datavorous/entrance-exam-dataset huggingface dataset:
5 g sulphur is present in 100 g of CS2.ΔTb of solution 0.954 and Kb is 4.88. The molecular formula of sulphur is

Options:
A S2
B S4
C S3
D S8

The correct answer is:
S8

Solution:
ÎTb=KbÃm

m=ÎTbKb
or W2Ã1000M2ÃW1=ÎTbKb
or 5Ã1000M2Ã100=0.9544.88
M2=256
Atomic weight of sulphur = 32
Number of sulphur atom =Molecular weightAtomic weight=25632=8
Hence Molecular formula =S8 .

Topic: Chemistry, Solutions, JEE Main
\( 5 \) girls and \( 10 \) boys sit a random in a row having \( 15 \) chairs numbered as \( 1 \) to \( 15 \). If the probability that the end seats are occupied by the girls and odd number of boys take seat between any two girls is \( \frac{20}{n} \cdot \) then find the value of \( \frac{3003 n}{10} \)

Options:

The correct answer is:
2

Solution:
Total number of arrangements =15!

Boys in these four gaps be2x+1, 2y+1, 2z+1 and 2t+1, then
2x+1+2y+1+2z+1+2t+1=10
⇒x+y+z+t=3
Where x, y, z, t are integers and 0≤x≤3,
0≤y≤3, 0≤z≤3, 0≤t≤3
∴ The number of ways of selecting positions for boys
= coefficient of x3 in 1+x+x2+x34
=coefficient of x3 in 1-x41-x4
= coefficient of x3 in 1-x44(1-x)-4=C36=20
∴ Number of arrangements of boys and girls with given condition = 20 × 20! × 5!

∴ Required probability =20×10!×5!15!=203003

Topic: Mathematics, Probability, JEE Main
5 L of an alkane requires 25 L of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is

Options:
A Isobutane
B Ethane
C Butane
D Propane

The correct answer is:
Propane

Solution:
Combustion of Hydrocarbon,
CxHy+x+y4O2→x CO2+y2H2O
∴ n∝v (P,T=Constant)
for 1 mole of CxHy=x+y4, O2 required
for 5 L of CxHy=x+y4 5 L, O2 required
∴ ratio of alkane and O2 is 5/25
So, x+y45=25
x+y4=5,
which is satisfied by C3H8.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2016 (09 Apr Online)
$5 \mathrm{~m}$ 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, JEE Main
$0.5 \mathrm{M}$ of $\mathrm{H}_2 \mathrm{SO}_4$ is diluted from 1 litre to 10 litre, normality of resulting solution is

Options:
A $1 \mathrm{~N}$
B $0.1 \mathrm{~N}$
C $10 \mathrm{~N}$
D $11 \mathrm{~N}$

The correct answer is:
$0.1 \mathrm{~N}$

Solution:
Molarity of $\mathrm{H}_2 \mathrm{SO}_4=0.5$
Normality of $\mathrm{H}_2 \mathrm{SO}_4\left(N_1\right)=0.5 \times 2=1$ $N_1 V_1=N_2 V_2$
$1 \times 1=N_2 \times 10$ or $\quad N_2=\frac{1}{10}=0.1 N$.

Topic: Chemistry, Solutions, JEE Main
\(0.05 \mathrm{M} \mathrm{CuSO}_4\) when treated with \(0.01 \mathrm{M} \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\) gives green colour solution of \(\mathrm{Cu}_2 \mathrm{Cr}_2 \mathrm{O}_7\). The two solutions are separated as shown below :
[SPM : Semi Permeable Membrane]
Due to osmosis :

Options:
A Molarity of \(\mathrm{CuSO}_4\) solution is lowered.
B Molarity of \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\) solution is lowered.
C Green colour formation observed on side Y.
D Green colour formation observed on side X.

The correct answer is:
Molarity of \(\mathrm{CuSO}_4\) solution is lowered.

Solution:
Only solvent Molecules are allowed to pass through the SPM.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (09 Apr Shift 1)
$5 \mathrm{~mL}$ of $0.4 \mathrm{~N} \mathrm{NaOH}$ is mixed with $20 \mathrm{~mL}$ of $0.1 \mathrm{~N} \mathrm{HCl}$. The $\mathrm{pH}$ of the resulting solution will be

Options:
A 7
B 8
C 5
D 6

The correct answer is:
7

Solution:
From the first law of thermodynamics, the total energy of the universe remains constant.
$$
\begin{aligned}
&\text { Milliequivalents of } \mathrm{NaOH}=\mathrm{N}_{1} \mathrm{~V}_{1} \\
&\qquad=0.4 \times 5=0.20 \\
&\text { Milliequivalents of } \mathrm{HCl}=\mathrm{N}_{2} \mathrm{~V}_{2} \\
&\qquad=0.1 \times 20=0.20
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, KCET, KCET 2012
5âml of As2S3 is mixed with distilled waterand 0.01âM solution of an electrolyte AB so that total volume is 10âml . it was found that all solution containing more than 5âml of AB coagulate within 5 min . What is the Flocculation value of AB for As2S3 sol?

Options:

The correct answer is:
5

Solution:
A minimum of 5âml of AB is required to coagulate the sol, the moles of AB in sol is =5Ã0.0110=0.005 moles this means that a minimum of 0.005 moles or 0.005Ã1000=5 milli moles are required for coagulation 1 litre of sol. flocculation value of AB for As2S3 sol = 5

Topic: Chemistry, Surface Chemistry, JEE Main
5 mol of Fe2(C2O4)3 is oxidised by x mol of K2Cr2O7 in acidic medium, calculate the value of X ?

Options:

The correct answer is:
5

Solution:
Eq of Fe2(C2O4)3 = eq of K2Cr2O7
5×6=mol×6
Mol of K2Cr2O7 = 5

Topic: Chemistry, Redox Reactions, JEE Main
5 moles of AB2 weigh 125×10-3 kg and 10 moles of A2B2 weigh 300×10-3 kg . The molar mass of A (MA) in kg mol-1 are:

Options:
A MA=5×10-3 and MB=10×10-3
B MA=25×10-3 and MB=50×10-3
C MA=50×10-3 and MB=25×10-3
D MA=10×10-3 and MB=5×10-3

The correct answer is:
MA=5×10-3 and MB=10×10-3

Solution:
Mass of 1 mole of AB2=125×10-3kg5=125 gm5
=25 gm

Similarly, mass of 1 mole of A2B2=300×10-3kg10=30 gm

Let molar mass of A is x and B is y

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2019 (12 Apr Shift 1)
5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV=28 J K-1, calculate ∆U and ∆pV for the process. (R=8.0 J K-1 mol-1)

Options:
A âU=14Â J;Â âpV=0.8Â J
B ∆U=14 kJ; ∆pV=18 kJ
C ∆U=14 kJ; ∆pV=4 kJ
D ∆U=2.8 kJ; ∆pV=0.8 kJ

The correct answer is:
∆U=14 kJ; ∆pV=4 kJ

Solution:
ΔU=nCVΔT
=5×28×100=14000 J
=14 kJ
Δ(PV)=nRΔT=5×8×100=4000 J=4 kJ

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2019 (08 Apr Shift 2)
0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10 m3 at 1000 K . Given, R is the gas constant in JK-1mol-1, x is:

Options:
A 4+R2R
B 2R4+R
C 4-R2R
D 2R4-R

The correct answer is:
4-R2R

Solution:
According to the ideal gas equation:
PV=nRT
200×10=(0.5+x)R×1000
2=0.5R+xR
x=2-0.5RR=4-R2R

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2019 (09 Jan Shift 1)
$\int 5 \sin x d x=$

Options:
A $5 \cos x+c$
B $-5 \cos x+c$
C $5 \sin x+c$
D $-5 \sin x+c$

The correct answer is:
$-5 \cos x+c$

Solution:
$\int 5 \sin x d x=-5 \cos x+c$

Topic: Mathematics, Indefinite Integration, JEE Main
5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is 156 cm, joined them. The variance in cm2 of the height of these six students is:

Options:
A 22
B 16
C 18
D 20

The correct answer is:
20

Solution:
Let 1st 5 students are
x1, x2, x3 ,x4 and x5
∴18=Ex2-Ex2
=x12+x22+x32+x42+x525-1502
⇒x12+x22+x32+x42+x52=112590
∴ new variance =x12+x22+....+x52+15626-(151)2 Since new mean=150×5+1566=151
=112590+243366-22801
=20

Topic: Mathematics, Statistics, JEE Main, JEE Main 2019 (09 Jan Shift 1)
∫616logex2logex2+logex2-44x+484dx is equal to

Options:
A 5
B 10
C 8
D 6

The correct answer is:
5

Solution:
Let ∫616logex2 dxlogex2+logex2-44x+484

∫616logex2logex2+logex-222dx .....(1)

We know

∫abfxdx=∫abfa+b-xdx

So, ∫616loge22-x2loge22-x2+loge22-22-x2dx

I=∫616loge22-x2logex2+loge22-x2dx ....(2)

Add (1) & (2)

2I=∫616dx=10,I=5

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (27 Aug Shift 1)
$6$ litres of an alkene require $27$ litres of oxygen at constant temperature and pressure for complete combustion. The alkene is :

Options:
A
Ethene

B
Propene

C
$1-$Butene

D
$2-$Butene

The correct answer is:
$2-$Butene

Solution:
No Solution Available

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2013 (25 Apr Online)
$6^{\text {th }}$ term in expansion of $\left(2 x^2-\frac{1}{3 x^2}\right)^{10}$

Options:
A $\frac{4580}{17}$
B $-\frac{896}{27}$
C $\frac{5580}{17}$
D None of these

The correct answer is:
$-\frac{896}{27}$

Solution:
Applying $T_{r+1}={ }^n C_r x^{n-r} a^r$ for $(x+a)^n$
Hence $T_6={ }^{10} C_5\left(2 x^2\right)^5\left(-\frac{1}{3 x^2}\right)^5$
$=-\frac{10!}{5!5!} 32 \times \frac{1}{243}=-\frac{896}{27}$

Topic: Mathematics, Binomial Theorem, JEE Main
0.6g of a solute is dissolved in 0.1 litre of a solvent which develops an osmotic pressure of $1.23 \mathrm{~atm}$ at $27^{\circ} \mathrm{C}$. The molecular mass of the substance is

Options:
A $149.5 \mathrm{gmole}^{-1}$
B $120 \mathrm{gmole}^{-1}$
C $430 \mathrm{~g} \mathrm{~mole}^{-1}$
D None of these

The correct answer is:
$120 \mathrm{gmole}^{-1}$

Solution:
$m=\frac{w R T}{P V}=\frac{0.6 \times 0.082 \times 300}{1.23 \times 0.1}=120 \mathrm{gmole}^{-1}$

Topic: Chemistry, Solutions, JEE Main
$\int \frac{d x}{7+5 \cos x}$ is equal to

Options:
A $\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{1}{\sqrt{3}} \tan \frac{x}{2}\right)+c$
B $\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{1}{\sqrt{6}} \tan \frac{x}{2}\right)+c$
C $\frac{1}{7} \tan ^{-1}\left(\tan \frac{x}{2}\right)+c$
D $\frac{1}{4} \tan ^{-1}\left(\tan \frac{x}{2}\right)+c$

The correct answer is:
$\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{1}{\sqrt{6}} \tan \frac{x}{2}\right)+c$

Solution:
$$
\text { Let } \begin{aligned}
I & =\int \frac{d x}{7+5 \cos x} \\
& =\int \frac{d x}{7\left(\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}\right)+5\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right)} \\
& =\int \frac{d x}{12 \cos ^2 \frac{x}{2}+2 \sin ^2 \frac{x}{2}} \\
& =\frac{1}{2} \int \frac{\sec ^2 \frac{x}{2}}{6+\tan ^2 \frac{x}{2}} d x
\end{aligned}
$$
Let $\tan \frac{x}{2}=z$
$$
\begin{aligned}
& \Rightarrow \frac{1}{2} \sec ^2 \frac{x}{2} d x=d z \\
& \therefore \quad I=\int \frac{d z}{6+z^2} \\
& =\frac{1}{\sqrt{6}} \tan ^{-1} \frac{z}{\sqrt{6}}+c \\
& =\frac{1}{\sqrt{6}} \tan ^{-1}\left(\frac{1}{\sqrt{6}} \tan \frac{x}{2}\right)+c \\
&
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main
$$
\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x=A \sqrt{7-6 x-x^2}+B \sin ^{-1}\left(\frac{x+3}{4}\right)+C
$$
(where $C$ is a constant of integration), then the ordered pair $(A, B)$ is equal to

Options:
A
$(-2,-1)$

B
$(2,-1)$

C
$(-2,1)$

D
$(2,1)$

The correct answer is:
$(-2,-1)$

Solution:
$\because 7-6 x-x^2=16-(x+3)^2$
and $\frac{d}{d x}\left(7-6 x-x^2\right)=-2 x-6$
So, $\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x=\int \frac{2 x+6}{\sqrt{7-6 x-x^2}} d x$
$$
-\int \frac{1}{\sqrt{16-(x+3)^2}} d x
$$
$$
=-2 \sqrt{7-6 x-x^2}-\sin ^{-1}\left(\frac{x+3}{4}\right)+C
$$
Therefore, $A=-2, \& B=-1$.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
0.7 grams of Na2CO3.xH2O were dissolved in water and the volume was made to 100 mL, 20 mL of this solution required 19.8 mL of N/10 HCl for complete neutralization. The value of x is
Report your answer by rounding it upto nearest whole number.

Options:

The correct answer is:
2

Solution:
Meq. of Na 2 CO 3 â xH 2 O in 20âmL=19.8Ã 1 10
â´ Meq. of Na 2 CO 3 â xH 2 Oâ Na 2 CO 3 â xH 2 O in 100âmL=19.8Ã 1 10 Ã5
â´wEÃ1000=19.8Ã110Ã5
or 0.7M2Ã1000=19.82
â´ M=141.41
23Ã2+12+3Ã16+18x=141.41
∴ x≅2

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
7 mole of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is
(Given R=8.3 J K-1 mol-1)

Options:
A 5810 J
B 3486 J
C 11620 J
D 6972 J

The correct answer is:
3486 J

Solution:
For the given process pressure is constant therefore, it is an isobaric process.

For a quasi-static process the change in internal energy of an ideal gas is independent of the nature of the process and is given by,

ΔU=nCvΔT

=n×3R2×ΔT

[molar heat capacity at constant volume for monatomic gas =3R2]

ΔU=7×32×8.3×40=3486 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (26 Jul Shift 1)
$7^{\text {th }}$ term of the sequence $\sqrt{2}, \sqrt{10}, 5 \sqrt{2}, \ldots \ldots$ is

Options:
A $125 \sqrt{10}$
B $25 \sqrt{2}$
C $125$
D $125 \sqrt{2}$

The correct answer is:
$125 \sqrt{2}$

Solution:
Given sequence is $\sqrt{2}, \sqrt{10}, \sqrt{50} \ldots \ldots$ Common ratio $r=\sqrt{5}$, first term $a=\sqrt{2}$, then $7^{\text {th }}$ term
$t_7=\sqrt{2}(\sqrt{5})^{7-1}=\sqrt{2}(\sqrt{5})^6=\sqrt{2}(5)^3=125 \sqrt{2}$

Topic: Mathematics, Sequences and Series, JEE Main
8 g of Cu66 undergoes radioactive decay and after 15 minutes only 1 g remains. The half-life, in minutes, is then

Options:
A 5
B 15/8
C 8/15
D 15 ln(2)

The correct answer is:
5

Solution:
Given,
m0=8 g
m=1g
Time of decay =15 min
lnMM0=-λt
⇒ ln18=-λ15
∴ λ=ln815
So, T1/2= ln 2λ
=15ln2ln8

=5

Topic: Physics, Nuclear Physics, NEET
8 g of NaOH is dissolved in 18g of H2O. Mole fraction of NaOH in solution and molality (in mol kg-1) of the solution respectively are:

Options:
A 0.167,11.11
B 0.167,22.20
C 0.2,11.11
D 0.2,22.20

The correct answer is:
0.167,11.11

Solution:
n1=840=0.2
n2=1818=1;
nt=n1+n2=0.2+1=1.2
Mole fraction of NaOH=n1nt=0.21.2=0.167
Molality =840×100018=11.11

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2019 (12 Jan Shift 2)
$8 \mathrm{~mol}$ of $A B_3(\mathrm{~g})$ are introduced into a $1.0 \mathrm{dm}^3$ vessel. If it dissociates as
$2 \mathrm{AB}_3(g) \rightleftharpoons A_2(g)+3 B_2(g)$. At equilibrium, $2 \mathrm{~mol}$ of $A_2$ are found to be present. The equilibrium constant of this reaction is

Options:
A
2

B
3

C
27

D
36

The correct answer is:
27

Solution:
$2 A B_3(g) \leftrightharpoons A_2(g)+3 B_2(g)$
at $t=0 \quad 8 \quad 0 \quad 0$
at eq. $(8-2 \times 2) \quad 2 \quad 3 \times 2$ $=4 \quad 2 \quad 6$
now $K_C=\frac{\left[A_2\right]\left[B_2\right]^3}{\left[A B_3\right]^2}=\frac{2 / 1 \times[6 / 1]^3}{[4 / 1]^2}=27$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2012 (12 May Online)
8-digit numbers are formed using the digits 1, 1, 2, $2,2,3,4,4$. The number of such numbers in which the odd digits do no occupy odd places, is:

Options:
A
160

B
120

C
60

D
48

The correct answer is:
120

Solution:
In 8 digits numbers, 4 places are odd places.
Also, in the given 8 digits, there are three odd digits 1,1 and 3.
No. of ways three odd digits arranged at
$$
\text { four even places }=\frac{4 P_3}{2 !}=\frac{4 !}{2 !}
$$
No. of ways the remaining five digits 2 , 2, 2, 4 and 4 arranged at remaining five places $=\frac{5 !}{3 ! 2 !}$
Hence, required number of 8 digits number $=\frac{4 !}{2 !} \times \frac{5 !}{3 ! 2 !}=120$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2014 (12 Apr Online)
9.3 g of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is 100% completed is _____×10-1 g.

(Given molar mass in g mol-1 N=14,O=16,C=12,H=1)

Options:

The correct answer is:
135

Solution:
Given, Aniline required. is 9.3 gms

Molar mass of aniline= 93 g mol-1

Molar mass of Acetanilide = 135 g mol-1

The reaction is given as,

Number of moles=Mass / Molar mass

nAcetanilide =nAniline

⇒m135=9.393

⇒m=13.5 g

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 2)
$99^{\text {th }}$ term of the series $2+7+14+23+34+\ldots \ldots$ is

Options:
A $9998$
B $9999$
C $10000$
D $100000$

The correct answer is:
$9998$

Solution:
Let $S=2+7+14+23+34+\ldots . .+T_n \ldots(i)$
and $S=2+7+14+\ldots \ldots \ldots \ldots \ldots .+T_{n-1}+T_n \ldots(ii)$
From (i) and (ii), we get
$0=2+\left[5+7+9+11 \ldots \ldots . .+T_n-T_{n-1}\right]-T_n$
$T_n=2+\left[\frac{n-1}{2}\{2 \times 5+(n-2) 2\}\right]$
$\Rightarrow T_n=2+(n-1)(n+3)$
Now put $n=99$
$\Rightarrow T_{99}=2+98 \times 102=9998$

Topic: Mathematics, Sequences and Series, JEE Main
$$
\text { 9. If } 25^{15} \text { is divided by } 13 \text {, the remainder is }
$$

Options:

None

Solution:
12

Topic: Mathematics, Binomial Theorem
$9 \mathrm{~kg}$ solution is poured into a glass U-tube as shown in the figure below. The tube's inner diameter is $2 \sqrt{\frac{\pi}{5}} \mathrm{~m}$ and the solution oscillates freely up and down about its position of equilibriwm $(x=0)$. The period of oscillation in seconds is $\left(1 \mathrm{~m}^3\right.$ of solution has a mass $\mu=900 \mathrm{~kg}, g=10 \mathrm{~m} / \mathrm{s}^2$, ignore frictional and swrface tension effects)

Options:
A 0.1
B 10
C $\sqrt{\pi}$
D 1

The correct answer is:
0.1

Solution:
Diameter of tube $=2 \sqrt{\frac{\pi}{5}} \mathrm{~m}$
$\therefore$ Radius of tube $(r)=\sqrt{\frac{\pi}{5}} \mathrm{~m}$
Mass of solution $(u)=9 \mathrm{~kg}$
We know that,
$$
\begin{aligned}
& \because \quad \operatorname{Mass}(u)=d \cdot v \\
& \therefore \quad u=d \pi r^2 L \\
&
\end{aligned}
$$
or,
$$
9=900 \times \pi \times\left(\sqrt{\frac{\pi}{5}}\right)^2 L
$$
or,
$$
9=900 \times \frac{\pi^2}{5} \times 1
$$
or,
$L=\frac{5}{100 \pi^2}$
Now, $\quad T=2 \pi \sqrt{\frac{L}{2 g}}$
From Eq. (i),
Or,
$$
\begin{aligned}
& T=2 \pi \sqrt{\frac{5}{200 \pi^2 \times 10}} \\
& T=2 \times \sqrt{\frac{1}{400}}=\frac{2}{20} \\
& T=0.1 \mathrm{~s}
\end{aligned}
$$

Topic: Physics, Oscillations, JEE Main
$\int \frac{3^x d x}{\sqrt{9^x-1}}$ is equal to

Options:
A $\frac{1}{\log 3} \log \left|3^x+\sqrt{9^x-1}\right|+c$
B $\frac{1}{\log 3} \log \left|3^x-\sqrt{9^x-1}\right|+c$
C $\frac{1}{\log 9} \log \left|3^x+\sqrt{9^x-1}\right|+c$
D $\frac{1}{\log 3} \log \left|9^x+\sqrt{9^x-1}\right|+c$

The correct answer is:
$\frac{1}{\log 3} \log \left|3^x+\sqrt{9^x-1}\right|+c$

Solution:
Let $I=\int \frac{3^x d x}{\sqrt{9^x-1}}=\int \frac{3^x}{\sqrt{3^{2 x}-1}} d x$
Let $3^x=z \Rightarrow 3^x \log 3 d x=d z$
$$
I=\frac{1}{\log 3} \int \frac{z}{\sqrt{z^2-1}}
$$
$\begin{aligned} & =\frac{1}{\log 3} \log \left\{z+\sqrt{z^2-1}\right\}+c \\ & =\frac{1}{\log 3} \log \left[3^x+\sqrt{9^x-1}\right]+c\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int_9^x \frac{f(y)}{y^2} d y=2 \sqrt{x}-6 \Rightarrow f(x)=$

Options:
A $\sqrt{x}$
B $x \sqrt{x}$
C $x^2 \sqrt{x}$
D $x+\sqrt{x}$

The correct answer is:
$x \sqrt{x}$

Solution:
Given, $\int_9^x \frac{f(y)}{y^2} d y=2 \sqrt{x}-6$
Differentiating both sides using Leibnitz rule,
$\frac{f(x)}{x^2}=\frac{2}{2 \sqrt{x}}$
$\Rightarrow f(x)=\frac{x^2}{\sqrt{x}}=x \sqrt{x}$

Topic: Mathematics, Definite Integration, JEE Main
10.0 mL of 0.05 M KMnO4 solution was consumed in a titration with 10.0 mL of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is ........ ×10-2 g/L. (Round off to the nearest integer)

Options:

The correct answer is:
1575

Solution:
neqKMnO4=neqH2C2O4·2H2O

or, 10×0.051000×5=10×M1000×2

∴ Conc. of oxalic acid solution =0.125M

=0.125×126 g/L=15.75 g/L

=1575×10-2 g/L

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (27 Jul Shift 2)
1022 gas molecules each of mass 10-26 kg collide with a surface (perpendicular to it) elastically per second over an area 1 m2 with a speed of 104 m s-1. The pressure exerted by the gas molecules will be

Options:
A 2 N m-2
B 4 N m-2
C 8 N m-2
D 16 N m-2

The correct answer is:
2 N m-2

Solution:
v=104 m s-1
m=10-26
n=1022
A=1 m2
Δp=2mnv
Δp=2×1022×10-26×104=2
P=FA=2 N m-2

Topic: Physics, Work Power Energy, JEE Main
${ }^{10} C_1+{ }^{10} C_3+{ }^{10} C_5+{ }^{10} C_7+{ }^{10} C_9=$

Options:
A $2^9$
B $2^{10}$
C $2^{10}-1$
D None of these

The correct answer is:
$2^9$

Solution:
We know that
$2^{n-1}={ }^n C_0+{ }^n C_2+{ }^n C_4+\ldots .={ }^n C_1+{ }^n C_3+{ }^n C_5+\ldots . .$
So, ${ }^{10} C_1+{ }^{10} C_3+{ }^{10} C_5+\ldots . .+{ }^{10} C_9=2^{10-1}=2^9$

Topic: Mathematics, Binomial Theorem, JEE Main
10 divisions on the main scale of a Vernier calliper coincide with 11 divisions on the Vernier scale. If each division on the main scale is of 5 units, the least count of the instrument is :

Options:
A 12
B 1011
C 5011
D 511

The correct answer is:
511

Solution:
Given:

10 MSD=11 VSD

⇒1 VSD=1011 MSD

LC=1MSD-1VSD

=1 MSD-1011 MSD

=1 MSD11

=511 units

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2024 (01 Feb Shift 1)
10 g of a gas at STP occupies a volume of 2 L. At what temperature will the volume be doubled, if pressure and amount of the gas remains constant?

Options:
A 273 K
B 546 K
C -273 K
D 546°C

The correct answer is:
546 K

Solution:
At constant pressure, the volume is proportional to the absolute temperature.

V1T1=V2T2

Here,

V1=2 L T1=273 K STPV2=4 L T2=?

Putting all the values, we get T2,

2273=4T2T2=546 K

Topic: Physics, Kinetic Theory of Gases, AP EAMCET, AP EAMCET 2020 (23 Sep Shift 1)
$10 \mathrm{~g}$ of hydrogen and $64 \mathrm{~g}$ of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be -

Options:
A $1 \mathrm{~mol}$
B $2 \mathrm{~mol}$
C $3 \mathrm{~mol}$
D $4 \mathrm{~mol}$

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

Solution:
$\mathrm{H}_2+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O}$
$2 \mathrm{~g} \quad 16 \mathrm{~g} \quad 18 \mathrm{~g}$
$10 \mathrm{~g} \mathrm{H}_2$ required $\mathrm{O}_2=80$ which is not present 64 g $\mathrm{O}_2$ required $8 \mathrm{~g}$ of $\mathrm{H}_2$ and $\mathrm{H}_2$ left $=2 \mathrm{~g}$. Thus, $\mathrm{O}_2$ is the limiting reactant and $\mathrm{H}_2$ is excess reactant.
Hence, $\mathrm{H}_2 \mathrm{O}$ formed from 64 of $\mathrm{O}_2$
$$
\begin{aligned}
& =\frac{18}{16} \times 64 \\
& =72 \mathrm{~g}=\frac{72}{18} \mathrm{~mole} \\
& =4 \mathrm{~mole}
\end{aligned}
$$

Topic: Chemistry, Redox Reactions, NEET, NEET 2009 (Mains)
$10 \mathrm{~g}$ of hydrogen and 64 of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be

Options:
A $2 \mathrm{~mol}$
B $3 \mathrm{~mol}$
C $4 \mathrm{~mol}$
D $1 \mathrm{~mol}$

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

Solution:
Key Idea
(i) $\mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O}$
(ii) Amount of water produced is decided by limited reactant (ie, the reactant which is used in small amount)
$$
\begin{aligned}
& \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O} \\
& 1 \mathrm{~mol} \quad \frac{1}{2} \mathrm{~mol} \quad 1 \mathrm{~mol} \\
& \frac{10}{2} \mathrm{~mol} \quad \frac{64}{32} \mathrm{~mol} \\
& =5 \mathrm{~mol}=2 \mathrm{~mol} \\
& \because \frac{1}{2} \mathrm{molO}_2 \text { gives }=1 \mathrm{~mol} \mathrm{H}_2 \mathrm{O} \\
& \therefore 2 \mathrm{~mol} \mathrm{O}_2 \text { will give }=1 \times 2 \times 2=4 \mathrm{~mol} .
\end{aligned}
$$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
10 g of MgCO3 decomposes on heating to 0.1 mole CO2 and 4 g MgO. The percent purity of MgCO3 is
(Given that atomic weights of Mg,âC and O are 24, 12 and 16 u)

Options:
A 44%
B 54%
C 74%
D 84%

The correct answer is:
84%

Solution:
MgC O 3 decomposes as below:
MgC O 3 âMgO+C O 2
mole ratio =1:1:1
Thus 0.1 mole of C O 2 will be produced by 0.1 mole of MgC O 3 .
Mass of MgC O 3 in 10âg sample =0.1Ã84=8.4 g
Percent purity = 8.4 10 Ã100=84%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel. Which reaction is left in excess and how much? (At, wt. Mg=24; O=16)

Options:
A Mg, 0.16 g
B O2,Â 0.16Â g
C Mg,Â 0.44Â g
D O2,Â O.28Â g

The correct answer is:
Mg, 0.16 g

Solution:
Mg+12O2→MgO
1.024 0.5632
0.512 0.074
Oxygen is limiting reagent so 0.074-x2=0
x=0.072
excess Mg=0.512-0.072 mole
mass of Mg is =1-0.7×12=0.16 gram.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
1.0 g of Mg is burnt with 0.28 g of O2 in a closed vessel. Which reactant is left in excess and how much?

Options:
A Mg, 5.8 g
B Mg, 0.58 g
C O2 , 0.24 g
D O2 , 2.4 g

The correct answer is:
Mg, 0.58 g

Solution:
Balanced reaction can be given as
2Mg24×2=48+O232→2MgO
32 g of O2 is required to burn 48 g Mg
So, 0.28 g O2 will be required for 4832×0.28 g Mg
=0.42 g of Mg
Thus, Mg will remain in excess =1-0.42=0.58 g

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET