AP EAMCET 2003 Chemistry Question Paper with Answer and Solution

(C) According to Stefan-Boltzmann Law,the power radiated by a star is given by $Q = \sigma A T^4 = \sigma (4 \pi r^2) T^4$,where $r$ is the radius and $T$ is the surface temperature.
Therefore,$Q \propto r^2 T^4$.
Given: $\frac{Q_{\text{star}}}{Q_{\text{sun}}} = 10,000$,$T_{\text{sun}} = 6000 \ K$,and $T_{\text{star}} = 2000 \ K$.
Using the ratio formula: $\frac{Q_{\text{star}}}{Q_{\text{sun}}} = \left( \frac{r_{\text{star}}}{r_{\text{sun}}} \right)^2 \times \left( \frac{T_{\text{star}}}{T_{\text{sun}}} \right)^4$.
Substituting the values: $10,000 = \left( \frac{r_{\text{star}}}{r_{\text{sun}}} \right)^2 \times \left( \frac{2000}{6000} \right)^4$.
$10,000 = \left( \frac{r_{\text{star}}}{r_{\text{sun}}} \right)^2 \times \left( \frac{1}{3} \right)^4$.
$10,000 = \left( \frac{r_{\text{star}}}{r_{\text{sun}}} \right)^2 \times \frac{1}{81}$.
$\left( \frac{r_{\text{star}}}{r_{\text{sun}}} \right)^2 = 10,000 \times 81 = 810,000$.
$\frac{r_{\text{star}}}{r_{\text{sun}}} = \sqrt{810,000} = 900$.
Thus,the ratio of the radii is $900:1$.

(B) Initially,when a $1.0\, kg$ mass is suspended,the spring stretches by $x = 5\, cm = 0.05\, m$. Using Hooke's Law,$F = kx$,where $F = mg$:
$k = \frac{mg}{x} = \frac{1.0 \times 10}{0.05} = 200\, N/m$.
Now,for a mass $M = 2.0\, kg$ suspended from the same spring,the angular frequency $\omega$ is given by:
$\omega = \sqrt{\frac{k}{M}} = \sqrt{\frac{200}{2}} = \sqrt{100} = 10\, rad/s$.
The block is pulled by an amplitude $A = 10\, cm = 0.1\, m$ and released. The maximum velocity $v_{\max}$ is given by:
$v_{\max} = A\omega = 0.1 \times 10 = 1\, m/s$.

(C) The maximum speed of an object in simple harmonic motion is given by the formula $v_{\max} = a\omega$,where $a$ is the amplitude and $\omega$ is the angular frequency.
We know that $\omega = \frac{2\pi}{T}$,where $T$ is the time period.
Substituting this into the formula,we get $v_{\max} = a \left( \frac{2\pi}{T} \right)$.
Rearranging for amplitude $a$,we have $a = \frac{v_{\max} \times T}{2\pi}$.
Given values are $v_{\max} = 15 \, cm/s$ and $T = 628 \, ms = 628 \times 10^{-3} \, s = 0.628 \, s$.
Using $\pi \approx 3.14$,we calculate $a = \frac{15 \times 0.628}{2 \times 3.14} = \frac{15 \times 0.628}{6.28} = \frac{15 \times 0.628}{10 \times 0.628} = \frac{15}{10} = 1.5 \, cm$.

(B) The frequency of the $n$-th harmonic for a string is given by $f_n = \frac{n}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.
Thus,$f_n = \frac{n}{2lr} \sqrt{\frac{T}{\pi \rho}}$.
The first overtone of $A$ is the second harmonic $(n=2)$,so $f_{A,2} = \frac{2}{2l_A r_A} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{l_A r_A} \sqrt{\frac{T}{\pi \rho}}$.
The second overtone of $B$ is the third harmonic $(n=3)$,so $f_{B,3} = \frac{3}{2l_B r_B} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_{A,2} = f_{B,3}$,we have $\frac{1}{l_A r_A} = \frac{3}{2l_B r_B}$.
Given $r_A = 2r_B$,we substitute this into the equation:
$\frac{1}{l_A (2r_B)} = \frac{3}{2l_B r_B} \Rightarrow \frac{1}{2l_A} = \frac{3}{2l_B}$.
Therefore,$\frac{l_A}{l_B} = \frac{1}{3}$.

(A) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l}\sqrt{\frac{T}{m}}$.
From this formula,we observe that $n \propto \frac{\sqrt{T}}{l}$.
Therefore,the ratio of the final frequency $n'$ to the initial frequency $n$ is $\frac{n'}{n} = \sqrt{\frac{T'}{T}} \times \frac{l}{l'}$.
Given that the length is shortened by $40\%$,the new length is $l' = l - 0.4l = 0.6l = \frac{3}{5}l$.
Given that the tension is increased by $44\%$,the new tension is $T' = T + 0.44T = 1.44T = \frac{144}{100}T$.
Substituting these values into the ratio formula:
$\frac{n'}{n} = \sqrt{\frac{1.44T}{T}} \times \frac{l}{0.6l} = \sqrt{1.44} \times \frac{1}{0.6} = 1.2 \times \frac{1}{0.6} = 2$.
Thus,the ratio of the final to the initial fundamental frequency is $2:1$.

(C) To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel with the galvanometer.
Given: Galvanometer resistance $G = 50 \,\Omega$,full-scale current $I_g = 0.05 \,A$,maximum current to be measured $I = 5 \,A$.
The formula for shunt resistance is $S = \frac{I_g \cdot G}{I - I_g}$.
Substituting the values: $S = \frac{0.05 \times 50}{5 - 0.05} = \frac{2.5}{4.95} = \frac{250}{495} = \frac{50}{99} \,\Omega$.
We know $S = \frac{\rho \cdot l}{A}$,where $\rho = 5 \times 10^{-7} \,\Omega \cdot m$ and $A = 2.97 \times 10^{-2} \,cm^2 = 2.97 \times 10^{-6} \,m^2$.
Rearranging for length $l$: $l = \frac{S \cdot A}{\rho} = \frac{50}{99} \times \frac{2.97 \times 10^{-6}}{5 \times 10^{-7}}$.
$l = \frac{50}{99} \times \frac{29.7 \times 10^{-7}}{5 \times 10^{-7}} = \frac{50}{99} \times 5.94 = \frac{297}{99} = 3 \,m$.

(A) The radius $r$ of a circular path of a charged particle moving in a uniform magnetic field $B$ is given by the formula $r = \frac{mv}{qB}$.
Given the ratios:
Mass ratio: $\frac{m_1}{m_2} = \frac{1}{1}$
Charge ratio: $\frac{q_1}{q_2} = \frac{1}{2}$
Speed ratio: $\frac{v_1}{v_2} = \frac{2}{3}$
Since the magnetic field $B$ is uniform and the same for both,the ratio of the radii is:
$\frac{r_1}{r_2} = \left( \frac{m_1}{m_2} \right) \times \left( \frac{v_1}{v_2} \right) \times \left( \frac{q_2}{q_1} \right)$
Substituting the given values:
$\frac{r_1}{r_2} = \left( \frac{1}{1} \right) \times \left( \frac{2}{3} \right) \times \left( \frac{2}{1} \right) = \frac{4}{3}$
Therefore,the ratio of the radii is $4:3$.

(D) Let $\alpha$ be the maximum value of the base angle for which light is totally reflected from the hypotenuse.
From the geometry of the right-angled prism,the angle of incidence $i$ at the hypotenuse is $(90^\circ - \alpha)$.
For total internal reflection to occur,the angle of incidence must be greater than or equal to the critical angle $C$,where $\sin C = \frac{1}{\mu}$.
Thus,for the limiting case (maximum angle $\alpha$):
$i = C$
$90^\circ - \alpha = C$
Taking the sine of both sides:
$\sin(90^\circ - \alpha) = \sin C$
$\cos \alpha = \frac{1}{\mu}$
Therefore,the maximum value of the base angle is:
$\alpha = \cos^{-1}\left(\frac{1}{\mu}\right)$

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

(D) Given expression: $225 + (3\omega + 8\omega^2)^2 + (3\omega^2 + 8\omega)^2$
Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 = -1 - \omega$.
Substitute $\omega^2$ in the terms:
$(3\omega + 8\omega^2) = 3\omega + 8(-1 - \omega) = 3\omega - 8 - 8\omega = -5\omega - 8$.
$(3\omega^2 + 8\omega) = 3(-1 - \omega) + 8\omega = -3 - 3\omega + 8\omega = 5\omega - 3$.
Now,expand the squares:
$(-5\omega - 8)^2 = 25\omega^2 + 80\omega + 64$.
$(5\omega - 3)^2 = 25\omega^2 - 30\omega + 9$.
Summing these: $25\omega^2 + 80\omega + 64 + 25\omega^2 - 30\omega + 9 = 50\omega^2 + 50\omega + 73$.
Using $50(\omega^2 + \omega) = 50(-1) = -50$:
Result $= 225 - 50 + 73 = 248$.

(C) Given the cubic equation $x^3 + 0x^2 + 4x + 1 = 0$,the roots are $\alpha, \beta, \gamma$.
From Vieta's formulas,we have:
$\alpha + \beta + \gamma = 0$
$\alpha\beta + \beta\gamma + \gamma\alpha = 4$
$\alpha\beta\gamma = -1$
Since $\alpha + \beta + \gamma = 0$,we can write $\alpha + \beta = -\gamma$,$\beta + \gamma = -\alpha$,and $\gamma + \alpha = -\beta$.
Substituting these into the expression:
$(\alpha + \beta)^{-1} + (\beta + \gamma)^{-1} + (\gamma + \alpha)^{-1} = \frac{1}{-\gamma} + \frac{1}{-\alpha} + \frac{1}{-\beta} = -(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma})$
$= -(\frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma})$
$= -(\frac{4}{-1}) = 4$.

(D) To find the number of solutions,we first calculate the determinant $\Delta$ of the coefficient matrix:
$\Delta = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & -3 \end{vmatrix}$
Expanding along the first row:
$\Delta = 2((-3)(-3) - (2)(4)) - 1((1)(-3) - (2)(1)) - 1((1)(4) - (-3)(1))$
$\Delta = 2(9 - 8) - 1(-3 - 2) - 1(4 + 3)$
$\Delta = 2(1) - 1(-5) - 1(7)$
$\Delta = 2 + 5 - 7 = 0$
Since $\Delta = 0$,the system either has no solution or infinitely many solutions.
Now,calculate $\Delta_x$ by replacing the first column with the constants $(7, 1, 5)$:
$\Delta_x = \begin{vmatrix} 7 & 1 & -1 \\ 1 & -3 & 2 \\ 5 & 4 & -3 \end{vmatrix} = 7(9 - 8) - 1(-3 - 10) - 1(4 + 15) = 7(1) - 1(-13) - 1(19) = 7 + 13 - 19 = 1 \neq 0$.
Since $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero,the system is inconsistent.
Therefore,the number of solutions is $0$.

(D) The period of $\sin(a\theta)$ is given by $\frac{2\pi}{|a|}$.
For the first term $\sin(\frac{\theta}{3})$,the period $T_1 = \frac{2\pi}{1/3} = 6\pi$.
For the second term $\cos(\frac{\theta}{2})$,the period $T_2 = \frac{2\pi}{1/2} = 4\pi$.
The period of the sum of two periodic functions is the $L.C.M.$ of their individual periods.
$L.C.M.$ of $6\pi$ and $4\pi$ is $12\pi$.
Therefore,the period of $f(\theta)$ is $12\pi$.

(A) The given equations of the tangents are $5x - 12y + 10 = 0$ and $12y - 5x + 16 = 0$.
Rewriting the second equation to match the form $ax + by + c = 0$,we get $5x - 12y - 16 = 0$.
Since the tangents are parallel,the distance between them is the diameter of the circle.
The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
Here,$a = 5$,$b = -12$,$c_1 = 10$,and $c_2 = -16$.
$d = \frac{|10 - (-16)|}{\sqrt{5^2 + (-12)^2}} = \frac{|26|}{\sqrt{25 + 144}} = \frac{26}{\sqrt{169}} = \frac{26}{13} = 2$.
Since the diameter $d = 2$,the radius $r = \frac{d}{2} = \frac{2}{2} = 1$.

(A) By definition,a parabola is the locus of a point $P(x, y)$ such that its distance from the focus $S(0, 0)$ is equal to its perpendicular distance from the directrix $x + y - 4 = 0$.
$SP = PM \implies SP^2 = PM^2$
$x^2 + y^2 = \left( \frac{x + y - 4}{\sqrt{1^2 + 1^2}} \right)^2$
$x^2 + y^2 = \frac{(x + y - 4)^2}{2}$
$2(x^2 + y^2) = x^2 + y^2 + 16 + 2xy - 8x - 8y$
$x^2 + y^2 - 2xy + 8x + 8y - 16 = 0$

(B) The given equation of the hyperbola is $x^2 - 2y^2 = 2$,which can be rewritten as $\frac{x^2}{2} - \frac{y^2}{1} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 2$ and $b^2 = 1$.
The product of the lengths of the perpendiculars drawn from any point on the hyperbola to its asymptotes is given by the formula $\frac{a^2 b^2}{a^2 + b^2}$.
Substituting the values,we get $\frac{2 \times 1}{2 + 1} = \frac{2}{3}$.

(B) Given the vector equation $r = (1 - p - q)a + pb + qc$.
We can rewrite this as $r = a - pa - qa + pb + qc$.
Rearranging the terms,we get $r = a + p(b - a) + q(c - a)$.
This equation is in the form $r = a + p(b - a) + q(c - a)$,where $p$ and $q$ are scalar parameters.
This represents a plane passing through the point with position vector $a$ and parallel to the vectors $(b - a)$ and $(c - a)$.
Since $a, b, c$ are non-coplanar,the vectors $(b - a)$ and $(c - a)$ are linearly independent,confirming that the equation represents a plane.

(B) The equation of any plane passing through $(1, 1, 1)$ is given by $a(x - 1) + b(y - 1) + c(z - 1) = 0 \dots (i)$.
Since the plane passes through $(1, -1, -1)$,we substitute these coordinates into $(i)$:
$a(1 - 1) + b(-1 - 1) + c(-1 - 1) = 0 \implies 0a - 2b - 2c = 0 \implies b + c = 0 \dots (ii)$.
The plane is perpendicular to $2x - y + z + 5 = 0$,so the normal vectors are perpendicular. The normal vector of the required plane is $\vec{n} = (a, b, c)$ and the normal of the given plane is $\vec{n_1} = (2, -1, 1)$.
Thus,$2a - b + c = 0 \dots (iii)$.
From $(ii)$,$c = -b$. Substituting this into $(iii)$:
$2a - b - b = 0 \implies 2a - 2b = 0 \implies a = b$.
Let $a = 1$,then $b = 1$ and $c = -1$.
Substituting these values into $(i)$:
$1(x - 1) + 1(y - 1) - 1(z - 1) = 0 \implies x - 1 + y - 1 - z + 1 = 0 \implies x + y - z - 1 = 0$.

(A) The intercept form of a plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Given that the plane is parallel to the $y$-axis,the $y$-intercept is infinite,which means the term involving $y$ is absent.
The $x$-intercept is $a = 4$ and the $z$-intercept is $c = 3$.
Substituting these values into the equation,we get $\frac{x}{4} + \frac{z}{3} = 1$.
Multiplying the entire equation by $12$,we obtain $3x + 4z = 12$.

(C) Given $f(x) = 2x + 3$ and $g(x) = x^2 + 7$.
We need to find $x$ such that $g(f(x)) = 8$.
First,compute $g(f(x)) = g(2x + 3) = (2x + 3)^2 + 7$.
Set the expression equal to $8$:
$(2x + 3)^2 + 7 = 8$
$(2x + 3)^2 = 1$
Taking the square root on both sides:
$2x + 3 = \pm 1$
Case $1$: $2x + 3 = 1 \implies 2x = -2 \implies x = -1$.
Case $2$: $2x + 3 = -1 \implies 2x = -4 \implies x = -2$.
Thus,the values of $x$ are $-1$ and $-2$.

(A) To find the angle between the curves $y = \sin x$ and $y = \cos x$,we first find their point of intersection by setting $\sin x = \cos x$.
This gives $\tan x = 1$,so $x = \pi/4$.
Let $m_1$ be the slope of the first curve $y = \sin x$. Then $m_1 = \frac{dy}{dx} = \cos x$. At $x = \pi/4$,$m_1 = \cos(\pi/4) = \frac{1}{\sqrt{2}}$.
Let $m_2$ be the slope of the second curve $y = \cos x$. Then $m_2 = \frac{dy}{dx} = -\sin x$. At $x = \pi/4$,$m_2 = -\sin(\pi/4) = -\frac{1}{\sqrt{2}}$.
The angle $\theta$ between the curves is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values: $\tan \theta = \left| \frac{\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}})}{1 + (\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})} \right| = \left| \frac{\frac{2}{\sqrt{2}}}{1 - \frac{1}{2}} \right| = \left| \frac{\sqrt{2}}{\frac{1}{2}} \right| = 2\sqrt{2}$.
Thus,$\theta = \tan^{-1}(2\sqrt{2})$.

(A) Given $x(y^3 - 1) = 1$,we have $\frac{1}{x} = y^3 - 1$.
Since $0 < y < 2^{1/3}$,we have $0 < y^3 < 2$,so $-1 < y^3 - 1 < 1$.
Thus,$|1/x| < 1$.
The given series is $S = 2 \left[ \frac{1}{x} + \frac{1}{3x^3} + \frac{1}{5x^5} + \dots \right]$.
Using the logarithmic series expansion $\ln \left( \frac{1+t}{1-t} \right) = 2 \left( t + \frac{t^3}{3} + \frac{t^5}{5} + \dots \right)$ for $|t| < 1$,where $t = 1/x$:
$S = \ln \left( \frac{1 + 1/x}{1 - 1/x} \right) = \ln \left( \frac{x+1}{x-1} \right)$.
Substituting $x = \frac{1}{y^3 - 1}$:
$S = \ln \left( \frac{\frac{1}{y^3 - 1} + 1}{\frac{1}{y^3 - 1} - 1} \right) = \ln \left( \frac{1 + y^3 - 1}{1 - (y^3 - 1)} \right) = \ln \left( \frac{y^3}{2 - y^3} \right)$.
Note: The provided option $(A)$ is $\ln \left( \frac{y^3}{y^3 - 2} \right)$,which is equivalent to $\ln \left( -\frac{y^3}{2 - y^3} \right)$. Given the domain constraints,the expression simplifies to $\ln \left( \frac{y^3}{2 - y^3} \right)$.

(B) The given equation of the hyperbola is $x^2 - 2y^2 = 2$,which can be written as $\frac{x^2}{2} - \frac{y^2}{1} = 1$.
Here,$a^2 = 2$ and $b^2 = 1$.
The equations of the asymptotes are $\frac{x}{\sqrt{2}} - y = 0$ and $\frac{x}{\sqrt{2}} + y = 0$.
Let $P(x_1, y_1)$ be any point on the hyperbola. The product of the lengths of the perpendiculars from $P$ to the asymptotes is given by the formula $\frac{a^2 b^2}{a^2 + b^2}$.
Substituting the values,we get $\frac{2 \times 1}{2 + 1} = \frac{2}{3}$.

(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:
$\text{Total moles} = n(N_2) + n(H_2) + n(NH_3) = 0.6 + 1.8 + 0.8 = 3.2 \text{ mol}$.

(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.

(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 a zero order reaction:
Rate $= K[A]^0 = K$
Integrating the rate law,$[A]_t = [A]_0 - Kt$.
At half-life $(t = t_{1/2})$,the concentration $[A]_t = [A]_0/2$.
Substituting these values: $[A]_0/2 = [A]_0 - K t_{1/2}$.
Rearranging gives $K t_{1/2} = [A]_0/2$,or $t_{1/2} = \frac{[A]_0}{2K}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = t_{1/2}$,$x = [A]_0$,and $m = 1/(2K)$,the plot of $t_{1/2}$ versus $[A]_0$ is a straight line passing through the origin with a slope of $1/(2K)$.

(B) For a first-order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{K}$.
Given $t_{1/2} = 10 \text{ minutes}$,the rate constant $K = \frac{0.693}{10} = 0.0693 \text{ min}^{-1}$.
The integrated rate law for a first-order reaction is $t = \frac{2.303}{K} \log \frac{[A]_0}{[A]}$.
Here,$[A] = 10 \% [A]_0 = 0.1 [A]_0$,so $\frac{[A]_0}{[A]} = 10$.
Substituting the values: $t = \frac{2.303}{0.0693} \log(10) = \frac{2.303}{0.0693} \times 1 \approx 33.2 \text{ minutes}$.

(A) From the given reactions:
$1$. $X + HCl \xrightarrow{AlCl_3} C_2H_5Cl$ implies $X$ is ethene $(C_2H_4)$.
$2$. $Y + HCl \xrightarrow{ZnCl_2} C_2H_5Cl$ implies $Y$ is ethanol $(C_2H_5OH)$ (Lucas test).
$3$. The conversion of $Y$ (ethanol) to $X$ (ethene) is a dehydration reaction.
$4$. Ethanol undergoes dehydration to ethene when heated with $Al_2O_3$ at $350^{\circ}C$.
Reaction: $\underset{(Y)}{C_2H_5OH} \xrightarrow{Al_2O_3, 350^{\circ}C} \underset{(X)}{C_2H_4} + H_2O$.

(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.