JEE Main 2016 Chemistry Question Paper with Answer and Solution

(A) Given the curve $y = 1 + \sqrt{4x - 3}$.
First,find the slope of the tangent by differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2\sqrt{4x - 3}} \times 4 = \frac{2}{\sqrt{4x - 3}}$.
Given the slope of the tangent is $\frac{2}{3}$,we set:
$\frac{2}{\sqrt{4x - 3}} = \frac{2}{3} \Rightarrow \sqrt{4x - 3} = 3$.
Squaring both sides,$4x - 3 = 9 \Rightarrow 4x = 12 \Rightarrow x = 3$.
Substituting $x = 3$ into the curve equation,$y = 1 + \sqrt{4(3) - 3} = 1 + \sqrt{9} = 4$.
So,the point $P$ is $(3, 4)$.
The slope of the normal at $P$ is the negative reciprocal of the tangent slope:
$m_{normal} = -\frac{1}{2/3} = -\frac{3}{2}$.
The equation of the normal at $(3, 4)$ is:
$y - 4 = -\frac{3}{2}(x - 3) \Rightarrow 2y - 8 = -3x + 9 \Rightarrow 3x + 2y - 17 = 0$.
Checking the options,for $(1, 7)$: $3(1) + 2(7) - 17 = 3 + 14 - 17 = 0$. Thus,the normal passes through $(1, 7)$.

(C) According to Henry's Law,the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid ($S \propto P$ or $S = K_H \times P$).
Given:
$P_1 = 500 \ torr$,$S_1 = 0.01 \ g \ L^{-1}$
$P_2 = 750 \ torr$,$S_2 = ?$
Using the relation $\frac{S_1}{P_1} = \frac{S_2}{P_2}$:
$\frac{0.01}{500} = \frac{S_2}{750}$
$S_2 = \frac{0.01 \times 750}{500}$
$S_2 = 0.01 \times 1.5 = 0.015 \ g \ L^{-1}$