TS EAMCET 2003 Chemistry Question Paper with Answer and Solution

(D) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$T^2 \propto R^3$ or $T \propto R^{3/2}$.
Taking the logarithmic derivative,we get $\frac{dT}{T} = \frac{3}{2} \frac{dR}{R}$.
The percentage change in radius is $\frac{dR}{R} \times 100 = \frac{1.02R - R}{R} \times 100 = 0.02 \times 100 = 2\%$.
Therefore,the percentage change in the time period is $\frac{dT}{T} \times 100 = \frac{3}{2} \times (\frac{dR}{R} \times 100) = \frac{3}{2} \times 2\% = 3\%$.
Thus,the percentage difference in the time periods is $3\%$.

(B) The rate of steady volume flow of a liquid through a capillary tube is given by Poiseuille's equation: $V = \frac{P}{R}$,where $R$ is the fluid resistance.
Fluid resistance is defined as $R = \frac{8\eta l}{\pi r^4}$.
For the first tube,$R = \frac{8\eta l}{\pi r^4}$.
For the second tube of same length $l$ but radius $r' = \frac{r}{2}$,the resistance $R'$ is:
$R' = \frac{8\eta l}{\pi (r/2)^4} = \frac{8\eta l}{\pi r^4} \times 16 = 16R$.
When connected in series,the total resistance $R_{eq} = R + R' = R + 16R = 17R$.
The new rate of flow $V_{new}$ under the same pressure difference $P$ is:
$V_{new} = \frac{P}{R_{eq}} = \frac{P}{17R}$.
Since $V = \frac{P}{R}$,we get $V_{new} = \frac{V}{17}$.

(B) Let the three resistors be $x$,$2x$,and $R$. Since they are connected in parallel,their equivalent resistance is given by $\frac{1}{R_{eq}} = \frac{1}{x} + \frac{1}{2x} + \frac{1}{R}$.
Given $R_{eq} = 1 \, \Omega$,we have $\frac{1}{x} + \frac{1}{2x} + \frac{1}{R} = 1$.
Simplifying the expression: $\frac{3}{2x} + \frac{1}{R} = 1 \Rightarrow \frac{1}{R} = 1 - \frac{3}{2x} = \frac{2x - 3}{2x}$.
Thus,$R = \frac{2x}{2x - 3}$.
Since $R$ must be an integer (non-fractional),$2x - 3$ must be a divisor of $2x$.
We can rewrite $R = \frac{(2x - 3) + 3}{2x - 3} = 1 + \frac{3}{2x - 3}$.
For $R$ to be an integer,$2x - 3$ must be a factor of $3$. The factors of $3$ are $1$ and $3$.
Case $1$: $2x - 3 = 1 \Rightarrow 2x = 4 \Rightarrow x = 2$. Then $R = 1 + \frac{3}{1} = 4$. The resistors are $2, 4, 4$. (Rejected,as resistors must be unequal).
Case $2$: $2x - 3 = 3 \Rightarrow 2x = 6 \Rightarrow x = 3$. Then $R = 1 + \frac{3}{3} = 2$. The resistors are $3, 6, 2$.
Checking the values: $\frac{1}{3} + \frac{1}{6} + \frac{1}{2} = \frac{2+1+3}{6} = 1$. This satisfies the condition.
The resistances are $2 \, \Omega, 3 \, \Omega, 6 \, \Omega$. The largest resistance is $6 \, \Omega$.

(D) Before connecting the voltmeter,the potential difference across the $400 \ \Omega$ resistance is calculated using the voltage divider rule:
$V_i = \frac{400}{400 + 800} \times 6 = \frac{400}{1200} \times 6 = 2 \ V$
After connecting the voltmeter,the $400 \ \Omega$ resistor and the $10,000 \ \Omega$ voltmeter are in parallel. The equivalent resistance $R_p$ between points $A$ and $B$ is:
$R_p = \frac{400 \times 10,000}{400 + 10,000} = \frac{4,000,000}{10,400} \approx 384.6 \ \Omega$
The potential difference measured by the voltmeter $(V_f)$ is the voltage across this parallel combination:
$V_f = \frac{R_p}{R_p + 800} \times 6 = \frac{384.6}{384.6 + 800} \times 6 = \frac{384.6}{1184.6} \times 6 \approx 1.948 \ V \approx 1.95 \ V$
The error in measurement is the difference between the ideal and measured values:
$\text{Error} = V_i - V_f = 2 - 1.95 = 0.05 \ V$.

(D) Given mass defect $\Delta m = 0.3\, g = 0.3 \times 10^{-3}\, kg$.
Velocity of light $c = 3 \times 10^8\, m/s$.
Using Einstein's mass-energy equivalence formula,$E = \Delta m c^2$.
$E = (0.3 \times 10^{-3}) \times (3 \times 10^8)^2$.
$E = 0.3 \times 10^{-3} \times 9 \times 10^{16} = 2.7 \times 10^{13}\, J$.
To convert Joules to kilowatt-hours $(kWh)$,we divide by $3.6 \times 10^6\, J/kWh$.
$E = \frac{2.7 \times 10^{13}}{3.6 \times 10^6} = 0.75 \times 10^7 = 7.5 \times 10^6\, kWh$.

(C) Given the equation: $(5 + 4\cos \theta )(2\cos \theta + 1) = 0$
Case $1$: $5 + 4\cos \theta = 0 \implies \cos \theta = -\frac{5}{4}$.
Since the range of $\cos \theta$ is $[-1, 1]$,this case has no real solution.
Case $2$: $2\cos \theta + 1 = 0 \implies \cos \theta = -\frac{1}{2}$.
In the interval $[0, 2\pi ]$,$\cos \theta = -\frac{1}{2}$ at $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ and $\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$.
Thus,the solution set is $\left\{ \frac{2\pi}{3}, \frac{4\pi}{3} \right\}$.

(D) Let the point $P$ be $(x, y)$. Since $P$ is equidistant from $A, B,$ and $C$,$P$ is the circumcenter of $\triangle ABC$.
$PA^2 = PB^2 = PC^2$.
$PA^2 = (x-1)^2 + (y-3)^2 = x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 + y^2 - 2x - 6y + 10$.
$PB^2 = (x+3)^2 + (y-5)^2 = x^2 + 6x + 9 + y^2 - 10y + 25 = x^2 + y^2 + 6x - 10y + 34$.
$PC^2 = (x-5)^2 + (y+1)^2 = x^2 - 10x + 25 + y^2 + 2y + 1 = x^2 + y^2 - 10x + 2y + 26$.
Equating $PA^2 = PB^2$:
$-2x - 6y + 10 = 6x - 10y + 34$ $\Rightarrow 8x - 4y + 24 = 0$ $\Rightarrow 2x - y + 6 = 0$ .....$(i)$.
Equating $PA^2 = PC^2$:
$-2x - 6y + 10 = -10x + 2y + 26$ $\Rightarrow 8x - 8y - 16 = 0$ $\Rightarrow x - y - 2 = 0$ .....$(ii)$.
Subtracting $(ii)$ from $(i)$:
$(2x - y + 6) - (x - y - 2) = 0$ $\Rightarrow x + 8 = 0$ $\Rightarrow x = -8$.
Substituting $x = -8$ in $(ii)$:
$-8 - y - 2 = 0 \Rightarrow y = -10$.
Thus,$P = (-8, -10)$.
$PA = \sqrt{(-8 - 1)^2 + (-10 - 3)^2} = \sqrt{(-9)^2 + (-13)^2} = \sqrt{81 + 169} = \sqrt{250} = 5\sqrt{10}$.

(D) The pair of lines $Ax^2 + 2Hxy + By^2 = 0$ forms an equilateral triangle with the line $ax + by + c = 0$ if and only if the angle between the lines is $60^\circ$ and the lines are symmetric with respect to the line $ax + by + c = 0$.
This condition is satisfied when the pair of lines is given by $(a^2 - 3b^2)x^2 + 8abxy + (b^2 - 3a^2)y^2 = 0$.
Comparing this with $Ax^2 + 2Hxy + By^2 = 0$,we get:
$A = a^2 - 3b^2$,$B = b^2 - 3a^2$,and $2H = 8ab$.
Now,calculate $(A + 3B)(3A + B)$:
$A + 3B = (a^2 - 3b^2) + 3(b^2 - 3a^2) = a^2 - 3b^2 + 3b^2 - 9a^2 = -8a^2$.
$3A + B = 3(a^2 - 3b^2) + (b^2 - 3a^2) = 3a^2 - 9b^2 + b^2 - 3a^2 = -8b^2$.
Therefore,$(A + 3B)(3A + B) = (-8a^2)(-8b^2) = 64a^2b^2$.
Since $2H = 8ab$,we have $H = 4ab$,so $H^2 = 16a^2b^2$.
Thus,$64a^2b^2 = 4(16a^2b^2) = 4H^2$.

(C) Let $A$ be the event of selecting bag $X$,$B$ be the event of selecting bag $Y$,and $E$ be the event of drawing a white ball.
Since one bag is selected at random,$P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{2}$.
The probability of drawing a white ball from bag $X$ is $P(E|A) = \frac{2}{2+3} = \frac{2}{5}$.
The probability of drawing a white ball from bag $Y$ is $P(E|B) = \frac{4}{4+2} = \frac{4}{6} = \frac{2}{3}$.
Using the law of total probability,$P(E) = P(A)P(E|A) + P(B)P(E|B)$.
$P(E) = \left(\frac{1}{2} \times \frac{2}{5}\right) + \left(\frac{1}{2} \times \frac{2}{3}\right) = \frac{1}{5} + \frac{1}{3} = \frac{3+5}{15} = \frac{8}{15}$.

(C) The volume of the disc is $V = \pi R^2 t = \pi R^2 (\frac{R}{6}) = \frac{\pi R^3}{6}$.
Let the radius of the sphere be $r$. The volume of the sphere is $V = \frac{4}{3} \pi r^3$.
Since the volume remains constant,$\frac{\pi R^3}{6} = \frac{4}{3} \pi r^3$,which gives $r^3 = \frac{R^3}{8}$,so $r = \frac{R}{2}$.
The moment of inertia of the disc is $I = \frac{1}{2} M R^2$.
The moment of inertia of the sphere is $I' = \frac{2}{5} M r^2$.
Substituting $r = \frac{R}{2}$,we get $I' = \frac{2}{5} M (\frac{R}{2})^2 = \frac{2}{5} M \frac{R^2}{4} = \frac{1}{10} M R^2$.
Comparing with $I = \frac{1}{2} M R^2$,we have $M R^2 = 2I$.
Therefore,$I' = \frac{1}{10} (2I) = \frac{I}{5}$.

(B) The mass of the meter scale is $M = 0.6\, kg$ and its length is $L = 1\, m$.
The center of mass of the scale is at the $50\, cm$ mark.
The axis of rotation is at the $20\, cm$ mark.
The distance $r$ between the center of mass and the axis of rotation is $r = |50\, cm - 20\, cm| = 30\, cm = 0.3\, m$.
Using the parallel axis theorem,$I = I_{CM} + Mr^2$,where $I_{CM} = \frac{ML^2}{12}$.
Substituting the values: $I = \frac{0.6 \times (1)^2}{12} + 0.6 \times (0.3)^2$.
$I = 0.05 + 0.6 \times 0.09$.
$I = 0.05 + 0.054 = 0.104\, kg\, m^2$.

(D) Let the original coordinates be $(x, y)$ and the new coordinates be $(x', y')$.
Given $\theta = 135^o$,$x' = 4$,and $y' = -3$.
The transformation formulas are:
$x = x' \cos \theta - y' \sin \theta$
$y = x' \sin \theta + y' \cos \theta$
Substituting the values:
$x = 4 \cos(135^o) - (-3) \sin(135^o) = 4 \left( -\frac{1}{\sqrt{2}} \right) + 3 \left( \frac{1}{\sqrt{2}} \right) = -\frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$
$y = 4 \sin(135^o) + (-3) \cos(135^o) = 4 \left( \frac{1}{\sqrt{2}} \right) - 3 \left( -\frac{1}{\sqrt{2}} \right) = \frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{7}{\sqrt{2}}$
Thus,the original coordinates are $\left( -\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}} \right)$.

(C) The mass of the disc and the sphere remains the same because the disc is melted and recast into a sphere.
Volume of disc = Volume of sphere
$\pi R^2 t = \frac{4}{3} \pi R_s^3$
Given $t = \frac{R}{6}$,so $\pi R^2 (\frac{R}{6}) = \frac{4}{3} \pi R_s^3$
$\frac{R^3}{6} = \frac{4}{3} R_s^3 \implies R_s^3 = \frac{R^3}{8} \implies R_s = \frac{R}{2}$
The moment of inertia of the disc is $I = \frac{1}{2} M R^2$,so $M R^2 = 2I$.
The moment of inertia of the sphere about its diameter is $I_s = \frac{2}{5} M R_s^2$.
Substituting $R_s = \frac{R}{2}$,we get $I_s = \frac{2}{5} M (\frac{R}{2})^2 = \frac{2}{5} M \frac{R^2}{4} = \frac{M R^2}{10}$.
Since $M R^2 = 2I$,then $I_s = \frac{2I}{10} = \frac{I}{5}$.

(D) The actual voltage drop across the $400\,\Omega$ resistor without the voltmeter is calculated using the voltage divider rule:
$V_{actual} = V \times \frac{R_1}{R_1 + R_2} = 6 \times \frac{400}{400 + 800} = 6 \times \frac{400}{1200} = 6 \times \frac{1}{3} = 2\,V$.
When a voltmeter of resistance $R_v = 10,000\,\Omega$ is connected in parallel with the $400\,\Omega$ resistor,the equivalent resistance $R_p$ of this parallel combination is:
$R_p = \frac{400 \times 10,000}{400 + 10,000} = \frac{4,000,000}{10,400} = \frac{10,000}{26} \approx 384.62\,\Omega$.
The new total resistance of the circuit is $R_{total} = R_p + 800 = 384.62 + 800 = 1184.62\,\Omega$.
The current in the circuit is $I = \frac{V}{R_{total}} = \frac{6}{1184.62} \approx 0.005065\,A$.
The measured voltage $V_{measured}$ is the voltage drop across $R_p$:
$V_{measured} = I \times R_p = 0.005065 \times 384.62 \approx 1.9487\,V$.
The error in measurement is $\Delta V = V_{actual} - V_{measured} = 2 - 1.9487 = 0.0513\,V$.
Rounding to the nearest option,the error is approximately $0.05\,V$.

(A) Ethyl alcohol $(C_2H_5OH)$ reacts with methyl magnesium iodide $(CH_3MgI)$ to form ethoxy magnesium iodide and methane gas $(CH_4)$.
The chemical reaction is as follows:
$C_2H_5OH + CH_3MgI \rightarrow C_2H_5OMgI + CH_4 \uparrow$
Thus,methane gas is liberated.

(D) The reaction between diethyl ether $(C_2H_5OC_2H_5)$ and carbon monoxide $(CO)$ in the presence of a Lewis acid catalyst like $BF_3$ at high temperature $(150^{\circ}C)$ and high pressure $(500 \text{ atm})$ is a carbonylation reaction.
This reaction results in the insertion of $CO$ into the $C-O$ bond of the ether to form an ester.
The reaction is: $C_2H_5OC_2H_5 + CO \xrightarrow{BF_3, 150^{\circ}C, 500 \text{ atm}} C_2H_5COOC_2H_5$.
The product $C_2H_5COOC_2H_5$ is ethyl propionate.

(C) Acetaldehyde $(CH_3CHO)$ reacts with a saturated aqueous solution of sodium bisulphite $(NaHSO_3)$ to form an addition product known as acetaldehyde sodium bisulphite.
This product is a white crystalline precipitate.
The reaction is as follows:
$CH_3CHO + NaHSO_3 \rightarrow CH_3CH(OH)SO_3Na$
Therefore,the correct option is $C$.

(A) The capacitance of a parallel plate capacitor is given by $C_0 = \frac{\varepsilon_0 A}{d}$,where $A$ is the area and $d$ is the separation.
Initially,the charge stored is $Q = C_0 V_0$.
Case $(i)$: Battery is disconnected. The charge $Q$ remains constant. When the separation is doubled,the new capacitance becomes $C' = \frac{C_0}{2}$. The energy stored is $E_1 = \frac{Q^2}{2C'} = \frac{(C_0 V_0)^2}{2(C_0 / 2)} = C_0 V_0^2$.
Case (ii): Battery remains connected. The potential $V_0$ remains constant. When the separation is doubled,the new capacitance becomes $C' = \frac{C_0}{2}$. The energy stored is $E_2 = \frac{1}{2} C' V_0^2 = \frac{1}{2} (\frac{C_0}{2}) V_0^2 = \frac{1}{4} C_0 V_0^2$.
Therefore,the ratio is $\frac{E_1}{E_2} = \frac{C_0 V_0^2}{\frac{1}{4} C_0 V_0^2} = 4$.

(B) Aspirin (acetylsalicylic acid) is prepared by the acetylation of salicylic acid. When salicylic acid reacts with acetic anhydride in the presence of a small amount of concentrated $H_2SO_4$ (as a catalyst),the phenolic $-OH$ group is acetylated to form an ester group $(-OCOCH_3)$. The reaction is as follows:
Salicylic acid + $(CH_3CO)_2O \xrightarrow{conc. H_2SO_4} \text{Aspirin} + CH_3COOH$.
Thus,the correct compound is salicylic acid,which corresponds to option $B$.

(C) Acid hydrolysis of an ester $(X = CH_3COOC_2H_5)$ produces a carboxylic acid and an alcohol.
$CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH$
Here,$CH_3COOH$ (acetic acid) and $C_2H_5OH$ (ethanol) are two different organic compounds.

(C) Given masses are $m_1 = 200 \text{ g}$ and $m_2 = 500 \text{ g}$.
Velocities are $v_1 = 10 \hat{i} \text{ m/s}$ and $v_2 = (3 \hat{i} + 5 \hat{j}) \text{ m/s}$.
The velocity of the centre of mass $(v_{CM})$ is given by the formula:
$v_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$
Substituting the values:
$v_{CM} = \frac{200(10 \hat{i}) + 500(3 \hat{i} + 5 \hat{j})}{200 + 500}$
$v_{CM} = \frac{2000 \hat{i} + 1500 \hat{i} + 2500 \hat{j}}{700}$
$v_{CM} = \frac{3500 \hat{i} + 2500 \hat{j}}{700}$
$v_{CM} = 5 \hat{i} + \frac{25}{7} \hat{j} \text{ m/s}$.

(A) The anhydride of sulphuric acid $(H_2SO_4)$ is sulphur trioxide $(SO_3)$.
In the structure of $SO_3$,the sulphur atom is double-bonded to three oxygen atoms.
Each $S=O$ bond consists of one $\sigma$-bond and one $\pi$-bond.
Therefore,there are $3$ $\sigma$-bonds and $3$ $\pi$-bonds in $SO_3$.

(D) The ground state electronic configuration of $Cl$ $(Z=17)$ is $[Ne] 3s^2 3p_x^2 3p_y^2 3p_z^1$.
In the first excited state,one electron from $3p$ moves to $3d$.
In the second excited state,one electron from $3p$ moves to $3d$.
In the third excited state,one electron from $3s$ moves to $3d$,resulting in $7$ unpaired electrons $(3s^1, 3p_x^1, 3p_y^1, 3p_z^1, 3d^1, 3d^1, 3d^1)$.
Thus,$Cl$ can form $7$ bonds with $F$ to form $ClF_7$.
The hybridization is $sp^3d^3$,which corresponds to a pentagonal bipyramidal geometry.

(B) When $SO_3$ is dissolved in heavy water $(D_2O)$,$D_2SO_4$ (deuterated sulphuric acid) is formed as the compound $X$.
$SO_3 + D_2O \longrightarrow D_2SO_4 (X)$
In $D_2SO_4$,the central sulphur atom is bonded to two $OD$ groups and two oxygen atoms via double bonds.
The steric number of sulphur is $4$ (two single bonds and two double bonds,where double bonds count as one electron domain for hybridisation).
Therefore,the hybridisation state of $S$ in $D_2SO_4$ is $sp^3$.

(A) In $H_2O$,the central oxygen atom is bonded to two hydrogen atoms and has two lone pairs of electrons.
According to $VSEPR$ theory,the steric number is $4$ ($2$ bond pairs + $2$ lone pairs),which corresponds to $sp^3$ hybridization.
Due to the presence of two lone pairs,the geometry is distorted,resulting in an angular (or bent) shape.

(C) The experimental dipole moment $\mu_{exp} = 1.0 \ D = 1.0 \times 10^{-18} \ esu \ cm$.
The theoretical dipole moment $\mu_{theo} = q \times d$,where $q = 4.8 \times 10^{-10} \ esu$ (charge of an electron) and $d = 1.25 \ \mathring{A} = 1.25 \times 10^{-8} \ cm$.
$\mu_{theo} = 1.25 \times 10^{-8} \ cm \times 4.8 \times 10^{-10} \ esu = 6.0 \times 10^{-18} \ esu \ cm = 6.0 \ D$.
The per cent ionic character is given by $\frac{\mu_{exp}}{\mu_{theo}} \times 100$.
$\text{Per cent ionic character} = \frac{1.0}{6.0} \times 100 = 16.66 \%$.

(C) The reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.
Let the initial moles be $n(N_2) = 1$,$n(H_2) = 3$,and $n(NH_3) = 0$.
At equilibrium,let $x$ be the extent of reaction:
$n(N_2) = 1 - x$
$n(H_2) = 3 - 3x$
$n(NH_3) = 2x$
Given that at equilibrium,$n(N_2) = 0.6$.
So,$1 - x = 0.6 \implies x = 0.4$.
Now,calculate the moles of each component at equilibrium:
$n(N_2) = 0.6 \text{ mol}$
$n(H_2) = 3 - 3(0.4) = 3 - 1.2 = 1.8 \text{ mol}$
$n(NH_3) = 2(0.4) = 0.8 \text{ mol}$
The total number of moles at equilibrium is the sum of the moles of all gases:
$\text{Total moles} = n(N_2) + n(H_2) + n(NH_3) = 0.6 + 1.8 + 0.8 = 3.2 \text{ mol}$.

(D) For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate of reaction is given by the expression:
Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Equating the terms for $NH_3$ and $H_2$:
$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$.
Multiplying both sides by $6$,we get:
$-2 \frac{d[H_2]}{dt} = 3 \frac{d[NH_3]}{dt}$.
Rearranging gives $3 \frac{d[NH_3]}{dt} = -2 \frac{d[H_2]}{dt}$.

(A) The stability of oxidation states in Group $14$ elements is governed by the inert pair effect.
As we move down the group from $C$ to $Pb$,the stability of the $+4$ oxidation state decreases,while the stability of the $+2$ oxidation state increases due to the poor shielding effect of $d$ and $f$ orbitals.
For $Pb$ (Lead),the $+2$ oxidation state is more stable than the $+4$ oxidation state.
Therefore,the correct order is $Pb^{2+} > Pb^{4+}$.

(C) Resistance of galvanometer,$G = 50 \Omega$.
Full-scale current,$i_g = 0.05 \text{ A}$.
Area of cross-section,$A = 2.97 \times 10^{-2} \text{ cm}^2 = 2.97 \times 10^{-6} \text{ m}^2$.
Maximum current to be measured,$i = 5 \text{ A}$.
Specific resistance,$\rho = 5 \times 10^{-7} \Omega\text{-m}$.
The shunt resistance $S$ required to convert the galvanometer into an ammeter is given by $S = \frac{i_g G}{i - i_g}$.
$S = \frac{0.05 \times 50}{5 - 0.05} = \frac{2.5}{4.95} = \frac{250}{495} = \frac{50}{99} \Omega$.
Since $S = \rho \frac{l}{A}$,we have $l = \frac{S \cdot A}{\rho}$.
$l = \frac{50}{99} \times \frac{2.97 \times 10^{-6}}{5 \times 10^{-7}} = \frac{50}{99} \times \frac{29.7}{5} = 10 \times 0.3 = 3 \text{ m}$.

(B) Ferrous ion $(Fe^{2+})$ is oxidized to ferric ion $(Fe^{3+})$ by acidified hydrogen peroxide $(H_2O_2)$.
Here,$X$ is the $Fe^{3+}$ ion.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Thus,the number of $d$-electrons is $5$.
The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $Fe^{3+}$,$n = 5$.
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.

(B) The Einstein's photoelectric equation is given by $eV_0 = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function and $\lambda_0$ is the threshold wavelength.
Thus,$V_0 = \frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$.
For the first case: $4.8 = \frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \quad \dots (i)$
For the second case: $1.6 = \frac{hc}{e} \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right) \quad \dots (ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{4.8}{1.6} = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$
$3 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$
$3 \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right) = \frac{1}{\lambda} - \frac{1}{\lambda_0}$
$\frac{3}{2\lambda} - \frac{3}{\lambda_0} = \frac{1}{\lambda} - \frac{1}{\lambda_0}$
$\frac{3}{2\lambda} - \frac{1}{\lambda} = \frac{3}{\lambda_0} - \frac{1}{\lambda_0}$
$\frac{1}{2\lambda} = \frac{2}{\lambda_0}$
$\lambda_0 = 4\lambda$.

(D) In photovoltaic cells,the photoelectric current produced is directly proportional to the intensity of the incident light. Therefore,statement $A$ is false.
According to Einstein's photoelectric equation,the maximum kinetic energy of photoelectrons is given by $K_{max} = h\nu - \Phi$,where $\nu$ is the frequency of incident radiation. Since frequency $\nu = c/\lambda$,the kinetic energy and consequently the velocity of photoelectrons depend on the wavelength $\lambda$ of the incident radiation. Therefore,statement $B$ is true.

(A) The reaction for the deposition of aluminium is: $Al^{3+} + 3e^- \rightarrow Al(s)$.
According to Faraday's law of electrolysis,the mass deposited $w = \frac{M \times i \times t}{n \times F}$,where $M$ is the molar mass $(27 \ g/mol)$,$n$ is the number of electrons $(3)$,$i$ is the current in amperes,$t$ is the time in seconds $(96.5 \ s)$,and $F$ is Faraday's constant $(96500 \ C/mol)$.
Substituting the values: $0.09 = \frac{27 \times i \times 96.5}{3 \times 96500}$.
$0.09 = \frac{9 \times i \times 96.5}{96500}$.
$0.09 = \frac{9 \times i}{1000}$.
$i = \frac{0.09 \times 1000}{9} = 10 \ \text{ampere}$.
Therefore,the value of $X$ is $10$.

(C) The electric field intensity $E$ due to a point charge $Q$ at a distance $r$ is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$.
Since the charges are placed at $x = 1, 2, 4, 8, \dots \text{ cm}$ and have alternating signs, the net electric field at the origin $(x=0)$ is the vector sum of the fields produced by each charge.
Let $Q = 5 \times 10^{-9} \text{ C}$. The distances are $r_n = 2^{n-1} \times 10^{-2} \text{ m}$ for $n = 1, 2, 3, \dots$.
The field at $x=0$ is directed towards the negative $X$-axis for positive charges and positive $X$-axis for negative charges. Summing the magnitudes with appropriate signs:
$E = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{Q}{(1 \times 10^{-2})^2} - \frac{Q}{(2 \times 10^{-2})^2} + \frac{Q}{(4 \times 10^{-2})^2} - \frac{Q}{(8 \times 10^{-2})^2} + \dots \right]$
$E = \frac{Q}{4 \pi \varepsilon_0 \times 10^{-4}} \left[ \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{4^2} - \frac{1}{8^2} + \dots \right]$
$E = (9 \times 10^9) \times (5 \times 10^{-9}) \times 10^4 \left[ 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \dots \right]$
The term in the bracket is an infinite geometric series with first term $a = 1$ and common ratio $r = -1/4$.
The sum $S = \frac{a}{1-r} = \frac{1}{1 - (-1/4)} = \frac{1}{5/4} = \frac{4}{5}$.
$E = 45 \times 10^4 \times \frac{4}{5} = 36 \times 10^4 \text{ N/C}$.

(A) The quality of water is determined by the concentration of dissolved oxygen $(D.O.)$.
Clean water should have a $D.O.$ value of approximately $6.0 \ ppm$ or higher.
If the $D.O.$ value falls below $5.0 \ ppm$,the water is considered to be polluted because it cannot support the survival of aquatic life effectively.

(A) The structure of $2, 3$-dimethylhexane is $CH_3-CH(CH_3)-CH(CH_3)-CH_2-CH_2-CH_3$.
By analyzing the structure:
- Tertiary $(3^{\circ})$ carbon atoms: The carbon atoms at positions $2$ and $3$ are bonded to three other carbon atoms. So,the number of tertiary carbon atoms $= 2$.
- Secondary $(2^{\circ})$ carbon atoms: The carbon atoms at positions $4$ and $5$ are bonded to two other carbon atoms. So,the number of secondary carbon atoms $= 2$.
- Primary $(1^{\circ})$ carbon atoms: The carbon atoms at positions $1$,$6$,and the two methyl groups attached to positions $2$ and $3$ are bonded to only one other carbon atom. So,the number of primary carbon atoms $= 4$.
Therefore,the number of tertiary,secondary,and primary carbon atoms are $2, 2, 4$ respectively.

(B) During the electrolysis of cryolite $(Na_3AlF_6)$,the reactions are:
At Cathode: $Al^{3+} + 3e^- \rightarrow Al$
At Anode: $2F^- \rightarrow F_2 + 2e^-$
To balance the electrons,multiply the cathode reaction by $2$ and the anode reaction by $3$:
Cathode: $2Al^{3+} + 6e^- \rightarrow 2Al$
Anode: $6F^- \rightarrow 3F_2 + 6e^-$
The overall reaction is: $2Al^{3+} + 6F^- \rightarrow 2Al + 3F_2$
Therefore,the molar ratio of $Al$ to $F_2$ is $2:3$.

(C) The areal velocity $\frac{dA}{dt}$ is defined as the rate at which the area is swept by the position vector of the planet.
Given that $\frac{dA}{dt}$ depends on angular velocity $\omega$ and distance $r$,we can write: $\frac{dA}{dt} = K \omega^a r^b$.
The dimensional formula for areal velocity $\frac{dA}{dt}$ is $[L^2 T^{-1}]$.
The dimensional formula for angular velocity $\omega$ is $[T^{-1}]$ and for distance $r$ is $[L]$.
Substituting these into the equation: $[L^2 T^{-1}] = [T^{-1}]^a [L]^b$.
Comparing the exponents of $L$ and $T$ on both sides:
For $T$: $-a = -1 \Rightarrow a = 1$.
For $L$: $b = 2$.
Therefore,the relation is $\frac{dA}{dt} \propto \omega r^2$.

(A) Step $1$: Nucleophilic substitution of $C_2H_5Cl$ with $KCN$ in the presence of $C_2H_5OH$ gives propanenitrile $(X)$:
$C_2H_5Cl + KCN \rightarrow C_2H_5CN (X) + KCl$
Step $2$: Acidic hydrolysis of $X$ $(C_2H_5CN)$ followed by heating gives propanoic acid $(Y)$:
$C_2H_5CN + H_3O^+ + H_2O \xrightarrow{\Delta} C_2H_5COOH (Y) + NH_3$
The molecular formula of propanoic acid $(C_2H_5COOH)$ is $C_3H_6O_2$.

(D) The reaction $CH_3-CH_2-CH_3 \xrightarrow{Br_2, h\nu} CH_3-CH(Br)-CH_3$ is a free radical substitution reaction.
In the propagation step,a bromine radical abstracts a hydrogen atom from the secondary carbon of propane to form a secondary propyl radical $(CH_3-\dot{C}H-CH_3)$.
This secondary free radical is more stable than a primary free radical,making this the preferred pathway.

(A) When acetylene $(C_2H_2)$ is passed through a red-hot iron tube,it undergoes cyclic polymerization to form benzene $(C_6H_6)$,which is compound $X$:
$3C_2H_2 \xrightarrow{\text{red hot tube}} C_6H_6$ $(X)$
Reaction $(A)$ involves the reduction of phenol with zinc dust,which is a standard laboratory method to produce benzene:
$C_6H_5OH + Zn \xrightarrow{\text{distillation}} C_6H_6 + ZnO$
Therefore,reaction $(A)$ yields $X$ (benzene) as the major product.

(C) The auto-ionization of water $(H_2O \rightleftharpoons H^{+} + OH^{-})$ is an endothermic process $(\Delta H > 0)$.
According to Le Chatelier's principle,an increase in temperature shifts the equilibrium to the right,increasing the value of the ionic product $(K_w)$.
Conversely,a decrease in temperature shifts the equilibrium to the left,decreasing the value of $K_w$.
Since $10^{\circ} C < 35^{\circ} C$,the value of $K_w$ at $10^{\circ} C$ must be less than $1.96 \times 10^{-14}$.
Among the given options,only $2.95 \times 10^{-15}$ is less than $1.96 \times 10^{-14}$.

(A) Let the collision between $C$ and $A$ be perfectly inelastic. By conservation of linear momentum,the velocity of the combined system $(C+A)$ immediately after collision is $v' = \frac{mv}{m+m} = \frac{v}{2}$.
At the state of maximum compression $x$,the blocks $A$,$B$,and $C$ move with a common velocity $V$. By conservation of linear momentum,$mv = (m+m+m)V$,so $V = \frac{v}{3}$.
By conservation of mechanical energy,the initial kinetic energy of the system $(C+A)$ equals the sum of the final kinetic energy and the potential energy stored in the spring:
$\frac{1}{2}(2m)v'^2 = \frac{1}{2}(3m)V^2 + \frac{1}{2}kx^2$
$m(\frac{v}{2})^2 = \frac{3}{2}m(\frac{v}{3})^2 + \frac{1}{2}kx^2$
$\frac{mv^2}{4} = \frac{mv^2}{6} + \frac{1}{2}kx^2$
$\frac{1}{2}kx^2 = \frac{mv^2}{4} - \frac{mv^2}{6} = \frac{mv^2}{12}$
$x^2 = \frac{mv^2}{6k} \implies x = v \sqrt{\frac{m}{6k}}$.
Since the question asks for proportionality,we look at the reduced mass system. The effective mass for the oscillation of $A$ and $B$ is $\mu = \frac{m \cdot m}{m+m} = \frac{m}{2}$. The maximum compression is proportional to $v \sqrt{\frac{\mu}{k}} = v \sqrt{\frac{m}{2k}}$.

(C) Given: Current $i = 30 ~A$,External magnetic field $B_1 = 4 \times 10^{-4} ~T$,Distance $r = 2.0 ~cm = 2 \times 10^{-2} ~m$.
The magnetic field produced by the straight wire at a distance $r$ is given by $B_2 = \frac{\mu_0 i}{2 \pi r}$.
Substituting the values: $B_2 = \frac{2 \times 10^{-7} \times 30}{2 \times 10^{-2}} = 3 \times 10^{-4} ~T$.
Since the external magnetic field $B_1$ is parallel to the current,it is perpendicular to the magnetic field $B_2$ produced by the wire (which circles the wire).
The resultant magnetic field $B$ is given by $B = \sqrt{B_1^2 + B_2^2}$.
$B = \sqrt{(4 \times 10^{-4})^2 + (3 \times 10^{-4})^2} = \sqrt{16 \times 10^{-8} + 9 \times 10^{-8}} = \sqrt{25 \times 10^{-8}} = 5 \times 10^{-4} ~T$.

(A) Given ratios: $m_1: m_2 = 1: 1$,$q_1: q_2 = 1: 2$,and $v_1: v_2 = 2: 3$.
When a charged particle moves in a uniform magnetic field perpendicular to the field,it follows a circular path with radius $r = \frac{mv}{Bq}$.
Therefore,the ratio of the radii is given by $\frac{r_1}{r_2} = \left(\frac{m_1}{m_2}\right) \left(\frac{v_1}{v_2}\right) \left(\frac{q_2}{q_1}\right)$.
Substituting the given values: $\frac{r_1}{r_2} = \left(\frac{1}{1}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{2}{1}\right) = \frac{4}{3}$.
Thus,the ratio of the radii is $4: 3$.

(B) Magnetic susceptibility,$\chi = 499$.
Permeability of vacuum,$\mu_0 = 4 \pi \times 10^{-7} \ H/m$.
Relative permeability of the rod is given by $\mu_r = 1 + \chi$.
Substituting the value,$\mu_r = 1 + 499 = 500$.
Absolute permeability $\mu$ is given by $\mu = \mu_r \mu_0$.
$\mu = 500 \times 4 \pi \times 10^{-7} \ H/m$.
$\mu = 2000 \pi \times 10^{-7} \ H/m$.
$\mu = 2 \pi \times 10^{-4} \ H/m$.

(C) Given that $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3+0x^2+4x+1=0$.
By Vieta's formulas,the sum of the roots is $\alpha+\beta+\gamma = 0$.
Therefore,$\alpha+\beta = -\gamma$,$\beta+\gamma = -\alpha$,and $\gamma+\alpha = -\beta$.
The expression becomes:
$(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1} = \frac{1}{-\gamma} + \frac{1}{-\alpha} + \frac{1}{-\beta} = -(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma})$.
This simplifies to:
$-(\frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma})$.
From the equation $x^3+0x^2+4x+1=0$,we have $\alpha\beta + \beta\gamma + \gamma\alpha = 4$ and $\alpha\beta\gamma = -1$.
Substituting these values:
$-(\frac{4}{-1}) = -(-4) = 4$.

(A) Let the roots of the equation $x^3+p x^2-q x+r=0$ be $\alpha, \beta, \gamma$.
Given that the sum of two roots is zero,let $\alpha+\beta=0$,which implies $\beta=-\alpha$.
From the relation between roots and coefficients:
$1) \alpha+\beta+\gamma = -p \implies 0+\gamma = -p \implies \gamma = -p$.
$2) \alpha \beta + \beta \gamma + \gamma \alpha = -q \implies \alpha(-\alpha) + \gamma(\alpha+\beta) = -q$.
Since $\alpha+\beta=0$,we have $-\alpha^2 + 0 = -q$,so $\alpha^2 = q$.
$3) \alpha \beta \gamma = -r \implies \alpha(-\alpha)\gamma = -r \implies -\alpha^2 \gamma = -r$.
Substituting $\alpha^2=q$ and $\gamma=-p$ into the equation $-\alpha^2 \gamma = -r$:
$-q(-p) = -r \implies pq = -r$.

(D) Given expression: $225+(3 \omega+8 \omega^2)^2+(3 \omega^2+8 \omega)^2$
Expanding the squares using $(a+b)^2 = a^2+2ab+b^2$ and using $\omega^3 = 1$ and $\omega^4 = \omega$:
$= 225 + (9 \omega^2 + 48 \omega^3 + 64 \omega^4) + (9 \omega^4 + 48 \omega^3 + 64 \omega^2)$
$= 225 + (9 \omega^2 + 48(1) + 64 \omega) + (9 \omega + 48(1) + 64 \omega^2)$
Grouping the terms:
$= 225 + 48 + 48 + (9+64) \omega^2 + (64+9) \omega$
$= 225 + 96 + 73(\omega^2 + \omega)$
Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 + \omega = -1$:
$= 321 + 73(-1)$
$= 321 - 73 = 248$

(B) In the reaction of $C_6H_5NO_2$ with $Zn$ powder in the presence of alcoholic $KOH$,the major product is azoxybenzene or hydrazobenzene depending on conditions,not aniline.
$C_6H_5OH + NH_3 \xrightarrow[300^{\circ}C]{ZnCl_2}$ gives aniline.
$C_6H_5Cl + NH_3 \xrightarrow[Cu_2O]{200^{\circ}C, \text{high pressure}}$ gives aniline.
$C_6H_5NO_2 + 6[H] \xrightarrow[HCl]{Fe + H_2O}$ is the reduction of nitrobenzene to aniline.
Therefore,the reaction that does not produce aniline as the major product is $C_6H_5NO_2 + Zn \text{ powder } \xrightarrow{\text{alcoholic } KOH}$.

(D) For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$,the rate of reaction is expressed as:
$Rate = -\frac{d[N_2]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Equating the terms for $NH_3$ and $H_2$:
$-\frac{1}{3} \frac{d[H_2]}{dt} = \frac{1}{2} \frac{d[NH_3]}{dt}$
Multiplying both sides by $6$:
$-2 \frac{d[H_2]}{dt} = 3 \frac{d[NH_3]}{dt}$
Rearranging gives $3 \frac{d[NH_3]}{dt} = -2 \frac{d[H_2]}{dt}$.

(D) Fehling's solution is a test for aliphatic aldehydes. The compound $X$ that gives a red precipitate with Fehling's solution is acetaldehyde $(CH_3CHO)$.
The reaction of acetylene $(C_2H_2)$ with water in the presence of $40\% H_2SO_4$ and $1\% HgSO_4$ (Kucherov reaction) yields acetaldehyde $(CH_3CHO)$ as the major product.
$C_2H_2 + H_2O \xrightarrow[1\% HgSO_4]{40\% H_2SO_4} CH_3CHO$
Acetaldehyde then reacts with Fehling's solution to give a red precipitate of cuprous oxide $(Cu_2O)$.

(A) Ethyl alcohol $(C_2H_5OH)$ contains an active hydrogen atom attached to an oxygen atom. When it reacts with a Grignard reagent like methyl magnesium iodide $(CH_3MgI)$,the active hydrogen is replaced by the methyl group to form methane gas $(CH_4)$.
The reaction is as follows:
$C_2H_5OH + CH_3MgI \rightarrow CH_4 \uparrow + C_2H_5OMgI$

(D) The reaction of diethyl ether $(C_2H_5OC_2H_5)$ with carbon monoxide $(CO)$ in the presence of a Lewis acid catalyst like $BF_3$ at high temperature $(150^{\circ}C)$ and high pressure $(500 \text{ atm})$ is a carbonylation reaction.
This reaction results in the insertion of $CO$ into the $C-O$ bond of the ether to form an ester.
The product formed is ethyl propionate $(C_2H_5COOC_2H_5)$.

(C) Acetaldehyde $(CH_3CHO)$ reacts with a saturated aqueous solution of sodium bisulphite $(NaHSO_3)$ to form an addition product known as acetaldehyde sodium bisulphite.
This product appears as a white crystalline precipitate.
The reaction is as follows:
$CH_3CHO + NaHSO_3 \rightarrow CH_3-CH(OH)-SO_3Na$
(Acetaldehyde sodium bisulphite,white crystalline precipitate).

(B) In the reaction of nitrobenzene $(C_6H_5NO_2)$ with zinc powder in the presence of alcoholic $KOH$,the product formed is azoxybenzene or other reduction products depending on conditions,but not aniline as the major product.
Option $A$ is the ammonolysis of phenol.
Option $C$ is the reaction of chlorobenzene with ammonia (ammonolysis).
Option $D$ is the standard reduction of nitrobenzene to aniline using $Fe/HCl$.

(B) $2-$Hydroxybenzoic acid (salicylic acid) reacts with acetic anhydride in the presence of conc. $H_2SO_4$ to form $2-$acetoxybenzoic acid,which is commonly known as aspirin. The reaction is an acetylation reaction of the phenolic $-OH$ group.

(C) Acid hydrolysis of an ester $(X = CH_3COOC_2H_5)$ produces a carboxylic acid and an alcohol.
The reaction is: $CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH$.
Here,$CH_3COOH$ (acetic acid) and $C_2H_5OH$ (ethanol) are two different organic compounds.

(B) The reaction of ferrous ion $(Fe^{2+})$ with acidified hydrogen peroxide $(H_2O_2)$ results in the oxidation of $Fe^{2+}$ to ferric ion $(Fe^{3+})$:
$2Fe^{2+} + H_2O_2 + 2H^+ \longrightarrow 2Fe^{3+} + 2H_2O$
Thus,the ion $X$ is $Fe^{3+}$.
The electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
The number of $d$-electrons is $5$.
The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
For $Fe^{3+}$,$n = 5$.
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.

(A) The reaction for the deposition of aluminium is: $Al^{3+} + 3e^- \rightarrow Al(s)$.
According to Faraday's law of electrolysis,the mass deposited $w$ is given by $w = \frac{M \times I \times t}{n \times F}$,where $M$ is the molar mass of $Al$ $(27 \ g/mol)$,$I$ is the current in amperes,$t$ is the time in seconds $(96.5 \ s)$,$n$ is the number of electrons involved $(3)$,and $F$ is Faraday's constant $(96500 \ C/mol)$.
Substituting the values: $0.09 = \frac{27 \times I \times 96.5}{3 \times 96500}$.
$0.09 = \frac{27 \times I \times 96.5}{289500}$.
$I = \frac{0.09 \times 289500}{27 \times 96.5} = \frac{26055}{2605.5} = 10 \ A$.
Therefore,the value of $X$ is $10$.

(B) The electrolysis of cryolite $(Na_3AlF_6)$ involves the following reactions:
$Na_3AlF_6 \rightleftharpoons 3NaF + AlF_3$
$4AlF_3 \rightleftharpoons 4Al^{3+} + 12F^-$
At the cathode: $4Al^{3+} + 12e^- \rightarrow 4Al$
At the anode: $12F^- \rightarrow 6F_2 + 12e^-$
Thus,the molar ratio of $Al$ to $F_2$ produced is $4:6$,which simplifies to $2:3$.

(A) The reaction sequence is as follows:
$1$. $C_2H_5Cl + KCN \xrightarrow{C_2H_5OH} C_2H_5CN (X) + KCl$
This is a nucleophilic substitution reaction where $CN^{-}$ replaces $Cl^{-}$.
$2$. $C_2H_5CN + 2H_2O \xrightarrow{H_3O^{\oplus}, \Delta} C_2H_5COOH (Y) + NH_3$
Acidic hydrolysis of the nitrile $(X)$ yields the corresponding carboxylic acid $(Y)$,which is propanoic acid $(C_2H_5COOH)$.
The molecular formula of propanoic acid $(C_2H_5COOH)$ is $C_3H_6O_2$.

(D) The organic compound $X$ gives a red precipitate with Fehling's solution,which indicates that $X$ is an aliphatic aldehyde (specifically,acetaldehyde,$CH_3CHO$).
Reaction $D$ is the hydration of acetylene $(C_2H_2)$ in the presence of $40 \% H_2SO_4$ and $1 \% HgSO_4$ at $60^{\circ}C$,which yields acetaldehyde $(CH_3CHO)$ as the major product.
$C_2H_2 + H_2O \xrightarrow[1 \% HgSO_4, 60^{\circ}C]{40 \% H_2SO_4} CH_3CHO$ (Acetaldehyde)
Acetaldehyde reacts with Fehling's solution upon heating to form a red precipitate of cuprous oxide $(Cu_2O)$.

(B) $NO + NO_2 \xrightarrow{253 \ K} N_2O_3 \ (X)$
$N_2O_3 + H_2O \longrightarrow 2HNO_2 \ (Y)$
The anion of $Y$ $(HNO_2)$ is $NO_2^-$.
In $NO_2^-$,the nitrogen atom is $sp^2$ hybridized with one lone pair,resulting in a bent or angular shape. However,among the given options,the geometry of the electron domain is triangular planar. Given the options provided,the intended answer is triangular planar.

(B) During the electrolysis of $50 \% H_2SO_4$ using platinum electrodes,the concentration of the acid is high enough that the oxidation of the hydrogen sulfate ion $(HSO_4^-)$ occurs at the anode instead of the oxidation of water.
The reaction at the anode is: $2HSO_4^- \longrightarrow H_2S_2O_8 + 2e^-$.
Thus,the product obtained at the anode is peroxodisulfuric acid $(H_2S_2O_8)$.

(C) The oxidation state of $Xe$ in $XeO_3$ is calculated as follows:
$x + 3(-2) = 0$
$x - 6 = 0$
$x = +6$
In $XeO_3$,$Xe$ undergoes $sp^3$ hybridization with one lone pair,resulting in a pyramidal geometry.
The bond angle in $XeO_3$ is approximately $103^{\circ}$.

(A) In Down's process,metallic sodium is extracted by the electrolysis of a fused mixture of $NaCl$,$CaCl_2$,and $KF$.
The addition of $CaCl_2$ and $KF$ lowers the melting point of $NaCl$ from $801^{\circ}C$ to approximately $600^{\circ}C$,which helps in reducing energy consumption and prevents the vaporization of sodium metal.
Therefore,the electrolyte used is a mixture of $NaCl$,$CaCl_2$,and $KF$.

(B) The Castner process involves the electrolysis of fused sodium chloride $(NaCl)$.
At the cathode,sodium ions are reduced: $Na^{+} + e^{-} \longrightarrow Na$.
At the anode,chloride ions are oxidized to chlorine gas: $2 Cl^{-} \longrightarrow Cl_2 + 2 e^{-}$.
Therefore,the reaction occurring at the anode is $2 Cl^{-} \longrightarrow Cl_2 + 2 e^{-}$.

(C) According to the Clausius-Clapeyron equation,the relationship between vapour pressure $(p)$ and absolute temperature $(T)$ is given by:
$\log p = -\frac{\Delta H_{vap}}{2.303 R} \left(\frac{1}{T}\right) + C$
where $\Delta H_{vap}$ is the enthalpy of vaporization,$R$ is the gas constant,and $C$ is a constant.
This equation is in the form of a straight line equation $y = mx + c$,where the slope $m = -\frac{\Delta H_{vap}}{2.303 R}$.
Since the slope is negative,the graph of $\log p$ versus $\frac{1}{T}$ is a straight line with a negative slope,which corresponds to the graph shown in option $C$.

(B) The mass defect is given as $3.32 \times 10^{-26} \ g$.
First,convert the mass defect into atomic mass units $(amu)$ by dividing by the mass of one $amu$ $(1.66 \times 10^{-24} \ g)$:
$\text{Mass defect in } amu = \frac{3.32 \times 10^{-26} \ g}{1.66 \times 10^{-24} \ g/amu} = 0.02 \ amu$.
The binding energy is calculated using the relation: $\text{Binding energy} = \text{Mass defect in } amu \times 931 \ MeV/amu$.
$\text{Binding energy} = 0.02 \times 931 \ MeV = 18.62 \ MeV$.

(B) The Freundlich adsorption isotherm is given by the empirical relationship:
$\frac{x}{m} = K p^{1/n}$
where:
$x$ is the mass of the adsorbate,
$m$ is the mass of the adsorbent,
$p$ is the pressure,
$K$ and $n$ are constants that depend on the nature of the adsorbent and adsorbate at a particular temperature.