AP EAMCET 2009 Chemistry Question Paper with Answer and Solution

(C) Let the radius of each small drop be $r$ and the radius of the big drop be $R$. Since the volume is conserved,the volume of $8$ small drops equals the volume of the big drop: $8 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$.
This simplifies to $R^3 = 8r^3$,which gives $R = 2r$.
The terminal velocity $V_T$ of a spherical drop is given by Stokes' Law as $V_T = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$.
Since the density $\rho$ and viscosity $\eta$ are constant,$V_T \propto r^2$.
Therefore,$\frac{V_{T_{big}}}{V_{T_{small}}} = (\frac{R}{r})^2 = (\frac{2r}{r})^2 = 4$.
Given $V_{T_{small}} = 6 \ cm \ s^{-1}$,the terminal velocity of the big drop is $V_{T_{big}} = 4 \times 6 = 24 \ cm \ s^{-1}$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the ejected photoelectron is given by:
$K_{max} = E - W_0 = \frac{1}{2}mv^2$
From this,the velocity $v$ of the photoelectron is:
$v = \sqrt{\frac{2(E - W_0)}{m}}$
When a charged particle of charge $e$ and mass $m$ moves with velocity $v$ perpendicular to a uniform magnetic field $B$,it experiences a magnetic Lorentz force which provides the necessary centripetal force for circular motion:
$evB = \frac{mv^2}{r}$
Solving for the radius $r$:
$r = \frac{mv}{eB}$
Substituting the expression for $v$:
$r = \frac{m}{eB} \sqrt{\frac{2(E - W_0)}{m}} = \frac{\sqrt{m^2 \cdot \frac{2(E - W_0)}{m}}}{eB} = \frac{\sqrt{2m(E - W_0)}}{eB}$

(D) The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
For material $X_1$,$N_1 = N_0 e^{-10\lambda t}$.
For material $X_2$,$N_2 = N_0 e^{-\lambda t}$.
Given the ratio $N_1/N_2 = 1/e$ at time $t$,we have:
$\frac{N_1}{N_2} = \frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = e^{-10\lambda t + \lambda t} = e^{-9\lambda t}$.
Setting this equal to $1/e = e^{-1}$,we get:
$e^{-9\lambda t} = e^{-1}$.
Equating the exponents:
$-9\lambda t = -1$.
Therefore,$t = 1/(9\lambda)$.

(C) The potential at point $P(0, 0, z)$ due to a point charge $Q$ at distance $r$ is given by $V = \frac{1}{4\pi \epsilon_0} \frac{Q}{r}$.
$1$. Potential at $P$ due to charge $+q$ located at $(0, 0, a)$:
The distance of $P$ from $(0, 0, a)$ is $r_1 = z - a$.
$V_1 = \frac{1}{4\pi \epsilon_0} \frac{q}{z - a}$.
$2$. Potential at $P$ due to charge $-q$ located at $(0, 0, -a)$:
The distance of $P$ from $(0, 0, -a)$ is $r_2 = z - (-a) = z + a$.
$V_2 = \frac{1}{4\pi \epsilon_0} \frac{-q}{z + a}$.
$3$. Total potential at $P$:
$V = V_1 + V_2 = \frac{q}{4\pi \epsilon_0} \left( \frac{1}{z - a} - \frac{1}{z + a} \right)$.
$V = \frac{q}{4\pi \epsilon_0} \left( \frac{(z + a) - (z - a)}{(z - a)(z + a)} \right)$.
$V = \frac{q}{4\pi \epsilon_0} \left( \frac{2a}{z^2 - a^2} \right) = \frac{2qa}{4\pi \epsilon_0 (z^2 - a^2)}$.

(D) According to Einstein's photoelectric equation,the kinetic energy $K$ of the ejected photoelectron is given by $K = E - W_0$.
Since $K = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2(E - W_0)}{m}}$.
When a charged particle of mass $m$ and charge $e$ moves with velocity $v$ perpendicular to a uniform magnetic field $B$,it experiences a magnetic Lorentz force that provides the necessary centripetal force for circular motion.
Thus,$evB = \frac{mv^2}{r}$.
Solving for the radius $r$,we get $r = \frac{mv}{eB}$.
Substituting the expression for $v$ into the equation for $r$:
$r = \frac{m}{eB} \sqrt{\frac{2(E - W_0)}{m}} = \frac{\sqrt{m^2 \cdot \frac{2(E - W_0)}{m}}}{eB} = \frac{\sqrt{2m(E - W_0)}}{eB}$.

(C) Step $1$: Conservation of linear momentum during the perfectly inelastic collision.
Let $m_1 = 0.02\, kg$ be the mass of the bullet and $u_1 = 250\, ms^{-1}$ be its initial velocity.
Let $m_2 = 0.23\, kg$ be the mass of the block and $u_2 = 0\, ms^{-1}$ be its initial velocity.
Let $v$ be the common velocity of the bullet and block after the collision.
According to the law of conservation of momentum:
$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$
$0.02 \times 250 + 0.23 \times 0 = (0.02 + 0.23) v$
$5 = 0.25 v$
$v = \frac{5}{0.25} = 20\, ms^{-1}$
Step $2$: Work-Energy theorem to find the coefficient of friction.
The kinetic energy of the combined system is dissipated by the work done by friction.
Initial kinetic energy $K = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 0.25 \times (20)^2 = 0.125 \times 400 = 50\, J$.
The work done by friction $W = f_k \cdot d = \mu N \cdot d = \mu (m_1 + m_2) g \cdot d$.
Equating work done to the initial kinetic energy:
$50 = \mu \times 0.25 \times 9.8 \times 40$
$50 = \mu \times 98$
$\mu = \frac{50}{98} \approx 0.51$.

(C) The reaction of ethanol $(C_2H_5OH)$ with chlorine $(Cl_2)$ proceeds in two steps:
$1$. Ethanol is oxidized to acetaldehyde $(CH_3CHO)$,which is $X$.
$CH_3CH_2OH + Cl_2 \rightarrow CH_3CHO + 2HCl$
$2$. Acetaldehyde then undergoes chlorination to form chloral $(CCl_3CHO)$,which is $Y$.
$CH_3CHO + 3Cl_2 \rightarrow CCl_3CHO + 3HCl$
Therefore,$X = CH_3CHO$ and $Y = CCl_3CHO$.

(A) The reaction sequence is: $\text{Alkyne}$ $\xrightarrow{H_2, \text{Lindlar's catalyst}} A$ $\xrightarrow{\text{Ozonolysis}} B \text{ (only)}$.
In the Wacker process,ethene $(CH_2=CH_2)$ is oxidized to acetaldehyde $(CH_3CHO)$,which is $B$.
Since ozonolysis of alkene $A$ yields $B$ $(CH_3CHO)$ only,$A$ must be $CH_3CH=CHCH_3$ (but$-2-$ene).
Lindlar's catalyst reduces an alkyne to a $cis$-alkene.
Therefore,the starting alkyne must be $CH_3-C \equiv C-CH_3$ (but$-2-$yne).

(D) The synthesis of crotonaldehyde from acetaldehyde occurs via an aldol condensation reaction.
First,two molecules of acetaldehyde undergo a nucleophilic addition reaction to form $3-$hydroxybutanal (aldol).
Then,the aldol undergoes an elimination reaction (dehydration) upon heating to form crotonaldehyde $(CH_3CH=CHCHO)$.
Since the overall process involves both a nucleophilic addition step and an elimination step,it is classified as a nucleophilic addition-elimination reaction.

(C) The impedance of an $L-C-R$ series circuit is given by $Z = \sqrt{(X_L - X_C)^2 + R^2}$,where $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$.
Given: $\omega = 70 \times 10^3 \text{ rad/s}$,$L = 100 \times 10^{-6} \text{ H}$,and $C = 1 \times 10^{-6} \text{ F}$.
Calculating inductive reactance: $X_L = \omega L = (70 \times 10^3) \times (100 \times 10^{-6}) = 7 \text{ } \Omega$.
Calculating capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{(70 \times 10^3) \times (1 \times 10^{-6})} = \frac{1}{70 \times 10^{-3}} = \frac{1000}{70} \approx 14.29 \text{ } \Omega$.
Since $X_C > X_L$,the net reactance is capacitive $(X_C - X_L > 0)$.
Therefore,the circuit behaves like a series $R-C$ circuit.

(A) In the Gattermann reaction,the diazonium group $(-N_2^+Cl^-)$ is replaced by a halogen atom (like $Cl$ or $Br$) using copper powder $(Cu)$ in the presence of the corresponding halogen acid ($HCl$ or $HBr$).
Specifically,the diazonium cation is attacked by the nucleophilic halide ion $(Cl^{\ominus})$ generated in the reaction mixture,facilitated by the copper catalyst.
Therefore,$\underline{X}$ is $Cl^{\ominus}$ and $\underline{Y}$ is $Cu / HCl$.

(D) codon is a sequence of $3$ nitrogenous bases in $DNA$ or $RNA$ that codes for a specific amino acid.
These amino acids are then linked together to form a polypeptide chain,which constitutes a protein.
Therefore,$\underline{A} = 3 \text{ bases}$,$\underline{B} = \text{amino acid}$,and $\underline{C} = \text{protein}$.

(C) Tripeptides are polymers of amino acids in which three individual amino acid units,called residues,are linked together by peptide bonds.
For three distinct amino acids,the number of possible tripeptides is given by the number of permutations of $3$ items taken $3$ at a time,which is $3! = 3 \times 2 \times 1 = 6$.
Let the amino acids be $G$ (Glycine),$A$ (Alanine),and $P$ (Phenylalanine). The possible sequences are:
$GAP, GPA, AGP, APG, PGA, PAG$.
Thus,there are $6$ different tripeptides that can be prepared.

(C) Reaction $(I)$: Dry distillation of calcium acetate gives acetone $(CH_3 COCH_3)$ and calcium carbonate $(CaCO_3)$. Thus,$\underline{A} = CH_3 COCH_3$.
Reaction $(II)$: Reduction of acetic acid with $HI$ and red phosphorus gives ethane $(C_2 H_6)$. Thus,$\underline{B} = C_2 H_6$.
Reaction $(III)$: Dehydration of acetic acid with $P_4 O_{10}$ gives acetic anhydride $((CH_3 CO)_2 O)$. Thus,$\underline{C} = (CH_3 CO)_2 O$.
Therefore,the correct sequence is $\underline{A} = CH_3 COCH_3$,$\underline{B} = C_2 H_6$,and $\underline{C} = (CH_3 CO)_2 O$.

(A) Let the mass of the first body be $m_1 = 5 \ kg$ and its initial velocity be $u$. Let the mass of the second body be $M$ and its initial velocity be $u_2 = 0$.
After the elastic collision,the first body moves with velocity $v_1 = \frac{u}{10}$ in the same direction. Let the velocity of the second body be $v_2$.
By the law of conservation of linear momentum:
$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
$5u + M(0) = 5 \left(\frac{u}{10}\right) + M v_2$
$5u = \frac{u}{2} + M v_2 \implies M v_2 = 4.5u$ --- $(i)$
For an elastic collision,the coefficient of restitution $e = 1$,so:
$v_2 - v_1 = e(u_1 - u_2)$
$v_2 - \frac{u}{10} = 1(u - 0)$
$v_2 = u + \frac{u}{10} = \frac{11u}{10}$ --- (ii)
Substituting $v_2$ from (ii) into $(i)$:
$M \left(\frac{11u}{10}\right) = 4.5u$
$M = \frac{4.5 \times 10}{11} = \frac{45}{11} \approx 4.09 \ kg$.

(C) Initial mass of the particle $= 4 M$. Initial velocity $= 0$. Therefore,initial momentum $= 0$.
After the explosion,the system consists of three pieces with masses $M$,$M$,and $2 M$. Let the velocities of the pieces with mass $M$ be $\vec{v}_x = 4 \hat{i} ~ms^{-1}$ and $\vec{v}_y = 6 \hat{j} ~ms^{-1}$. Let the velocity of the piece with mass $2 M$ be $\vec{v}$.
According to the law of conservation of linear momentum:
$\vec{P}_{initial} = \vec{P}_{final}$
$0 = M(4 \hat{i}) + M(6 \hat{j}) + 2 M \vec{v}$
$0 = 4 M \hat{i} + 6 M \hat{j} + 2 M \vec{v}$
Dividing by $2 M$:
$0 = 2 \hat{i} + 3 \hat{j} + \vec{v}$
$\vec{v} = -2 \hat{i} - 3 \hat{j} ~ms^{-1}$
The magnitude of the velocity is:
$|\vec{v}| = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} ~ms^{-1}$.

(C) Let the mass of the bullet be $m_1 = 0.02 \ kg$ and its initial velocity be $u_1 = 250 \ ms^{-1}$.
Let the mass of the block be $m_2 = 0.23 \ kg$ and its initial velocity be $u_2 = 0 \ ms^{-1}$.
By the law of conservation of linear momentum,the momentum before impact equals the momentum after impact:
$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$
$0.02 \times 250 + 0.23 \times 0 = (0.02 + 0.23)v$
$5 = 0.25v$
$v = \frac{5}{0.25} = 20 \ ms^{-1}$
Now,the combined mass $M = m_1 + m_2 = 0.25 \ kg$ moves with initial velocity $v = 20 \ ms^{-1}$ and comes to rest after covering a distance $d = 40 \ m$ due to friction.
The work done by friction equals the change in kinetic energy:
$W = \Delta K$
$f_k \cdot d = \frac{1}{2} M v^2$
$(\mu M g) \cdot d = \frac{1}{2} M v^2$
$\mu = \frac{v^2}{2gd} = \frac{20^2}{2 \times 9.8 \times 40}$
$\mu = \frac{400}{784} \approx 0.51$

(C) Initially,the particle of mass $4 m$ is at rest,so its initial momentum is zero.
Let the velocity of the heavier mass $(2 m)$ be $\vec{v} = v_x \hat{i} + v_y \hat{j}$.
According to the law of conservation of linear momentum,the total final momentum must be zero:
$\vec{P}_{initial} = \vec{P}_{final} = 0$
$m(4 \hat{i}) + m(6 \hat{j}) + 2m(v_x \hat{i} + v_y \hat{j}) = 0$
Dividing by $m$:
$4 \hat{i} + 6 \hat{j} + 2v_x \hat{i} + 2v_y \hat{j} = 0$
Equating the components:
$4 + 2v_x = 0 \Rightarrow v_x = -2 ms^{-1}$
$6 + 2v_y = 0 \Rightarrow v_y = -3 ms^{-1}$
The magnitude of the velocity of the heavier mass is:
$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} ms^{-1}$

(B) Sulphuric anhydride is $SO_3$. In its structure,the central sulphur atom is $sp^2$ hybridized.
It forms three $\sigma$ bonds with three oxygen atoms.
Out of the three $\pi$ bonds,one is formed by $p \pi-p \pi$ overlap (between $S$ and $O$) and the other two are formed by $p \pi-d \pi$ overlap (between $S$ and $O$).
Thus,the molecule contains $3 \sigma$,$1 p \pi-p \pi$,and $2 p \pi-d \pi$ bonds.

(A) In $XeO_3$,the Xenon atom is $sp^3$ hybridized with one lone pair. It forms $3$ double bonds with oxygen atoms. Each double bond consists of one $\sigma$ bond and one $p\pi-d\pi$ $\pi$ bond. Thus,there are $3$ $p\pi-d\pi$ $\pi$ bonds.
In $XeO_4$,the Xenon atom is $sp^3$ hybridized with no lone pair. It forms $4$ double bonds with oxygen atoms. Each double bond consists of one $\sigma$ bond and one $p\pi-d\pi$ $\pi$ bond. Thus,there are $4$ $p\pi-d\pi$ $\pi$ bonds.
Therefore,the number of $p\pi-d\pi$ bonds in $XeO_3$ and $XeO_4$ are $3$ and $4$ respectively.

(C) The dipole moment is given by the formula $\mu = \delta \times d$,where $\delta$ is the magnitude of the electric charge and $d$ is the bond length.
Therefore,the fraction of charge is $\delta = \frac{\mu}{d}$.
For the ratio of charges in $HCl$ and $HI$:
$\frac{\delta_{HCl}}{\delta_{HI}} = \frac{\mu_{HCl}}{d_{HCl}} \times \frac{d_{HI}}{\mu_{HI}}$
Substituting the given values:
$\frac{\delta_{HCl}}{\delta_{HI}} = \frac{1.03 \times 1.6}{1.3 \times 0.38} = \frac{1.648}{0.494} \approx 3.33 : 1$.
Thus,the ratio is $3.3 : 1$.

(C) For the reaction $N_{2(g)} + 2 O_{2(g)} \rightleftharpoons 2 NO_{2(g)}$,the equilibrium constant is $K_1 = 100$.
The expression is $K_1 = \frac{[NO_2]^2}{[N_2][O_2]^2} = 100$.
For the reaction $NO_{2(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}$,the equilibrium constant is $K_2 = \frac{[N_2]^{1/2} [O_2]}{[NO_2]}$.
Comparing $K_2$ with $K_1$,we see that $K_2 = \sqrt{\frac{1}{K_1}}$.
Therefore,$K_2 = \sqrt{\frac{1}{100}} = \frac{1}{10} = 0.1$.

(C) Dopamine is a neurotransmitter produced in several areas of the brain. $A$ deficiency of dopamine in the brain is associated with Parkinson's disease. The $IUPAC$ name of dopamine is $4-(2-aminoethyl)benzene-1,2-diol$ (commonly known as $2-(3,4-dihydroxyphenyl)ethylamine$). Its structure consists of a benzene ring with two hydroxyl groups at the $3$ and $4$ positions and an ethylamine chain at the $1$ position. The correct structure is represented by option $C$.

(C) The first ionisation energy $(IE_1)$ generally increases across a period from left to right due to an increase in effective nuclear charge. For the second period elements,the expected order is $Li < Be < B < C < N < O < F < Ne$.
However,there are exceptions due to stable electronic configurations.
The actual order for the given elements is $B < Be < O < N$.
$1$. The $IE_1$ of $B$ $(1s^2, 2s^2 2p^1)$ is lower than $Be$ $(1s^2, 2s^2)$ because the electron in $B$ is removed from a $2p$ orbital,which is more shielded and further from the nucleus than the $2s$ orbital of $Be$.
$2$. The $IE_1$ of $O$ $(1s^2, 2s^2 2p^4)$ is lower than $N$ $(1s^2, 2s^2 2p^3)$ because $N$ has a stable half-filled $2p$ subshell,which requires more energy to remove an electron.

(C) Let the internal resistance of the voltmeter be $R$. The voltmeter is connected in parallel with the $50 \text{ k}\Omega$ resistor between points $B$ and $C$.
The equivalent resistance $R'$ across $B$ and $C$ is given by:
$\frac{1}{R'} = \frac{1}{R} + \frac{1}{50 \text{ k}\Omega} = \frac{50 \text{ k}\Omega + R}{50 R \text{ k}\Omega}$
$R' = \frac{50 R}{50 + R} \text{ k}\Omega$
The total resistance of the circuit is $R_{total} = 50 \text{ k}\Omega + R' = 50 + \frac{50 R}{50 + R} = \frac{2500 + 100 R}{50 + R} \text{ k}\Omega$.
The total current $I$ in the circuit is:
$I = \frac{V}{R_{total}} = \frac{100 \text{ V}}{\frac{2500 + 100 R}{50 + R} \text{ k}\Omega} = \frac{100(50 + R)}{2500 + 100 R} \text{ mA}$.
The voltage across $B$ and $C$ is given as $V_{BC} = \frac{100}{3} \text{ V}$.
Using Ohm's law,$V_{BC} = I \cdot R'$:
$\frac{100}{3} = \left( \frac{100(50 + R)}{2500 + 100 R} \right) \cdot \left( \frac{50 R}{50 + R} \right)$
$\frac{1}{3} = \frac{50 R}{2500 + 100 R}$
$2500 + 100 R = 150 R$
$50 R = 2500$
$R = 50 \text{ k}\Omega$.

(C) The paramagnetic property of an ion is directly proportional to the number of unpaired electrons present in its $d$-orbitals.
$Cu^{2+} = [Ar] 3d^9$,number of unpaired electrons = $1$.
$V^{2+} = [Ar] 3d^3$,number of unpaired electrons = $3$.
$Cr^{2+} = [Ar] 3d^4$,number of unpaired electrons = $4$.
$Mn^{2+} = [Ar] 3d^5$,number of unpaired electrons = $5$.
Comparing the number of unpaired electrons $(1 < 3 < 4 < 5)$,the correct order of increasing paramagnetic property is $Cu^{2+} < V^{2+} < Cr^{2+} < Mn^{2+}$.

(A) According to Einstein's photoelectric equation,the kinetic energy $K$ of the ejected photoelectron is given by:
$K = E - W_0 = \frac{1}{2}mv^2$
From this,the velocity $v$ of the electron is:
$v = \sqrt{\frac{2(E - W_0)}{m}}$
When a charged particle of mass $m$ and charge $e$ moves in a uniform magnetic field $B$ perpendicular to its velocity,it experiences a magnetic Lorentz force which provides the necessary centripetal force for circular motion:
$evB = \frac{mv^2}{r}$
Rearranging for the radius $r$:
$r = \frac{mv}{eB}$
Substituting the expression for $v$:
$r = \frac{m}{eB} \sqrt{\frac{2(E - W_0)}{m}} = \frac{\sqrt{2m(E - W_0)}}{eB}$

(B) The work function $W_0 = 3.31 \times 10^{-19} \, J$.
The wavelength of incident radiation is $\lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, m$.
According to Einstein's photoelectric equation, the energy of the incident photon $E$ is given by $E = W_0 + KE_{max}$.
First, calculate the energy of the incident photon:
$E = \frac{hc}{\lambda} = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{5000 \times 10^{-10}} = \frac{19.86 \times 10^{-26}}{5 \times 10^{-7}} = 3.972 \times 10^{-19} \, J$.
Now, calculate the maximum kinetic energy $KE_{max}$:
$KE_{max} = E - W_0 = 3.972 \times 10^{-19} \, J - 3.31 \times 10^{-19} \, J = 0.662 \times 10^{-19} \, J$.
To convert this energy into electron-volts $(eV)$:
$KE_{max} (eV) = \frac{0.662 \times 10^{-19} \, J}{1.6 \times 10^{-19} \, J/eV} = 0.41375 \, eV \approx 0.41 \, eV$.

(A) The cell representation is $Ag|Ag^{+}|AgCl|Cl^{-}|Cl_2, Pt$.
The half-cell reactions are:
Anode: $Ag \rightarrow Ag^{+} + e^{-}$
Cathode: $AgCl + e^{-} \rightarrow Ag(s) + Cl^{-}$
Net cell reaction: $AgCl \rightarrow Ag^{+} + Cl^{-}$
The standard Gibbs free energy change for the reaction is:
$\Delta G_{\text{reaction}}^{\circ} = \Sigma \Delta G_f^{\circ}(\text{products}) - \Sigma \Delta G_f^{\circ}(\text{reactants})$
$\Delta G_{\text{reaction}}^{\circ} = [\Delta G_f^{\circ}(Ag^{+}) + \Delta G_f^{\circ}(Cl^{-})] - [\Delta G_f^{\circ}(AgCl)]$
$\Delta G_{\text{reaction}}^{\circ} = [78 + (-129)] - (-109) \ kJ / mol$
$\Delta G_{\text{reaction}}^{\circ} = -51 + 109 = +58 \ kJ / mol$
Using the relation $\Delta G^{\circ} = -n F E_{\text{cell}}^{\circ}$,where $n = 1$ and $F = 96500 \ C / mol$:
$58 \times 10^3 \ J / mol = -1 \times 96500 \ C / mol \times E_{\text{cell}}^{\circ}$
$E_{\text{cell}}^{\circ} = \frac{-58000}{96500} \ V \approx -0.60 \ V$

(B) According to Kohlrausch's law of independent migration of ions:
$\lambda_{m, Ba(OH)_2}^{\infty} = \lambda_{m, Ba^{2+}}^{\infty} + 2\lambda_{m, OH^-}^{\infty}$
We can express this using the given electrolytes:
$\lambda_{m, Ba(OH)_2}^{\infty} = \lambda_{m, BaCl_2}^{\infty} + 2\lambda_{m, NaOH}^{\infty} - 2\lambda_{m, NaCl}^{\infty}$
Substituting the given values:
$\lambda_{m, Ba(OH)_2}^{\infty} = 280 \times 10^{-4} + 2(248 \times 10^{-4}) - 2(126 \times 10^{-4})$
$= (280 + 496 - 252) \times 10^{-4}$
$= 524 \times 10^{-4} \ S \ m^2 \ mol^{-1}$

(A) Given:
Linear charge density,$\lambda = \frac{1}{3} ~Cm^{-1}$
Distance,$r = 18 ~cm = 18 \times 10^{-2} ~m$
Permittivity of free space,$\varepsilon_0 \approx 8.85 \times 10^{-12} ~C^2 N^{-1} m^{-2}$
The electric field intensity $E$ at a distance $r$ from an infinitely long thin straight wire is given by the formula:
$E = \frac{\lambda}{2 \pi \varepsilon_0 r}$
We can rewrite this as:
$E = \frac{2 \lambda}{4 \pi \varepsilon_0 r} = 2k \frac{\lambda}{r}$
where $k = \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 ~Nm^2 C^{-2}$.
Substituting the values:
$E = 2 \times (9 \times 10^9) \times \frac{(1/3)}{18 \times 10^{-2}}$
$E = 18 \times 10^9 \times \frac{1}{3 \times 18 \times 10^{-2}}$
$E = \frac{18 \times 10^9}{54 \times 10^{-2}}$
$E = \frac{1}{3} \times 10^{11} ~NC^{-1}$
$E \approx 0.33 \times 10^{11} ~NC^{-1}$

(C) The electric potential $V$ at a point due to a system of charges is the algebraic sum of the potentials due to individual charges.
Potential at point $P(0, 0, z)$ due to charge $+q$ at $(0, 0, a)$ is:
$V_1 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{(z - a)}$
Potential at point $P(0, 0, z)$ due to charge $-q$ at $(0, 0, -a)$ is:
$V_2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{-q}{(z - (-a))} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{-q}{(z + a)}$
Total potential $V$ at point $P$ is:
$V = V_1 + V_2$
$V = \frac{q}{4 \pi \varepsilon_0} \left( \frac{1}{z - a} - \frac{1}{z + a} \right)$
$V = \frac{q}{4 \pi \varepsilon_0} \left( \frac{(z + a) - (z - a)}{(z - a)(z + a)} \right)$
$V = \frac{q}{4 \pi \varepsilon_0} \left( \frac{2a}{z^2 - a^2} \right)$
$V = \frac{2qa}{4 \pi \varepsilon_0(z^2 - a^2)}$

(A) The thermosphere is the fourth layer of the Earth's atmosphere,located above the mesosphere.
In this region,the air is very thin.
The thermosphere includes the ionosphere,which is characterized by the presence of charged particles.
Due to high-energy solar radiation,molecules in this layer undergo ionization,resulting in the formation of species such as $O_2^{+}$,$O^{+}$,and $NO^{+}$.

(C) The formula for specific rotation is $[\alpha] = \frac{\alpha}{l \times c}$,where $\alpha$ is the observed rotation in degrees,$l$ is the path length in decimeters $(dm)$,and $c$ is the concentration in $g/mL$.
Given: $\alpha = -1.2^{\circ}$,$l = 5 \ cm = 0.5 \ dm$,and $c = 6.15 \ g / 100 \ mL = 0.0615 \ g/mL$.
Substituting these values: $[\alpha] = \frac{-1.2}{0.5 \times 0.0615} = \frac{-1.2}{0.03075} = -39^{\circ}$.

(B) According to the law of conservation of energy,the total energy at the surface of the earth is equal to the total energy at the maximum height $h$.
At the surface: $E_i = K + U = \frac{1}{2}mv^2 - \frac{GMm}{R}$
At maximum height $h$: $E_f = K + U = 0 - \frac{GMm}{R+h}$
Equating $E_i = E_f$: $\frac{1}{2}mv^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$
Given $v = \frac{v_e}{2} = \frac{1}{2} \sqrt{\frac{2GM}{R}}$,so $v^2 = \frac{GM}{2R}$.
Substituting $v^2$ into the energy equation: $\frac{1}{2}m(\frac{GM}{2R}) - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$\frac{GMm}{4R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$-\frac{3GMm}{4R} = - \frac{GMm}{R+h}$
$\frac{3}{4R} = \frac{1}{R+h} \Rightarrow 3(R+h) = 4R$
$3R + 3h = 4R \Rightarrow 3h = R \Rightarrow h = \frac{R}{3}$

(B) The reaction of ethyl chloride $(C_2H_5Cl)$ with silver cyanide $(AgCN)$ is a nucleophilic substitution reaction.
$AgCN$ is a covalent compound,and the nitrogen atom acts as the nucleophile.
Therefore,the major product $\underline{X}$ is ethyl isocyanide $(C_2H_5NC)$.
In ethyl isocyanide $(C_2H_5-N \equiv C)$,the nitrogen atom is directly linked to the ethyl carbon.
Thus,statement $(III)$ is correct.

(A) To determine the structure of the alkene that undergoes ozonolysis,place the carbonyl oxygen atoms of the products face to face and replace them with a double bond $(C=C)$.
Acetaldehyde is $CH_3CHO$ and acetone is $(CH_3)_2CO$.
By removing the oxygen atoms and joining the carbon atoms with a double bond,we get:
$CH_3-CH=C(CH_3)_2$.
The $IUPAC$ name of this alkene is $2-$methyl$-2-$butene.

(B) The mixture of acetic acid $(CH_3COOH)$ and potassium acetate $(CH_3COOK)$ forms an acidic buffer.
Using the Henderson-Hasselbalch equation: $pH = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right)$.
First,calculate $pK_a$: $pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74$.
Let the concentration of potassium acetate be $M$. The number of moles of acid $= 0.1 \times 20 = 2 \text{ mmol}$. The number of moles of salt $= M \times 50 = 50M \text{ mmol}$.
Substituting the values into the equation: $4.8 = 4.74 + \log\left(\frac{50M}{2}\right)$.
$0.06 = \log(25M)$.
Taking antilog on both sides: $10^{0.06} = 25M$.
Since $10^{0.06} \approx 1.15$,we have $25M = 1.15$.
$M = \frac{1.15}{25} = 0.046 \text{ M}$.
Rounding to the nearest provided option,$M \approx 0.04 \text{ M}$.

(D) Buffer capacity,$\beta = \frac{d C_{HA}}{d_{pH}}$,where $d C_{HA}$ is the number of moles of acid added per liter and $d_{pH}$ is the change in $pH$.
First,calculate the number of moles of acetic acid added: $n = \frac{0.12 \ g}{60 \ g/mol} = 0.002 \ mol$.
Next,calculate the concentration of acid added per liter $(d C_{HA})$: $d C_{HA} = \frac{0.002 \ mol}{0.250 \ L} = 0.008 \ mol/L$.
Given the change in $pH$ $(d_{pH})$ is $0.02$.
Therefore,$\beta = \frac{0.008}{0.02} = 0.4$.

(C) The magnetic field at the centre $O$ is the vector sum of the magnetic field due to the straight wire and the magnetic field due to the circular loop.
$1$. Magnetic field due to the straight wire at distance $R$ from the centre $O$:
$B_1 = \frac{\mu_0 I}{2 \pi R}$ (directed outwards,perpendicular to the plane).
$2$. Magnetic field due to the circular loop of radius $R$ at its centre $O$:
$B_2 = \frac{\mu_0 I}{2 R}$ (directed outwards,perpendicular to the plane).
Since both magnetic fields are in the same direction (outwards),the resultant magnetic field $B$ is:
$B = B_1 + B_2$
$B = \frac{\mu_0 I}{2 \pi R} + \frac{\mu_0 I}{2 R}$
$B = \frac{\mu_0 I}{2 \pi R} (1 + \pi)$
Thus,the magnitude of the magnetic induction at the centre is $\frac{\mu_0 I}{2 \pi R}(\pi+1)$.

(C) The magnetic field at the centre of a circular loop of radius $R$ with $N$ turns carrying current $I$ is given by $B = \frac{\mu_0 N I}{2 R}$.
For a single loop $(N=1)$,$B = \frac{\mu_0 I}{2 R}$. Since the length of the wire $l = 2 \pi R$,we have $R = \frac{l}{2 \pi}$.
Thus,$B = \frac{\mu_0 I}{2 (l / 2 \pi)} = \frac{\mu_0 I \pi}{l}$.
When the same wire of length $l$ is bent into a double loop $(N=2)$,the new radius $R'$ is given by $l = 2 (2 \pi R')$,so $R' = \frac{l}{4 \pi} = \frac{R}{2}$.
The new magnetic field $B'$ at the centre is $B' = \frac{\mu_0 N' I}{2 R'} = \frac{\mu_0 (2) I}{2 (R / 2)} = \frac{2 \mu_0 I}{R} = 4 \left( \frac{\mu_0 I}{2 R} \right) = 4 B$.

(B) The frequency of oscillation $v$ in a vibration magnetometer is given by $v = \frac{1}{2\pi} \sqrt{\frac{MB}{I}}$.
When magnets are placed together,the effective magnetic moment $M_{eff}$ is $(M_A + M_B)$ for similar poles and $(M_A - M_B)$ for opposite poles.
Let $v_s = 20 \text{ oscillations/min}$ (similar poles) and $v_d = 15 \text{ oscillations/min}$ (opposite poles).
Since $v \propto \sqrt{M}$,we have $\frac{v_s}{v_d} = \sqrt{\frac{M_A + M_B}{M_A - M_B}}$.
Squaring both sides: $\left(\frac{20}{15}\right)^2 = \frac{M_A + M_B}{M_A - M_B} \Rightarrow \left(\frac{4}{3}\right)^2 = \frac{M_A + M_B}{M_A - M_B}$.
$\frac{16}{9} = \frac{M_A + M_B}{M_A - M_B}$.
Applying componendo and dividendo: $\frac{M_A}{M_B} = \frac{16+9}{16-9} = \frac{25}{7}$.
Thus,the ratio $M_A : M_B = 25:7$.

(C) Given,$\alpha, \beta, \gamma$ are the roots of $x^3+4x+1=0$.
From Vieta's formulas: $\alpha+\beta+\gamma=0$,$\alpha\beta+\beta\gamma+\gamma\alpha=4$,and $\alpha\beta\gamma=-1$.
Since $\alpha+\beta+\gamma=0$,we have $\beta+\gamma=-\alpha$,$\gamma+\alpha=-\beta$,and $\alpha+\beta=-\gamma$.
The roots of the new equation are $y_1 = \frac{\alpha^2}{-\alpha} = -\alpha$,$y_2 = \frac{\beta^2}{-\beta} = -\beta$,and $y_3 = \frac{\gamma^2}{-\gamma} = -\gamma$.
Let $y = -x$,then $x = -y$.
Substituting $x = -y$ into the original equation $x^3+4x+1=0$:
$(-y)^3+4(-y)+1=0
\implies -y^3-4y+1=0
\implies y^3+4y-1=0$.
Thus,the required equation is $x^3+4x-1=0$.

(C) Given,$f(x)=2 x^4-13 x^2+a x+b$ is divisible by $x^2-3 x+2 = (x-2)(x-1)$.
Since $f(x)$ is divisible by $(x-2)$ and $(x-1)$,we must have $f(2)=0$ and $f(1)=0$.
For $f(2)=0$:
$2(2)^4-13(2)^2+a(2)+b=0$
$2(16)-13(4)+2a+b=0$
$32-52+2a+b=0$
$2a+b=20$ ... $(i)$
For $f(1)=0$:
$2(1)^4-13(1)^2+a(1)+b=0$
$2-13+a+b=0$
$a+b=11$ ... $(ii)$
Subtracting Eq. $(ii)$ from Eq. $(i)$:
$(2a+b)-(a+b)=20-11$
$a=9$
Substituting $a=9$ in Eq. $(ii)$:
$9+b=11$
$b=2$
Thus,$(a, b) = (9, 2)$.

(C) Let $z = x + iy$.
Given,$\left|\frac{z+2i}{2z+i}\right| < 1$.
This implies $|z+2i| < |2z+i|$.
Squaring both sides,we get $|z+2i|^2 < |2z+i|^2$.
Substituting $z = x + iy$,we have $|x + i(y+2)|^2 < |2x + i(2y+1)|^2$.
$x^2 + (y+2)^2 < (2x)^2 + (2y+1)^2$.
$x^2 + y^2 + 4y + 4 < 4x^2 + 4y^2 + 4y + 1$.
$4 - 1 < 4x^2 - x^2 + 4y^2 - y^2$.
$3 < 3x^2 + 3y^2$.
Dividing by $3$,we get $x^2 + y^2 > 1$.

(D) Let the subscripts $1$ denote steel and $2$ denote brass.
Given ratios: $\frac{l_1}{l_2} = a$,$\frac{r_1}{r_2} = b$,$\frac{Y_1}{Y_2} = c$.
From the free body diagram,the tension in the steel wire is $F_1 = 2g$ and the tension in the brass wire is $F_2 = 2g + 2g = 4g$.
The formula for elongation is $\Delta l = \frac{F l}{A Y} = \frac{F l}{\pi r^2 Y}$.
For steel: $\Delta l_1 = \frac{F_1 l_1}{\pi r_1^2 Y_1} = \frac{2g l_1}{\pi r_1^2 Y_1}$.
For brass: $\Delta l_2 = \frac{F_2 l_2}{\pi r_2^2 Y_2} = \frac{4g l_2}{\pi r_2^2 Y_2}$.
The ratio of increase in lengths of brass to steel is $\frac{\Delta l_2}{\Delta l_1} = \frac{4g l_2}{\pi r_2^2 Y_2} \cdot \frac{\pi r_1^2 Y_1}{2g l_1}$.
Substituting the given ratios: $\frac{\Delta l_2}{\Delta l_1} = \left(\frac{4}{2}\right) \cdot \left(\frac{l_2}{l_1}\right) \cdot \left(\frac{r_1}{r_2}\right)^2 \cdot \left(\frac{Y_1}{Y_2}\right) = 2 \cdot \left(\frac{1}{a}\right) \cdot b^2 \cdot c = \frac{2 b^2 c}{a}$.
Wait,re-evaluating the question: the ratio of steel to brass is $a, b, c$. Let's re-calculate $\frac{\Delta l_2}{\Delta l_1} = \frac{4g l_2}{\pi r_2^2 Y_2} \cdot \frac{\pi r_1^2 Y_1}{2g l_1} = 2 \cdot \frac{1}{a} \cdot b^2 \cdot c = \frac{2 b^2 c}{a}$.
Looking at the options,if the question asks for $\frac{\Delta l_1}{\Delta l_2}$,then $\frac{\Delta l_1}{\Delta l_2} = \frac{a}{2 b^2 c}$,which matches option $D$.

(A) Let at time $t$,the position of $A$ be $\overrightarrow{r}_A = (0,0) + (0,2)t = (0, 2t)$.
Let at time $t$,the position of $B$ be $\overrightarrow{r}_B = (0,10) + (2,0)t = (2t, 10)$.
The displacement vector between them is $\overrightarrow{r}_{BA} = \overrightarrow{r}_B - \overrightarrow{r}_A = (2t - 0, 10 - 2t) = (2t, 10 - 2t)$.
The square of the distance $D^2$ between them is given by $f(t) = (2t)^2 + (10 - 2t)^2$.
$f(t) = 4t^2 + 100 + 4t^2 - 40t = 8t^2 - 40t + 100$.
To find the time for the minimum distance,we differentiate $f(t)$ with respect to $t$ and set it to zero:
$\frac{df}{dt} = 16t - 40 = 0$.
$16t = 40 \implies t = \frac{40}{16} = 2.5 \text{ s}$.
Since the second derivative $\frac{d^2f}{dt^2} = 16 > 0$,the distance is indeed minimum at $t = 2.5 \text{ s}$.

(A) $1$. Kinetic energy $(KE)$ and momentum $(p)$:
$KE = \frac{p^2}{2m}$. Since $m$ is constant,$KE \propto p^2$. The graph of $KE$ versus $p^2$ is a straight line passing through the origin.
$2$. Kinetic energy $(KE)$ and time $(t)$:
$v_x = u \cos \theta$ and $v_y = u \sin \theta - gt$.
$KE = \frac{1}{2}m(v_x^2 + v_y^2) = \frac{1}{2}m(u^2 \cos^2 \theta + (u \sin \theta - gt)^2)$.
This is a quadratic equation in $t$,representing a parabola.
$3$. Kinetic energy $(KE)$ and vertical displacement $(y)$:
Using $v_y^2 = u_y^2 - 2gy$,we have $KE = \frac{1}{2}m(v_x^2 + v_y^2) = \frac{1}{2}m(u^2 \cos^2 \theta + u^2 \sin^2 \theta - 2gy) = \frac{1}{2}mu^2 - mgy$.
This is a linear equation of the form $KE = -mgy + C$,which represents a straight line with a negative slope.
$4$. Analysis of Graph $(A)$:
The graph $(A)$ shows a $V$-shaped curve for $KE$ versus $y$. However,the relationship derived is $KE = \frac{1}{2}mu^2 - mgy$,which is a linear relationship,not a $V$-shaped curve. Therefore,graph $(A)$ does not represent the variation of $KE$ correctly.

(C) Let the body reach the maximum height $C$ at time $T$. Due to symmetry,the time taken to go from $C$ to $B$ is the same as the time taken to go from $B$ to $C$. Let this time be $t^{\prime}$.
Thus,$t_2 = T + t^{\prime}$ and $t_1 = T - t^{\prime}$.
Adding these equations: $t_1 + t_2 = 2T$,so $T = \frac{t_1 + t_2}{2}$.
Subtracting these equations: $t_2 - t_1 = 2t^{\prime}$,so $t^{\prime} = \frac{t_2 - t_1}{2}$.
The maximum height $H_{\max}$ is the distance covered in time $T$ starting from rest at point $C$ (downward motion).
$H_{\max} = \frac{1}{2} g T^2 = \frac{1}{2} g \left( \frac{t_1 + t_2}{2} \right)^2 = \frac{g(t_1 + t_2)^2}{8}$.

(D) The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
For material $X_1$,$N_1 = N_0 e^{-10 \lambda t}$.
For material $X_2$,$N_2 = N_0 e^{-\lambda t}$.
We are given the ratio $\frac{N_1}{N_2} = \frac{1}{e} = e^{-1}$.
Substituting the expressions:
$\frac{N_0 e^{-10 \lambda t}}{N_0 e^{-\lambda t}} = e^{-1}$
$e^{-10 \lambda t + \lambda t} = e^{-1}$
$e^{-9 \lambda t} = e^{-1}$
Equating the exponents:
$-9 \lambda t = -1$
$t = \frac{1}{9 \lambda}$.

(C) The rate law for the reaction is given by $\text{Rate} = k[A][B]^2$.
Given values are:
$k = 5.0 \times 10^{-6} \ mol^{-2} \ L^2 \ s^{-1}$
$[A] = 0.05 \ mol \ L^{-1} = 5 \times 10^{-2} \ mol \ L^{-1}$
$[B] = 0.1 \ mol \ L^{-1} = 1 \times 10^{-1} \ mol \ L^{-1}$
Substituting these values into the rate equation:
$\text{Rate} = (5.0 \times 10^{-6}) \times (0.05) \times (0.1)^2$
$\text{Rate} = 5.0 \times 10^{-6} \times 5 \times 10^{-2} \times 1 \times 10^{-2}$
$\text{Rate} = 25 \times 10^{-10} \ mol \ L^{-1} \ s^{-1}$
$\text{Rate} = 2.50 \times 10^{-9} \ mol \ L^{-1} \ s^{-1}$

(C) The rate law is given by: $\text{Rate} = k[A][B]^2$.
Given values are: $k = 5.0 \times 10^{-6} \, mol^{-2} \, L^2 \, s^{-1}$,$[A] = 0.05 \, mol \, L^{-1}$,and $[B] = 0.1 \, mol \, L^{-1}$.
Substituting these values into the rate equation:
$\text{Rate} = (5.0 \times 10^{-6}) \times (0.05) \times (0.1)^2$
$= (5.0 \times 10^{-6}) \times (5.0 \times 10^{-2}) \times (1.0 \times 10^{-2})$
$= 25.0 \times 10^{-10} \, mol \, L^{-1} \, s^{-1}$
$= 2.50 \times 10^{-9} \, mol \, L^{-1} \, s^{-1}$.

(B) The correct matches are as follows:
$(A)$ Feldspar is $KAlSi_3O_8$ $(IV)$.
$(B)$ Asbestos is $CaMg_3(SiO_3)_4$ $(V)$.
$(C)$ Pyrargyrite is $[Ag_3SbS_3]$ $(I)$.
$(D)$ Diaspore is $Al_2O_3 \cdot H_2O$ $(II)$.
Therefore,the correct sequence is $A-IV, B-V, C-I, D-II$.

(C) The reaction of ethanol $(C_2H_5OH)$ with chlorine $(Cl_2)$ proceeds as follows:
$1$. Ethanol is oxidized by chlorine to form acetaldehyde $(CH_3CHO)$,which is $X$.
$CH_3CH_2OH + Cl_2 \rightarrow CH_3CHO + 2HCl$
$2$. Acetaldehyde then reacts with further chlorine to form chloral $(CCl_3CHO)$,which is $Y$.
$CH_3CHO + 3Cl_2 \rightarrow CCl_3CHO + 3HCl$
Thus,$X$ is $CH_3CHO$ and $Y$ is $CCl_3CHO$.

(D) The synthesis of crotonaldehyde from acetaldehyde occurs in two steps:
$1$. Aldol condensation of two molecules of acetaldehyde $(CH_3CHO)$ in the presence of dilute $NaOH$ to form $3-$hydroxybutanal (aldol). This step is a nucleophilic addition reaction.
$2$. Dehydration (elimination of water) of $3-$hydroxybutanal upon heating to form crotonaldehyde $(CH_3CH=CHCHO)$.
Since the overall process involves both nucleophilic addition and elimination,it is classified as a nucleophilic addition-elimination reaction.

(A) In the Gattermann reaction,the diazonium group $(-N_2^+Cl^-)$ is replaced by a chlorine atom $(-Cl)$ using copper powder $(Cu)$ in the presence of hydrochloric acid $(HCl)$.
Specifically,the nucleophilic species involved in the substitution is the chloride ion $(Cl^{\ominus})$,and the reagents used are $Cu$ powder and $HCl$.
Therefore,$\underline{X} = Cl^{\ominus}$ and $\underline{Y} = Cu / HCl$.

(D) codon is a specific sequence of $3$ adjacent bases on a strand of $DNA$ or $RNA$.
It specifies a particular amino acid that is to be incorporated into a protein chain during the process of translation.
Therefore,$\underline{A} = 3$ bases,$\underline{B} = \text{amino acid}$,and $\underline{C} = \text{protein}$.

(C) Tripeptides are polymers of amino acids in which three individual amino acid units,called residues,are linked together by amide bonds (peptide bonds).
Since we have $3$ distinct amino acids (glycine,alanine,and phenylalanine),the number of possible tripeptides is given by the number of permutations of these $3$ distinct items.
The number of arrangements is $3! = 3 \times 2 \times 1 = 6$.
Therefore,there are $6$ different ways to link these amino acids.

(C) Reaction $(I)$: Dry distillation of calcium acetate gives acetone.
$(CH_3 COO)_2 Ca \xrightarrow{\Delta} CH_3 COCH_3 + CaCO_3$
So,$\underline{A} = CH_3 COCH_3$.
Reaction $(II)$: Reduction of acetic acid with $HI$ and red phosphorus gives ethane.
$CH_3 COOH + 6HI \xrightarrow{\text{Red P}} CH_3 CH_3 + 3I_2 + 2H_2 O$
So,$\underline{B} = C_2 H_6$.
Reaction $(III)$: Dehydration of acetic acid with $P_4 O_{10}$ gives acetic anhydride.
$2 CH_3 COOH \xrightarrow{P_4 O_{10}} (CH_3 CO)_2 O + H_2 O$
So,$\underline{C} = (CH_3 CO)_2 O$.
Therefore,the correct sequence is $\underline{A} = CH_3 COCH_3$,$\underline{B} = C_2 H_6$,$\underline{C} = (CH_3 CO)_2 O$.

(A) In $XeO_3$,the central atom $Xe$ is $sp^3$ hybridized with one lone pair. It forms $3$ double bonds with oxygen atoms. Each double bond consists of one $\sigma$ bond and one $p \pi-d \pi$ $\pi$ bond. Thus,there are $3$ $p \pi-d \pi$ $\pi$ bonds.
In $XeO_4$,the central atom $Xe$ is $sp^3$ hybridized with no lone pair. It forms $4$ double bonds with oxygen atoms. Each double bond consists of one $\sigma$ bond and one $p \pi-d \pi$ $\pi$ bond. Thus,there are $4$ $p \pi-d \pi$ $\pi$ bonds.
Therefore,the number of $p \pi-d \pi$ $\pi$ bonds in $XeO_3$ and $XeO_4$ are $3$ and $4$ respectively.

(C) Dopamine is a neurotransmitter produced in several areas of the brain. $A$ deficiency of dopamine is associated with Parkinson's disease. The $IUPAC$ name of dopamine is $2-(3,4-\text{dihydroxyphenyl})\text{ethylamine}$. Its structure consists of a benzene ring with two hydroxyl groups at the $3$ and $4$ positions and an ethylamine side chain $(-CH_2CH_2NH_2)$ at the $1$ position. This corresponds to the structure shown in option $C$.

(C) Paramagnetic property depends upon the number of unpaired electrons. Higher the number of unpaired electrons,higher the paramagnetic property will be.
$Cu^{2+} = [Ar] 3d^9$,number of unpaired electrons $= 1$
$V^{2+} = [Ar] 3d^3$,number of unpaired electrons $= 3$
$Cr^{2+} = [Ar] 3d^4$,number of unpaired electrons $= 4$
$Mn^{2+} = [Ar] 3d^5$,number of unpaired electrons $= 5$
Hence,the correct order of increasing paramagnetic property is $Cu^{2+} < V^{2+} < Cr^{2+} < Mn^{2+}$.

(A) The cell reaction is: $Ag(s) + AgCl(s) \rightarrow Ag^{+}(aq) + Cl^{-}(aq) + Ag(s)$
Simplified net cell reaction: $AgCl(s) \rightarrow Ag^{+}(aq) + Cl^{-}(aq)$
Calculate $\Delta G^{\circ}_{reaction}$:
$\Delta G^{\circ}_{reaction} = [\Delta G_f^{\circ}(Ag^{+}) + \Delta G_f^{\circ}(Cl^{-})] - [\Delta G_f^{\circ}(AgCl)]$
$\Delta G^{\circ}_{reaction} = [78 + (-129)] - (-109) \ kJ/mol$
$\Delta G^{\circ}_{reaction} = -51 + 109 = 58 \ kJ/mol = 58000 \ J/mol$
Using the relation $\Delta G^{\circ} = -nFE^{\circ}_{cell}$:
Here,$n = 1$ (as $Ag \rightarrow Ag^{+} + e^{-}$ and $AgCl + e^{-} \rightarrow Ag + Cl^{-}$).
$58000 = -1 \times 96500 \times E^{\circ}_{cell}$
$E^{\circ}_{cell} = -\frac{58000}{96500} \approx -0.60 \ V$

(B) According to Kohlrausch's law of independent migration of ions:
$\lambda_m^{\circ} Ba(OH)_2 = \lambda_m^{\circ} BaCl_2 + 2 \lambda_m^{\circ} NaOH - 2 \lambda_m^{\circ} NaCl$
Substituting the given values:
$\lambda_m^{\circ} Ba(OH)_2 = (280 \times 10^{-4}) + 2(248 \times 10^{-4}) - 2(126 \times 10^{-4})$
$\lambda_m^{\circ} Ba(OH)_2 = (280 + 496 - 252) \times 10^{-4} \ S \ m^2 \ mol^{-1}$
$\lambda_m^{\circ} Ba(OH)_2 = 524 \times 10^{-4} \ S \ m^2 \ mol^{-1}$

(A) The correct matches are:
$(A)$ Feldspar: $KAlSi_3O_8$ $(IV)$
$(B)$ Asbestos: $CaMg_3(SiO_3)_4$ $(V)$
$(C)$ Pyrargyrite: $Ag_3SbS_3$ $(I)$
$(D)$ Diaspore: $Al_2O_3 \cdot H_2O$ $(II)$
Therefore,the correct sequence is $A-IV, B-V, C-I, D-II$.

(B) The reaction of an alkyl halide $(C_2H_5Cl)$ with $AgCN$ is a nucleophilic substitution reaction.
$AgCN$ is a covalent compound,and the nitrogen atom acts as the nucleophilic center.
Therefore,the reaction proceeds as follows:
$C_2H_5Cl + AgCN \xrightarrow{EtOH / H_2O} C_2H_5-NC + AgCl$
Here,$\underline{X}$ is ethyl isocyanide $(C_2H_5NC)$.
In ethyl isocyanide,the nitrogen atom is directly linked to the ethyl carbon $(C_2H_5-N \equiv C)$.
Thus,statement $(III)$ is correct.

(C) The presence of $P-H$ bonds in phosphorus oxyacids is determined by their structures:
$1$. $H_3PO_4$ (Orthophosphoric acid): Contains three $P-OH$ bonds and one $P=O$ bond. No $P-H$ bond.
$2$. $H_3PO_3$ (Phosphorous acid): Contains two $P-OH$ bonds,one $P=O$ bond,and one $P-H$ bond.
$3$. $H_3PO_2$ (Hypophosphorous acid): Contains one $P-OH$ bond,one $P=O$ bond,and two $P-H$ bonds.
Therefore,the pair $H_3PO_3$ and $H_3PO_2$ contains $P-H$ bonds.

(B) The reaction of fluorine with dilute $NaOH$ is given by:
$2F_2 + 2NaOH \rightarrow 2NaF + OF_2 + H_2O$
Thus,the gaseous product $A$ is oxygen difluoride $(OF_2)$.
In $OF_2$,the central oxygen atom is $sp^3$ hybridized.
It has two bond pairs and two lone pairs of electrons.
Due to the strong repulsion between the two lone pairs of electrons on the oxygen atom,the bond angle is compressed from the ideal tetrahedral angle of $109^{\circ} 28^{\prime}$ to $103^{\circ}$.

(B) The density formula is given by $d = \frac{M \times Z}{N_A \times a^3}$, where $Z$ is the number of atoms per unit cell.
Rearranging for $Z$: $Z = \frac{d \times N_A \times a^3}{M}$.
Given: $d = 8.92 \ g \ cm^{-3}$, $M = 63.55 \ g \ mol^{-1}$, $a = 362 \ pm = 362 \times 10^{-10} \ cm$, and $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
Substituting the values:
$Z = \frac{8.92 \times 6.022 \times 10^{23} \times (362 \times 10^{-10})^3}{63.55} \approx 4$.
Since $Z = 4$, the unit cell is face-centred cubic $(FCC)$.

(A) The freezing point of a substance is defined as the temperature at which the solid and liquid phases of the substance are in equilibrium.
In the context of the depression in freezing point experiment,the equilibrium is established between the molecules of the liquid solvent (present in the solution) and the solid solvent (which separates out upon freezing).

(B) $As_2S_3$ is a negatively charged sol.
According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the valency of the ion carrying a charge opposite to that of the colloidal particles.
Since $As_2S_3$ is a negative sol,the positive ions (cations) are responsible for coagulation.
The coagulating power increases with the increase in the valency of the cation.
The valencies of the cations in the given options are:
$K^+$ $(KCl)$,$Mg^{2+}$ $(MgSO_4)$,$Al^{3+}$ $(AlCl_3)$.
Since $Al^{3+}$ has the highest valency $(+3)$,it is the most effective in causing the coagulation of the $As_2S_3$ sol.