datavorous/entrance-exam-dataset huggingface dataset:
$2 \mathrm{~moles}$ of the ideal gas at $27^{\circ} \mathrm{C}$ temperature is expanded reversibly from $2 \mathrm{~L}$ to $20 \mathrm{~L}$. Find the entropy change $\left(\mathrm{R}=2 \mathrm{cal} \mathrm{mol}^{-1}\right.$ $\left.\mathrm{K}^{-1}\right)$ :

Options:
A 92.1
B 0
C 4
D 9.2

The correct answer is:
9.2

Solution:
$\begin{aligned}
& \Delta \mathrm{S}=2.303 \mathrm{n} \mathrm{R} \log _{10} \frac{\mathrm{V}_2}{\mathrm{~V}_1} \\
& =2.303 \times 2 \times 2 \times \log _{10} \frac{20}{2} \\
& =2.303 \times 2 \times 2 \times 1 \\
& =9.212 \mathrm{cal}
\end{aligned}$

Topic: Chemistry, Thermodynamics (C), JEE Main
2N2O5(g)→4NO2(g)+O2(g)
What is the ratio of the rate of decomposition of N2O5 to rate of formation of O2?

Options:
A 1:2
B 2:1
C 1:4
D 4:1

The correct answer is:
2:1

Solution:
As per unique value of reaction rate:
- 1 2 d dt N 2 O 5 = 1 1 d dt O 2
∴ d dt N 2 O 5 d dt O 2 = 2 : 1 ​

Topic: Chemistry, Chemical Kinetics, JEE Main
2n<n! is true for where nâN

Options:
A n<4
B n≥4
C n<3
D None of these

The correct answer is:
n≥4

Solution:
P n : 2n<n !
P 1 ,âP 2 ,âP 3 are not true.
P 4 : 16<24 is true
P K : K!>2K is true for some KâN, K>4
To prove P(K+1): K+1!>2K+1 is true
Multiplying both sides of P K by (K+1), we get,
K+1!>2KK+1>2K.2(as K>4)
Hence by PMI, P n is true â nâN, nâ¥4

Topic: Mathematics, Mathematical Induction, JEE Main
2NOClg⇌2NOg+Cl2g

In an experiment, 2.0 moles of NOCl was placed in a one-litre flask and the concentration of NO after equilibrium established, was found to be 0.4 mol/L. The equilibrium constant at 30 °C is_____×10-4.

Options:

The correct answer is:
125

Solution:
2NOClg⇌2NOg+Cl2g

t=02 M

t=teq2-xMxMx2M

∵ x=0.4 M

∴ NOCleq=1.6 M

NOeq=0.4 M

Cl2eq=0.2 M

⇒Kc=NO2Cl2NOCl2=0.420.21.62

Kc=322.56×10-3

Kc=12.5×10-3

Kc=125×10-4

Integer answer is 125

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2022 (27 Jun Shift 1)
2O3g⇌3O2g

At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is -......J mol-1. (Nearest integer)

[Given: ln 1.35=0.3 and R=8.3 J K-1 mol-1]

Options:

The correct answer is:
747

Solution:
2O3g⇌3O2g

Initially 1 mole 0

1-0.5 32×0.5

0.5 mole 1.52=0.75 mole

Kp=PO23PO32=0.751.2530.51.252=353252

=0.630.42=0.2160.16=1.35

ΔG°=-RTlnKp

=-8.3×300ln1.35

=-8.3×300×0.3

=-747J/mole

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2022 (24 Jun Shift 1)
$2 p$ orbitals have

Options:
A $n=1, I=2$
B $n=1, I=0$
C $n=2, I=1$
D $n=2, I=0$

The correct answer is:
$n=2, I=1$

Solution:
In $2 p$-orbital, 2 denotes principal quantum number $(n)$ and $p$ denotes azimuthal quantum number $(I=1)$.

Topic: Chemistry, Structure of Atom, JEE Main
\( 2 \pi-\left(\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{16}{65}\right) \) is equal to :

Options:
A \( \frac{\pi}{2} \)
B \( \frac{5 \pi}{4} \)
C \( \frac{3 \pi}{2} \)
D \( \frac{7 \pi}{4} \)

The correct answer is:
\( \frac{3 \pi}{2} \)

Solution:
sin-145+sin-1513+sin-11665

=tan-143+tan-1512+tan-11663

=tan-143+5121-43·512+tan-11663

=tan-16316+tan-11663=π2

∴2π-π2=3π2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
2 players A,B tosses a fair coin in cyclic order A,A,B,A,A,B… till a head appears. If the probability that A gets head first is p, then 24p is equal to

Options:

The correct answer is:
28

Solution:
Required probability =A+A-A+A-A-B-A+A-A-B-A-A+A-A-B-A-A-B-A
=12+122+124+125+127+128+1210+…
=12+124+127+1210+…+122+125+128+…
=121-18+141-18
=47+27=67=p

Hence, 24p=24×76=28

Topic: Mathematics, Probability, JEE Main
2 RCl + Si → 5 7 0 K Cu powder R 2 SiCl 2 → H 2 O R 2 Si(OH) 2 → Polymerisation X , then X will be

Options:
A Linear long chain silicone
B Cyclic silicone
C Cross linked silicone
D None of these

The correct answer is:
Linear long chain silicone

Solution:
Structure of X will be
Linear long chain polymer

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
2π-sin-145+sin-1513+sin-11665 is equal to :

Options:
A π2
B 5π4
C 3π2
D 7π4

The correct answer is:
3π2

Solution:
sin-145+sin-1513+sin-11665

=tan-143+tan-1512+tan-11663

=tan-143+5121-43·512+tan-11663

=tan-16316+tan-11663=π2

∴2π-π2=3π2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2020 (03 Sep Shift 1)
$2 \sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{24}{25}=$

Options:
A $\frac{\pi}{2}$
B $\frac{2 \pi}{3}$
C $\frac{5 \pi}{3}$
D None of these

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

Solution:
$\begin{aligned} & 2 \sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{24}{25}=\sin ^{-1} 2 \times \frac{3}{5} \sqrt{1-\frac{9}{25}}+\cos ^{-1} \frac{24}{25} \\ & \quad=\sin ^{-1} \frac{24}{25}+\cos ^{-1} \frac{24}{25}=\frac{\pi}{2}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$2 \sin A \cos ^3 A-2 \sin ^3 A \cos A=$

Options:
A $\sin 4 A$
B $\frac{1}{2} \sin 4 A$
C $\frac{1}{4} \sin 4 A$
D None of these

The correct answer is:
$\frac{1}{2} \sin 4 A$

Solution:
$\begin{aligned}
& 2 \sin A \cos ^3 A-2 \sin ^3 A \cos A \quad=2 \sin A \cos A\left(\cos ^2 A-\sin ^2 A\right) \\
& =2 \sin A \cos A \cos 2 A=\sin 2 A \cos 2 A=\frac{1}{2} \sin 4 A
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
2SO2( g)+O2( g)⇌2SO3( g)
In an equilibrium mixture, the partial pressures are
PSO3=43kPa;PO2=530 Pa and
PSO2=45kPa. The equilibrium constant
KP=_____×10-2. (Nearest integer)

Options:

The correct answer is:
172

Solution:
2SO2( g)+O2( g)=2SO3( g)
KP=pSO3( g)2pSO2g2×pO2( g)
=43×4345×45×0.530
=172.28×10-2 kPa-1
≃172 × 10-2
Ans is 172

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2021 (20 Jul Shift 1)
$2 \tan ^{-1}\left(\frac{1}{3}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=$

Options:
A $\frac{\pi}{4}$
B $0$
C $\tan ^{-1}\left(\frac{5}{4}\right)$
D $\frac{\pi}{2}$

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

Solution:
$\begin{aligned} & 2 \tan ^{-1}\left(\frac{1}{3}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1} \frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^2}+\cos ^{-1}\left(\frac{3}{5}\right) \\ & =\tan ^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right)+\cos ^{-1}\left(\frac{3}{5}\right) \\ & =\tan ^{-1}\left(\frac{3}{4}\right)+\cot ^{-1}\left(\frac{3}{4}\right) \\ & =\frac{\pi}{2}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=$

Options:
A $\tan ^{-1}\left(\frac{49}{29}\right)$
B $\frac{\pi}{2}$
C $0$
D $\frac{\pi}{4}$

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

Solution:
$\begin{aligned} & 2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\tan ^{-1}\left(\frac{2(1 / 3)}{1-(1 / 9)}\right)+\tan ^{-1}\left(\frac{1}{7}\right) \\ & \quad=\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{1}{7}=\tan ^{-1}\left(\frac{(3 / 4)+(1 / 7)}{1-(3 / 4) \times(1 / 7)}\right) \\ & \quad=\tan ^{-1}\left(\frac{25}{25}\right)=\frac{\pi}{4}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$2 \tan ^{-1}\left[\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right]=$

Options:
A $\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)$
B $\cos ^{-1}\left(\frac{a+b \cos \theta}{a \cos \theta+b}\right)$
C $\cos ^{-1}\left(\frac{a \cos \theta}{a+b \cos \theta}\right)$
D $\cos ^{-1}\left(\frac{b \cos \theta}{a \cos \theta+b}\right)$

The correct answer is:
$\cos ^{-1}\left(\frac{a \cos \theta+b}{a+b \cos \theta}\right)$

Solution:
$\begin{aligned} & 2 \tan ^{-1}\left[\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right]=\cos ^{-1}\left[\frac{1-\left(\frac{a-b}{a+b}\right) \tan ^2 \frac{\theta}{2}}{1+\left(\frac{a-b}{a+b}\right) \tan ^2 \frac{\theta}{2}}\right] \\ & \left(2 \tan ^{-1} x=\cos ^{-1} \frac{1-x^2}{1+x^2}\right) \\ & =\cos ^{-1}\left[\frac{(a+b)-(a-b) \tan ^2 \frac{\theta}{2}}{(a+b)+(a-b) \tan ^2 \frac{\theta}{2}}\right] \\ & =\cos ^{-1}\left[\frac{a\left(1-\tan ^2 \frac{\theta}{2}\right)+b\left(1+\tan ^2 \frac{\theta}{2}\right)}{a\left(1+\tan ^2 \frac{\theta}{2}\right)+b\left(1-\tan ^2 \frac{\theta}{2}\right)}\right] \\ & =\cos ^{-1}\left[\frac{\frac{a\left(1-\tan ^2 \frac{\theta}{2}\right)}{1+\tan ^2 \frac{\theta}{2}}+b}{a+b\left(\frac{1-\tan ^2 \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}\right)}\right]=\cos ^{-1}\left[\frac{a \cos \theta+b}{a+b \cos \theta}\right] \\ & \end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
2Zn+O2→2ZnO;ΔG°=-616 J
2Zn+S2→2ZnS;ΔG°=-293J
S2+2O2→2SO2;ΔG°=-408J
ΔG° for the following reaction
2ZnS+3O2→2ZnO+2SO2 is

Options:
A -731J
B -1317J
C -501J
D +731J

The correct answer is:
-731J

Solution:
For the reactions,
2ZnS→2Zn+S2;ΔG°=+293J ...(i)
2Zn+O2→2ZnO;ΔG°=-616J ...(ii)
S2+2O2→2SO2;ΔG°=-408J ...(iii)
The ΔG° for the reaction,
2ZnS+3O2→2ZnO+2SO2
can be obtained by adding equations. (i), (ii) and (iii)
So, ΔG°=293-616-408=-731J

Topic: Chemistry, Thermodynamics (C), JEE Main
2-bromopentane is heated with potassium ethoxide in ethanol. The major product is

Options:
A trans-pent-2-ene
B 2-ethoxypentane
C pent-1-ene
D 3-ethoxypentane

The correct answer is:
trans-pent-2-ene

Solution:
When 2-bromopentane is heated with
potassium ethoxide, we get trans-pent-2-ene.
Because when alkyl halides react in ethanol with potassium ethoxide, it give trans product as major product due to more symmetrical nature as compared to cis-isomer. The reaction occurs as follows:
(It shows elimination reaction.)
Hence, trans-2-pantane is the main product and option (a) is the correct answer.

Topic: Chemistry, Haloalkanes and Haloarenes, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 2)
2-celled pollen grains of angiosperm represent:

Options:
A Microspores
B Partially developed male gametophyte
C Mature male gametophyte
D Male gametangium

The correct answer is:
Partially developed male gametophyte

Solution:
The 3-celled pollen grain of Angiosperm is considered to be mature male gametophyte where the terminal position has more of cytoplasm and tube-nucleus. The two male gametes remain in the back side of tube nucleus. Nearly 40% of Angiosperms release their pollen grain in 3-celled stage while rest of the 60% of angiosperm release pollen grain in 2-celled stage. The 2-celled stage is referred to as partially developed male gametophyte.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
2-chlorobutane is heated with alcoholic $\mathrm{NaOH}$, the product formed in larger amount is

Options:
A 1-Butene
B 1-Butyne
C 2-Butene
D All of these

The correct answer is:
2-Butene

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
2-Methylbut-1-ene reacts with mercuric acetate in presence of water to form a product, which on reduction with NaBH4 yields

Options:
A 2-Methylbutan-2-ol
B 2-Methylbutan-1-ol
C 3-Methylbutan-2-ol
D none of the above

The correct answer is:
2-Methylbutan-2-ol

Solution:
This is an example of oxymercuration-demercuration and gives alcohol corresponding to Markovnikov's addition of water to carbon-carbon double bond.

Topic: Chemistry, Hydrocarbons, JEE Main
2-Methylbutan-2-ol can be obtained by the acid catalyzed hydration of

Options:
A CH3CH2CH=CH2
B CH3CH=CHCH2
C CH32C=CHCH3
D Either of the three

The correct answer is:
CH32C=CHCH3

Solution:
Proceed backward i.e., remove a molecule of water to get the alkene.

Topic: Chemistry, Hydrocarbons, 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, Hydrocarbons, JEE Main
2-Methylpent-2-ene on reductive ozonlysis will give

Options:
A Propanal only
B Propanal and ethanal
C Propanone & propanal
D Propan-2-ol and ethanal

The correct answer is:
Propanone & propanal

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
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, JEE Main
2RCl + Si → 570 K Cu powder R 2 SiCl 2 → H 2 O R 2 Si(OH) 2 → Polymerisation (A), (A)

Options:
A cyclic silicone
B cross linked silicone
C linear silicone
D none of these

The correct answer is:
linear silicone

Solution:

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
$30^{\circ} 30^t$ is equal to

Options:
A $\left(\frac{41}{2}\right)^{\circ}$
B $61^{\circ}$
C $\left(\frac{61}{2}\right)^{\circ}$
D None of these

The correct answer is:
$\left(\frac{61}{2}\right)^{\circ}$

Solution:
We know that $30^{\prime}=\left(\frac{1}{2}\right)^2$
$30^{\circ}+\left(\frac{1}{2}\right)^{\circ}=\left(\frac{61}{2}\right)^{\circ}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\int_{-3 \pi / 2}^{-\pi / 2}\left[(x+\pi)^3+\cos ^2(x+3 \pi)\right] d x$ is equal to

Options:
A
$\frac{\pi^4}{32}$

B
$\frac{\pi^4}{32}+\frac{\pi}{2}$

C
$\frac{\pi}{2}$

D
$\frac{\pi}{4}-1$

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

Solution:
$$
I=\int_{-3 \pi / 2}^{-\pi / 2}\left[(x+\pi)^3+\cos ^2(x+3 \pi)\right] d x
$$
Put $\mathrm{x}+\pi=\mathrm{t}$
$$
\begin{aligned}
& I=\int_{-\pi / 2}^{\pi / 2}\left[t^3+\cos ^2 t\right] d t=2 \int_0^{\pi / 2} \cos ^2 t d t \\
& =\int_0^{\pi / 2}(1+\cos 2 t) d t=\frac{\pi}{2}+0 .
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2006
∫324334489-4x2dx is equal to

Options:
A π3
B π2
C π6
D 2π

The correct answer is:
2π

Solution:
Given integration is

∫324334489-4x2dx

We know that

∫dxa2-x2=sin-1xa+C

So, the equation can be re-written as

∫324334482322-x2dx=482×sin-12x3324334

=24×sin-123×334-sin-123×324

=24×sin-132-sin-112

=24×π3-π4=2π

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (24 Jan Shift 2)
$\int \frac{\sin x d x}{3+4 \cos ^2 x}=$

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

The correct answer is:
$\frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+c$

Solution:
$\begin{aligned} & I=\int \frac{\sin x}{3+4 \cos ^2 x} d x \\ & \text { Put } \cos x=t \Rightarrow-\sin x d x=d t \\ & =\frac{-1}{4} \int \frac{d t}{t^2+\binom{\sqrt{3}}{2}^2} \\ & I=-\frac{1}{\text { 4. } \frac{\sqrt{3}}{2}} \cdot \tan ^{-1} \frac{t}{\left(\frac{\sqrt{3}}{2}\right)}+c=\frac{-1}{2 \sqrt{3}} \tan ^{-1} \frac{2 t}{\sqrt{3}}+c \\ & \Rightarrow I=\frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+c \\ & \end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
$\frac{1}{\sqrt[3]{6-3 x}}=$

Options:
A $6^{1 / 3}\left[1+\frac{x}{6}+\frac{2 x^2}{6^2}+\ldots.\right]$
B $6^{-1 / 3}\left[1+\frac{x}{6}+\frac{2 x^2}{6^2}+\ldots.\right]$
C $6^{1 / 3}\left[1-\frac{x}{6}+\frac{2 x^2}{6^2}-\ldots.\right]$
D $6^{-1 / 3}\left[1-\frac{x}{6}+\frac{2 x^2}{6^2}-\ldots\right]$

The correct answer is:
$6^{-1 / 3}\left[1+\frac{x}{6}+\frac{2 x^2}{6^2}+\ldots.\right]$

Solution:
$\begin{aligned} & \frac{1}{(6-3 x)^{1 / 3}}=(6-3 x)^{-1 / 3}=6^{-1 / 3}\left[1-\frac{x}{2}\right]^{-1 / 3} \\ & =6^{-1 / 3}\left[1+\left(-\frac{1}{3}\right)\left(-\frac{x}{2}\right) x+\frac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{2.1}\left(-\frac{x}{2}\right)^2+\ldots .\right] \\ & =6^{-1 / 3}\left[1+\frac{x}{6}+\frac{2 x^2}{6^2}+\ldots .\right]\end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main
3×722+2×1022-44 when divided by 18 leaves the remainder

Options:

The correct answer is:
15

Solution:
Given that: 3×722+2×1022-44

=3×(6+1)22+2×(9+1)22-44

As we know that (1+x)n=1+x(C1n+C2nx+.....)=1+kx ; k,x∈I

=18λ+3+18μ+2-44 ; λ,μ∈I
=18δ-39 ;δ∈I
=18β+15

⇒remainder =15

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (27 Aug Shift 2)
3. \( 88 \mathrm{~g} \) of a mixture of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) and \( \mathrm{CaCO}_{3} \) is heated to a constant mass of \( 3.00 \mathrm{~g} \). What was the mass of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) (in
grams) in the mixture?

Options:

The correct answer is:
1.88

Solution:
Na2CO3 does not decompose on heating.

CaCO3 decomposes to form CaO and CO2.

CaCO3⟶ΔCaO+CO2

∴ Loss in mass is due to loss of CO2 gas

∴ (3.88-3.00=0.88g) of CO2 is lost during heating.

0.88 g CO2=0.8844=0.02 mol CO2

From the reaction stoichiometry,

0.02 mol CO2≡0.02 mol CaCO3

0.02 mol CaCO3=0.02×100=2.00g CaCO3

∴ Mass of Na2CO3 in the mixture =3.88-2.00 g=1.88 g

Topic: Chemistry, s Block Elements, JEE Main
$3 \lambda \vec{c}+2 \mu(\vec{a} \times \vec{b})=0$ then

Options:
A
$3 \lambda+2 \mu=0$

B
$3 \lambda=2 \mu$

C
$\lambda=\mu$

D
$\lambda+\mu=0$

The correct answer is:
$3 \lambda=2 \mu$

Solution:
No Solution Available

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2002
3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) in a flask. After an hour, it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is:

Options:
A 54 mg
B 18 mg
C 36 mg
D 42 mg

The correct answer is:
18 mg

Solution:
Milliequivalents (Meqs) of CH3COOH (initial)=50×0.06=3
Milliequivalents of CH3COOH (final)=50×0.042=2.1
CH3COOH adsorbed=3-2.1=0.9 Meqs
=9×10-1×60 g/mol×10-3g
= 5 4 0 × 1 0 - 4 = 0.054 g
=54 mg
Amount of acetic acid adsorbed per gram of charcoal=543=18 mg/g of charcoal.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2015 (04 Apr)
3 infinitely long thin wires each carrying current i in the same direction, are in the x-y plane in a gravity-free space. The central wire is along the y-axis while the other two are along with x=âd. If the central wire is displaced along the z-direction by a small amount & released, the wire executes the simple harmonic motion. If the linear density of the wire is Î», find the frequency of oscillation.

Options:
A i2ÏdÎ¼0ÏÎ»
B i2πdπλμ0
C 2μdiπλμ0
D 2μdiμ0μλ

The correct answer is:
i2πdμ0πλ

Solution:
When wire (2) is displaced by x along y-axis force is taken per unit length
F→2=F→12+F→23
sinceF12=F23=μ04π2i2d2+x2
F2=2F12sinθ
F=2μ04π2i2d2+x2×xd2+x2=μ0πi2xd2+x2
x<<d
FR=-μ0πi2xd2 comparing with FR=-Kx⇒K=μ0πi2x2d2
n=12πKm=12πμ0i2πd2×λ=i2πdμ0πλ

Topic: Physics, Magnetic Effects of Current, JEE Main
3-Methylbutan-2-ol+HI→ΔX
Identify the position of carbon in the main chain to which nucleophile is added in the product 'X'?

Options:

The correct answer is:
2

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
\( 3 \mathrm{~g} \) of acetic acid is added to \( 250 \mathrm{~mL} \) of \( 0.1 \mathrm{M} \mathrm{HCl} \) and the solution made up to \( 500 \mathrm{~mL} \). To \( 20 \mathrm{~mL} \) of this solution \( \frac{1}{2} \mathrm{~mL} \) of \( 5 \mathrm{M} \mathrm{NaOH} \) is added. The \( \mathrm{pH} \) of the solution is [Given: pKa of acetic acid \( =4.75 \), molar mass of acetic acid \( 60 \mathrm{~g} / \mathrm{mol}, \log 3=0.4771 \), Neglect any changes in volume]

Options:

The correct answer is:
5.22

Solution:
millimole of acetic acid in 20mL=2
millimole of HCl in 20mL=1
millimole of NaOH=2.5

pH=PKa+log⁡3/22
=4.74+log⁡3
=4.74+0.48=5.22

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is

Options:
A zero
B two
C four
D six

The correct answer is:
four

Solution:
In the presence of peroxide anti-Markownikoff's product form.
E⨁ attacks C-where hindrance is less.
2 chiral centres are present in the product.
Number of stereoisomers 2n=22=4.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2017 (02 Apr)
3-Methylhex-2-ene on reaction with HBr in presence of peroxide forms an addition product (A). The number of possible stereoisomers for ' A ' is ________.

Options:

The correct answer is:
4

Solution:
The hydrogen bromide addition in peroxides follows Anti Markovnikov rule. Anti Markovnikov addition reaction is found to follow a free radical mechanism. The peroxide compound involved helps in the generation of free radicals. The Anti-Markovnikov rule defines regiochemistry in which the substituent is attached to a less substituted carbon instead of the more substituted carbon.

The reaction of 3-Methylhex-2-ene with HBr in presence of peroxide, takes place as follows,

2 chiral carbons, here, so, total number of stereoisomers will be 2n=22=4

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (27 Jan Shift 1)
3 moles of an ideal gas at a temperature of $27^{\circ} \mathrm{C}$ are mixed with 2 moles of an ideal gas at a temperature $227^{\circ} \mathrm{C}$, determine the equilibrium temperature of the mixture, assuming no loss of energy.

Options:
A $327^{\circ} \mathrm{C}$
B $107^{\circ} \mathrm{C}$
C $318^{\circ} \mathrm{C}$
D $410^{\circ} \mathrm{C}$

The correct answer is:
$107^{\circ} \mathrm{C}$

Solution:
Energy possessed by the ideal gas at $27^{\circ} \mathrm{C}$ is
$$
E_{1}=3\left(\frac{3}{2} R \times 300\right)=\frac{2700 R}{2}
$$
Energy possessed by the ideal gas at $227^{\circ} \mathrm{C}$ is
$$
\mathrm{E}_{2}=2\left(\frac{3 \mathrm{R}}{2} \times 500\right)=1500 \mathrm{R}
$$
If $\mathrm{T}$ be the equilibrium temperature, of the mixture, then its energy will be
$$
\mathrm{E}_{\mathrm{m}}=5\left(\frac{3 \mathrm{R} \mathrm{T}}{2}\right)
$$
Since, energy remains conserved,
$$
\mathrm{E}_{\mathrm{m}}=\mathrm{E}_{1}+\mathrm{E}_{2}
$$
or $\quad 5\left(\frac{3 R T}{2}\right)=\frac{2700 R}{2}+1500 R$
or $\quad \mathrm{T}=380 \mathrm{~K}$ or $107^{\circ} \mathrm{C}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
3 moles of an ideal monoatomic gas performs $A B C D A$ cyclic process as shown in figure below. The gas temperatures are $T_A=400 \mathrm{~K}$, $T_B=800 \mathrm{~K}, T_C=2400 \mathrm{~K}$ and $T_D=1200 \mathrm{~K}$. The work done by the gas is (approximately) $(R=8.314 \mathrm{~J} / \mathrm{mol} \mathrm{K})$

Options:
A $10 \mathrm{~kJ}$
B $20 \mathrm{~kJ}$
C $40 \mathrm{~kJ}$
D $100 \mathrm{~kJ}$

The correct answer is:
$20 \mathrm{~kJ}$

Solution:
Processes $A$ to $B$ and $C$ to $D$ are parts of straight line graphs of form $y=m x$.
Also,
$p=\frac{\mu R}{V} T(\mu=3)$
$p \propto T$
So, volume remains constant for the graphs $A B$ and $C D$.
So, no work is done during processes for $A$ to $B$ and $C$ to $D$.
$W_{A B}=W_{C D}=0$
and $\quad W_{B C}=p_2\left(V_C-V_B\right)$
$=\mu R\left(T_C-T_B\right)$
$=3 R(2400-800)$
$=3 R \times 1600$
$=4800 R$
$W_{D A}=p_1\left(V_A-V_D\right)$
$=\mu R\left(T_A-T_D\right)$
$=3 R(400-1200)$
$=-2400 R$
Work done in complete cycle
$W=W_{A B}+W_{B C}+W_{C D}+W_{D A}$
$=0+4800 R+0+(-2400) R$
$=2400 R$
$=19.944 \mathrm{~J}=20 \mathrm{~kJ}$

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2010