datavorous/entrance-exam-dataset huggingface dataset:
∫-π2π2sinx-xdx=, where . denotes GIF

Options:
A 3(1-cos1)+sin2-sin1
B cos2-sin2
C 3(1-cos1)+cos2-sin1
D 0

The correct answer is:
3(1-cos1)+sin2-sin1

Solution:
I=∫-π2π2sinx-xdx, where · denotes G.I.F.

⇒I=∫-π2-1sin2+xdx+∫-10sin1+xdx+∫01sinxdx+∫1π2sinx-1dx

⇒I=-cosx+2-π2-1-cosx+1-10-cosx01-cosx-11π2

⇒I=-cos1-cos-π2+2-cos1-cos0-cos1-cos0-cosπ2-1-cos0

⇒I=-cos1+sin2-cos1+1-cos1+1-sin1+1

⇒I=31-cos1+sin2-sin1

Topic: Mathematics, Definite Integration, JEE Main
2, 2-dichloropropane $\stackrel{A q . \mathrm{KOH}}{\longrightarrow} A$
$\frac{\text { Clemmensen }}{\text { . reduction }} B,B$ is

Options:
A propanol
B propene
C propane
D ethane

The correct answer is:
propane

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
$\int_{2}^{3} \frac{x}{x^{2}-1} d x=$

Options:
A $\left(\frac{-1}{2}\right) \log \left(\frac{8}{3}\right)$
B $\left(\frac{1}{2}\right) \log \left(\frac{8}{3}\right)$
C $\left(\frac{-1}{3}\right) \log \left(\frac{8}{3}\right)$
D $\left(\frac{1}{3}\right) \log \left(\frac{8}{3}\right)$

The correct answer is:
$\left(\frac{1}{2}\right) \log \left(\frac{8}{3}\right)$

Solution:
$\int \frac{1}{x^{2}-1} d x$
$=\int_{2}^{1} \frac{(x-1)+1}{(x-1)(x+1)} d x-\int_{2}^{1} \frac{d x}{x+1}+\int_{2}^{1}\left(\frac{1}{x-1}-\frac{1}{x+1}\right) \frac{1}{2} d x$
$=|\log | x+1||_{2}^{1}+\frac{1}{2}\left[\left.\log \left|\frac{x-1}{x+1}\right|\right|_{2} ^{1}\right.$
$=(\log 4-\log 3)+\frac{1}{2}\left|\log \left(\frac{2}{4}\right)-\log \left(\frac{1}{3}\right)\right|$
$\left.=\log \left(\frac{4}{3}\right)+\frac{1}{2} \mid \frac{\log \left(\frac{1}{2}\right)}{1}\right)-\log \left(\frac{4}{3}\right)+\log \left|\frac{3}{2}\right|^{2}$
$=\log \left(\frac{4}{3} x \sqrt{\frac{3}{2}}\right)^{1}-\log \left(\frac{8}{3}\right)^{2} \mid-\frac{1}{2} \log \left(\frac{8}{3}\right)$
This problem con ulso be nolved an follows :
$\begin{array}{l}
\int_{0}^{1} \frac{x}{x^{2}-1} d x \\
-\int_{2}^{1} \frac{x}{(x-1)(x+1)} d x-\frac{1}{2} \int_{0}^{1}\left|\frac{1}{x-1}+\frac{1}{x+1}\right| d x
\end{array}$
$=\frac{1}{2}\left\{[\log (x-1)]_{2}^{3}+[\log (x+1)]_{2}^{3}\right\}$
$=\frac{1}{2}\left\{[\log [(x-1)(x+1)]]_{2}^{3}\right\}=\frac{1}{2}\left[\log \left(x^{2}-1\right)\right]_{2}^{3}$
$=\frac{1}{2}[\log (9-1)-\log (4-1)]=\frac{1}{2} \log \left(\frac{8}{3}\right)$

Topic: Mathematics, Definite Integration, JEE Main
$\int_2^3 \frac{d x}{x^2-x}$ is equal to

Options:
A $\log \frac{2}{3}$
B $\log \frac{4}{3}$
C $\log \frac{8}{3}$
D $\log \frac{1}{4}$

The correct answer is:
$\log \frac{4}{3}$

Solution:
We have,
$$
\begin{aligned}
\int_2^3 \frac{d x}{x^2-x} & =\int_2^3 \frac{1}{x(x-1)} d x \\
& =\int_2^3\left[-\frac{1}{x}+\frac{1}{x-1}\right] d x \\
& =\left[\log \frac{x-1}{x}\right]_2^3=\log \frac{2}{3}-\log \frac{1}{2}=\log \frac{4}{3}
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main
$\sqrt{-2} \sqrt{-3}=$

Options:
A $\sqrt{6}$
B $-\sqrt{6}$
C $i \sqrt{6}$
D None of these

The correct answer is:
$-\sqrt{6}$

Solution:
$\sqrt{-2} \sqrt{-3}=i \sqrt{2} i \sqrt{3}=i^2 \sqrt{6}=-\sqrt{6}$

Topic: Mathematics, Complex Number, JEE Main
$2+4+7+11+16+\ldots \ldots$ to $n$ terms $=$

Options:
A $\frac{1}{6}\left(n^2+3 n+8\right)$
B $\frac{n}{6}\left(n^2+3 n+8\right)$
C $\frac{1}{6}\left(n^2-3 n+8\right)$
D $\frac{n}{6}\left(n^2-3 n+8\right)$

The correct answer is:
$\frac{n}{6}\left(n^2+3 n+8\right)$

Solution:
We have $S=2+4+7+11+16+\ldots .+T_n$
Again $S=2+4+7+11+\ldots \ldots+T_{n-1}+T_n$
Subtracting, we get
$0=2+\left\{2+3+4+5+\ldots . .\left(T_n-T_{n-1}\right)\right\}-T_n$
$T_n=2+\frac{1}{2}(n-1)(4+\{n-2) 1\}=\frac{1}{2}\left(n^2+n+2\right)$
Now, $S=\Sigma T_n=\frac{1}{2} \Sigma\left(n^2+n+2\right)=\frac{1}{2}\left(\Sigma n^2+\Sigma n+2 \Sigma 1\right)$
$=\frac{1}{2}\left\{\frac{1}{6} n(n+1)(2 n+1)+\frac{1}{2} n(n+1)+2 n\right\}$
$=\frac{n}{12}\{(n+1)(2 n+1+3)+12\}$
$=\frac{n}{6}\{(n+1)(n+2)+6\}=\frac{n}{6}\left(n^2+3 n+8\right)$

Topic: Mathematics, Sequences and Series, JEE Main
2, 4-DNP test can be used to identify :

Options:
A aldehyde
B amine
C halogens
D ether

The correct answer is:
aldehyde

Solution:
2, 4-DNP test is useful for the identification of carbonyl compounds.

The 2,4-dinitrophenylhydrazine can be used as a process for the identification of aldehydes and ketones.

The formed 2,4-dinitrophenylhydrazone is a solid that is purified and the melting point is determined. The comparison of a melting point with a table of known values could identify the compound as aldehyde or ketone

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2021 (26 Feb Shift 2)
0.2+0.22+0.222+â¦ upto n terms is equal to

Options:
A 29-2811-10-n
B n191-10-n
C 29n-191-10-n
D 29

The correct answer is:
29n-191-10-n

Solution:
0.2+0.22+0.222+...n terms
=20.1+0.11+0.111+…n terms
=2110+11100+1111000+… n terms
=29910+99100+9991000+… n  terms
=291-110+1-1100+1-11000+… n  terms
=29n-110+1100+11000+…n
=29n-1101-110n1-110
=29n-110×109.10n-110n
=29n-191-10-n

Topic: Mathematics, Sequences and Series, JEE Main
$\int \frac{d x}{\left(2 a x+x^2\right)^{\frac{3}{2}}}=$

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

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

Solution:
We have,
$$
\begin{aligned}
& \int \frac{1}{\left(2 a x+x^2\right)^{3 / 2}} d x=\int \frac{1}{x^3\left(\frac{2 a}{x}+1\right)^{3 / 2}} d x \\
& \text { Put } \begin{aligned}
\frac{2 a}{x}+1 & =t \Rightarrow x=\frac{2 a}{t-1} \Rightarrow \frac{-2 a}{x^2} d x=d t \\
& =-\frac{1}{2 a} \int \frac{t-1}{2 a(t)^{3 / 2}} d t=-\frac{1}{4 a^2} \int \frac{t-1}{t^{3 / 2}} d t
\end{aligned}
\end{aligned}
$$
$$
\begin{aligned}
& =-\frac{1}{4 a^2} \int\left(t^{-1 / 2}-t^{-3 / 2}\right) d t=-\frac{1}{4 a^2}\left[2 t^{1 / 2}+2 t^{-1 / 2}\right]+c \\
& =-\frac{1}{2 a^2}\left(t^{1 / 2}+t^{-1 / 2}\right)+c=-\frac{1}{2 a^2}\left(\frac{t+1}{\sqrt{t}}\right)+c
\end{aligned}
$$
Put $t=\frac{2 a}{x}+1$
$$
\begin{aligned}
& =-\frac{1}{2 a^2}\left(\frac{\frac{2 a}{x}+1+1}{\sqrt{\frac{2 a}{x}+1}}\right)+c=-\frac{1}{a^2}\left(\frac{\frac{a}{x}+1}{\sqrt{\frac{2 a}{x}+1}}\right)+c \\
& =-\frac{1}{a^2}\left(\frac{\frac{a+x}{x}}{\sqrt{\frac{2 a+x}{x}}}\right)+c=-\frac{1}{a^2}\left(\frac{a+x}{\sqrt{2 a x+x^2}}\right)+c
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main
$2 \mathrm{CH}_3 \mathrm{COOH} \xrightarrow{-\mathrm{MeO} \mathrm{C}} A$, product ' $A$ ' in the reaction is

Options:
A $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$
B $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OH}$
C $\mathrm{CH}_3 \mathrm{COCH}_3$
D

The correct answer is:
$\mathrm{CH}_3 \mathrm{COCH}_3$

Solution:
No Solution Available

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$2 \mathrm{CH}_3 \mathrm{COOH} \xrightarrow[300^{\circ} \mathrm{C}]{\mathrm{MnO}} A$, product ' $A$ ' in the reaction is

Options:
A $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$
B $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OH}$
C $\mathrm{CH}_3 \mathrm{COCH}_3$
D

The correct answer is:
$\mathrm{CH}_3 \mathrm{COCH}_3$

Solution:
$2 \mathrm{CH}_3 \mathrm{COOH} \xrightarrow[300^{\circ} \mathrm{C}]{Mn \mathrm{O}} \mathrm{CH}_3 \mathrm{COCH}_3+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
2- chloro - 2 - methylpentane on reaction with sodium methoxide in methanol yields:
(i)
(ii)
(iii)

Options:
A iii only
B i and ii
C i and iii
D All of these

The correct answer is:
All of these

Solution:
When given reactant reacts with the sodium methoxide solution then both substitution and elimination reactions are possible.

Substitution reaction gives the following product through SN1 reaction.

E1 will give the following products

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2016 (03 Apr)
2-chlorobutane+Cl2→C4H8Cl2 (isomers)
Total number of optically active isomers shown by C4H8Cl2, obtained in the above reaction is ________.

Options:

The correct answer is:
6

Solution:
Optical activity is the ability of a substance to rotate the plane of polarisation of a beam of light that is passed through it.A compound to be said as optically active is only if it has to have an asymmetric chiral carbon centre (the carbon is said to be chiral if they have different substituents attached).

The optically active substances that rotate the plane of polarisation of the light towards the right are known as right-handed or dextrorotatory. It can be Indicated by including (+) or ‘D’ before the compound name.

Substances that rotate the plane of polarisation of the light toward the left are known as left-handed or laevorotatory. It can be Indicated by including (-) or ‘L’ before the compound name

The reaction is as follows,

2-chlorobutane+Cl2→C4H8Cl2

C4H8Cl2 will give four isomers of which one is optically inactive. Their structures are as follows,

Here only two, i.e., II and IV are optically active. Since each optically active compound has two enantiomers, therefore, total 6 isomeric compounds are possible.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2024 (30 Jan Shift 2)
$2 \cos ^{-1} \sqrt{\frac{1+x}{2}}=\frac{\pi}{2}$, then $x=$

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

The correct answer is:
$0$

Solution:
Given equation is
$2 \cos ^{-1} \sqrt{\left(\frac{1+x}{2}\right)}=\frac{\pi}{2}$
$\begin{aligned}
& \Rightarrow \cos ^{-1} \sqrt{\left(\frac{1+x}{2}\right)}=\frac{\pi}{4} \Rightarrow \cos \frac{\pi}{4}=\frac{\sqrt{1+x}}{\sqrt{2}} \\
& \Rightarrow \frac{1}{\sqrt{2}}=\frac{\sqrt{1+x}}{\sqrt{2}} \Rightarrow 1=\sqrt{1+x} \Rightarrow x=0 \\
&
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
2cos45°+cos56°+cos58°-cos66°2cos28°cos29°sin33°=

Options:
A 2
B 22
C 22
D 42

The correct answer is:
22

Solution:
Let

A=2cos45°+cos56°+cos58°-cos66°2cos28°cos29°sin33°

⇒A=1-cos66°+cos56°+cos58°2cos28°cos29°sin90-57°

∵cos45°=12

⇒A=1-1+2sin233°+2cos57°cos1°2cos28°cos29°sin90-57°

∵cosC+cosD=2cosC+D2cosC-D2cos2A=1-2sin2A

⇒A=2cos257°+2cos57°cos1°2cos28°cos29°cos57°

⇒A=2cos57°+cos1°cos28°cos29°

⇒A=22cos57°+cos1°2cos28°cos29°

⇒A=22cos57°+cos1°cos57°+cos1°

∵2cosAcosB=cosA+B+cosA-B

⇒A=22

Topic: Mathematics, Trigonometric Equations, JEE Main
2 dice are thrown. Suppose a random variable X is assigned a value 2k, if the sum on the dice is equal to k, then the expected value of X is

Options:
A 10
B 12
C 14
D 509

The correct answer is:
14

Solution:
k	2	3	4	5	6	7	8	9	10	11	12

Pk	136	236	336	436	536	636	536	436	336	236	136

k= sum on dice
Expected value =136×4+236×6+336×8+436×10
+536×12+636×14+536×16+436×18
+336×20+236×22+136×24
=50436=14

Topic: Mathematics, Probability, JEE Main
0.02 equivalent of Ag was deposited in an electrolysis experiment. If same quantity of a electricity is passed through a gold solution, 1.314 g of gold is deposited. Find oxidation state of the gold. (Atomic mass of Au = 197)

Options:

The correct answer is:
3

Solution:
Ag++e-→Ag1 mol 1eq.
0.02 q. of Ag = 0.02 F = 0.02 × 96500 = 1930 C
1.314 Au is deposited by 0.02 F
197 is deposited by 0.021.314×197=3F

Topic: Chemistry, Electrochemistry, JEE Main
2 g of a non-volatile non-electrolyte solute is disolved in 200 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1:8. The elevation in boiling points of A and B are in the ratio xyx:y. The value of y is____(Nearest Integer)

Options:

The correct answer is:
8

Solution:
Given : KbAKbB=18

∴ ΔTBAΔTBB=KbA·mKbB·m=18=xy

∵ xy=18

∴ y=8 (nearest integer)

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (27 Jun Shift 1)
$2 \mathrm{~g}$ of a radioactive sample having half-life of 15 days was synthesised on 1st Jan 2009. The amount of the sample left behind on 1st March, 2009 (including both the days) is

Options:
A $0 \mathrm{~g}$
B $0.125 \mathrm{~g}$
C $1 \mathrm{~g}$
D $0.5 \mathrm{~g}$

The correct answer is:
$0.125 \mathrm{~g}$

Solution:
$\begin{array}{llrl}\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}} & & \\ \text { Given, } & \mathrm{N}_{0} =2 \mathrm{~g} \\ \mathrm{t}_{1 / 2} =15 \text { days } \\ \Rightarrow & \mathrm{T} =60 \text { days } \\ & \mathrm{n} =\frac{60}{15}=4 \\ & \mathrm{~N} =2\left(\frac{1}{2}\right)^{4} \\ \text { or } & \mathrm{N} =0.125 \mathrm{~g}\end{array}$

Topic: Chemistry, Chemical Kinetics, JEE Main
0.2 g sample of benzoic acid, C6H5COOH is titrated with 0.12 M Ba OH2 solution. What volume of Ba(OH)2 solution is required to reach the equivalence point?
Molar mass of benzoic acid is 122 g mol-1 .

Options:
A 6.83 mL
B 13.6 mL
C 17.6 mL
D 35.2 mL

The correct answer is:
6.83 mL

Solution:
Number of moles of benzoic acid =0.2122=1.64×10-3
Neutralization takes place as
2C6H5COOH+Ba(OH)2→(C6H5COO)2Ba+2H2O
Number of moles of Ba(OH)2 required =12×1.64×10-3
MV1000=12×1.64×10-3
0.12×V1000=12×1.64×10-3
V = 6.83 mL

Topic: Chemistry, Redox Reactions, JEE Main
$2 \mathrm{gm}$ of $\mathrm{O}_2$ at $27^{\circ} \mathrm{C}$ and $760 \mathrm{~mm}$ of $\mathrm{Hg}$ pressure has volume

Options:
A 1.4 lit.
B $2.8 \mathrm{lit}$.
C $11.2 \mathrm{lit}$.
D 22.4 lit.

The correct answer is:
1.4 lit.

Solution:
Molecular Mass of O2=22.4 L at NTP
$32 \mathrm{~g}$ of $O 2=22.4 \mathrm{~L}$
$2 \mathrm{~g}$ ofO2 $=(22.4 / 22) \times 2=1.4$ lit

Topic: Chemistry, States of Matter, JEE Main
→ 2)  H 2 O 1)  HNO 3 A Major ; Product A is

Options:
A
B
C
D All of these

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main
2-hexene→ii H2Oi O3Products

The two products formed in above reaction are

Options:
A Butanal and acetaldehyde
B Butanoic acid and acetaldehyde
C Butanal and acetic acid
D Butanoic acid and acetic acid

The correct answer is:
Butanoic acid and acetic acid

Solution:
Ozone will cleave carbon-carbon double bonds to give ketones/carboxylic acids after oxidative workup. Here in this reaction Cleavage occurs at unsaturation of hexene and Butanoic acid and acetic acid is formed.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2023 (12 Apr Shift 1)
$2-$ Hexyne gives trans$-2-$Hexene on treatment with :

Options:
A
$\mathrm{Pt} / \mathrm{H}_2$

B
$\mathrm{Li} / \mathrm{NH}_3$

C
$\mathrm{Pd} / \mathrm{BaSO}_4$

D
$\mathrm{LiAlH}_4$

The correct answer is:
$\mathrm{Li} / \mathrm{NH}_3$

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2012 (Offline)
$2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})$
The equilibrium constant of the above reaction is $6.4$ at $300 \mathrm{~K}$. If $0.25$ mole each of $\mathrm{H}_{2}$ and $\mathrm{I}_{2}$ are added to the system, the equilibrium constant will be

Options:
A $6.4$
B $0.8$
C $3.2$
D $1.6$

The correct answer is:
$6.4$

Solution:
The value of equilibrium constant remains constant for a given reaction at constant temperature.

Topic: Chemistry, Chemical Equilibrium, KCET, KCET 2009
$\frac{1-2 i}{2+i}+\frac{4-i}{3+2 i}=$

Options:
A $\frac{24}{13}+\frac{10}{13} i$
B $\frac{24}{13}-\frac{10}{13} i$
C $\frac{10}{13}+\frac{24}{13} i$
D $\frac{10}{13}-\frac{24}{13} i$

The correct answer is:
$\frac{10}{13}-\frac{24}{13} i$

Solution:
$\begin{aligned} & \frac{1-2 i}{2+i}+\frac{4-i}{3+2 i}=\frac{(1-2 i)(3+2 i)+(4-i)(2+i)}{(2+i)(3+2 i)} \\ & =\frac{50-120 i}{65}=\frac{10}{13}-\frac{24}{13} i\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
$\frac{1+7 i}{(2-i)^2}=$

Options:
A $\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$
B $\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$
C $\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$
D None of these

The correct answer is:
$\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$

Solution:
$\frac{1+7 i}{(2-i)^2}=\frac{(1+7 i)}{(3-4 i)} \frac{(3+4 i)}{(3+4 i)}=\frac{-25+25 i}{25}=-1+i$
Let $z=x+i y=-1+i$
$\therefore r \cos \theta=-1$ and $r \sin \theta=1 \quad \therefore \theta=\frac{3 \pi}{4}$ and $r=\sqrt{2}$ Thus $\frac{1+7 i}{(2-i)^2}=\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]$
Aliter :$\left|\frac{1+7 i}{(2-i)^2}\right|=\left|\frac{1+7 i}{3-4 i}\right|=\sqrt{2}$
and $\arg \left(\frac{1+7 i}{3-4 i}\right)=\tan ^{-1} 7-\tan ^{-1}\left(-\frac{4}{3}\right)$
$=\tan ^{-1} 7+\tan ^{-1} \frac{4}{3}=\frac{3 \pi}{4}$
$\therefore \frac{1+7 i}{(2-i)^2}=\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$

Topic: Mathematics, Complex Number, JEE Main
2IO3-+xI-+12H+→6I2+6H2 What is the value of x?

Options:
A 2
B 12
C 10
D 6

The correct answer is:
10

Solution:
Assign the oxidation number to all the elements in the reaction,

2I+5O3-+xI--1+12H++1→6I20+6H2O

2IO3-+10e-→I2

2I-→I2+2e-×5

2IO3-+10I-+12H+→6I2+6H2O

Therefore, the value of x=10 in the reaction.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2023 (08 Apr Shift 1)
2 kg of a monoatomic gas is at a pressure of 4×104 N m-2. The density of the gas is 8 kg m-3. What is the order of energy of the gas due to its thermal motion?

Options:
A 105 J
B 106 J
C 104 J
D 103 J

The correct answer is:
104 J

Solution:
Internal energy is the energy due to thermal motion.

The internal energy of an ideal gas given by
U=f2nRT (for n moles)
U=f2PV=f2×P×Mρ (Since, f=3 for a monoatomic gas.)
U=32×4×104×28
=1.5×104J
So, the order of energy due to thermal motion is 104

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2019 (10 Jan Shift 2)
2 kg of ice at -20Â°C is mixed with 5 kg of water at 20Â°C. The water content of the final mixture is (Latent heat of ice =80 kcal kg-1, the specific heat of water =1 kcalkg-1Â°C-1 and specific heat of ice =0.5 kcal kgâ1Â°Câ1 )

Options:
A 7 kg
B 6 kg
C 4 kg
D 3 kg

The correct answer is:
6 kg

Solution:
Here, m, s and θ1 are the mass, specific heat and temperature difference for water respectively. Similarly, M, S and θ2 are the corresponding quantities for ice. Here, L is the latent heat of fusion.

We have, the heat lost by water =msθ, heat gained by ice =MSθ+ML.
From the principle of calorimetry, we have, the heat lost by water=heat gained by ice.

Hence, msθ1=MSθ2+ML.

5×1×20-0=2×0.5×20+x×80
x=1 kg
1 kg of ice melts to form water.
Hence, final amount of water 5 kg+1 kg=6 kg.

Topic: Physics, Thermodynamics, JEE Main
2 L of 0.2MH2SO4 is reacted with 2 L of 0.1MNaOH solution, the molarity of the resulting product Na2SO4 in the solution is____millimolar.

Options:

The correct answer is:
25

Solution:
H2SO4+2NaOH→Na2SO4+2H2O

0.4mole0.2mole-0.3mol-0.1mol

Molarity=Number of molesVolume in litres

Molarity of Na2SO4 is 0.14=0.025M

=25×10-3M

Molarity of the resulting product Na2SO4 in the solution is 25 millimoles

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (28 Jul Shift 2)
2 loudspeakers are emitting sound waves of wavelength Î» with an initial phase difference of Ï2. At what minimum distance from O on line AB will one hear maxima?

Options:
A 25Â Î»
B 100Î»15
C 50Â Î»
D 25 Î»3

The correct answer is:
25 Î»3

Solution:
Δϕ=2nπ
⇒π2+2πλdsin⁡θ=2nπ
2πλ dsin⁡θ=2n-12π
sinθ= 2n-12λ2d=12×λ2×3λ=112
⇒y100λ2=112
144y2=100λ2
y≈100λ12=25λ3

Topic: Physics, Waves and Sound, JEE Main
\( 2 \mathrm{~kg} \) of a monoatomic gas is at a pressure of \( 4 \times 10^{4} \mathrm{~N} \mathrm{~m}^{-2} \). The density of the gas is \( 8 \mathrm{~kg} \mathrm{~m}^{-3} \). What is the order of energy of the gas due to its thermal motion?

Options:
A \( 10^{5} \mathrm{~J} \)
B \( 10^{6} \mathrm{~J} \)
C \( 10^{4} \mathrm{~J} \)
D \( 10^{3} \mathrm{~J} \)

The correct answer is:
\( 10^{4} \mathrm{~J} \)

Solution:
Internal energy is the energy due to thermal motion.

The internal energy of an ideal gas given by
U=f2nRT (for n moles)
U=f2PV=f2×P×Mρ (Since, f=3 for a monoatomic gas.)
U=32×4×104×28
=1.5×104J
So, the order of energy due to thermal motion is 104

Topic: Physics, Kinetic Theory of Gases, JEE Main
2 methylbutane on reacting with bromine in the presence of sunlight gives mainly

Options:
A
1 – bromo -2 - methylbutane

B
2 – bromo -2 - methylbutane

C
2 – bromo -3 - methylbutane

D
1 – bromo -3 – methylbutane

The correct answer is:
2 – bromo -2 - methylbutane

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2005
$2 \mathrm{MnO}_4^{-}+5 \mathrm{H}_2 \mathrm{O}_2+6 \mathrm{H}^{+} \rightarrow 2 \mathrm{Z}+5 \mathrm{O}_2+8 \mathrm{H}_2 \mathrm{O}$. In this reaction $\mathrm{Z}$ is

Options:
A $\mathrm{Mn}^{+2}$
B $M n^{+4}$
C $\mathrm{MnO}_2$
D $M n$

The correct answer is:
$\mathrm{Mn}^{+2}$

Solution:
Hydrogen peroxide reduces \(\mathrm{Mn}(\mathrm{V}\)II) to \(\mathrm{Mn}\) (II) in acidic medium.
Thus \(\mathrm{Z}\) is \(\mathrm{Mn}^{2+}\).
Hence, option A is correct.

Topic: Chemistry, Redox Reactions, JEE Main
2MnO4-+bC2O42-+cH+→xMn2++yCO2+zH2O
If the above equation is balanced with integer coefficients, the value of c is _______ . (Round off to the Nearest Integer).

Options:

The correct answer is:
16

Solution:
Writing the half reaction

Reduction half reaction

MnO4-→Mn2+

balancing oxygen

MnO4-→Mn2++4H2O

balancing Hydrogen

8H++MnO4-→Mn2++4H2O

balancing charge

5e-+8H++MnO4-→Mn2++4H2O

Oxidation half

C2O42-→CO2
Balancing carbon

C2O42-→2CO2

Balancing charge

C2O42-→2CO2+2e-

Net equation

16H++2MnO4-+5C2O42-→10CO2+2Mn2++8H2O

So c=16

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (16 Mar Shift 1)
2MnO4-+bI-+cH2O→xI2+yMnO2+zOH-

If the above equation is balanced with integer coefficients, the value of z is ________.

Options:

The correct answer is:
08

Solution:
Reduction Half reaction

2MnO4-→2MnO2

2MnO4-+4H2O+6e-→2MnO2+8OH-

Oxidation Half reaction

2I-→I2+2e-

6I-→3I2+6e-

Adding oxidation half and reduction half, net reaction is

2MnO4-+6I-+4H2O→3I2+2MnO2+8OH-

⇒z=8

⇒Ans 8

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2024 (30 Jan Shift 1)
0.2 molar solution of formic acid is ionized $3.2 \%$. Its ionization constant is

Options:
A $9.6 \times 10^{-3}$
B $2.1 \times 10^{-4}$
C $1.25 \times 10^{-6}$
D $4.8 \times 10^{-5}$

The correct answer is:
$2.1 \times 10^{-4}$

Solution:
$K_a=C \alpha^2=0.2 \times\left(\frac{32}{100}\right)^2=2.048 \times 10^{-4}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
0.02 mole of CoNH35BrCl2 and 0.02 mole of CoNH35ClSO4 are present in 200 cc of a solution X. The number of moles of the precipitates Y and Z that are formed when the solution X is treated with excess silver nitrate and excess barium chloride are respectively

Options:
A 0.02, 0.02
B 0.01, 0.02
C 0.02, 0.04
D 0.04, 0.02

The correct answer is:
0.04, 0.02

Solution:
[Co(NH3)5Br]Cl21 mole0.02 mole+2AgNO32 moles→[Co(NH3)5Br](NO3)21 mole+2AgCl(ppt.)(Y)2 moles0.02×2=0.04 mole
[Co(NH3)5Cl]SO41 mole+BaCl20.02 moles→[Co(NH3)5Cl]Cl21 mole+BaSO4(ppt.)(Z)0.02 moles

Topic: Chemistry, Coordination Compounds, JEE Main
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, JEE Main
2 moles of solid benzoic acid undergoes combustion at 300 K .
C 6 H 5 COOH s  +  15 2   O 2 g  → 7  CO 2 g  + 3  H 2 O l .
The value of ΔH–ΔU value is

Options:
A -1.247 kJ
B -2.494 kJ
C +2.494Â kJ
D +1.247Â kJ

The correct answer is:
-2.494 kJ

Solution:
C 6 H 5 COOH s + 15 2 O 2 g â7â CO 2 g +3â H 2 O l
ÎHâÎU=ÎngRT
=â12(8.314)Ã3001000
=â1.247âkJ for two moles it is −2.494 kJ

Topic: Chemistry, Thermodynamics (C), JEE Main