datavorous/entrance-exam-dataset huggingface dataset:
$$
\lim _{x \rightarrow 2}\left(\frac{\sqrt{1-\cos \{2(x-2)\}}}{x-2}\right)

Options:
A
equals $\sqrt{2}$

B
equals $-\sqrt{2}$

C
equals $\frac{1}{\sqrt{2}}$

D
does not exist

The correct answer is:
does not exist

Solution:
$$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{\sqrt{2 \sin ^2(x-2)}}{x-2} \\
& \lim _{x \rightarrow 2} \frac{\sqrt{2}|\sin (x-2)|}{x-2} \\
& \text { R.H.L. }=\sqrt{2}, \text { L.H.L. }=-\sqrt{2}
\end{aligned}
$$
Limit does not exist.

Topic: Mathematics, Limits, JEE Main, JEE Main 2011
\(\lim _{x \rightarrow \pi} \frac{1-\sin x / 2}{\left(\cos \frac{x}{2}\right)\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)}=\)

Options:
A \(\frac{3}{\sqrt{2}}\)
B \(\frac{-1}{\sqrt{2}}\)
C \(\frac{1}{\sqrt{2}}\)
D \(\frac{5}{\sqrt{2}}\)

The correct answer is:
\(\frac{1}{\sqrt{2}}\)

Solution:
\(\lim _{x \rightarrow \pi} \frac{1-\sin \frac{x}{2}}{\left(\cos \frac{x}{2}\right)\left(\cos \frac{x}{4}-\sin \frac{x}{4}\right)}\)
Let \(\quad x=\pi+h, x \rightarrow \pi, h \rightarrow 0\)
\(\begin{gathered}
=\lim _{\substack{x \rightarrow \pi \\
h \rightarrow 0}} \frac{1-\sin \left(\frac{\pi}{2}+\frac{h}{2}\right)}{\cos \left(\frac{\pi}{2}+\frac{h}{2}\right)\left(\cos \left(\frac{\pi+h}{4}\right)-\sin \left(\frac{\pi+h}{4}\right)\right)} \\
=\lim _{h \rightarrow 0} \frac{1-\cos \frac{h}{2}}{-\sin \frac{h}{2} \cdot\left(\cos \frac{\pi}{4} \cos \frac{h}{4}-\sin \frac{\pi}{4} \sin \frac{h}{4}-\sin \frac{\pi}{4} \cdot \cos \frac{h}{4}-\cos \frac{\pi}{4} \cdot \sin \frac{h}{4}\right)}
\end{gathered}\)
\(\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\sqrt{2}\left(1-\cos \frac{h}{2}\right)}{-\sin \frac{h}{2}\left\{\cos \frac{h}{4}-\sin \frac{h}{4}-\cos \frac{h}{4}-\sin \frac{h}{4}\right\}} \\ & =\lim _{h \rightarrow 0} \frac{\sqrt{2}\left(1-\cos \frac{h}{2}\right)}{\sin \frac{h}{2} \cdot\left(-2 \sin \frac{h}{4}\right)} \\ & =\lim _{h \rightarrow 0} \frac{\sqrt{2} \cdot 2 \sin ^2 \frac{h}{4}}{2 \cdot \sin \frac{h}{2} \times \sin \frac{h}{4}} \\ & =\lim _{h \rightarrow 0} \sqrt{2} \frac{\sin \frac{h}{4}}{\sin \frac{h}{2}}=\sqrt{2} \lim _{h \rightarrow 0} \frac{\sin \frac{h}{4}}{4\left(\frac{h}{4}\right)} \cdot \frac{2(h / 2)}{\sin \frac{h}{2}} \\ & =\sqrt{2} \times \frac{2}{4} \cdot \lim _{h \rightarrow 0}\left(\frac{\sin \frac{h}{4}}{\frac{h}{4}}\right)\left(\frac{\frac{h}{2}}{\sin \frac{h}{2}}\right) \\ & =\sqrt{2} \times \frac{2}{4} \times 1 \times 1=\frac{1}{\sqrt{2}}\end{aligned}\)

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow \infty}\left(\frac{x+2}{x+1}\right)^{x+3}$

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

The correct answer is:
$e$

Solution:
Let $A=\lim _{x \rightarrow \infty}\left(\frac{x+2}{x+1}\right)^{x+3}$
$\begin{array}{l}
=\lim _{x \rightarrow \infty}\left(1+\frac{1}{x+1}\right)^{x+3}=\lim _{x \rightarrow \infty}\left[\left(1+\frac{1}{x+1}\right)^{x+1}\right]^{\frac{(x+3)}{(x+1)}}=e \\
\left\{\because \lim _{x \rightarrow \infty}\left(1+\frac{1}{x+1}\right)^{x+1}=e\right. \text { and } \left.\lim _{x \rightarrow \infty} \frac{(x+3)}{(x+1)}=\lim _{x \rightarrow \infty} \frac{\{1+(3 / x)\}}{\{1+(1 / x)\}}=1\right\} .
\end{array}$

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow \infty} \frac{\sqrt{x^2+a^2}-\sqrt{x^2+b^2}}{\sqrt{x^2+c^2}-\sqrt{x^2+d^2}}=$

Options:
A $\frac{a^2-b^2}{c^2-d^2}$
B $\frac{a^2+b^2}{c^2-d^2}$
C $\frac{a^2+b^2}{c^2+d^2}$
D None of these

The correct answer is:
$\frac{a^2-b^2}{c^2-d^2}$

Solution:
\(\lim _{x \rightarrow \infty}\left[\frac{\sqrt{x^2+a^2}-\sqrt{x^2+b^2}}{\sqrt{x^2+c^2}-\sqrt{x^2+d^2}}\right]\)
Rationalising the numerator and the denominator :
\(\begin{aligned}
& \lim _{x \rightarrow \infty}\left[\frac{\left(\sqrt{x^2+a^2}-\sqrt{x^2+b^2}\right)}{\left(\sqrt{x^2+c^2}-\sqrt{x^2+d^2}\right)} \times \frac{\left(\sqrt{x^2+c^2}+\sqrt{x^2+d^2}\right)}{\left(\sqrt{x^2+c^2}+\sqrt{x^2+d^2}\right)} \times \frac{\left(\sqrt{x^2+a^2}+\sqrt{x^2+b^2}\right)}{\left(\sqrt{x^2+a^2}+\sqrt{x^2+b^2}\right)}\right] \\
& =\lim _{x \rightarrow \infty}\left[\frac{\left(\sqrt{x^2+a^2}-\sqrt{x^2+b^2}\right)\left(\sqrt{x^2+a^2}+\sqrt{x^2+b^2}\right)\left(\sqrt{x^2+c^2}+\sqrt{x^2+d^2}\right)}{\left(\sqrt{x^2+c^2}-\sqrt{x^2+d^2}\right)\left(\sqrt{x^2+c^2}+\sqrt{x^2+d^2}\right)\left(\sqrt{x^2+a^2}+\sqrt{x^2+b^2}\right)}\right] \\
& =\lim _{x \rightarrow \infty} \frac{\left(x^2+a^2\right)-\left(x^2+b^2\right)}{\left(x^2+c^2\right)-\left(x^2+d^2\right)} \times\left(\frac{\sqrt{x^2+c^2}+\sqrt{x^2+d^2}}{\sqrt{x^2+a^2}+\sqrt{x^2+b^2}}\right) \\
& =\lim _{x \rightarrow \infty}\left(\frac{a^2-b^2}{c^2-d^2}\right)\left(\frac{\sqrt{x^2+c^2}+\sqrt{x^2+d^2}}{\sqrt{x^2+a^2}+\sqrt{x^2+b^2}}\right)
\end{aligned}\)
Dividing the numerator and the denominator by x :
\(\begin{aligned}
& \lim _{x \rightarrow \infty}\left(\frac{a^2-b^2}{c^2-d^2}\right)\left(\frac{\sqrt{1+\frac{c^2}{x^2}}+\sqrt{1+\frac{d^2}{x^2}}}{\sqrt{1+\frac{1}{x^2}}+\sqrt{1+\frac{b^2}{x^2}}}\right) \\
& A s x \rightarrow \infty, \frac{1}{x}, \frac{1}{x^2} \rightarrow 0 \\
& =\left(\frac{a^2-b^2}{c^2-d^2}\right)\left(\frac{\sqrt{1}+\sqrt{1}}{\sqrt{1}+\sqrt{1}}\right) \\
& =\frac{a^2-b^2}{c^2-d^2}
\end{aligned}\)

Topic: Mathematics, Limits, JEE Main
$\lim _{x \rightarrow \alpha} \frac{\sin x-\sin \alpha}{x-\alpha}=$

Options:
A $0$
B $1$
C $\sin \alpha$
D $\cos \alpha$

The correct answer is:
$\cos \alpha$

Solution:
$\lim _{x \rightarrow \alpha} \frac{\cos x}{1}=\cos \alpha$, (Apply L-Hospital's rule)

Topic: Mathematics, Limits, JEE Main
Lime reacts exothermally with water to give ‘A’ which has low solubility in water. Aqueous solution of ‘A’ is often used for the test of CO2, a test in which insoluble B is formed. If B is further reacted with CO2 then soluble compound is formed. ‘A' is

Options:
A Quick lime
B Slaked lime
C White lime
D Lime water

The correct answer is:
Slaked lime

Solution:
CaOLime→Limited waterCaOH2

The process is known as slaking of lime and the product is slaked lime. It is an exothermic process.

CaOH2aq solution is known as lime water and used for test of CO2.

CaOH2aq→CO2CaCO3CaCO3+CO2excess+H2O→CaHCO32

Milkiness is due to the formation of calcium carbonate and disappearance of milkiness is due to the formation of calcium bicarbonate.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Limit $\lim _{x \rightarrow \infty} x^{2} \int_{0}^{x} e^{t^{3}-x^{3}}$ dt equals

Options:
A $\frac{1}{3}$
B 2
C $\infty$
D $\frac{2}{3}$

The correct answer is:
$\frac{1}{3}$

Solution:
$$
\lim _{x \rightarrow \infty} \frac{x^{2} \int_{0}^{x} e^{x^{3}} d t}{e^{x^{3}}}
$$
Apply L Hospital
$$
\begin{array}{l}
\lim _{x \rightarrow \infty} \frac{2 x \int_{0}^{x} e^{t^{3}}+x^{2} e^{x^{3}}}{3 x^{2} e^{x^{3}}} \\
\lim _{x \rightarrow \infty} \frac{2 \int_{0}^{x} e^{t^{3}}+x e^{x^{3}}}{3 x e^{x^{3}}} \\
\lim _{x \rightarrow \infty} \frac{2 e^{x^{3}}+e^{x^{3}}+3 x^{3} e^{x^{3}}}{3 e^{x^{3}}+9 x^{3} e^{x^{3}}} \\
\lim _{x \rightarrow \infty} \frac{3+3 x^{3}}{3+9 x^{3}}=\frac{1}{3}
\end{array}
$$

Topic: Mathematics, Limits, JEE Main
Limiting compositions of $f$-block hydrides are

Options:
A $\mathrm{MH}_2$ and $M \mathrm{H}_3$
B $\mathrm{MH}_3$ and $\mathrm{MH}_5$
C $M \mathrm{H}_2$ and $M \mathrm{H}_8$
D $\mathrm{MH}_2$ and $M \mathrm{H}_6$

The correct answer is:
$\mathrm{MH}_2$ and $M \mathrm{H}_3$

Solution:
Limiting composition of $f$ block hydrides are $\mathrm{MH}_2$ and $\mathrm{MH}_3$.

Topic: Chemistry, Hydrogen, JEE Main
Line L1â¡3x-4y+1=0 touches the circles C1 and C2. Centres of C1 and C2 are A11,2 and A23,1 respectively, then identify the INCORRECT statement from the following statements.

Options:
A L1 is direct common tangent of these circles
B L1 is transverse common tangent to these circles
C Radius of circle C1 is 45 units
D Radius of circle C2 is 65 units

The correct answer is:
L1 is direct common tangent of these circles

Solution:
As A1 & A2 lie on different sides of L1 hence, L1 is transverse common tangent
Radius of C1=3-8+15=45
Radius of C2=9-4+15=65

Topic: Mathematics, Circle, JEE Main
Line L has intercepts a and b on the coordinate axes, when the axes are rotated through a given angle keeping the origin fixed, the same line has intercepts p and q, then

Options:
A a 2 + b 2 = p 2 + q 2
B 1 a 2 + 1 b 2 = 1 p 2 + 1 q 2
C a 2 + p 2 = b 2 + q 2
D 1 a 2 + 1 p 2 = 1 b 2 + 1 q 2

The correct answer is:
1 a 2 + 1 b 2 = 1 p 2 + 1 q 2

Solution:
Equation of L is x a + y b = 1 and its distance from the origin is 1 1 a 2 + 1 b 2
Let the axis be rotated through an angle θ and let X, Y be the new coordinates of any point P x, y in the plane, then new equation of L will be
X p + Y q = 1 and its distance from the origin is 1 1 p 2 + 1 q 2
As the origin is fixed, distance from origin will be same in both cases
1 a 2 + 1 b 2 = 1 p 2 + 1 q 2

Topic: Mathematics, Straight Lines, JEE Main
Line PQ is parallel to y-axis and moment of inertia of a rigid body about PQ line is given by I=2x2-12x+27, where x is in meter and I is in kg m2. The minimum value of I is:

Options:
A 27 kg m2
B 11Â kgÂ m2
C 17Â kgÂ m2
D 9Â kgÂ m2

The correct answer is:
9 kg m2

Solution:
I=2x2-12x+27
dIdx =4x-12
For minimum or maximum
dIdx =0âx=3
d2Idx2=4. Hence I is minimum at x=3
I=9 kg m2

Topic: Physics, Rotational Motion, JEE Main
Line x+1λ=y-1=z+2-4 is perpendicular to 2x+2y-8z+5=0, then λ is

Options:
A 1
B -4
C -5
D -3

The correct answer is:
1

Solution:
When line x-x1a1=y-y1b1=z-z1c1 is perpendicular to plane ax+by+cz+d=0, then

a1a=b1b=c1c

∴ λ2=12=-4-8

⇒λ=1

Hence, option (a) is correct.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Lines are drawn parallel to the line 4x-3y+2=0, at a distance 35 units from the origin. Then which one of the following points lies on any of these lines?

Options:
A 14,-13
B -14,23
C -14,-23
D 14,13

The correct answer is:
-14,23

Solution:
The equation of a line parallel to the line ax+by+c=0 is ax+by+k=0.

Thus, the equation of a line parallel to 4x-3y+2=0 is 4x-3y+c=0.

The distance of a line ax+by+c=0 from origin is ca2+b2+c2.

Given the line 4x-3y+c=0 is at a distance of 35 units from origin

⇒c16+9=35

⇒c5=35

⇒c=±3.

Hence, the equation of the lines are 4x-3y+3=0 and 4x-3y-3=0.

Clearly point -14,23 satisfy the given equation 4x-3y+3=0.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Lines L1 & L2 are rotating in an anticlockwise direction about the points Aâ2,0 and B2,0 respectively in such a way that the speed of angle of rotation of line L2 is double as that of L1. Initially equations of both lines are y=0. If the angle of rotation of line L2 varies between 0 to Ï2, then the locus of the point of intersection P of lines L1 & L2 is part of a circle whose radius is equal to

Options:
A 2 units
B 4 units
C 6 units
D 8 units

The correct answer is:
4 units

Solution:
Locus of P is part of circle whose center is 2,0 & radius is equal to 4 units

Topic: Mathematics, Circle, JEE Main
Liquid helium is stored at its boiling point (4.2 K) in a spherical can, separated by a vacuum space from a surrounding shield which is maintained at the temperature of liquid nitrogen (77 K). If the can is 0.1m in radius and is blacked on the outside so that it acts as a black body, how much helium boils away per hour?
(Latent heat of vapourization is 21 kJ/kg)

Options:
A 43 g/h
B 43 kg/h
C 4.3 g/h
D 43 x 10-3 g/h

The correct answer is:
43 g/h

Solution:
Heat absorbed per second by liquid helium α A = T 0 4 - T 4
Heat required to boil liquid helium away = dm dt L
dm dt = σ A T 0 4 - T 4 L
= 6.67 × 1 0 - 8 × 4 × 3.14 × 0.1 2 7 7 4 - 4.2 4 2 1 × 1 0 3
= 1.19 × 1 0 - 5 kg/s
= 4.3 × 1 0 - 2 kg/h = 4 3 g/h

Topic: Physics, Thermodynamics, JEE Main
Liquid M and liquid N form an ideal solution. The vapour pressures of pure liquids M and N are 450 and 700 mmHg, respectively, at the same temperature. Then correct statements is:
( xM=Mole fraction of 'M' in solution;
xN=Mole fraction of 'N'in solution;
yM=Mole fraction of 'M' in vapour phase;
yN=Mole fraction of 'N' in vapour phase; )

Options:
A xMxN>yMyN
B xMxN=yMyN
C xMxN<yMyN
D xM-yM<xN-yN

The correct answer is:
xMxN>yMyN

Solution:
PMo is given as=450 mm of Hg.

PNo is given as=700 mm of Hg.

We know that PM=xM×P MO→ Raoult’s law.

PN=xN×P NO→ Raoult’s law.

From Dalton’s law PM=yM×PT

PN=yM×PT

So,

xM×P Mo=yM×PT …(i)

xM×PNo=yN×PT …(ii)

(i) (ii) xMxN×P MoP No = yMyN

xMxN=yMyN×750450

xMxN>yMyN

Topic: Chemistry, Solutions, JEE Main, JEE Main 2019 (09 Apr Shift 1)
Liquid oxygen at 50 K is heated to 300 K at a constant pressure of 1 atm. The rate of heating is constant. Among the following graphs that represent the variation of temperature with time is

Options:
A
B
C
D

The correct answer is:

Solution:
Temperature of liquid oxygen will first increase in the same phase. Then, phase change ( liquid to gas) will take place. During which temperature will remain constant. After that temperature of oxygen in gaseous state will further increase.

Topic: Physics, Thermodynamics, JEE Main
Liquid oxygen at $50 \mathrm{~K}$ is heated to $300 \mathrm{~K}$ at constant pressure of 1 atm. The rate of heating is constant. Which one of the following graphs represents the variation of temperature with time?

Options:
A

B

C

D

The correct answer is:

Solution:
Graph (a) shows the variation of temperature with time. At first temperature will increase then there will be state change from liquid to gas.

Topic: Physics, Thermal Properties of Matter, JEE Main
Liquid oxygen remains suspended between two pole faces of a magnet because it is

Options:
A diamagnetic
B paramagnetic
C ferromagnetic
D antiferromagnetic.

The correct answer is:
paramagnetic

Solution:
Molecular bonding of $\mathrm{O}_2$ based on orbitals show that $\mathrm{O}_2$ is paramagnetic, which is true experimentally.

Topic: Physics, Magnetic Properties of Matter, AIIMS, AIIMS 2004
Liquids A and B are at 30   o C and 20   o C , respectively. When mixed in equal masses, the temperature of the mixture is found to be 26   o C , The specific heats of A and B are in the ratio of m:n , where m and n are integers, then find minimum value of m+n.

Options:

The correct answer is:
5

Solution:
At thermal equilibrium
heat rejected by one liquid = heat accepted by other liquid
msA30â26=msB26â20
â4sA=6sB
â´ sAsB=64=32 = m n âm+n=3+2=5

Topic: Physics, Thermal Properties of Matter, JEE Main
Liquids A and B form an ideal solution. At $30^{\circ} \mathrm{C}$, the total vapour pressure of a solution containing $1 \mathrm{~mol}$ of A and $2 \mathrm{~mol}$ of B is $250 \mathrm{~mm} \mathrm{Hg}$. The total vapour pressure becomes $300 \mathrm{~mm} \mathrm{Hg}$ when 1 more mol of $\mathrm{A}$ is added to the first solution. The vapour pressures of pure $\mathrm{A}$ and $\mathrm{B}$ at the same temperature are

Options:
A
$150,450 \mathrm{mmHg}$

B
$125,150 \mathrm{~mm} \mathrm{Hg}$

C
$450,150 \mathrm{~mm} \mathrm{Hg}$

D
$250,300 \mathrm{~mm} \mathrm{Hg}$

The correct answer is:
$450,150 \mathrm{~mm} \mathrm{Hg}$

Solution:
Let vapour pressure of $A=P_A^0$
Vapour pressure of $B=P_B^0$
In first solution,
Mole fraction of $A\left(x_A\right)=\frac{1}{1+2}=\frac{1}{3}$
Mole fraction of $B\left(x_B\right)=\frac{2}{1+2}=\frac{2}{3}$
According to Raoult's law,
Total vapour pressure
$$
\begin{aligned}
& =250=P_A^0 x_A+P_B^0 x_B \\
& 250=\frac{1}{3} P_A^0+\frac{2}{3} P_B^0
\end{aligned}
$$
In second solution
Mole fraction of $A\left(x_A\right)=\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}$
Mole fraction of $B\left(x_B\right)=\frac{2}{4}=\frac{1}{2}$
$\therefore$ Total vapour pressure
$$
\begin{aligned}
& =300=P_A^0 x_A+P_B^0 x_B \\
& 300=\frac{1}{2} P_A^0+\frac{1}{2} P_B^0
\end{aligned}
$$
Multiplying equation (i) by $\frac{1}{2}$ and equation (ii) by $\frac{1}{3}$
$$
\begin{aligned}
& \frac{1}{6} P_A^0+\frac{2}{6} P_B^0=125 \\
& \frac{1}{6} P_A^0+\frac{1}{6} P_B^0=100 \\
& \frac{1}{6} P_B^0=25 \\
& P_B^0=25 \times 6=150 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$
On substituting value of $P_B^0$ in equation
(ii) we get
$$
\begin{aligned}
& 300=P_A^0 \times \frac{1}{2}+150 \times \frac{1}{2} \\
& P_A^0=450 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main
Liquids A and B form an ideal solution. At $30^{\circ} \mathrm{C}$, the total vapour pressure of a solution containing $1 \mathrm{~mol}$ of A and $2 \mathrm{~mol}$ of B is $250 \mathrm{~mm} \mathrm{Hg}$. The total vapour pressure becomes $300 \mathrm{~mm} \mathrm{Hg}$ when 1 more mol of $\mathrm{A}$ is added to the first solution. The vapour pressures of pure $\mathrm{A}$ and $\mathrm{B}$ at the same temperature are

Options:
A
$150,450 \mathrm{mmHg}$

B
$125,150 \mathrm{~mm} \mathrm{Hg}$

C
$450,150 \mathrm{~mm} \mathrm{Hg}$

D
$250,300 \mathrm{~mm} \mathrm{Hg}$

The correct answer is:
$450,150 \mathrm{~mm} \mathrm{Hg}$

Solution:
Let vapour pressure of $A=P_A^0$
Vapour pressure of $B=P_B^0$
In first solution,
Mole fraction of $A\left(x_A\right)=\frac{1}{1+2}=\frac{1}{3}$
Mole fraction of $B\left(x_B\right)=\frac{2}{1+2}=\frac{2}{3}$
According to Raoult's law,
Total vapour pressure
$$
\begin{aligned}
& =250=P_A^0 x_A+P_B^0 x_B \\
& 250=\frac{1}{3} P_A^0+\frac{2}{3} P_B^0
\end{aligned}
$$
In second solution
Mole fraction of $A\left(x_A\right)=\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}$
Mole fraction of $B\left(x_B\right)=\frac{2}{4}=\frac{1}{2}$
$\therefore$ Total vapour pressure
$$
\begin{aligned}
& =300=P_A^0 x_A+P_B^0 x_B \\
& 300=\frac{1}{2} P_A^0+\frac{1}{2} P_B^0
\end{aligned}
$$
Multiplying equation (i) by $\frac{1}{2}$ and equation (ii) by $\frac{1}{3}$
$$
\begin{aligned}
& \frac{1}{6} P_A^0+\frac{2}{6} P_B^0=125 \\
& \frac{1}{6} P_A^0+\frac{1}{6} P_B^0=100 \\
& \frac{1}{6} P_B^0=25 \\
& P_B^0=25 \times 6=150 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$
On substituting value of $P_B^0$ in equation
(ii) we get
$$
\begin{aligned}
& 300=P_A^0 \times \frac{1}{2}+150 \times \frac{1}{2} \\
& P_A^0=450 \mathrm{~mm} \mathrm{Hg}
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2012 (19 May Online)
Liquids A and B form an ideal solution in the entire composition range. At $350 \mathrm{~K}$, the vapor pressures of pure A and pure B are $7 \times 10^{3} \mathrm{~Pa}$ and $12 \times 10^{3} \mathrm{~Pa}$, respectively. The composition of the vapour is in equilibrium with a solution containing 40 mole percent of $\mathrm{A}$ at this temperature is:

Options:
A $x_{\mathrm{A}}=0.37 ; x_{\mathrm{B}}=0.63$
B $x_{\mathrm{A}}=0.28 ; x_{\mathrm{B}}=0.72$
C $x_{\mathrm{A}}=0.4 ; x_{\mathrm{B}}=0.6$
D $x_{\mathrm{A}}=0.76 ; x_{\mathrm{B}}=0.24$

The correct answer is:
$x_{\mathrm{A}}=0.28 ; x_{\mathrm{B}}=0.72$

Solution:
$\mathrm{P}_{\mathrm{A}}^{\circ}=7 \times 10^{3}$

$\mathrm{P}_{\mathrm{B}}^{\circ}=12 \times 10^{3}$

$x_{\mathrm{A}}^{\prime}=0.4 ; x_{\mathrm{B}}^{\prime}=1-0.4$

$x_{\mathrm{B}}^{\prime}=0.6$

$\mathrm{P}_{\text {total }}=\mathrm{P}_{\mathrm{A}^{\circ}}^{x_{\mathrm{A}}^{\prime}}+\mathrm{P}_{\mathrm{B}}^{\circ} x_{\mathrm{B}}^{\prime}$

$=7 \times 10^{3} \times 0.4+12 \times 10^{3} \times 0.6$

$=(7 \times 0.4+12 \times 0.6) \times 10^{3}=10^{4}$

$x_{\mathrm{A}}=\frac{\mathrm{P}_{\mathrm{A}^{\circ}}^{x^{\prime}} \mathrm{A}}{\mathrm{P}_{\text {total }}}=\frac{7 \times 10^{3} \times 0.4}{10^{4}}$

$\therefore x_{\mathrm{A}}=0.28, x_{\mathrm{B}}=1-0.28=0.72$

Topic: Chemistry, Solutions, JEE Main
Liquids A and B form an ideal solution in the entire composition range. At 350K, the vapour pressure of pure A and pure B are 7×103Pa and 12×103Pa , respectively. The composition of the vapour in equilibrium with a solution containing 40 mole percent of A at this temperature is:

Options:
A xA=0.4;xB=0.6
B xA=0.76;xB=0.24
C xA=0.28;xB=0.72
D xA=0.37;xB=0.63

The correct answer is:
xA=0.28;xB=0.72

Solution:
Since, PA0=7Ã103Pa;âPB0=12Ã103Pa and XA=0.4, X B =1â0.4=0.6
According to Raoult's Law: PT=PA0XA+PB0XB
PT=7Ã1030.4+12Ã1030.6
PT=10Ã103Pa
Now, PA0XA=PTxA
xA=PA0XAPT=7Ã103Ã0.4104â0.28
xB=1-xAââ1-0.28â0.72

Topic: Chemistry, Solutions, JEE Main, JEE Main 2019 (10 Jan Shift 1)
List 10 modern-day animals and using the internet resources link it to a corresponding ancient fossil. Name both.

Options:

None

Solution:
(i) Cockroach, Limulus (king crab), Neopilina, Latimaria (Fish) are fossil that has remain unchanged over years.
(ii) Trilobites - fossil arthropods
(iii) Lung fishes - connecting link between fishes and amphibians
(iv) Peripatus - connecting link between annelids and arthropods
(v) Woody mammoth - ice fossils
(vi) Gastropods - mould and cast fossil
(vii) Giant elk - amber fossil of asphalt
(viii) Dinosaur footprint-imprints

Topic: Biology, Evolution
List five main groups of natural plant growth regulators. Write a note on discovery, physiological functions and agricultural/horticultural applications of any one of them.

Options:

None

Solution:
The five main groups of natural growth regulators are
(a) auxins
(b) gibberellins
(c) cytokinins
(d) ethylene
(e) abscisic acid
Gibberellins
Discovery : - Theyare another kind of promotory PGR. There are more than 100 gibberellins reported from different organisms such as fungi and higher plants. They are denoted as $\mathrm{GA}_1, \mathrm{GA}_2, \mathrm{GA}_3$. E. Kurosawa reported the symptoms of the disease in infected rice seedings when they were treated with filtrates of the fungus. Gibberalla fujikuroi caused, 'bakane' (foolish seedling) a disease of rice seedlings. The Physiological functions
(i) They cause an increase in length of axis is used to increase the length of grapes stalk.
(ii) Gibberellins cause fruit like apple to elongate and improve its shape.
(iii) They also delay senescence. Thus the fruits can be left on the tree longer so as to extend the market period.
Agricultural Applications
(i) Spraying sugarcane crop with gibberellins increases the length of stem. Thus increasing the yield as much as 20 tonnes per acre.
(ii) Spraying juvenile conifers with GAs hastens the maturity period, thus leading to early seed production.
(iii) Gibberellins also promotes bolting (internode elongation just prior to flowering) in beet, cabbages and many plants with rosette habit.

Topic: Biology, Plant - Growth and Development
List -I

(Drug)

List-II
(Class of Drug)

a	Furacin	i	Antibiotic

b	Arsphenamine	ii	Tranquilizers

c	Dimetone	iii	Antiseptic

d	Valium	iv

Synthetic antihistamines

Choose the most appropriate match :

Options:
A (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
B (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
C (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
D (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

The correct answer is:
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

Solution:
→ furacine acts as Antiseptic
→ Arsphenamine also known as salvarsan acts as antibiotic
→ Dimetone is synthetic histamine
→ valium is a Tranqulizer

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2021 (27 Jul Shift 1)
List-I	List-II

(a) MI of the rod (length L, Mass M, about an axis ⊥ to the rod passing through the midpoint)	(i) 8ML23

(b) MI of the rod (length L, Mass 2M, about an axis ⊥ to the rod passing through one of its end)	(ii) ML23

(c) MI of the rod (length 2L, Mass M, about an axis ⊥ to the rod passing through its midpoint)	(iii) ML212

(d) MI of the rod (Length 2L, Mass 2M, about an axis ⊥ to the rod passing through one of its end)	(iv) 2ML23

Choose the correct answer from the options given below:

Options:
A (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)
B (a)-(ii), (b)-(i), (c)- (iii), (d)-(iv)
C (a)-(iii), (b)-(iv), (c)- (ii), (d)-(i)
D (a) - (iii), (b)-(iv), (c)-(i), (d)-(ii)

The correct answer is:
(a)-(iii), (b)-(iv), (c)- (ii), (d)-(i)

Solution:

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (27 Jul Shift 1)