boiling point of the solution at 1 atm is _____ K. (Rounded-off to the nearest integer) [Given Kb for H2O=0.52 K kg mol-1]

Options:

The correct answer is:
375

Solution:
ΔTb=iKbm

=1+4α×0.52×1

=3.4×0.52×1=1.768

Tb=1.768+373.15=374.918 K

=375 K

Hence answer is 375

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (25 Feb Shift 1)
1 mole of a gas with $\gamma=7 / 5$ is mixed with 1 mole of a gas with $\gamma=5 / 3$, then the value of $\gamma$ for the resulting mixture is

Options:
A
7/5

B
2/5

C
24/16

D
12/7

The correct answer is:
24/16

Solution:
$\gamma$ for resulting mixture should be in between $7 / 5$ and $5 / 3$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2002
1 Mole of CO2 gas at 300 K expanded under the reversible adiabatic condition such that its volume becomes 27 times. The magnitude of work done inâkJ/mol is:
(Given Î³=1.33 and Cv=25.10âJmolâ1Kâ1âforâCO2 )
report your answer by rounding it up to nearest whole number

Options:

The correct answer is:
5

Solution:
Number of moles =1
T1=300 K
V2=27V1
TV1Î³-1=T2V2Î³-1
T1T2=V2V1râ1
T2=30012713
T2=100 K
Adiabatic condition, q=0;ââÎE=w=nCvT2âT1
w=1Ã25.10100â300=-5020âJ/mole
w=-5.02ââkJ/mole

Topic: Chemistry, Thermodynamics (C), JEE Main
0.001 mole of $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_3\right)\left(\mathrm{SO}_4\right)\right]$ was passed through a cation exchanger and the acid coming out of it required 20 mL of 0.1 M NaOH for neutralisation. Thus, the complex is

Options:
A $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_3\right)\right] \mathrm{SO}_4$
B $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{SO}_4\right)\right] \mathrm{NO}_3$
C $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\right] \mathrm{NO}_3 \cdot \mathrm{SO}_4$
D None of the above

The correct answer is:
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_3\right)\right] \mathrm{SO}_4$

Solution:
Millimoles of acid required for neutralisation
$\begin{aligned} & =20 \times 0.1=2 \text { millimoles } \\ & =0.002 \text { moles }\end{aligned}$
If 0.001 mole is neutralising 0.002 moles of base, thus, 0.001 mole of base remains unreacted. Hence, acid coming out should be basic.
$\begin{aligned} & {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_3\right)\right] \mathrm{SO}_4 \longrightarrow} \\ & {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_3\right]^{2+}+\mathrm{SO}_4^{2-}\right.}\end{aligned}$

Topic: Chemistry, Coordination Compounds, JIPMER, JIPMER 2015
1 mole of equimolar mixture of Fe 2 (C 2 O 4 ) 3 and FeC 2 O 4 requires X moles of KMnO 4 in acid medium for complete reaciton. The value of X is:

Options:
A 0.9
B 0.6
C 1.2
D 0.8

The correct answer is:
0.9

Solution:
Fe 2 C 2 O 4 3 â¶ 2 Fe 3 + + 6 CO 2 + 6 e - n Factor = 6 Fe C 2 O 4 â¶ Fe 3 + + 2 CO 2 + 3 e - n Factor = 3
Meqs of Fe 2 (C 2 O 4 ) 3  +  Fe(C 2 O 4 )  = 0.5 × 6 + 0.5 × 3
= 4.5 Eqs.
= 4.5 Eqs.of KMnO 4
=  4.5 5  = 0.9moles of KMnO 4 .

Topic: Chemistry, Redox Reactions, JEE Main
1 mole of N2O4(g) at 300 K is kept in a closed container under 1 atm . It is heated to 600 K when 20% by mole of N2O4(g) decomposes to NO2(g). The resultant pressure is

Options:
A 1.2 atm
B 2.4 atm
C 2.0 atm
D 1.0 atm

The correct answer is:
2.4 atm

Solution:
N 2 O 4 (g) ⇌ 2NO 2 (g) Initial mole 1 0 At equi. (1-0.20) mole 0.40 mole
n total â=â1.2
P total â=â n total RT V â=â2.4âatm
as P 2 P 1 â=â n 2 T 2 n 1 T 1

Topic: Chemistry, Chemical Equilibrium, JEE Main
1 mole of PbS is oxidised by X moles of O3 to get Y moles of O2. X+Y=

Options:

The correct answer is:
8

Solution:
Given that, 1 mole of PbS is oxidised by X moles of O3 to get Y moles of O2.

The balanced chemical equation can be written as,

PbS+4O3→PbSO4+4O2

Here x = 4, y = 4

Hence,

X+Y=4+4=8

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 2)
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, JEE Main
1 mole of rigid diatomic gas performs a work of Q5 when heat Q is supplied to it. The molar heat capacity of the gas during this transformation is xR8. The value of x is
[R universal gas constant]

Options:

The correct answer is:
25

Solution:
Q=ΔU+W

Q=ΔU+Q5

ΔU=4Q5

nCvΔT=45nCΔT

C=258R

x=25

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (26 Feb Shift 2)
(1) $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}), \mathrm{K}_1$
(2) $\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}), \mathrm{K}_2$
(3) $\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g}), \mathrm{K}_3$
The equation for the equilibrium constant of the reaction
$$
2 \mathrm{NH}_3(\mathrm{~g})+\frac{5}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \text {, }
$$
$\left(K_4\right)$ in terms of $\mathrm{K}_1, \mathrm{~K}_2$ and $\mathrm{K}_3$ is :

Options:
A
$\frac{\mathrm{K}_1 \cdot \mathrm{K}_2}{\mathrm{~K}_3}$

B
$\frac{\mathrm{K}_1 \cdot \mathrm{K}_3^2}{\mathrm{~K}_2}$

C
$\mathrm{K}_1 \mathrm{~K}_2 \mathrm{~K}_3$

D
$\frac{\mathrm{K}_2 \cdot \mathrm{K}_3^3}{\mathrm{~K}_1}$

The correct answer is:
$\frac{\mathrm{K}_2 \cdot \mathrm{K}_3^3}{\mathrm{~K}_1}$

Solution:
To calculate the value of $\mathrm{K}_4$ in the given equation we should apply :
eqn. (2) $+$ eqn. $(3) \times 3-$ eqn. (1)
hence $\mathrm{K}_4=\frac{\mathrm{K}_2 \mathrm{~K}_3^3}{\mathrm{~K}_1}$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2013 (23 Apr Online)
$(1+\Lambda)^{n} f(a)$ is equal to

Options:
A $f(a+h)$
B $f(a+2 h)$
C $f(a+n h)$
D $f(a+(n-1) h)$

The correct answer is:
$f(a+n h)$

Solution:
$\begin{aligned}(1+\Delta)^{n} f(a) &=E^{n} f(a) \\ &=f(a+n h) \end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
1% of 1012 Hz of a satellite link was used for telephony. The number of channels or subscribers if each channel is of 8 kHz are

Options:
A 2.5×107
B 1.25×106
C 2.5×108
D 1.25×108

The correct answer is:
1.25×106

Solution:
According to the question
1% of 1012 Hz=1010 Hz
Number of channels =10108Ã103=1.25Ã106

Topic: Physics, Communication System, JEE Main
$\int \frac{d x}{1-\sin x}=$

Options:
A $x+\cos x+c$
B $1+\sin x+c$
C $\sec x-\tan x+c$
D $\sec x+\tan x+c$

The correct answer is:
$\sec x+\tan x+c$

Solution:
$\begin{aligned} & \int \frac{d x}{1-\sin x}=\int \frac{(1+\sin x)}{1-\sin ^2 x} d x \\ & =\int \sec ^2 x d x+\int \tan x \cdot \sec x d x=\tan x+\sec x+c .\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
1st member of ketone family and its next homologue reacts with NH2OH to form oxime. The number of products (including stereoisomers) formed are:

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Products are:

product 2 shows GI
Hence, total number of products = 3.

Topic: Chemistry, General Organic Chemistry, JEE Main
$1 \%$ solution of $\mathrm{Ca}\left(\mathrm{NO}_3\right)_2$ has freezing point

Options:
A $0^{\circ} \mathrm{C}$
B Less than $0^{\circ} \mathrm{C}$
C Greater than $0^{\circ} \mathrm{C}$
D None of the above

The correct answer is:
Less than $0^{\circ} \mathrm{C}$

Solution:
Freezing point is lowered on addition of solute in it..

Topic: Chemistry, Solutions, JEE Main
1. The function $f(x)=x^3-6 x^2+a x+b$ satisfy the conditions of Rolle's theorem in $[1,3]$. The values of $a$ and $b$ are

Options:
A $11. -6$
B $-6,11$
C $-11,6$
D $6,-11$

The correct answer is:
$11. -6$

Solution:
$f(1)=f(3) \Rightarrow a+b-5=3 a+b-27 \Rightarrow a=11$, which is given in option (1) only.

Topic: Mathematics, Application of Derivatives, JEE Main
$\int \frac{e^{\tan ^{-1} x}}{1+x^2} d x=$

Options:
A $\log \left(1+x^2\right)+c$
B $\log e^{\tan ^{-1} x}+c$
C $e^{\tan ^{-1} x}+c$
D $\tan ^{-1} e^{\tan ^{-1} x}+c$

The correct answer is:
$e^{\tan ^{-1} x}+c$

Solution:
Putting $\quad t=\tan ^{-1} x \Rightarrow d t=\frac{1}{1+x^2} d x$, we get
$\int \frac{e^{\tan ^{-1} x}}{1+x^2} d x=\int e^t d t=e^t+c=e^{\tan ^{-1} x}+c .$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int \frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^{2}}} d x=$

Options:
A $\frac{2}{5}\left(\sin ^{-1} x\right)^{\frac{5}{2}}+c$
B $\frac{2}{5}\left(\cos ^{-1} x\right)^{\frac{5}{2}}+c$
C $\frac{5}{2}\left(\cos ^{-1} x\right)^{\frac{5}{2}}+c$
D $\frac{5}{2}\left(\sin ^{-1} x\right)^{\frac{5}{2}}+c$

The correct answer is:
$\frac{2}{5}\left(\sin ^{-1} x\right)^{\frac{5}{2}}+c$

Solution:
Let $I=\int \frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^{2}}} d x$
Put $\sin ^{-1} x=t \Rightarrow \frac{1}{\sqrt{1-x^{2}}} d x=d t$
$\begin{aligned} \therefore I &=\int t^{\frac{3}{2}} d t \\ &=\frac{t^{\frac{5}{2}}}{\left(\frac{5}{2}\right)}=\frac{2}{5} t^{\frac{5}{2}}+c=\frac{2}{5}\left(\sin ^{-1} x\right)^{\frac{5}{2}}+c \end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x$ 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, JEE Main
$\int \frac{x+1}{\sqrt{1+x^2}} d x=$

Options:
A $\sqrt{1+x^2}+\tan ^{-1} x+c$
B $\sqrt{1+x^2}-\log \left\{x+\sqrt{1+x^2}\right\}+c$
C $\sqrt{1+x^2}+\log \left\{x+\sqrt{1+x^2}\right\}+c$
D $\sqrt{1+x^2}+\log (\sec x+\tan x)+c$

The correct answer is:
$\sqrt{1+x^2}+\log \left\{x+\sqrt{1+x^2}\right\}+c$

Solution:
$\int \frac{x+1}{\sqrt{x^2+1}} d x=\int \frac{x}{\sqrt{x^2+1}} d x+\int \frac{1}{\sqrt{x^2+1}} d x$
Put $x^2+1=t \Rightarrow 2 x d x=d t$, then it reduce to $\frac{1}{2} \int \frac{d t}{t^{1 / 2}}+\int \frac{1}{\sqrt{x^2+1}} d x=\frac{1}{2} \cdot 2 \cdot t^{1 / 2}+\log \left(x+\sqrt{x^2+1}\right)+c$
$=\left(x^2+1\right)^{1 / 2}+\log \left(x+\sqrt{x^2+1}\right)+c .$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^2+c$, then $k$ is equal to

Options:
A $\frac{1}{50}$
B $-\frac{1}{50}$
C $\frac{1}{100}$
D $-\frac{1}{100}$

The correct answer is:
$\frac{1}{100}$

Solution:
Let $I=\int x^{49} \frac{\tan ^{-1}\left(x^{50}\right)}{1+\left(x^{50}\right)^2} d x$
Let $x^{50}=t \Rightarrow 50 x^{49} d x=d t$
$\therefore \quad I=\frac{1}{50} \int \frac{\tan ^{-1} t}{1+t^2} d t$
Let $\tan ^{-1} t=u \Rightarrow \frac{1}{1+t^2} d t=d u$
$\begin{aligned}
\therefore \quad I & =\frac{1}{50} \int u d u=\frac{u^2}{100}+c \\
& =\frac{\left(\tan ^{-1} x^{50}\right)^2}{100}+c
\end{aligned}$
But $I=k\left(\tan ^{-1} x^{50}\right)^2+c$(given)
$\begin{array}{rlrl}\Rightarrow & k\left(\tan ^{-1} x^{50}\right)^2+c & =\frac{1}{100}\left(\tan ^{-1} x^{50}\right)^2+c \\ \Rightarrow & & k & =\frac{1}{100}\end{array}$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x$ is equal to :

Options:
A $(1+x) e^{x+x^{-1}}+C$
B $(x-1) e^{x+x^{-1}}+C$
C $-x e^{x+x^{-1}}+C$
D $x e^{x+x^{-1}}+C$

The correct answer is:
$x e^{x+x^{-1}}+C$

Solution:
$\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x$
$=\int e^{x+x^{-1}} d x+\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x$
$=e^{x+x^{-1}} \int d x-\int\left[\frac{d}{d x}\left(e^{x+x^{-1}}\right)\right] x d x$ $+\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x$
$=x e^{x+x^{-1}}-\int\left(1-\frac{1}{x^2}\right) x e^{x+x^{-1}} d x$ $+\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x$
$=x e^{x+x^{-1}}-\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x$
$+\int\left(x-\frac{1}{x}\right) e^{x+x^{-1}} d x+C$
$=x e^{x+x^{-1}}+C$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int \frac{d x}{1+x+x^2+x^3}=$

Options:
A $\log \sqrt{1+x}-\frac{1}{2} \log \sqrt{1+x^2}+\frac{1}{2} \tan ^{-1} x+c$
B $\log \sqrt{1+x}-\log \sqrt{1+x^2}+\tan ^{-1} x+c$
C $\log \sqrt{1+x^2}-\log \sqrt{1+x}+\frac{1}{2} \tan ^{-1} x+c$
D $\log \sqrt{1+x}+\tan ^{-1} x+\log \sqrt{1+x^2}+c$

The correct answer is:
$\log \sqrt{1+x}-\frac{1}{2} \log \sqrt{1+x^2}+\frac{1}{2} \tan ^{-1} x+c$

Solution:
$\begin{aligned} & \int \frac{d x}{1+x+x^2+x^3}=\int \frac{d x}{(1+x)\left(1+x^2\right)} \\
& =\frac{1}{2} \int \frac{1}{1+x^2} d x+\frac{1}{2} \int \frac{1}{1+x} d x-\frac{1}{2} \int \frac{x}{1+x^2} d x\end{aligned}$
\(\Rightarrow\) As, assume,
\(\frac{1}{(1+x)\left(1+x^2\right)}=\frac{A}{(1+x)}+\frac{B x+c}{\left(1+x^2\right)}\)
By solving it,
\(\begin{aligned}
& A=\frac{1}{2}, B=\frac{-1}{2}, \\
& c=\frac{1}{2} \\
& =\frac{1}{2} \tan ^{-1} x+\log \sqrt{1+x}-\frac{1}{2} \log \sqrt{1+x^2}+c
\end{aligned}\)

Topic: Mathematics, Indefinite Integration, JEE Main
$\int \frac{d x}{\sqrt{1+x}+\sqrt{x}}=$

Options:
A $\frac{2}{3}(1+x)^{2 / 3}-\frac{2}{3} x^{2 / 3}+c$
B $\frac{3}{2}(1+x)^{2 / 3}+\frac{3}{2} x^{2 / 3}+c$
C $\frac{3}{2}(1+x)^{3 / 2}+\frac{3}{2} x^{3 / 2}+c$
D $\frac{2}{3}(1+x)^{3 / 2}-\frac{2}{3} x^{3 / 2}+c$

The correct answer is:
$\frac{2}{3}(1+x)^{3 / 2}-\frac{2}{3} x^{3 / 2}+c$

Solution:
$\begin{aligned} & \int \frac{d x}{\sqrt{1+x}+\sqrt{x}}=\int\left[\frac{(x+1)-x}{\sqrt{1+x}+\sqrt{x}}\right] d x \\ & \int(\sqrt{x+1}-\sqrt{x}) d x=\frac{(x+1)^{3 / 2}}{3 / 2}-\frac{x^{3 / 2}}{3 / 2}+c \\ & =\frac{2}{3}\left[(x+1)^{3 / 2}-x^{3 / 2}\right]+c=\frac{2}{3}(x+1)^{3 / 2}-\frac{2}{3} x^{3 / 2}+c .\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int_1^x \frac{\log x^2}{x} d x=$

Options:
A $(\log x)^2$
B $\frac{1}{2}(\log x)^2$
C $\frac{\log x^2}{2}$
D None of these

The correct answer is:
$(\log x)^2$

Solution:
$\begin{aligned} & I=\int_1^x \frac{2 \log x}{x} d x \\ & \text { Let } \log x=t \Rightarrow \frac{d x}{x}=d t \\ & \therefore I=2 \int_0^{\log x} t d t=2\left[\frac{t^2}{2}\right]_0^{\log x}=(\log x)^2\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
1-Bromo-3-chloro cyclobutane when treated with 2-equivalents of Na in presence of ether which of the following product will be formed?

Options:
A
B
C
D

The correct answer is:

Solution:
It is an internal Wurtz reaction. The given dihalo compound forms a bicyclo compound.

Topic: Chemistry, Hydrocarbons, JEE Main
1-bromo-3-chlorocyclobutane is treated with two equivalents of Na in the presence of dry ether. Which of the following will be formed?

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
1-Bromo-2,2-dimethylcyclohexane on treatment with methanol gives

Options:
A
B
C
D all of these

The correct answer is:
all of these

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
1-Methylcyclohexene on addition of HCl produces

Options:
A 1-chloro-1-methylcyclohexane
B (±)-trans-2-chloro-1-methylcyclohexane
C (±) cis-2-chloro-1-methylcyclohexane
D 1-chloro-2-methylcyclohexane

The correct answer is:
1-chloro-1-methylcyclohexane

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
1-Methylethylene oxide on reaction with an excess of HBr produces

Options:
A
B
C
D

The correct answer is:

Solution:
This reaction occurs by S N 1 mechanism as follows

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
1st polaroid 2nd polaroid
Find the final intensity of light \( \left(I^{\prime \prime}\right) \), if the angle between the axes of two polaroids is \( 60^{\circ} \)

Options:
A \( \frac{3 I_{0}}{2} \)
B \( \frac{I_{0}}{2} \)
C \( \frac{I_{0}}{4} \)
D \( \frac{I_{0}}{8} \)

The correct answer is:
\( \frac{I_{0}}{8} \)

Solution:
From first polaroid the unpolarised light will become polarized with half the intensity.
So, I′=I02
From second polaroid
I′′=I′cos2⁡θ=I02cos2⁡60=I0 2 14=I08

Topic: Physics, Wave Optics, JEE Main
$20^{2-3 x^2}=(40 \sqrt{5})^{3 x^2-2}$, then $x$ is equal to

Options:
A $\pm \sqrt{\frac{3}{2}}$
B $\pm \sqrt{\frac{2}{3}}$
C $\pm \sqrt{\frac{4}{3}}$
D $\pm \sqrt{\frac{5}{4}}$

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

Solution:
We have,
$(20)^{2-3 x^2}=(40 \sqrt{5})^{3 x^2-2}$
On taking log on both sides, we get
$\left(2-3 x^2\right) \log 20=\left(3 x^2-2\right) \log 40 \sqrt{5}$
$\begin{aligned} & \Rightarrow\left(2-3 x^2\right)[2 \log 2+\log 5] \\ & =\left(3 x^2-2\right)\left(3 \log 2+\frac{3}{2} \log 5\right) \\ & \Rightarrow\left(3 x^2-2\right)\left[3 \log 2+\frac{3}{2} \log 5+2 \log 2+\log 5\right]=0 \\ & \Rightarrow\left(3 x^2-2\right)=0 \Rightarrow x^2=\frac{2}{3} \\ & \Rightarrow x= \pm \frac{\sqrt{2}}{3}\end{aligned}$

Topic: Mathematics, Quadratic Equation, JEE Main
2.0 g of H2 gas is adsorbed on 2.5 g of platinum powder at 300 K and 1 bar pressure. The volume of the gas adsorbed per gram of the adsorbent is mL

(Given : R=0.083 L bar K-1 mol-1)

Options:

The correct answer is:
9960

Solution:
VH2 = nRTPVH2g=220.0833001

∴ Volume of H2g adsorbed on 2.5 Pt=24.9 L

∴ Volume of H2g adsorbed on 1 g of Pt=24.92.5=9.96 L=9960 ml

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2022 (26 Jun Shift 1)
$\left(\frac{-1+i \sqrt{3}}{2}\right)^{20}+\left(\frac{-1-i \sqrt{3}}{2}\right)^{20}=$

Options:
A $20 \sqrt{3} i$
B $1$
C $\frac{1}{2^{19}}$
D $-1$

The correct answer is:
$-1$

Solution:
$\begin{aligned} & \text { As } \frac{-1+i \sqrt{3}}{2}=\omega \text { and } \frac{-1-i \sqrt{3}}{2}=\omega^2 \\ & \therefore(\omega)^{20}+\left(\omega^2\right)^{20}=\omega^{18} \cdot \omega^2+\omega^{39} \cdot \omega=\omega^2+\omega=-1\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
$\sqrt{2+\sqrt{2+2 \cos 4 \theta}}=$

Options:
A $\cos \theta$
B $\sin \theta$
C $2 \cos \theta$
D $2 \sin \theta$

The correct answer is:
$2 \cos \theta$

Solution:
$\begin{aligned}
& \sqrt{2+\sqrt{2+2 \cos 4 \theta}}=\sqrt{2+\sqrt{2.2 \cos ^2 2 \theta}} \\
& =\sqrt{2+2 \cos 2 \theta}=\sqrt{4 \cos ^2 \theta}=2 \cos \theta
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\frac{2}{2 !}+\frac{2+4}{3 !}+\frac{2+4+6}{4 !}+\ldots$ is equal to

Options:
A $e$
B $e^{-1}$
C $e^{-2}$
D $e^{-3}$

The correct answer is:
$e$

Solution:
We have,
$\begin{aligned} T_n & =\frac{2+4+6+\ldots+2 n}{(n+1) !} \\ T_n & =\frac{\sum 2 n}{(n+1) !}=\frac{2 \frac{n(n+1)}{3}}{(n+1) n(n-1) !} \\ & =\frac{1}{(n-1) !}\end{aligned}$
$\begin{aligned} \therefore \quad S & =T_1+T_2+T_3+\ldots \\ & =1+1+\frac{1}{2 !}+\frac{1}{3 !}+\ldots=e\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main
2.2 g of nitrous oxide N2O gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1 mL to 167.75 mL. The change in internal energy of the process, △U is '-x'J. The value of 'x' is _____.

[nearest integer]

(Given: atomic mass of N=14 g mol-1 and of O=16 g mol-1. Molar heat capacity of N2O is 100 JK-1 mol-1)

Options:

The correct answer is:
72

Solution:
Molecular weight of nitrous oxide \(\left(\mathrm{N}_2 \mathrm{O}\right)=44 \mathrm{~g} / \mathrm{mol}\)
Number of moles of \(\mathrm{N}_2 \mathrm{O}=\frac{2.2}{44} \quad 0.05 \mathrm{~mol}\)
Specific heat capacity of \(\mathrm{N}_2 \mathrm{O}=880 \mathrm{~J} / \mathrm{g} . \mathrm{K}\)
\(\therefore\) Heat release on cooling from \(310 \mathrm{~K}\) to \(270 \mathrm{~K}\)
\(q=-2.2 \times 880 \times 40 \mathrm{~K}\)
\(q=-77.44 \mathrm{KJ}\)
Negative sign indicates heat released.
\(\therefore\) Work done by surrounding on system to compares the gas from \(217.1 \mathrm{ml}\) to \(167.75 \mathrm{ml}\) at constant pressure.
It means work done by surrounding is irreversible.
\(\begin{aligned} & \therefore W= P \times \Delta V \\ & =1.01 \times 10^5 \times(167.75 \quad 217.1 \mathrm{ml} l) \\ & =1.01 \times 10^5 \times 0.049 J \\ & W=4.95 k . J \end{aligned}\)
Using 1st Law of thermodynamics
\(\begin{aligned} & \Delta U=q+W \\ & \Delta U=-77.44 K J + 4.95 K J \\ & \Delta U=-72.49 K J \end{aligned}\).
$\therefore x = 72.$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2022 (29 Jun Shift 2)
$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta$ is equal to

Options:
A $0$
B $1$
C $2$
D $-1$

The correct answer is:
$0$

Solution:
Let $I=\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta$
Let $\quad f(\theta)=\log \left(\frac{2-\sin \theta}{2+\sin \theta}\right)$
$\Rightarrow \quad f(-\theta)=\log \left(\frac{2+\sin \theta}{2-\sin \theta}\right)=-\log \left(\frac{2-\sin \theta}{2+\sin \theta}\right)$
$\Rightarrow \quad f(-\theta)=-f(\theta)$
$\therefore$ It is an odd function
$\therefore \quad I=\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta=0$.

Topic: Mathematics, Definite Integration, JEE Main
â«-Ï2Ï2logâ¡2-sinâ¡ x2+sinâ¡ xdx=

Options:
A 1
B 3
C 2
D 0

The correct answer is:
0

Solution:
Let I=â«-Ï2Ï2logâ¡ 2-sinâ¡ x2+sinâ¡ xdx
As given function f( x )=log( 2âsinx 2+sinx ) is odd.
âI= â« f( x )dx=0

Topic: Mathematics, Definite Integration, JEE Main
$\int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x=$

Options:
A $\frac{2}{15}$
B $\frac{4}{15}$
C $\frac{6}{15}$
D $\frac{8}{15}$

The correct answer is:
$\frac{4}{15}$

Solution:
$\begin{aligned} & \int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x \\ & =\int_{-\pi / 2}^{\pi / 2} \sin ^3 x \cos ^2 x d x+\int_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^3 x d x \\ & =0+2 \int_0^{\pi / 2} \sin ^2 x \cos ^3 x d x=0+2 \times \frac{2}{15}=\frac{4}{15}\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_{-\pi / 2}^{\pi / 2} \sin ^2 x d x=$

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

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

Solution:
$\int_{-\pi / 2}^{\pi / 2} \sin ^2 x d x=2 \int_0^{\pi / 2} \sin ^2 x d x=2 \frac{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)}{2 \Gamma\left(\frac{2+2}{2}\right)}=\frac{\pi}{2}$

Topic: Mathematics, Definite Integration, JEE Main