AP EAMCET 2013 Chemistry Question Paper with Answer and Solution

(A) The number of radial nodes is calculated using the formula: $\text{Radial nodes} = (n - l - 1)$.
For $3s$ orbital: $n = 3, l = 0$.
$\text{Radial nodes} = 3 - 0 - 1 = 2$.
For $2p$ orbital: $n = 2, l = 1$.
$\text{Radial nodes} = 2 - 1 - 1 = 0$.
Therefore,the number of radial nodes for $3s$ and $2p$ orbitals are $2$ and $0$ respectively.

(A) The number of radial nodes is calculated using the formula: $\text{Radial nodes} = n - l - 1$.
For $3s$ orbital: $n = 3$,$l = 0$. Therefore,$\text{Radial nodes} = 3 - 0 - 1 = 2$.
For $2p$ orbital: $n = 2$,$l = 1$. Therefore,$\text{Radial nodes} = 2 - 1 - 1 = 0$.
Thus,the number of radial nodes for $3s$ and $2p$ orbitals are $2$ and $0$ respectively.

(B) When phenol reacts with $Br_2$ in the presence of a non-polar solvent like $CS_2$ at low temperature $(0^{\circ} C)$,the reaction is limited to mono-substitution.
This occurs because the non-polar solvent does not facilitate the ionization of phenol into the highly reactive phenoxide ion,which is responsible for poly-substitution in polar solvents like water.
Therefore,the major products formed are $o$-bromophenol and $p$-bromophenol.

(D) The resonant frequency $f$ of a series $LCR$ circuit is given by the formula:
$f = \frac{1}{2 \pi \sqrt{LC}}$
From this expression,we can see that the resonant frequency is inversely proportional to the square root of the inductance $L$ when the capacitance $C$ is kept constant:
$f \propto \frac{1}{\sqrt{L}}$
Let the initial resonant frequency be $f_0$ with inductance $L$. The new resonant frequency $f'$ with new inductance $L' = \sqrt{3} L$ is:
$\frac{f'}{f_0} = \sqrt{\frac{L}{L'}} = \sqrt{\frac{L}{\sqrt{3} L}} = \sqrt{\frac{1}{\sqrt{3}}} = \left(\frac{1}{3^{1/2}}\right)^{1/2} = \left(\frac{1}{3}\right)^{1/4}$
Therefore,the new resonant frequency is $f' = \left(\frac{1}{3}\right)^{1/4} f_0$.
Note that the change in resistance $R$ does not affect the resonant frequency of an $LCR$ circuit.

(B) The initial capacitance of the parallel plate capacitor with air as the medium is given by $C = \frac{\varepsilon_0 A}{d}$.
When a metal plate of thickness $t = d/2$ is inserted between the plates,the effective distance between the plates becomes $d_{eff} = d - t = d - d/2 = d/2$.
Since the inserted plate is a metal,it acts as a short circuit across the region it occupies. The effective capacitance $C^{\prime}$ of the system is equivalent to a parallel plate capacitor with an air gap of $d/2$.
Thus,$C^{\prime} = \frac{\varepsilon_0 A}{d - d/2} = \frac{\varepsilon_0 A}{d/2} = \frac{2 \varepsilon_0 A}{d}$.
Comparing this with the initial capacitance $C$,we get $C^{\prime} = 2C$.
Therefore,the ratio $\frac{C^{\prime}}{C} = 2$.

(B) Let $M = 2.9 \ kg$ and $m = 0.1 \ kg$. The combined mass is $M' = M + m = 3.0 \ kg$.
Using the law of conservation of momentum for the collision:
$m v = (M + m) V$
$0.1 \times 150 = 3.0 \times V$
$V = 5 \ m/s$ (velocity of the combined mass just after collision).
Now,using the law of conservation of energy between the lowest point and the point where the string makes an angle $\theta = 60^{\circ}$:
$\frac{1}{2} M' V^2 = \frac{1}{2} M' v_{\theta}^2 + M' g l(1 - \cos \theta)$
$v_{\theta}^2 = V^2 - 2gl(1 - \cos 60^{\circ})$
$v_{\theta}^2 = 5^2 - 2 \times 10 \times 0.5 \times (1 - 0.5) = 25 - 5 = 20 \ m^2/s^2$.
The tension $T$ in the string at angle $\theta$ is given by:
$T - M' g \cos \theta = \frac{M' v_{\theta}^2}{l}$
$T = M' g \cos \theta + \frac{M' v_{\theta}^2}{l}$
$T = 3.0 \times 10 \times \cos 60^{\circ} + \frac{3.0 \times 20}{0.5}$
$T = 30 \times 0.5 + 120 = 15 + 120 = 135 \ N$.

(A) The velocity of the centre of mass is given by the formula:
$v_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$
Given: $m_1 = 4 ~kg$,$m_2 = 5 ~kg$.
Velocity $v_1$ is along the east direction (let this be $\hat{i}$): $v_1 = 5 \hat{i} ~m/s$.
Velocity $v_2$ is along the north direction (let this be $\hat{j}$): $v_2 = 3 \hat{j} ~m/s$.
Substituting the values:
$v_{CM} = \frac{4(5 \hat{i}) + 5(3 \hat{j})}{4 + 5}$
$v_{CM} = \frac{20 \hat{i} + 15 \hat{j}}{9} = \frac{20}{9} \hat{i} + \frac{15}{9} \hat{j}$
The magnitude of the velocity of the centre of mass is:
$|v_{CM}| = \sqrt{\left(\frac{20}{9}\right)^2 + \left(\frac{15}{9}\right)^2}$
$|v_{CM}| = \sqrt{\frac{400 + 225}{81}} = \sqrt{\frac{625}{81}}$
$|v_{CM}| = \frac{25}{9} ~m/s$

(B) Lateral overlap of atomic orbitals (like $p-p$,$p-d$,or $d-d$) results in the formation of $\pi$ bonds,whereas axial overlap results in $\sigma$ bonds.

(C) Let us analyze each molecule:
$1$. $PCl_5$: The central atom $P$ has $5$ bond pairs and $0$ lone pairs. The hybridisation is $sp^3d$ and the shape is trigonal bipyramidal.
$2$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ has $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons. The hybridisation is $dsp^2$ and the shape is square planar.
$3$. $SF_6$: The central atom $S$ has $6$ bond pairs and $0$ lone pairs. The hybridisation is $sp^3d^2$ and the shape is octahedral. This is correct.
$4$. $IF_3$: The central atom $I$ has $3$ bond pairs and $2$ lone pairs. The hybridisation is $sp^3d$ and the shape is bent $T$-shaped.
Therefore,the correct set is $(c)$.

(D) Phenacetin is a derivative of $p$-aminophenol and is used as an analgesic (painkiller). Its chemical structure is $N$-($4$-ethoxyphenyl)acetamide. Among the given options,the structure corresponding to $p$-ethoxyacetanilide is option $D$.

(C) For the $4$th period,the principal quantum number $n = 4$.
The orbitals being filled are $4s$,$3d$,and $4p$.
The number of electrons that can be accommodated in these orbitals is $2$ (for $4s$) $+ 10$ (for $3d$) $+ 6$ (for $4p$) $= 18$.
Therefore,the number of elements in the $4$th period is $18$.

(D) The correct option is $(d)$.
Prussian blue is ferric ferrocyanide,which has the chemical formula $Fe_4[Fe(CN)_6]_3$.
It is formed by the reaction of ferric ions $(Fe^{3+})$ with ferrocyanide ions $([Fe(CN)_6]^{4-})$.

(A) The wavelength region $490-500 \ nm$ corresponds to the blue-green part of the visible spectrum.
According to the colour wheel,the complementary colour of blue-green is red.
When a substance absorbs a specific wavelength of light,the colour observed is the complementary colour of the absorbed radiation.

(D) The sensitivity of a galvanometer is given by the ratio of the current through the galvanometer $(i_g)$ to the total current $(i)$.
Given that the sensitivity is decreased by $\frac{1}{40}$ times,we have $\frac{i_g}{i} = \frac{1}{40}$.
The formula for the current through a galvanometer with a shunt resistance $S$ and galvanometer resistance $G$ is $i_g = \frac{S}{S+G} i$.
Substituting the given values,where $S = 10 \Omega$ and $G = x$:
$\frac{1}{40} = \frac{10}{10+x}$.
Cross-multiplying gives $10 + x = 400$.
Therefore,$x = 400 - 10 = 390 \Omega$.

(A) When a current $I$ flows through a conductor with a temperature difference $\Delta T$ between its ends,the heat developed per unit time is governed by the Thomson effect. The heat produced per unit time (Thomson heat) is given by the formula $H = \sigma I \Delta T$,where $\sigma$ is the Thomson coefficient. Therefore,the heat developed per unit time is directly proportional to both the current $I$ and the temperature difference $\Delta T$.

(C) Let the potential at junction $P$ be $V$. According to Kirchhoff's Current Law $(KCL)$ at junction $P$,the sum of currents entering the junction equals the sum of currents leaving the junction.
$\frac{24 - V}{3} = \frac{V - 10}{2} + \frac{V - 9}{1}$
Multiply the entire equation by $6$ to clear the denominators:
$2(24 - V) = 3(V - 10) + 6(V - 9)$
$48 - 2V = 3V - 30 + 6V - 54$
$48 - 2V = 9V - 84$
$11V = 132$
$V = 12 \,V$
Now,calculate the current $I$ flowing through the $3 \,\Omega$ resistor:
$I = \frac{24 - V}{3} = \frac{24 - 12}{3} = \frac{12}{3} = 4 \,A$
Thus,the current $I$ is $4 \,A$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE_{max})$ is given by:
$KE_{max} = \frac{hc}{\lambda} - \phi_0$
Given:
Work function $\phi_0 = 2 \text{ eV} = 2 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-19} \text{ J}$
Wavelength $\lambda = 3000 \text{ Å} = 3000 \times 10^{-10} \text{ m} = 3 \times 10^{-7} \text{ m}$
Energy of incident photon $E = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{3 \times 10^{-7}} \text{ J}$
$E = 6.6 \times 10^{-19} \text{ J}$
Now,$KE_{max} = E - \phi_0 = 6.6 \times 10^{-19} \text{ J} - 3.2 \times 10^{-19} \text{ J} = 3.4 \times 10^{-19} \text{ J}$

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $E$ of an emitted electron is given by $E = \frac{hc}{\lambda} - W$,where $W$ is the work function of the metal.
For the first case: $E_1 = \frac{hc}{\lambda_1} - W$ $(i)$
For the second case: $E_2 = \frac{hc}{\lambda_2} - W$ (ii)
Subtracting equation (ii) from equation $(i)$:
$E_1 - E_2 = \left(\frac{hc}{\lambda_1} - W\right) - \left(\frac{hc}{\lambda_2} - W\right)$
$E_1 - E_2 = hc \left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)$
$E_1 - E_2 = hc \left(\frac{\lambda_2 - \lambda_1}{\lambda_1 \lambda_2}\right)$
Rearranging the terms to solve for $hc$:
$hc = \frac{(E_1 - E_2) \lambda_1 \lambda_2}{\lambda_2 - \lambda_1}$

(B) During the electrolysis of an aqueous solution of copper sulphate $(CuSO_4)$ using copper electrodes,both $Cu^{2+}$ and $H^{+}$ ions move towards the cathode.
Since the reduction potential of $Cu^{2+}$ $(+0.34 \ V)$ is higher than that of $H^{+}$ $(0.00 \ V)$,$Cu^{2+}$ ions are preferentially reduced at the cathode.
The reaction at the cathode is:
$Cu^{2+}_{(aq)} + 2e^- \longrightarrow Cu_{(s)}$

(C) The extent of charge of a lead accumulator is determined by the specific gravity of the $H_2SO_4$ solution.
During discharge,$H_2SO_4$ is consumed,and its density decreases.
During charging,$H_2SO_4$ is regenerated,and its density increases.

(B) The most serious effect of the depletion of the ozone layer is that the $UV$ rays coming from the sun can pass through the stratosphere and reach the surface of the earth. It has been found that with an increase in the exposure to $UV$ rays,the chance for the occurrence of skin cancer increases. Also,exposure of the eye to $UV$ rays damages the cornea and lens of the eye,which may cause cataract and even blindness.

(D) Pyridinium chlorochromate $(PCC)$ is formed by the reaction of pyridine with chlorochromic acid,which is obtained by dissolving chromium trioxide $(CrO_3)$ in aqueous $HCl$.
The chemical formula for the pyridinium cation is $[C_5 H_5 NH]^+$ and the chlorochromate anion is $[CrO_3 Cl]^-$.
Therefore,the structure of $PCC$ is represented as $[C_5 H_5 NH]^+ [CrO_3 Cl]^-$,which corresponds to $C_5 H_5 \stackrel{\oplus}{N} HCrO_3 Cl^{\ominus}$.

(A) The $-NH_2$ group is a strong electron-donating group due to its lone pair of electrons,which it donates to the benzene ring via the resonance effect ($+M$ effect).
This increases the electron density at the ortho and para positions of the benzene ring.
When an electrophile attacks at the ortho or para position,the resulting arenium ion is stabilized by the resonance donation of the lone pair from the nitrogen atom.
Therefore,both the assertion and the reason are correct,and the reason is the correct explanation for the assertion.

(A) Enantiomers are stereoisomers that are non-superimposable mirror images of each other.
They possess identical physical properties such as melting point,boiling point,density,and refractive index in an achiral environment.
However,they differ in their interaction with plane-polarized light.
One enantiomer rotates the plane of plane-polarized light to the right (dextrorotatory,denoted by $+$),while the other rotates it to the left (levorotatory,denoted by $-$) to the same extent.
Therefore,they differ in the sign of their specific rotation.

(D) In the extraction of silver by the cyanide process (also known as the Mac-Arthur Forrest process),the silver compound is dissolved in a $NaCN$ solution in the presence of air to form a soluble complex salt.
Subsequently,silver is precipitated from this complex salt by the addition of $Zn$,as $Zn$ is more reactive and displaces the $Ag$.
$Na_2S_2O_3$ is not used in this process.
The chemical reactions are:
$Ag_2S + 4NaCN \rightleftharpoons 2Na[Ag(CN)_2] + Na_2S$
$2Na_2S + 2O_2 \longrightarrow 2Na_2SO_4$
$2Na[Ag(CN)_2] + Zn \longrightarrow Na_2[Zn(CN)_4] + 2Ag \downarrow$

(B) In reaction $I$,the reaction of $n$-propanol with $PBr_3$ is a nucleophilic substitution reaction ($S_N2$ mechanism) which converts the alcohol group into a bromide,yielding $n$-propyl bromide $(CH_3CH_2CH_2Br)$.
In reaction $II$,the reaction of propene with $HBr$ in the presence of benzoyl peroxide $((C_6H_5COO)_2)$ proceeds via the anti-Markovnikov addition mechanism (peroxide effect or Kharasch effect). This results in the formation of $n$-propyl bromide $(CH_3CH_2CH_2Br)$ as the major product.
Therefore,$X = CH_3CH_2CH_2Br$ and $Y = CH_3CH_2CH_2Br$.

(C) The decomposition of a compound by the application of heat is known as pyrolysis.
Specifically,the pyrolysis of higher alkanes into a mixture of lower alkanes,alkenes,and hydrogen is also referred to as cracking.

(B) For titration of $Na_2CO_3$ and $NaHCO_3$ with $HCl$ using phenolphthalein,the reaction is: $Na_2CO_3 + HCl \longrightarrow NaHCO_3 + NaCl$.
At the phenolphthalein end point,$20 \ mL$ of $0.1 \ M$ $HCl$ is consumed,which corresponds to half the $Na_2CO_3$.
$Moles \ of \ Na_2CO_3 = Moles \ of \ HCl = 20 \ mL \times 0.1 \ M = 2 \ mmol$.
Concentration of $Na_2CO_3 = \frac{2 \ mmol}{20 \ mL} = 0.1 \ M$.
Using methyl orange,both $Na_2CO_3$ and $NaHCO_3$ are neutralized.
Total $HCl$ consumed = $60 \ mL \times 0.1 \ M = 6 \ mmol$.
$HCl$ consumed by $Na_2CO_3$ (complete) = $2 \times (2 \ mmol) = 4 \ mmol$.
$HCl$ consumed by $NaHCO_3$ = $6 \ mmol - 4 \ mmol = 2 \ mmol$.
Concentration of $NaHCO_3 = \frac{2 \ mmol}{20 \ mL} = 0.1 \ M$.

(A) Statement $1$ is correct because heavy water $(D_2O)$ slows down metabolic rates and is harmful to living organisms.
Statement $2$ is incorrect because the reaction of aluminum carbide with heavy water produces deuterated methane $(CD_4)$,not acetylene: $Al_4C_3 + 12D_2O \rightarrow 4Al(OD)_3 + 3CD_4$.
Statement $3$ is correct as $BaCl_2 \cdot 2D_2O$ is a known example of an interstitial deuterate where $D_2O$ molecules occupy interstitial sites in the crystal lattice.
Therefore,the correct statements are $1$ and $3$.

(A) The acidity of a carboxylic acid is inversely proportional to its $pK_a$ value.
$A$ higher $pK_a$ value corresponds to a lower $K_a$ value,which means the acid is weaker.
Comparing the given values: $1.28 < 2.56 < 4.76 < 4.89$.
Since $4.89$ is the highest $pK_a$ value,the carboxylic acid with $pK_a = 4.89$ is the weakest acid.

(B) The mixture of $NH_4OH$ (a weak base) and $NH_4Cl$ (its salt with a strong acid) forms a basic buffer solution.
For a basic buffer,the $pOH$ is calculated using the Henderson-Hasselbalch equation:
$pOH = pK_b + \log \frac{[Salt]}{[Base]}$
Given:
$pK_b = 4.8$
$[Salt] = 0.2 \ M$
$[Base] = 0.02 \ M$
Substituting the values:
$pOH = 4.8 + \log \left( \frac{0.2}{0.02} \right)$
$pOH = 4.8 + \log(10)$
$pOH = 4.8 + 1 = 5.8$
Since $pH + pOH = 14$ at $25^{\circ}C$:
$pH = 14 - 5.8 = 8.2$

(A) In a deflection magnetometer,the magnetic field due to the magnet $F$ and the horizontal component of the earth's magnetic field $B_H$ are perpendicular to each other.
Therefore,the net magnetic field acting on the needle is $B_{net} = \sqrt{F^2 + B_H^2}$.
The time period of oscillation is given by $T = 2\pi \sqrt{\frac{I}{m B_{net}}}$,where $I$ is the moment of inertia and $m$ is the magnetic moment of the needle.
Thus,$T = 2\pi \sqrt{\frac{I}{m \sqrt{F^2 + B_H^2}}}$.
When the magnet is removed,the field acting on the needle is only $B_H$,so the time period is $T_0 = 2\pi \sqrt{\frac{I}{m B_H}}$.
From the principle of the deflection magnetometer,we have $\frac{F}{B_H} = \tan \theta$,which implies $F = B_H \tan \theta$.
Substituting $F$ into the expression for $T$:
$T = 2\pi \sqrt{\frac{I}{m \sqrt{(B_H \tan \theta)^2 + B_H^2}}} = 2\pi \sqrt{\frac{I}{m B_H \sqrt{\tan^2 \theta + 1}}} = 2\pi \sqrt{\frac{I}{m B_H \sec \theta}}$.
Comparing $T$ and $T_0$:
$\frac{T}{T_0} = \sqrt{\frac{1}{\sec \theta}} = \sqrt{\cos \theta}$.
Squaring both sides,we get $T^2 = T_0^2 \cos \theta$.

(C) Given the quadratic equation: $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$.
Let the roots of the equation be $\alpha$ and $\beta$.
From the properties of quadratic equations,the sum of the roots is $\alpha + \beta = \frac{-(-b)}{\sqrt{2}} = \frac{b}{\sqrt{2}}$.
The product of the roots is $\alpha\beta = \frac{8 - 2\sqrt{5}}{\sqrt{2}}$.
The harmonic mean $(HM)$ of two roots is given by $HM = \frac{2\alpha\beta}{\alpha + \beta}$.
Given $HM = 4$,we have:
$\frac{2\alpha\beta}{\alpha + \beta} = 4$
Substituting the values of $\alpha + \beta$ and $\alpha\beta$:
$\frac{2 \left( \frac{8 - 2\sqrt{5}}{\sqrt{2}} \right)}{\frac{b}{\sqrt{2}}} = 4$
$\frac{2(8 - 2\sqrt{5})}{b} = 4$
$2(8 - 2\sqrt{5}) = 4b$
$8 - 2\sqrt{5} = 2b$
$b = 4 - \sqrt{5}$.

(A) Given: Tension $F = 20 ~N$,Area $A = 0.01 ~cm^2 = 10^{-6} ~m^2$,Young's modulus $Y = 1.1 \times 10^{11} ~N/m^2$,Poisson's ratio $\sigma = 0.32$.
The longitudinal strain is given by $\frac{\Delta l}{l} = \frac{F}{AY} = \frac{20}{10^{-6} \times 1.1 \times 10^{11}} = \frac{20}{1.1 \times 10^5} \approx 1.818 \times 10^{-4}$.
Poisson's ratio is defined as $\sigma = -\frac{\Delta r/r}{\Delta l/l}$. Thus,the lateral strain is $\frac{\Delta r}{r} = -\sigma \frac{\Delta l}{l} = -0.32 \times 1.818 \times 10^{-4} \approx -5.818 \times 10^{-5}$.
The area $A = \pi r^2$,so $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.
The decrease in area is $\Delta A = 2 \times A \times \left( \sigma \frac{\Delta l}{l} \right) = 2 \times 10^{-6} \times 0.32 \times 1.818 \times 10^{-4} \approx 1.16 \times 10^{-10} ~m^2$.
Converting to $cm^2$: $1.16 \times 10^{-10} \times (10^2 ~cm)^2 = 1.16 \times 10^{-6} ~cm^2$.

(C) The average velocity is defined as the total displacement divided by the total time taken.
Let the initial position be $(0, 0)$. The highest point is reached at time $t = \frac{v \sin \theta}{g}$.
At this time,the horizontal displacement is $x = (v \cos \theta) t = \frac{v^2 \sin \theta \cos \theta}{g} = \frac{R}{2}$,and the vertical displacement is $y = H = \frac{v^2 \sin^2 \theta}{2g}$.
The displacement vector is $\vec{s} = \frac{R}{2} \hat{i} + H \hat{j}$.
The magnitude of displacement is $|\vec{s}| = \sqrt{(\frac{R}{2})^2 + H^2}$.
Substituting $R = \frac{v^2 \sin 2\theta}{g} = \frac{2v^2 \sin \theta \cos \theta}{g}$ and $H = \frac{v^2 \sin^2 \theta}{2g}$:
$|\vec{s}| = \sqrt{(\frac{v^2 \sin \theta \cos \theta}{g})^2 + (\frac{v^2 \sin^2 \theta}{2g})^2} = \frac{v^2 \sin \theta}{g} \sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{4}} = \frac{v^2 \sin \theta}{2g} \sqrt{4 \cos^2 \theta + \sin^2 \theta} = \frac{v^2 \sin \theta}{2g} \sqrt{3 \cos^2 \theta + 1}$.
The time taken is $t = \frac{v \sin \theta}{g}$.
Therefore,average velocity $v_{av} = \frac{|\vec{s}|}{t} = \frac{\frac{v^2 \sin \theta}{2g} \sqrt{3 \cos^2 \theta + 1}}{\frac{v \sin \theta}{g}} = \frac{v}{2} \sqrt{1 + 3 \cos^2 \theta}$.

(B) The relation between the nuclear radius $(R)$ and the mass number $(A)$ is given by the formula: $R = R_0 A^{1/3}$.
Therefore,the ratio of the radii of two nuclei is given by: $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3}$.
Given: $R_1 = 6 \text{ fermi}$,$A_1 = 125$,and $A_2 = 27$.
Substituting the values into the formula:
$\frac{6}{R_2} = \left(\frac{125}{27}\right)^{1/3} = \frac{5}{3}$.
Solving for $R_2$:
$R_2 = \frac{6 \times 3}{5} = \frac{18}{5} = 3.6 \text{ fermi}$.
Since $1 \text{ fermi} = 10^{-15} \text{ m}$,we have:
$R_2 = 3.6 \times 10^{-15} \text{ m}$.

(B) The maximum velocity of a particle in simple harmonic motion is given by $V_{\max} = A\omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
For a spring-mass system,$\omega = \sqrt{\frac{K}{m}}$.
Thus,$V_{\max} = A\sqrt{\frac{K}{m}}$.
Given for particle $A$: mass $m_A = m$,force constant $K_A = K_1$,amplitude $A_A$.
Given for particle $B$: mass $m_B = 2m$,force constant $K_B = K_2$,amplitude $A_B$.
Since $(V_{\max})_A = (V_{\max})_B$,we have:
$A_A \sqrt{\frac{K_1}{m}} = A_B \sqrt{\frac{K_2}{2m}}$
$\frac{A_A}{A_B} = \sqrt{\frac{K_2}{2m} \cdot \frac{m}{K_1}}$
$\frac{A_A}{A_B} = \sqrt{\frac{K_2}{2K_1}}$.

(A) Diborane $(B_2H_6)$ reacts with $HCl$ in the presence of $AlCl_3$ as a catalyst to undergo hydrochlorination,which liberates $H_2$ gas.
The chemical reaction is: $B_2H_6 + HCl \xrightarrow{AlCl_3} B_2H_5Cl + H_2 \uparrow$.

(C) In three-dimensional silicates,all four corners of the $SiO_4^{4-}$ tetrahedral unit are shared with other tetrahedra.
This sharing of all four oxygen atoms results in a three-dimensional network structure.
Examples of such structures include silica $(SiO_2)$ in its various forms like quartz,tridymite,and cristobalite.

(C) The chemical formula of pyrophosphoric acid is $H_4P_2O_7$.
In this structure,phosphorus is in the $+5$ oxidation state.
It contains four $P-OH$ bonds,two $P=O$ bonds,and one $P-O-P$ linkage.
It does not contain any $P-H$ bonds.
Since it has four $P-OH$ groups,it is a tetrabasic acid.
Orthophosphoric acid $(H_3PO_4)$ is prepared by dissolving $P_4O_{10}$ in water: $P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4$.
Therefore,the statement that pyrophosphoric acid contains two $P-H$ bonds is incorrect.

(D) Sodium thiosulphate is oxidised by moist $Cl_2$ (chlorine water) to form sodium sulphate,hydrochloric acid,and sulphur.
The balanced chemical equation is:
$Na_2S_2O_3 + 4Cl_2 + 5H_2O \longrightarrow Na_2SO_4 + 8HCl + S$
Here,$X$ is sulphur $(S)$.

(D) Liquid $Xe$ (Xenon) is used in bubble chambers for the detection of $\gamma$-photons and neutral mesons due to its high density and high atomic number,which increases the probability of interaction with high-energy particles.

(D) Given,$\sin \theta + \cos \theta = p$ $(i)$
and $\sin^3 \theta + \cos^3 \theta = q$ $(ii)$
We know that $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.
So,$(\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta) = q$
$p(1 - \sin \theta \cos \theta) = q$ (since $\sin^2 \theta + \cos^2 \theta = 1$)
$1 - \sin \theta \cos \theta = \frac{q}{p}$
$\sin \theta \cos \theta = 1 - \frac{q}{p} = \frac{p - q}{p}$ $(iii)$
Now,square equation $(i)$:
$(\sin \theta + \cos \theta)^2 = p^2$
$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = p^2$
$1 + 2 \sin \theta \cos \theta = p^2$
Substitute $\sin \theta \cos \theta$ from $(iii)$:
$1 + 2 \left( \frac{p - q}{p} \right) = p^2$
$p + 2p - 2q = p^3$
$3p - 2q = p^3$
$p^3 - 3p = -2q$
$p(p^2 - 3) = -2q$

(A) Given,$\tan (\pi \cos \theta)=\cot (\pi \sin \theta)$
$\Rightarrow \tan (\pi \cos \theta)=\tan \left(\frac{\pi}{2}-\pi \sin \theta\right)$
$\Rightarrow \pi \cos \theta=\frac{\pi}{2}-\pi \sin \theta$
$\Rightarrow \cos \theta+\sin \theta=\frac{1}{2}$
Dividing both sides by $\sqrt{2}$:
$\Rightarrow \frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta=\frac{1}{2 \sqrt{2}}$
$\Rightarrow \cos \theta \cos \frac{\pi}{4}+\sin \theta \sin \frac{\pi}{4}=\frac{1}{2 \sqrt{2}}$
$\Rightarrow \cos \left(\theta-\frac{\pi}{4}\right)=\frac{1}{2 \sqrt{2}}$

(A) Comparing the given equation $x^2-5xy+py^2+3x-8y+2=0$ with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=-\frac{5}{2}, b=p, g=\frac{3}{2}, f=-4, c=2$.
The equation represents a pair of straight lines if $abc+2fgh-af^2-bg^2-ch^2=0$.
Substituting the values: $1(p)(2) + 2(-4)(\frac{3}{2})(-\frac{5}{2}) - 1(-4)^2 - p(\frac{3}{2})^2 - 2(-\frac{5}{2})^2 = 0$.
$2p + 30 - 16 - \frac{9p}{4} - \frac{25}{2} = 0$.
Multiplying by $4$: $8p + 120 - 64 - 9p - 50 = 0$ $\Rightarrow -p + 6 = 0$ $\Rightarrow p=6$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
$\tan \theta = \left| \frac{2\sqrt{(-\frac{5}{2})^2 - 1(6)}}{1+6} \right| = \left| \frac{2\sqrt{\frac{25}{4}-6}}{7} \right| = \left| \frac{2\sqrt{\frac{1}{4}}}{7} \right| = \frac{2(\frac{1}{2})}{7} = \frac{1}{7}$.
Since $\tan \theta = \frac{1}{7}$,we can form a right-angled triangle with opposite side $1$ and adjacent side $7$. The hypotenuse is $\sqrt{1^2+7^2} = \sqrt{50}$.
Therefore,$\sin \theta = \frac{1}{\sqrt{50}}$.

(C) The point of intersection $(x_0, y_0)$ of the pair of straight lines given by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is found by solving $\frac{\partial}{\partial x} = 2ax + 2hy + 2g = 0$ and $\frac{\partial}{\partial y} = 2hx + 2by + 2f = 0$.
Solving these equations,we get $x_0 = \frac{hf-bg}{ab-h^2}$ and $y_0 = \frac{gh-af}{ab-h^2}$.
The square of the distance from the origin $(0,0)$ is $D^2 = x_0^2 + y_0^2$.
Using the condition for a pair of straight lines $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$,we can simplify $x_0^2 + y_0^2$ to $\frac{c(a+b)-f^2-g^2}{ab-h^2}$.

(C) The point of intersection $(x_0, y_0)$ of the pair of straight lines given by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is found by solving $\frac{\partial}{\partial x} = 0$ and $\frac{\partial}{\partial y} = 0$.
These partial derivatives are $2ax + 2hy + 2g = 0$ and $2hx + 2by + 2f = 0$.
Solving these,we get $x_0 = \frac{hf-bg}{ab-h^2}$ and $y_0 = \frac{gh-af}{ab-h^2}$.
The square of the distance from the origin $(0,0)$ is $D^2 = x_0^2 + y_0^2 = \left(\frac{hf-bg}{ab-h^2}\right)^2 + \left(\frac{gh-af}{ab-h^2}\right)^2$.
Using the condition for a pair of lines $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$,we simplify the expression to $D^2 = \frac{c(a+b)-f^2-g^2}{ab-h^2}$.

(A) Let $m_1$ and $m_2$ be the slopes of the lines.
Given that $m_1+m_2$ is the arithmetic mean of $4$ and $9$,so $m_1+m_2 = \frac{4+9}{2} = \frac{13}{2}$.
Given that $m_1 m_2$ is the geometric mean of $4$ and $9$,so $m_1 m_2 = \sqrt{4 \times 9} = \sqrt{36} = 6$.
The equation of a pair of lines passing through the origin is given by $y^2 - (m_1+m_2)xy + (m_1 m_2)x^2 = 0$.
Substituting the values,we get $y^2 - \frac{13}{2}xy + 6x^2 = 0$.
Multiplying by $2$,we get $2y^2 - 13xy + 12x^2 = 0$,which is $12x^2 - 13xy + 2y^2 = 0$.

(A) The given equation of the circle is $4x^2 + 4y^2 - 12x - 12y + 9 = 0$.
Dividing the entire equation by $4$,we get:
$x^2 + y^2 - 3x - 3y + \frac{9}{4} = 0$
Rearranging the terms to complete the square:
$(x^2 - 3x) + (y^2 - 3y) = -\frac{9}{4}$
Adding $(\frac{3}{2})^2$ to both sides for both $x$ and $y$ terms:
$(x^2 - 3x + \frac{9}{4}) + (y^2 - 3y + \frac{9}{4}) = -\frac{9}{4} + \frac{9}{4} + \frac{9}{4}$
$(x - \frac{3}{2})^2 + (y - \frac{3}{2})^2 = (\frac{3}{2})^2$
Comparing this with the standard form of a circle $(x - h)^2 + (y - k)^2 = r^2$,we have the centre $(h, k) = (\frac{3}{2}, \frac{3}{2})$ and the radius $r = \frac{3}{2}$.
Since the distance of the centre from both axes is equal to the radius (i.e.,$|h| = |k| = r = \frac{3}{2}$),the circle touches both the axes.

(C) The length of the tangent from a point $(h, k)$ to a circle $x^2+y^2+2gx+2fy+c=0$ is given by $\sqrt{h^2+k^2+2gh+2fk+c}$.
For the circle $x^2+y^2-16=0$,the length of the tangent is $L_1 = \sqrt{h^2+k^2-16}$.
For the circle $x^2+y^2+2x+2y=0$,the length of the tangent is $L_2 = \sqrt{h^2+k^2+2h+2k}$.
According to the problem,$L_1 = 2L_2$.
Squaring both sides,we get $L_1^2 = 4L_2^2$.
$h^2+k^2-16 = 4(h^2+k^2+2h+2k)$.
$h^2+k^2-16 = 4h^2+4k^2+8h+8k$.
Rearranging the terms,we get $3h^2+3k^2+8h+8k+16=0$.

(B) $1$. For the formation of $X$: Benzoic acid reacts with $B_2H_6$ followed by hydrolysis $(H_3O^+)$ to undergo reduction,yielding benzyl alcohol $(C_6H_5CH_2OH)$ as $X$.
$2$. For the formation of $Y$: Benzoic acid reacts with $SOCl_2$ to form benzoyl chloride $(C_6H_5COCl)$,which then undergoes Rosenmund reduction using $H_2/Pd-BaSO_4$ to yield benzaldehyde $(C_6H_5CHO)$ as $Y$.
$3$. Therefore,$X$ is benzyl alcohol and $Y$ is benzaldehyde.

(C) When the vapours of a primary alcohol are passed over heated copper at $573 \ K$,dehydrogenation takes place and an aldehyde is formed: $R-CH_2OH \xrightarrow{Cu/573 \ K} R-CHO + H_2$.
In the case of tertiary alcohols,dehydration takes place to form an alkene: $(CH_3)_3C-OH \xrightarrow{Cu/573 \ K} (CH_3)_2C=CH_2 + H_2O$.
Thus,$X$ is an aldehyde $(RCHO)$ and $Y$ is an alkene $((CH_3)_2C=CH_2)$.

(B) $1$. Reaction with $Zn$ dust and heat: $p-Cresol$ $(p-methylphenol)$ reacts with $Zn$ dust to undergo reduction,where the $-OH$ group is removed and replaced by a hydrogen atom,resulting in the formation of $Toluene$ $(X)$.
$2$. Reaction with $(CH_3CO)_2O$ followed by $H^+$: $p-Cresol$ reacts with acetic anhydride $(CH_3CO)_2O$ to undergo acetylation of the phenolic $-OH$ group,forming $p-acetoxy-toluene$ $(Y)$.

(D) The reaction of phenol with $CHCl_3$ and aqueous $NaOH$ followed by acidification is the Reimer-Tiemann reaction,which yields $2$-hydroxybenzaldehyde (salicylaldehyde) as product $X$.
The reaction of phenol with $NaOH$ followed by $CO_2$ and then acidification is the Kolbe-Schmitt reaction,which yields $2$-hydroxybenzoic acid (salicylic acid) as product $Y$.
Therefore,$X$ is $2$-hydroxybenzaldehyde and $Y$ is $2$-hydroxybenzoic acid.

(B) The reaction of toluene with $Cl_2$ in the presence of $hv$ (photochemical chlorination) leads to the formation of benzal chloride $(C_6H_5CHCl_2)$ as the intermediate $X$,which upon hydrolysis at $373 \ K$ gives benzaldehyde.
The reaction of toluene with chromyl chloride $(CrO_2Cl_2)$ in $CS_2$ is the Etard reaction. This reaction proceeds through the formation of a brown chromium complex intermediate $Y$,which is $C_6H_5CH[OCr(OH)Cl_2]_2$. This complex upon acidic hydrolysis $(H_3O^+)$ yields benzaldehyde.
Therefore,$X = C_6H_5CHCl_2$ and $Y = C_6H_5CH[OCr(OH)Cl_2]_2$.

(C) The given reaction is the Cannizzaro reaction. Benzaldehyde $(C_6H_5CHO)$ does not have an $\alpha$-hydrogen atom,so it undergoes a self-oxidation and reduction (disproportionation) reaction in the presence of a concentrated base like $NaOH$.
One molecule of benzaldehyde is reduced to benzyl alcohol $(C_6H_5CH_2OH)$,and another molecule is oxidized to sodium benzoate $(C_6H_5COONa)$.
Thus,the products '$X$' and '$Y$' are benzyl alcohol and sodium benzoate.

(C) The reaction of aniline with $CHCl_3$ and $KOH$ (carbylamine reaction) produces phenyl isocyanide $(C_6H_5NC)$ as $X$.
The reaction of aniline with $NaNO_2/HCl$ followed by $CuCN/KCN$ (Sandmeyer reaction) produces phenyl cyanide $(C_6H_5CN)$ as $Y$.
Therefore,$X$ is phenyl isocyanide and $Y$ is phenyl cyanide.

(A) $(i)$ Aniline is highly reactive towards electrophilic substitution. When treated with $Br_2$ in $H_2O$,it undergoes rapid poly-substitution to form $2, 4, 6-$tribromoaniline as the major product $(X)$.
$(ii)$ To control the reactivity of the $-NH_2$ group,it is first acetylated using acetic anhydride $(CH_3CO)_2O$ to form acetanilide. The $-NHCOCH_3$ group is less activating than the $-NH_2$ group,which restricts the bromination to the $p-$position,yielding $p-$bromoacetanilide as the major product $(Y)$.

(C) The reduction of nitrobenzene depends on the medium used:
$(i)$ In neutral medium,using $Zn$ dust and $NH_4Cl$ solution,nitrobenzene is reduced to phenylhydroxylamine $(C_6H_5NHOH)$. Thus,$X$ is phenylhydroxylamine.
(ii) In alkaline medium,using $Zn$ and $KOH/C_2H_5OH$,the reduction proceeds further to form hydrazobenzene $(C_6H_5NH-NHC_6H_5)$. Thus,$Y$ is hydrazobenzene.
Therefore,the correct option is $C$.

(A) In $DNA$,the nitrogenous bases pair specifically through hydrogen bonds.
$G$ (guanine) pairs with $C$ (cytosine) via $3$ hydrogen bonds $(G \equiv C)$.
$A$ (adenine) pairs with $T$ (thymine) via $2$ hydrogen bonds $(A = T)$.
Therefore,the number of hydrogen bonds between guanine and cytosine is $3$,and between adenine and thymine is $2$.

(D) For a first order reaction,the integrated rate equation is: $\ln(a-x) = -Kt + \ln a$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln(a-x)$,$x = \text{time}$,slope $m = -K$,and intercept $c = \ln a$.
From the given graph:
Intercept $c = -2.303 = \ln a$.
Therefore,$a = e^{-2.303} \approx 10^{-1} \ mol \ L^{-1}$.
Slope $m = -(10)^{-2} = -K$.
Therefore,$K = 10^{-2} \ s^{-1}$.
Thus,the rate constant is $10^{-2} \ s^{-1}$ and the initial concentration is $10^{-1} \ mol \ L^{-1}$.

(D) The given reaction is a Wurtz-Fittig reaction,which involves the coupling of an aryl halide with an alkyl halide in the presence of sodium metal and dry ether to form an alkylbenzene.
The reaction is:
$C_6H_5Cl + CH_3CH_2CH_2CH_2Cl + 2Na \xrightarrow{\text{dry ether}} C_6H_5-CH_2CH_2CH_2CH_3 + 2NaCl$
The product $X$ formed is $n$-butylbenzene (or simply butylbenzene).

(D) When phenol reacts with bromine water $(Br_2/H_2O)$,it undergoes electrophilic substitution at all ortho and para positions due to the strong activating effect of the $-OH$ group,resulting in the formation of $2,4,6$-tribromophenol $(X)$.
When phenol reacts with concentrated nitric acid $(Conc. HNO_3)$,it undergoes nitration to form $2,4,6$-trinitrophenol,commonly known as picric acid $(Y)$.

(B) In Whytlaw-Gray's method for the preparation of fluorine,the copper diaphragm is used to prevent the mixing of $H_2$ and $F_2$ gases liberated at the cathode and anode,respectively.
The reactions in the electrolytic cell are:
$KHF_2 \rightarrow KF + HF$
$KF \rightarrow K^+ + F^-$
At cathode:
$K^+ + e^- \rightarrow K$
$K + HF \rightarrow KF + H$
$2H \rightarrow H_2$
At anode:
$F^- \rightarrow F + e^-$
$2F \rightarrow F_2$

(A) The classification of colloidal systems based on the physical state of the dispersed phase and dispersion medium is as follows:
$(A)$ Solid dispersed in liquid is called a $Sol$ $(IV)$.
$(B)$ Liquid dispersed in liquid is called an $Emulsion$ $(I)$.
$(C)$ Gas dispersed in liquid is called $Foam$ $(II)$.
$(D)$ Liquid dispersed in solid is called a $Gel$ $(III)$.
Therefore,the correct match is $A-IV, B-I, C-II, D-III$.

(B) When phenol is treated with $Br_2$ in the presence of a non-polar solvent like $CS_2$ at low temperature $(0^{\circ} C)$,the reaction is controlled to yield mono-substituted products.
In a non-polar solvent,the ionization of phenol to the highly reactive phenoxide ion is suppressed.
Consequently,the electrophilic substitution occurs only once,leading to the formation of a mixture of $o$-bromophenol and $p$-bromophenol.

(C) The reduction of nitrobenzene depends on the medium used:
$1$. In neutral medium $(Zn/NH_4Cl)$: Nitrobenzene is reduced to phenylhydroxylamine $(C_6H_5NHOH)$. Thus,$X$ is phenylhydroxylamine.
$2$. In alkaline medium $(Zn + KOH/C_2H_5OH)$: Nitrobenzene undergoes reduction to form azoxybenzene,azobenzene,and finally hydrazobenzene $(C_6H_5NH-NHC_6H_5)$. Thus,$Y$ is hydrazobenzene.
Therefore,the correct option is $C$.

(B) In the $DNA$ double helix,the nitrogenous bases pair specifically through hydrogen bonds.
Guanine $(G)$ pairs with cytosine $(C)$ via three hydrogen bonds $(G \equiv C)$.
Adenine $(A)$ pairs with thymine $(T)$ via two hydrogen bonds $(A = T)$.
Therefore,the number of hydrogen bonds between guanine and cytosine is $3$,and between adenine and thymine is $2$.

(C) Let us analyze each molecule:
$1$. $PCl_5$: The central atom $P$ has $5$ bond pairs and $0$ lone pairs. Hybridisation is $sp^3d$ and the shape is trigonal bipyramidal.
$2$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ has $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons. Hybridisation is $dsp^2$ and the shape is square planar.
$3$. $SF_6$: The central atom $S$ has $6$ bond pairs and $0$ lone pairs. Hybridisation is $sp^3d^2$ and the shape is octahedral. This is the correct set.
$4$. $IF_3$: The central atom $I$ has $3$ bond pairs and $2$ lone pairs. Hybridisation is $sp^3d$ and the shape is bent $T$-shaped.

(D) Phenacetin is a derivative of $p-$aminophenol and is used as an analgesic (painkiller). Its chemical structure is $N-(4-ethoxyphenyl)acetamide$. Comparing this with the given options,the structure in option $D$ corresponds to phenacetin,which features an ethoxy group $(-OC_2H_5)$ at the para position relative to the acetamido group $(-NHCOCH_3)$.

(D) Ferric salts (such as $FeCl_3$) react with potassium ferrocyanide to form Prussian blue,which is ferric ferrocyanide.
The chemical reaction is:
$4 FeCl_3 + 3 K_4[Fe(CN)_6] \longrightarrow Fe_4[Fe(CN)_6]_3 + 12 KCl$
The product $Fe_4[Fe(CN)_6]_3$ is known as Prussian blue.

(A) The colour observed is the complementary colour of the light absorbed by the compound.
According to the colour wheel, the wavelength range $490-500 \,nm$ corresponds to the blue-green region of the visible spectrum.
The complementary colour of blue-green is red.
Therefore, the compound will appear red.

(B) During the electrolysis of an aqueous solution of copper sulphate using copper electrodes,both $Cu^{2+}$ and $H^{+}$ ions move towards the cathode.
Since the reduction potential of $Cu^{2+}$ is higher than that of $H^{+}$,$Cu^{2+}$ ions are reduced in preference to $H^{+}$ ions.
Thus,copper metal is deposited at the cathode according to the following reaction:
$Cu^{2+}_{(aq)} + 2e^{-} \longrightarrow Cu_{(s)}$

(C) The extent of charge of a lead accumulator is determined by the specific gravity of the $H_2SO_4$ solution.
During discharge,$H_2SO_4$ is consumed,and its density (specific gravity) decreases.
During charging,$H_2SO_4$ is regenerated,and its density increases.
$A$ fully charged battery typically has a specific gravity of about $1.25$ to $1.30$.

(D) Pyridinium chlorochromate $(PCC)$ is formed by the reaction of pyridine with chlorochromic acid,which is obtained by dissolving chromium trioxide $(CrO_3)$ in aqueous $HCl$.
The chemical formula of pyridine is $C_5H_5N$.
When pyridine is protonated,it forms the pyridinium cation,$C_5H_5NH^{\oplus}$.
The chlorochromate anion is $CrO_3Cl^{\ominus}$.
Therefore,the structure of $PCC$ is $C_5H_5NH^{\oplus} CrO_3Cl^{\ominus}$.

(A) The $-NH_2$ group of aniline is a strong electron-donating group due to the $+M$ (mesomeric) effect.
This effect increases the electron density at the ortho and para positions of the benzene ring,making them more nucleophilic and thus more susceptible to electrophilic attack.
During electrophilic aromatic substitution,the intermediate arenium ion (sigma complex) formed by the attack of an electrophile at the ortho or para positions is stabilized by the resonance donation of the lone pair of electrons from the nitrogen atom of the $-NH_2$ group.
Therefore,both the assertion and the reason are correct,and the reason correctly explains why the $-NH_2$ group is ortho,para directing.

(D) In the extraction of silver by the cyanide process (also known as the Mac-Arthur Forrest process),the silver ore is treated with a dilute solution of $NaCN$ in the presence of air to form a soluble complex,$Na[Ag(CN)_2]$.
The chemical reaction is: $Ag_2S + 4NaCN \rightleftharpoons 2Na[Ag(CN)_2] + Na_2S$.
Air is passed through the solution to oxidize $Na_2S$ to $Na_2SO_4$,preventing the reverse reaction.
Finally,$Zn$ is added to the solution to displace silver from the complex: $2Na[Ag(CN)_2] + Zn \longrightarrow Na_2[Zn(CN)_4] + 2Ag \downarrow$.
$Na_2S_2O_3$ is not used in this process.

(B) For titration of a mixture of $Na_2CO_3$ and $NaHCO_3$ against $HCl$:
$1$. With phenolphthalein indicator,the end point corresponds to the conversion of $Na_2CO_3$ to $NaHCO_3$:
$Na_2CO_3 + HCl \rightarrow NaHCO_3 + NaCl$
$Moles \ of \ HCl = Moles \ of \ Na_2CO_3 = 20 \ mL \times 0.1 \ M = 2 \ mmol$.
Concentration of $Na_2CO_3 = \frac{2 \ mmol}{20 \ mL} = 0.1 \ M$.
$2$. With methyl orange indicator,the end point corresponds to the complete neutralization of $Na_2CO_3$ and $NaHCO_3$:
$Na_2CO_3 + 2HCl \rightarrow 2NaCl + CO_2 + H_2O$
$NaHCO_3 + HCl \rightarrow NaCl + CO_2 + H_2O$
Total $HCl$ consumed = $60 \ mL \times 0.1 \ M = 6 \ mmol$.
$HCl$ consumed by $Na_2CO_3$ (complete) = $2 \times 2 \ mmol = 4 \ mmol$.
$HCl$ consumed by $NaHCO_3$ = $6 \ mmol - 4 \ mmol = 2 \ mmol$.
Concentration of $NaHCO_3 = \frac{2 \ mmol}{20 \ mL} = 0.1 \ M$.
Thus,the concentrations are $0.1 \ M$ and $0.1 \ M$.

(A) The acidity of a carboxylic acid is inversely proportional to its $pK_a$ value.
$pK_a = -\log(K_a)$.
$A$ higher $pK_a$ value indicates a lower $K_a$ value,which means the acid is weaker.
Comparing the given values: $1.28 < 2.56 < 4.76 < 4.89$.
Since $4.89$ is the highest $pK_a$ value,the carboxylic acid with $pK_a = 4.89$ is the weakest acid.

(C) The chemical formula of pyrophosphoric acid is $H_4P_2O_7$.
It has a structure with one $P-O-P$ linkage,four $P-OH$ bonds,and two $P=O$ bonds.
Since it has four $P-OH$ groups,it is a tetrabasic acid.
It does not contain any $P-H$ bonds.
Orthophosphoric acid $(H_3PO_4)$ is prepared by dissolving $P_4O_{10}$ in water: $P_4O_{10} + 6H_2O \longrightarrow 4H_3PO_4$.
Therefore,the statement that pyrophosphoric acid contains two $P-H$ bonds is incorrect.

(B) In Whytlaw-Gray's method for the preparation of fluorine,the copper diaphragm is used to prevent the mixing of $H_2$ and $F_2$ gases liberated at the cathode and anode,respectively.
The reactions in the electrolytic cell are as follows:
$KHF_2 \rightarrow KF + HF$
$KF \rightarrow K^+ + F^-$
At the cathode:
$K^+ + e^- \rightarrow K$
$K + HF \rightarrow KF + H$
$2H \rightarrow H_2$
At the anode:
$F^- \rightarrow F + e^-$
$2F \rightarrow F_2$

(D) Liquid xenon $(Xe)$ is used in bubble chambers for the detection of $\gamma$-photons and neutral mesons due to its high density and high atomic number,which increases the probability of interaction with these particles.

(B) The polymers which disintegrate by themselves during a certain period of time by enzymatic hydrolysis and to some extent by oxidation are known as biodegradable polymers.
Examples include:
$1$. Poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate $(PHBV)$,which is used in orthopaedic devices and in controlled drug release.
$2$. Poly(glycolic acid)-poly(lactic acid),commonly known as $Dexon$ (or dextron),which is used for stitching of wounds after surgery.

(C) When chloroform $(CHCl_3)$ is heated with an aqueous solution of sodium hydroxide $(NaOH)$,it undergoes hydrolysis.
The reaction is: $CHCl_3 + 4NaOH \rightarrow HCOONa + 3NaCl + 2H_2O$.
The product formed is sodium formate $(HCOONa)$.

(A) In a close packed structure ($hcp$ or $ccp$):
$(I)$ The number of octahedral voids is equal to the number of particles $(N)$ present in the close packing.
$(II)$ The number of tetrahedral voids is equal to $2N$,where $N$ is the number of particles.
Given that there are $X$ atoms,the number of octahedral voids is $X$ and the number of tetrahedral voids is $2X$.

(D) According to Raoult's law: $\frac{p^{\circ}-p_s}{p^{\circ}} = \frac{n_2}{n_1+n_2}$
Where $p^{\circ}$ is the vapour pressure of pure water at $100^{\circ}C = 760 \ mm \ Hg$.
$p_s$ is the vapour pressure of the solution.
Moles of solute (glucose),$n_2 = \frac{18 \ g}{180 \ g/mol} = 0.1 \ mol$.
Moles of solvent (water),$n_1 = \frac{180 \ g}{18 \ g/mol} = 10 \ mol$.
Substituting the values: $\frac{760 - p_s}{760} = \frac{0.1}{10 + 0.1} = \frac{0.1}{10.1}$.
$760 - p_s = 760 \times \frac{0.1}{10.1} = \frac{76}{10.1} \approx 7.524$.
$p_s = 760 - 7.524 = 752.476 \ mm \ Hg \approx 752.4 \ mm \ Hg$.

(A) Two solutions are isotonic if they have the same molar concentration of particles (osmotic pressure).
For $0.15 \ M \ NaCl$: $NaCl$ dissociates into $2$ ions ($Na^+$ and $Cl^-$). The concentration of particles $= 0.15 \times 2 = 0.30 \ M$.
For $0.1 \ M \ Na_2SO_4$: $Na_2SO_4$ dissociates into $3$ ions ($2Na^+$ and $SO_4^{2-}$). The concentration of particles $= 0.1 \times 3 = 0.30 \ M$.
Since both solutions have the same concentration of particles $(0.30 \ M)$,they are isotonic.

(A) The classification of colloidal systems based on the dispersed phase and dispersion medium is as follows:
$(A)$ Solid dispersed in liquid is called a Sol $(IV)$.
$(B)$ Liquid dispersed in liquid is called an Emulsion $(I)$.
$(C)$ Gas dispersed in liquid is called a Foam $(II)$.
$(D)$ Liquid dispersed in solid is called a Gel $(III)$.
Therefore,the correct matching is $(A-IV, B-I, C-II, D-III)$.