MHT CET 2020 Chemistry Question Paper with Answer and Solution

(D) The total entropy change is given by $\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}$.
Given $\Delta S_{\text{sys}} = 15 \ JK^{-1}$.
The entropy change of the surroundings is calculated as $\Delta S_{\text{surr}} = -\frac{\Delta H_{\text{sys}}}{T}$.
Since $\Delta H_{\text{sys}} = -29.8 \ kJ = -29800 \ J$,we have $\Delta S_{\text{surr}} = -\frac{-29800 \ J}{298 \ K} = 100 \ JK^{-1}$.
Therefore,$\Delta S_{\text{total}} = 15 \ JK^{-1} + 100 \ JK^{-1} = 115 \ JK^{-1}$.

(B) The total entropy change is given by $\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{sur}}$.
Given $\Delta S_{\text{sys}} = \Delta S^{\circ} = 15 \ J \ K^{-1}$.
The entropy change of the surroundings is $\Delta S_{\text{sur}} = -\frac{\Delta H_{\text{sys}}}{T}$.
Since $\Delta H_{\text{sys}} = -25 \ kJ = -25000 \ J$,we have $\Delta S_{\text{sur}} = -\frac{-25000 \ J}{300 \ K} = 83.33 \ J \ K^{-1}$.
Therefore,$\Delta S_{\text{total}} = 15 \ J \ K^{-1} + 83.33 \ J \ K^{-1} = 98.33 \ J \ K^{-1}$.

(B) According to the $3^{rd}$ law of thermodynamics,the entropy of a perfectly crystalline substance is zero at $T = 0 \ K$.
If a substance is not perfectly crystalline (e.g.,it has defects or frozen-in disorder),its entropy at $T = 0 \ K$ is greater than zero.
This remaining entropy at absolute zero is known as residual entropy.

(C) The condition for spontaneity is given by the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
For the reaction to be at equilibrium,$\Delta G = 0$,which implies $T = \frac{\Delta H}{\Delta S}$.
Given: $\Delta H = -30 \ kJ = -30000 \ J$ and $\Delta S = -45 \ J \ K^{-1}$.
Substituting the values: $T = \frac{-30000 \ J}{-45 \ J \ K^{-1}} = 666.67 \ K$.
Since both $\Delta H$ and $\Delta S$ are negative,the reaction is spontaneous at temperatures below $666.67 \ K$ and becomes non-spontaneous at temperatures above $666.67 \ K$.

(C) For a reaction to be spontaneous,the Gibbs free energy change must be negative,i.e.,$\Delta G < 0$.
According to the Gibbs-Helmholtz equation,$\Delta G = \Delta H - T\Delta S$.
If $\Delta H < 0$ (exothermic) and $\Delta S > 0$ (increase in entropy),then $\Delta G$ will always be negative at all temperatures because $\Delta H$ is negative and $-T\Delta S$ is also negative.
Therefore,the condition $\Delta S > 0, \Delta H < 0, \Delta G < 0$ represents a spontaneous reaction at all temperatures.

(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -RT \ln K$.
Given values: $K = 20$,$T = 300 \ K$,$R = 8 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.
Substituting the values into the formula: $\Delta G^{\circ} = -(8 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) \times (300 \ K) \times \ln(20)$.
Since $\ln(20) \approx 2.996$,we have: $\Delta G^{\circ} = -2.4 \times 2.996 \approx -7.19 \ kJ \ mol^{-1}$.
Therefore,the correct value is $-7.19 \ kJ \ mol^{-1}$.

(B) The spontaneity of a reaction is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
For a reaction to be spontaneous at all temperatures,$\Delta G$ must be negative at all values of $T$.
If $\Delta H$ is negative (exothermic) and $\Delta S$ is positive (increase in entropy),then $\Delta G = (\text{negative}) - T(\text{positive})$.
Since $T$ is always positive in Kelvin,$-T\Delta S$ will always be negative.
Thus,$\Delta G$ will always be negative regardless of the temperature.

(C) Let $x$ be the number of bacteria at time $t$.
$\frac{dx}{dt} \propto x \Rightarrow \frac{dx}{dt} = kx \Rightarrow \int \frac{dx}{x} = \int k \cdot dt$
$\log x = kt + c$
Initially,i.e.,when $t = 0$,let $x = x_{0}$.
$\log x_{0} = k \times 0 + c \Rightarrow c = \log x_{0}$
$\log x - \log x_{0} = kt \Rightarrow \log \left(\frac{x}{x_{0}}\right) = kt$ ... $(1)$
Since the number doubles in $4 \text{ hours}$,i.e.,$t = 4$ and $x = 2x_{0}$:
$\log \left(\frac{2x_{0}}{x_{0}}\right) = 4k \Rightarrow \log 2 = 4k \Rightarrow k = \frac{1}{4} \log 2$
When $t = 12$,substituting $k$ into equation $(1)$:
$\log \left(\frac{x}{x_{0}}\right) = 12 \times \left(\frac{1}{4} \log 2\right) = 3 \log 2 = \log 2^{3} = \log 8$
Therefore,$\frac{x}{x_{0}} = 8 \Rightarrow x = 8x_{0}$.
Given the original number is $N$,the number after $12 \text{ hours}$ will be $8N$.

(A) Let $y = f(\tan x)$ and $z = g(\sec x)$. We need to find $\frac{dy}{dz} = \frac{dy/dx}{dz/dx}$.
First,find the derivative of $y$ with respect to $x$:
$\frac{dy}{dx} = f^{\prime}(\tan x) \cdot \sec^2 x$.
Next,find the derivative of $z$ with respect to $x$:
$\frac{dz}{dx} = g^{\prime}(\sec x) \cdot \sec x \tan x$.
Now,calculate the ratio at $x = \frac{\pi}{4}$:
$\frac{dy}{dz} = \frac{f^{\prime}(\tan x) \cdot \sec^2 x}{g^{\prime}(\sec x) \cdot \sec x \tan x} = \frac{f^{\prime}(\tan x) \cdot \sec x}{g^{\prime}(\sec x) \cdot \tan x}$.
At $x = \frac{\pi}{4}$,$\tan x = 1$ and $\sec x = \sqrt{2}$:
$\frac{dy}{dz} = \frac{f^{\prime}(1) \cdot \sqrt{2}}{g^{\prime}(\sqrt{2}) \cdot 1} = \frac{2 \cdot \sqrt{2}}{4 \cdot 1} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.

(D) Let $I = \int \frac{dx}{\cos x \sqrt{\cos 2x}}$.
Divide the numerator and denominator by $\cos x$ inside the square root or manipulate the expression as follows:
$I = \int \frac{dx}{\cos x \sqrt{\cos^2 x - \sin^2 x}} = \int \frac{dx}{\cos x \cdot \cos x \sqrt{1 - \tan^2 x}} = \int \frac{\sec^2 x}{\sqrt{1 - \tan^2 x}} dx$.
Let $\tan x = t$,then $\sec^2 x \, dx = dt$.
Substituting these into the integral,we get:
$I = \int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + c$.
Replacing $t$ with $\tan x$,we get:
$I = \sin^{-1}(\tan x) + c$.

(A) We need to find the value of $\tan \left[\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right]$.
Let $\cos ^{-1}\left(\frac{4}{5}\right) = \theta$. Then $\cos \theta = \frac{4}{5}$.
Since $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}$,we have $\sin \theta = \frac{3}{5}$.
Therefore,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{4/5} = \frac{3}{4}$.
So,$\theta = \tan ^{-1}\left(\frac{3}{4}\right)$.
Substituting this into the expression,we get $\tan \left[\tan ^{-1}\left(\frac{3}{4}\right) + \tan ^{-1}\left(\frac{2}{3}\right)\right]$.
Using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left(\frac{x+y}{1-xy}\right)$ for $xy < 1$:
$\tan ^{-1}\left(\frac{3}{4}\right) + \tan ^{-1}\left(\frac{2}{3}\right) = \tan ^{-1} \left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \left(\frac{3}{4}\right)\left(\frac{2}{3}\right)}\right) = \tan ^{-1} \left(\frac{\frac{9+8}{12}}{1 - \frac{6}{12}}\right) = \tan ^{-1} \left(\frac{17/12}{6/12}\right) = \tan ^{-1} \left(\frac{17}{6}\right)$.
Finally,$\tan \left[\tan ^{-1} \left(\frac{17}{6}\right)\right] = \frac{17}{6}$.

(D) The implication $p \rightarrow (\sim p \vee q)$ is false only when the antecedent is true and the consequent is false.
Therefore,$p$ must be $T$ and $(\sim p \vee q)$ must be $F$.
Since $p$ is $T$,$\sim p$ is $F$.
For the disjunction $(\sim p \vee q)$ to be $F$,both $\sim p$ and $q$ must be $F$.
Thus,$q$ must be $F$.
Hence,the truth values of $p$ and $q$ are $T$ and $F$ respectively.

(C) To find the negation of the statement pattern $\sim p \vee (q \rightarrow \sim r)$,we apply the negation operator $\sim$ to the entire expression:
$\sim [\sim p \vee (q \rightarrow \sim r)]$
Using De Morgan's Law,$\sim (A \vee B) \equiv \sim A \wedge \sim B$:
$\equiv \sim (\sim p) \wedge \sim (q \rightarrow \sim r)$
$\equiv p \wedge \sim (\sim q \vee \sim r)$
Using De Morgan's Law again,$\sim (\sim q \vee \sim r) \equiv q \wedge r$:
$\equiv p \wedge (q \wedge r)$
Thus,the correct option is $C$.

(C) The lines pass through the origin and make an angle of $30^{\circ}$ with the $Y$-axis.
The angles these lines make with the positive $X$-axis are $90^{\circ}-30^{\circ}=60^{\circ}$ and $90^{\circ}+30^{\circ}=120^{\circ}$.
The slopes of the lines are $m_{1}=\tan 60^{\circ}=\sqrt{3}$ and $m_{2}=\tan 120^{\circ}=-\sqrt{3}$.
The equations of the lines are $y=\sqrt{3}x$ and $y=-\sqrt{3}x$,which can be written as $\sqrt{3}x-y=0$ and $\sqrt{3}x+y=0$.
The joint equation is $(\sqrt{3}x-y)(\sqrt{3}x+y)=0$.
Expanding this,we get $(\sqrt{3}x)^{2}-y^{2}=0$,which simplifies to $3x^{2}-y^{2}=0$.

(B) Since $f(x)$ is the p.d.f. of a random variable $X$,the total area under the curve must be $1$:
$\int_{0}^{1} K(x - x^2) dx = 1$
$K \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = 1$
$K \left( \frac{1}{2} - \frac{1}{3} \right) = 1 \Rightarrow K \left( \frac{1}{6} \right) = 1 \Rightarrow K = 6$
Now,we calculate $P(X < \frac{1}{2})$:
$P(X < \frac{1}{2}) = \int_{0}^{1/2} 6(x - x^2) dx$
$= 6 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1/2}$
$= 6 \left( \frac{(1/2)^2}{2} - \frac{(1/2)^3}{3} \right)$
$= 6 \left( \frac{1/4}{2} - \frac{1/8}{3} \right) = 6 \left( \frac{1}{8} - \frac{1}{24} \right)$
$= 6 \left( \frac{3 - 1}{24} \right) = 6 \left( \frac{2}{24} \right) = 6 \left( \frac{1}{12} \right) = \frac{1}{2}$

(A) Given: Area $= 10 \sqrt{3} \text{ cm}^2$,$\angle B = 60^{\circ}$,Perimeter $P = a + b + c = 20 \text{ cm}$.
Using the area formula: $\text{Area} = \frac{1}{2} ac \sin B$.
$10 \sqrt{3} = \frac{1}{2} ac \sin 60^{\circ} \implies 10 \sqrt{3} = \frac{1}{2} ac \left( \frac{\sqrt{3}}{2} \right) \implies ac = 40$.
Using the Law of Cosines: $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
$\cos 60^{\circ} = \frac{a^2 + c^2 - b^2}{2(40)} \implies \frac{1}{2} = \frac{(a+c)^2 - 2ac - b^2}{80}$.
Since $a+c = 20 - b$,we have $40 = (20-b)^2 - 2(40) - b^2$.
$40 = 400 - 40b + b^2 - 80 - b^2$.
$40 = 320 - 40b$.
$40b = 280 \implies b = 7$.
Thus,$\ell(AC) = b = 7 \text{ cm}$.

(C) Let the first line be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = \lambda$. Then any point on this line is $(2\lambda+1, 3\lambda-1, 4\lambda+1)$.
Let the second line be $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1} = \mu$. Then any point on this line is $(\mu+3, 2\mu+k, \mu)$.
If the lines intersect,there exist $\lambda$ and $\mu$ such that:
$2\lambda+1 = \mu+3 \implies 2\lambda - \mu = 2$ (Equation $1$)
$4\lambda+1 = \mu \implies 4\lambda - \mu = -1$ (Equation $2$)
Subtracting Equation $1$ from Equation $2$: $(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2 \implies 2\lambda = -3 \implies \lambda = -\frac{3}{2}$.
Substituting $\lambda = -\frac{3}{2}$ into Equation $1$: $2(-\frac{3}{2}) - \mu = 2 \implies -3 - \mu = 2 \implies \mu = -5$.
Now,equate the $y$-coordinates: $3\lambda - 1 = 2\mu + k$.
Substitute $\lambda = -\frac{3}{2}$ and $\mu = -5$: $3(-\frac{3}{2}) - 1 = 2(-5) + k$.
$-\frac{9}{2} - 1 = -10 + k \implies -\frac{11}{2} = -10 + k$.
$k = 10 - \frac{11}{2} = \frac{20-11}{2} = \frac{9}{2}$.

(B) For two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ to intersect,the condition is $\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right| = 0$.
Substituting the given values: $(x_1, y_1, z_1) = (1, -1, 1)$,$(a_1, b_1, c_1) = (2, 3, 4)$,$(x_2, y_2, z_2) = (3, k, 0)$,and $(a_2, b_2, c_2) = (1, 2, 1)$.
The determinant becomes: $\left|\begin{array}{ccc} 3-1 & k-(-1) & 0-1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right| = 0$.
$\left|\begin{array}{ccc} 2 & k+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right| = 0$.
Expanding the determinant: $2(3(1) - 4(2)) - (k+1)(2(1) - 4(1)) - 1(2(2) - 3(1)) = 0$.
$2(3-8) - (k+1)(2-4) - 1(4-3) = 0$.
$2(-5) - (k+1)(-2) - 1(1) = 0$.
$-10 + 2k + 2 - 1 = 0$.
$2k - 9 = 0$.
$k = \frac{9}{2}$.

(B) Let the two lines be $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $L_2: \frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$.
The lines intersect if the shortest distance between them is $0$. This condition is equivalent to the scalar triple product of the vector connecting a point on each line and the direction vectors of the lines being $0$.
Let $P_1 = (1, -1, 1)$ be a point on $L_1$ and $P_2 = (3, k, 0)$ be a point on $L_2$.
The direction vectors are $\vec{v_1} = (2, 3, 4)$ and $\vec{v_2} = (1, 2, 1)$.
The condition for intersection is given by the determinant:
$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$
Substituting the values:
$\begin{vmatrix} 3-1 & k-(-1) & 0-1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0$
$\begin{vmatrix} 2 & k+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2(3(1) - 4(2)) - (k+1)(2(1) - 4(1)) - 1(2(2) - 3(1)) = 0$
$2(3-8) - (k+1)(2-4) - 1(4-3) = 0$
$2(-5) - (k+1)(-2) - 1(1) = 0$
$-10 + 2k + 2 - 1 = 0$
$2k - 9 = 0$
$k = \frac{9}{2}$

(B) The given equation is $\cot x = \sqrt{3}$,which is equivalent to $\tan x = \frac{1}{\sqrt{3}}$.
We know that $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$.
Since $\tan x$ is positive in the first and third quadrants,the principal solutions lie in the interval $[0, 2\pi)$.
The first solution is $x = \frac{\pi}{6}$.
The second solution in the third quadrant is $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.
Thus,the principal solutions are $\frac{\pi}{6}$ and $\frac{7\pi}{6}$.

(C) Given that $\sin (y+z-x), \sin (z+x-y), \sin (x+y-z)$ are in $AP$.
Therefore,$2 \sin (z+x-y) = \sin (y+z-x) + \sin (x+y-z)$.
Using the sum-to-product formula $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$,we get:
$2 \sin (z+x-y) = 2 \sin \frac{(y+z-x) + (x+y-z)}{2} \cos \frac{(y+z-x) - (x+y-z)}{2}$
$2 \sin (z+x-y) = 2 \sin y \cos (z-x)$.
Since $\sin (z+x-y) = \sin (z-(y-x)) = \sin z \cos (y-x) - \cos z \sin (y-x)$,this approach is complex.
Alternatively,using $\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$:
$\sin (z+x-y) - \sin (y+z-x) = \sin (x+y-z) - \sin (z+x-y)$
$2 \cos z \sin (x-y) = 2 \cos x \sin (y-z)$
$\cos z (\sin x \cos y - \cos x \sin y) = \cos x (\sin y \cos z - \cos y \sin z)$
$\cos z \sin x \cos y - \cos z \cos x \sin y = \cos x \sin y \cos z - \cos x \cos y \sin z$
$2 \cos z \cos x \sin y = \cos x \cos y \sin z + \sin x \cos y \cos z$
Dividing by $\cos x \cos y \cos z$:
$2 \tan y = \tan z + \tan x$.

(B) We use the identity $\cot \theta = \frac{1 - \tan^2(\theta/2)}{2 \tan(\theta/2)}$ or $\cot \theta = \frac{1}{\tan \theta}$.
Starting with the expression: $\tan A + 2 \tan 2A + 4 \tan 4A + 8 \cot 8A$
$= \tan A + 2 \tan 2A + 4 \tan 4A + 8 \left( \frac{1 - \tan^2 4A}{2 \tan 4A} \right)$
$= \tan A + 2 \tan 2A + 4 \tan 4A + \frac{4(1 - \tan^2 4A)}{\tan 4A}$
$= \tan A + 2 \tan 2A + \frac{4 \tan^2 4A + 4 - 4 \tan^2 4A}{\tan 4A}$
$= \tan A + 2 \tan 2A + \frac{4}{\tan 4A} = \tan A + 2 \tan 2A + 4 \cot 4A$
$= \tan A + 2 \tan 2A + 4 \left( \frac{1 - \tan^2 2A}{2 \tan 2A} \right)$
$= \tan A + 2 \tan 2A + \frac{2(1 - \tan^2 2A)}{\tan 2A}$
$= \tan A + \frac{2 \tan^2 2A + 2 - 2 \tan^2 2A}{\tan 2A} = \tan A + \frac{2}{\tan 2A}$
$= \tan A + 2 \cot 2A = \tan A + 2 \left( \frac{1 - \tan^2 A}{2 \tan A} \right)$
$= \tan A + \frac{1 - \tan^2 A}{\tan A} = \frac{\tan^2 A + 1 - \tan^2 A}{\tan A} = \frac{1}{\tan A} = \cot A$

(B) The impedance of an $L-R$ circuit is given by $Z_1 = \sqrt{R^2 + X_L^2}$.
When a capacitor is connected in series,the circuit becomes an $L-C-R$ circuit.
The impedance of an $L-C-R$ circuit is given by $Z_2 = \sqrt{R^2 + (X_L - X_C)^2}$.
Since $(X_L - X_C)^2 < X_L^2$ (assuming the circuit is not at resonance where $X_L = X_C$),the total impedance $Z_2$ is less than $Z_1$.
According to Ohm's law for $AC$ circuits,$I = \frac{V}{Z}$.
Since the impedance $Z$ decreases,the alternating current $I$ flowing in the circuit increases.

(C) The impedance of an $LR$ circuit is given by $Z_{LR} = \sqrt{R^2 + X_L^2}$.
When a capacitor is connected in series,the circuit becomes an $LCR$ circuit.
The impedance of an $LCR$ circuit is given by $Z_{LCR} = \sqrt{R^2 + (X_L - X_C)^2}$.
Since $(X_L - X_C)^2 < X_L^2$ (assuming $X_C$ is not zero),the total impedance $Z$ of the circuit decreases.
According to Ohm's law for $AC$ circuits,$I = \frac{V}{Z}$.
Since $I \propto \frac{1}{Z}$,a decrease in impedance $Z$ leads to an increase in the alternating current $I$ flowing through the circuit.

(A) The impedance of an $LR$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$.
When a capacitor is connected in series,the circuit becomes an $LCR$ circuit.
The new impedance is $Z' = \sqrt{R^2 + (X_L - X_C)^2}$.
Since $(X_L - X_C)^2 < X_L^2$ (assuming $X_C < 2X_L$),the impedance $Z'$ decreases compared to $Z$.
According to Ohm's law for $AC$ circuits,$I = \frac{V}{Z}$.
Since the impedance $Z$ decreases,the alternating current $I$ flowing in the circuit increases.

(D) The energy levels for a Hydrogen atom are given by $n=1$ (ground state),$n=2$ (first excited state),$n=3$ (second excited state),and $n=4$ (third excited state).
For the first transition from the third excited state $(n=4)$ to the second excited state $(n=3)$:
$\frac{1}{\lambda_{1}} = R \left( \frac{1}{3^{2}} - \frac{1}{4^{2}} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16-9}{144} \right) = R \left( \frac{7}{144} \right)$.
For the second transition from the second excited state $(n=3)$ to the first excited state $(n=2)$:
$\frac{1}{\lambda_{2}} = R \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = R \left( \frac{5}{36} \right)$.
Taking the ratio $\frac{\lambda_{1}}{\lambda_{2}}$:
$\frac{\lambda_{1}}{\lambda_{2}} = \frac{1/\lambda_{2}}{1/\lambda_{1}} = \frac{R(5/36)}{R(7/144)} = \frac{5}{36} \times \frac{144}{7} = \frac{5 \times 4}{7} = \frac{20}{7}$.

(B) When a dielectric material is inserted between the plates of a capacitor,it gets polarized. The induced electric field $E_i$ inside the dielectric opposes the external electric field $E_0$ produced by the charges on the plates. The net electric field becomes $E = E_0 - E_i$. Since the potential difference $V$ is related to the electric field by $V = E \cdot d$,the reduction in the net electric field leads to a reduction in the effective potential difference between the plates. This reduction in potential,while the charge $Q$ remains constant,results in an increase in the capacitance $C = Q/V$.

(A) Let the distances of masses $m_{1}$ and $m_{2}$ from the centre of mass be $r_{1}$ and $r_{2}$ respectively. By the definition of the centre of mass,$m_{1}r_{1} = m_{2}r_{2}$.
If the second particle $(m_{2})$ is moved by a distance $d$ towards the centre of mass,its new distance becomes $(r_{2} - d)$.
To keep the centre of mass unchanged,let the first particle $(m_{1})$ be moved by a distance $d'$ towards the centre of mass,so its new distance becomes $(r_{1} - d')$.
The new condition for the centre of mass is $m_{1}(r_{1} - d') = m_{2}(r_{2} - d)$.
Expanding this,we get $m_{1}r_{1} - m_{1}d' = m_{2}r_{2} - m_{2}d$.
Since $m_{1}r_{1} = m_{2}r_{2}$,these terms cancel out,leaving $-m_{1}d' = -m_{2}d$.
Thus,$d' = \frac{m_{2}}{m_{1}} d$.
Since the displacement is towards the centre of mass for both to maintain the balance,the first particle must move by $\frac{m_{2}}{m_{1}} d$ towards the centre of mass.

(A) The circuit consists of two batteries connected in opposition and a resistor in series.
The net electromotive force $(EMF)$ of the circuit is $E_{net} = 200 \, V - 10 \, V = 190 \, V$.
The total resistance of the circuit is $R = 38 \, \Omega$.
According to Ohm's law, the current $i$ flowing through the circuit is given by:
$i = \frac{E_{net}}{R} = \frac{190 \, V}{38 \, \Omega} = 5 \, A$.

(B) To convert a galvanometer into a voltmeter,a high resistance $R$ is connected in series with the galvanometer.
Let $I_g$ be the full-scale deflection current of the galvanometer.
Given,$V_g = I_g G$,so $I_g = \frac{V_g}{G}$.
The total resistance of the voltmeter is $(G + R)$.
For a range $V$,the total voltage is $V = I_g(G + R)$.
Substituting $I_g = \frac{V_g}{G}$,we get $V = \frac{V_g}{G}(G + R)$.
Rearranging for $R$: $\frac{V}{V_g} = \frac{G + R}{G} = 1 + \frac{R}{G}$.
Therefore,$\frac{R}{G} = \frac{V}{V_g} - 1$.
$R = G(\frac{V}{V_g} - 1)$.

(B) The energy of each photon is given by $E = \frac{hc}{\lambda}$.
If $n$ photons are emitted in time $t$,the total energy emitted is $E_{total} = \frac{nhc}{\lambda}$.
Power $P$ is defined as the energy emitted per unit time:
$P = \frac{E_{total}}{t} = \frac{nhc}{\lambda t}$.
Rearranging the formula to solve for $n$:
$n = \frac{P \lambda t}{hc}$.

(D) The magnetic flux linked with an inductor is given by $\phi = LI$,where $L$ is the self-inductance.
From this,we can write $L = \frac{\phi}{I}$.
This expression represents the slope of the $\phi$ versus $I$ graph.
Since the slope is $\frac{\phi}{I}$,the inductor with the steepest slope will have the largest value of self-inductance $L$.
Looking at the graph,the line $A$ has the maximum slope compared to $B, C$ and $D$.
Therefore,inductor $A$ has the largest self-inductance.

(A) The total resistance of the circuit is $R_{total} = R + \frac{R}{2} = \frac{3R}{2}$.
According to Faraday's law of electromagnetic induction,the magnitude of induced electromotive force $(EMF)$ is $e = n \frac{|\Delta \phi|}{\Delta t} = n \frac{|\phi_{1} - \phi_{2}|}{t}$.
The induced current $i$ is given by $i = \frac{e}{R_{total}}$.
Substituting the values,we get $i = \frac{n |\phi_{1} - \phi_{2}| / t}{3R/2} = \frac{2n |\phi_{1} - \phi_{2}|}{3Rt}$.
Thus,the induced current is $\frac{2n(\phi_{1} - \phi_{2})}{3Rt}$.

(A) The relationship between magnetic flux $(\phi)$ and current $(I)$ for an inductor is given by $\phi = L I$,where $L$ is the self-inductance.
From this,we can write $L = \frac{\phi}{I}$.
In a graph of $\phi$ versus $I$,the slope of the line is given by $\frac{\phi}{I}$,which represents the self-inductance $L$.
$A$ steeper slope indicates a larger value of $L$.
Comparing the slopes of the lines $A, B, C,$ and $D$,line $A$ has the maximum slope.
Therefore,inductor $A$ has the largest self-inductance.

(D) The relationship between magnetic flux $\phi$ and current $I$ is given by $\phi = LI$,where $L$ is the self-inductance of the inductor.
From this equation,the self-inductance $L$ is the slope of the $\phi$ versus $I$ graph,i.e.,$L = \frac{\phi}{I}$.
The slope of the line represents the value of $L$.
$A$ steeper slope indicates a larger value of $L$,while a shallower slope indicates a smaller value of $L$.
Comparing the slopes of the four lines $A, B, C,$ and $D$,the line $D$ has the smallest slope.
Therefore,the inductor $D$ has the smallest value of self-inductance.

(C) Light is an electromagnetic wave in nature. Electromagnetic waves do not require any material medium for their propagation and can travel through a vacuum at a speed of approximately $3 \times 10^8 \ m/s$. Therefore,the statement that light requires a material medium for propagation is incorrect.

(C) Let the original length of the wire be $L$ and the cross-sectional area be $A$. Young's modulus $Y$ is given by $Y = \frac{F L}{A \Delta l}$,where $\Delta l$ is the extension.
Case $1$: Force $F_1 = 1.5 \text{ kg-wt}$. The total length is $a = L + \Delta l_1$,so $\Delta l_1 = a - L$.
Thus,$Y = \frac{1.5 L}{A (a - L)} \implies a - L = \frac{1.5 L}{Y A} \implies a = L (1 + \frac{1.5}{Y A}) \quad \dots (1)$
Case $2$: Force $F_2 = 2.5 \text{ kg-wt}$. The total length is $b = L + \Delta l_2$,so $\Delta l_2 = b - L$.
Thus,$Y = \frac{2.5 L}{A (b - L)} \implies b - L = \frac{2.5 L}{Y A} \implies b = L (1 + \frac{2.5}{Y A}) \quad \dots (2)$
From $(1)$,$\frac{1.5}{Y A} = \frac{a}{L} - 1 = \frac{a - L}{L}$.
From $(2)$,$\frac{2.5}{Y A} = \frac{b}{L} - 1 = \frac{b - L}{L}$.
Dividing the two equations: $\frac{2.5}{1.5} = \frac{b - L}{a - L} \implies \frac{5}{3} = \frac{b - L}{a - L}$.
$5(a - L) = 3(b - L) \implies 5a - 5L = 3b - 3L$.
$5a - 3b = 2L \implies L = \frac{5a - 3b}{2} = 2.5a - 1.5b$.

(D) The focal length $f$ of a lens is related to the refractive index $\mu$ of the material by the Lens Maker's Formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
From this formula,it is clear that $f \propto \frac{1}{\mu - 1}$.
According to Cauchy's equation,the refractive index $\mu$ is inversely proportional to the square of the wavelength $\lambda$ $(\mu \approx A + \frac{B}{\lambda^2})$.
Therefore,as the wavelength $\lambda$ increases,the refractive index $\mu$ decreases.
Among the given colors,red light has the maximum wavelength $(\lambda_{red} > \lambda_{yellow} > \lambda_{blue} > \lambda_{violet})$.
Since red light has the longest wavelength,it has the minimum refractive index $\mu$ for the lens material.
Consequently,the focal length $f$ will be maximum for red light.

(A) According to the lens maker's formula:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Given that the lens is a convex lens with both faces having the same radius of curvature,we have $R_1 = R$ and $R_2 = -R$.
The focal length $f = 20 \text{ cm}$ and refractive index $\mu = 1.55$.
Substituting these values into the formula:
$\frac{1}{20} = (1.55 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right)$
$\frac{1}{20} = 0.55 \times \left( \frac{1}{R} + \frac{1}{R} \right)$
$\frac{1}{20} = 0.55 \times \frac{2}{R}$
$\frac{1}{20} = \frac{1.1}{R}$
$R = 1.1 \times 20 = 22 \text{ cm}$
Therefore,the radius of curvature is $22 \text{ cm}$.

(C) In a resonance tube experiment,the resonance lengths $l_1$ and $l_2$ are related to the wavelength $\lambda$ and end correction $e$ as follows:
$l_1 + e = \frac{\lambda}{4}$
$l_2 + e = \frac{3\lambda}{4}$
Dividing the second equation by the first,we get:
$\frac{l_2 + e}{l_1 + e} = 3$
$l_2 + e = 3(l_1 + e)$
$l_2 + e = 3l_1 + 3e$
$l_2 - 3l_1 = 2e$
$e = \frac{l_2 - 3l_1}{2}$
Given $l_1 = 12 ~cm$ and $l_2 = 38 ~cm$:
$e = \frac{38 - 3(12)}{2} = \frac{38 - 36}{2} = \frac{2}{2} = 1 ~cm$.

(C) Given: $\frac{g_{m}}{g_{e}} = \frac{1}{6}$ and $\frac{\rho_{e}}{\rho_{m}} = \frac{5}{3}$.
We know that the acceleration due to gravity is given by $g = \frac{GM}{R^2}$.
Since $M = \rho \times V = \rho \times \frac{4}{3} \pi R^3$,we have $g = \frac{G \times \frac{4}{3} \pi R^3 \rho}{R^2} = \frac{4}{3} \pi G R \rho$.
Therefore,the ratio $\frac{g_{m}}{g_{e}} = \frac{R_{m} \rho_{m}}{R_{e} \rho_{e}}$.
Substituting the given values: $\frac{1}{6} = \left(\frac{R_{m}}{R_{e}}\right) \times \left(\frac{\rho_{m}}{\rho_{e}}\right)$.
Since $\frac{\rho_{e}}{\rho_{m}} = \frac{5}{3}$,then $\frac{\rho_{m}}{\rho_{e}} = \frac{3}{5}$.
So,$\frac{1}{6} = \left(\frac{R_{m}}{R_{e}}\right) \times \left(\frac{3}{5}\right)$.
$\frac{R_{m}}{R_{e}} = \frac{1}{6} \times \frac{5}{3} = \frac{5}{18}$.
Thus,$R_{m} = \left(\frac{5}{18}\right) R_{e}$.

(C) The root-mean-square (rms) velocity of gas molecules is given by the formula:
$v_{rms} = \sqrt{\frac{3kT}{m}}$
From this expression,we can see that $v_{rms} \propto \sqrt{T}$ for a given gas.
In an isothermal process,the temperature $(T)$ of the system remains constant.
Since the temperature does not change,the r.m.s. velocity of the gas molecules remains the same.

(A) The root mean square (r.m.s.) velocity of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
In an isothermal process,the temperature $T$ of the gas remains constant.
Since $v_{rms}$ depends only on the temperature $T$ (for a given gas),if $T$ is constant,the r.m.s. velocity $v_{rms}$ will also remain constant.
Therefore,when a gas is compressed isothermally,the r.m.s. velocity of the molecules remains the same.

(C) The magnetic field $B$ at the centre of a circular arc of radius $r$ carrying current $I$ that subtends an angle $\theta$ at the centre is given by the formula:
$B = \frac{\mu_{0} I \theta}{4 \pi r}$
Here,the angle subtended is $\theta = \frac{\pi}{16}$.
Substituting the value of $\theta$ into the formula:
$B = \frac{\mu_{0} I}{4 \pi r} \times \left( \frac{\pi}{16} \right)$
$B = \frac{\mu_{0} I \pi}{64 \pi r}$
$B = \frac{\mu_{0} I}{64 r}$
Thus,the correct option is $C$.

(C) The kinetic energy of the electron is given by $K = eV$,where '$e$' is the charge of the electron and '$V$' is the accelerating potential.
In the first case,$K_1 = eV_1 = eV$.
In the second case,$K_2 = eV_2 = e(2V) = 2eV$.
Since $K = \frac{1}{2}mv^2$,we have $v = \sqrt{\frac{2K}{m}}$.
Thus,the ratio of velocities is $\frac{v_2}{v_1} = \sqrt{\frac{K_2}{K_1}} = \sqrt{\frac{2eV}{eV}} = \sqrt{2}$.
The magnetic force on a moving charge is given by $F = evB \sin(\theta)$. Since the electron enters the transverse magnetic field,$\theta = 90^\circ$ and $\sin(90^\circ) = 1$,so $F = evB$.
Since '$e$' and '$B$' are constant,$F \propto v$.
Therefore,$\frac{F_2}{F_1} = \frac{v_2}{v_1} = \sqrt{2}$.
Hence,$F_2 = \sqrt{2}F$.

(B) The magnetic field produced by a circular current-carrying loop along its axis is directed along the axis itself.
Since the electron is projected along the axis,its velocity vector $\vec{v}$ is parallel or antiparallel to the magnetic field vector $\vec{B}$.
The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$,which has a magnitude $F = qvB \sin \theta$.
Here,the angle $\theta$ between the velocity and the magnetic field is $0^{\circ}$ or $180^{\circ}$.
Since $\sin(0^{\circ}) = 0$ and $\sin(180^{\circ}) = 0$,the magnetic force $F = 0$.
Therefore,the electron experiences no force.

(B) Let the mass of the lighter stone be $m_1 = m$ and its radius be $r_1 = r$. Let its tangential speed be $v_1$.
Let the mass of the heavier stone be $m_2 = 3m$ and its radius be $r_2 = \frac{r}{3}$. Let its tangential speed be $v_2$.
Given that $v_1 = n v_2$.
The centripetal force is given by $F = \frac{mv^2}{r}$.
Equating the centripetal forces for both stones: $F_1 = F_2$.
$\frac{m_1 v_1^2}{r_1} = \frac{m_2 v_2^2}{r_2}$.
Substituting the given values: $\frac{m (n v_2)^2}{r} = \frac{3m v_2^2}{r/3}$.
$\frac{m n^2 v_2^2}{r} = \frac{9m v_2^2}{r}$.
Canceling common terms $m, v_2^2,$ and $r$ from both sides,we get $n^2 = 9$.
Therefore,$n = 3$.

(C) The angular momentum $\overrightarrow{L}$ of a particle moving in a circular path is given by $\overrightarrow{L} = \overrightarrow{r} \times \overrightarrow{p} = \overrightarrow{r} \times m\overrightarrow{v}$.
Since the particle moves in a circular path,the position vector $\overrightarrow{r}$ and the velocity vector $\overrightarrow{v}$ are always perpendicular to each other.
The direction of the angular momentum vector $\overrightarrow{L}$ is determined by the right-hand rule,which is perpendicular to the plane of the circular motion.
As the speed $v$ decreases,the magnitude of the angular momentum $|\overrightarrow{L}| = mvr$ decreases.
However,because the particle remains in the same plane of motion,the direction of the angular momentum vector remains constant.

(D) The activity $R$ of a radioactive sample is defined as the rate of decay,given by $R = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of undecayed atoms.
The decay constant $\lambda$ is related to the half-life $T$ by the formula $\lambda = \frac{\ln 2}{T}$.
Substituting this into the activity formula,we get $R = \frac{(\ln 2) N}{T}$.
For the two elements,the activities are:
$R_{1} = \frac{(\ln 2) N_{1}}{T_{1}}$
$R_{2} = \frac{(\ln 2) N_{2}}{T_{2}}$
The ratio of their activities is:
$\frac{R_{1}}{R_{2}} = \frac{(\ln 2) N_{1} / T_{1}}{(\ln 2) N_{2} / T_{2}} = \frac{N_{1}}{T_{1}} \times \frac{T_{2}}{N_{2}} = \frac{N_{1} T_{2}}{N_{2} T_{1}}$.

(A) For a body of mass $m$ oscillating with amplitude $A$ and angular frequency $\omega$,the maximum velocity is given by $v_{max} = A\omega$.
Given that the masses are equal $(m_A = m_B = m)$ and the force constants are $k_1$ and $k_2$,the angular frequencies are $\omega_1 = \sqrt{\frac{k_1}{m}}$ and $\omega_2 = \sqrt{\frac{k_2}{m}}$.
Since the maximum velocities are equal,we have $A_1 \omega_1 = A_2 \omega_2$.
Substituting the expressions for $\omega_1$ and $\omega_2$:
$A_1 \sqrt{\frac{k_1}{m}} = A_2 \sqrt{\frac{k_2}{m}}$.
Squaring both sides and simplifying:
$A_1^2 \frac{k_1}{m} = A_2^2 \frac{k_2}{m} \implies \frac{A_1^2}{A_2^2} = \frac{k_2}{k_1}$.
Taking the square root,the ratio of the amplitudes is $\frac{A_1}{A_2} = \sqrt{\frac{k_2}{k_1}}$.

(B) In $[CoF_{6}]^{3-}$,the oxidation state of $Co$ is $+3$.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^{6}$.
Since $F^{-}$ is a weak field ligand $(WFL)$,it does not cause pairing of electrons in the $3d$ orbitals.
Therefore,the $Co^{3+}$ ion uses one $4s$,three $4p$,and two $4d$ orbitals to form six hybrid orbitals.
This results in $sp^{3}d^{2}$ hybridisation.
The geometry associated with $sp^{3}d^{2}$ hybridisation is octahedral.

(C) The primary valence of a central metal atom in a coordination compound corresponds to its oxidation state.
In $CoCl_{3}$,the oxidation state of $Co$ is calculated as follows:
$x + 3(-1) = 0$
$x - 3 = 0$
$x = +3$
Therefore,the primary valence of $Co$ is $3$.

(A) According to Werner's coordination theory,the primary valence corresponds to the oxidation state of the central metal atom.
It is ionisable and is satisfied by negative ions.
Therefore,it is also known as the ionisable valence.

(B) To find the oxidation number of $Fe$ in $K_3[Fe(CN)_6]$,let the oxidation number of $Fe$ be $x$.
The oxidation number of $K$ is $+1$ and the oxidation number of the $CN^-$ ligand is $-1$.
The sum of the oxidation numbers of all atoms in a neutral complex is $0$.
Therefore,$3(+1) + x + 6(-1) = 0$.
$3 + x - 6 = 0$.
$x - 3 = 0$.
$x = +3$.
Thus,the oxidation number of $Fe$ is $+3$.

(C) The chemical formula for the manganate ion is $MnO_4^{2-}$.
In $MnO_4^{2-}$,the oxidation state of $Mn$ is $+6$. The electronic configuration of $Mn^{6+}$ is $[Ar] 3d^1$.
Due to the presence of one unpaired electron in the $3d$ orbital,the manganate ion is paramagnetic.
The ionic charge of the manganate ion is $-2$.

(D) The chemical formula for Cuproammonium sulphate is $[Cu(NH_{3})_{4}]SO_{4} \cdot H_{2}O$.
In the complex ion $[Cu(NH_{3})_{4}]^{2+}$,the oxidation state of $Cu$ is $+2$.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^{9}$.
Due to the presence of $NH_{3}$ (a strong field ligand),the complex undergoes $dsp^{2}$ hybridization,resulting in a square planar geometry.
Since there is one unpaired electron in the $3d$ orbital,the complex is paramagnetic.

(B) The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$.
In the $+3$ oxidation state, $Cr^{3+}$ loses three electrons to form $[Ar] 3d^3$.
This configuration contains $n = 3$ unpaired electrons.
The effective magnetic moment $\mu$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=3$, we get $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \ BM$.

(C) In the structure of copper$(II)$ sulfate pentahydrate,$[Cu(H_{2}O)_{4}]SO_{4} \cdot H_{2}O$,four water molecules are directly coordinated to the $Cu^{2+}$ ion.
The fifth water molecule is held in the crystal lattice by hydrogen bonding between the $SO_{4}^{2-}$ ion and the coordinated water molecules.

(D) According to Werner's theory,the secondary valence of a central metal ion is equal to its coordination number.
In the complex $[Co(NH_3)_4 Cl_2]^+$,the $Co^{3+}$ ion is bonded to $4$ $NH_3$ molecules and $2$ $Cl^-$ ions.
Therefore,the coordination number $= 4 + 2 = 6$.
Thus,the secondary valence of $Co^{3+}$ is $6$.

(D) In the complex $[Pt(NH_{3})_{6}]^{4+}$:
$I$. Let the oxidation state of $Pt$ be $x$. Since $NH_{3}$ is a neutral ligand,$x + 6(0) = +4$,which gives $x = +4$.
$II$. The coordination number is the number of ligand donor atoms bonded to the central metal ion. Since there are $6$ $NH_{3}$ ligands,each donating one lone pair,the coordination number is $6$.

(B) The electronic configuration of $Zn$ $(Z=30)$ is $[Ar] 3d^{10} 4s^2$.
In the $+2$ oxidation state,$Zn^{2+}$ has the configuration $[Ar] 3d^{10}$.
Since the $3d$ subshell is completely filled,there are no unpaired electrons.
Therefore,$Zn^{2+}$ compounds are colourless due to the absence of $d-d$ transitions.

(D) The splitting power of ligands is determined by the spectrochemical series.
According to the spectrochemical series,the order of field strength is: $S^{2-} < OH^{-} < NCS^{-} < CO$.
$CO$ is a strong field ligand and causes the maximum splitting of $d$-orbitals.
Therefore,$CO$ has the highest splitting power.

(A) In the $3d-$ series elements,the density and atomic mass increase from $Sc$ to $Zn$ across the period.
Since $Sc$ $(Z = 21)$ is the first element in the $3d-$ series,it has the lowest atomic mass and the lowest density among the given options.

(B) Most transition elements are very hard due to strong metallic bonding involving $d$-electrons.
However,elements like $Zn, Cd,$ and $Hg$ have completely filled $d$-orbitals ($d^{10}$ configuration),which results in weak metallic bonding.
Therefore,$Zn$ is a soft element compared to $Co, W,$ and $Mo$.

(A) Babbitt metal is a soft alloy primarily composed of tin.
It typically contains $89.3 \%$ tin,$7.1 \%$ antimony,and $3.6 \%$ copper.

(C) The highest oxidation state exhibited by any transition element is $+8$.
This is observed in elements like Ruthenium $(Ru)$ and Osmium $(Os)$ in their tetroxides,such as $RuO_4$ and $OsO_4$.

(C) $(C)$
Across the $3d-$ series, the density of the elements generally increases with an increase in atomic number due to the increase in atomic mass and the decrease in atomic radius.
Therefore, the correct decreasing order of density is $Ni > Fe > Cr > V$.

(D) When tungsten $(W)$ metal is exposed to oxygen gas $(O_2)$,it undergoes surface adsorption followed by oxidation to form tungsten trioxide $(WO_3)$.

(B) The correct statement is that actinoid contraction is greater than lanthanoid contraction.
This is because the $5f$ electrons have poorer shielding effect compared to $4f$ electrons,leading to a greater effective nuclear charge and more significant contraction in the actinoid series.

(D) The reaction of lanthanoids $(Ln)$ with oxygen produces the stable oxide $(Ln_2O_3)$:
$4Ln + 3O_2 \rightarrow 2Ln_2O_3$ (Compound $A$)
When the lanthanoid oxide $(Ln_2O_3)$ reacts with excess $CO_2$,it forms the corresponding carbonate $(Ln_2(CO_3)_3)$:
$Ln_2O_3 + 3CO_2 \rightarrow Ln_2(CO_3)_3$ (Compound $B$)
Therefore,the formula of compound $(B)$ is $Ln_2(CO_3)_3$.

(B) Chemical twins are pairs of elements that exhibit very similar chemical properties due to the lanthanoid contraction.
$Nb-Ta$, $Mo-W$, and $Tc-Re$ are examples of chemical twins.
$Zr-Hf$ (Hafnium) are chemical twins, but $Zr-Rf$ (Rutherfordium) are not, as $Rf$ belongs to the actinoid series and has different chemical characteristics.

(C) In the lanthanoid series,the atomic size decreases from left to right due to lanthanoid contraction.
As the atomic number increases,the shielding effect of $4f$ electrons is poor,leading to an increase in effective nuclear charge.
Among the given elements ($Ce$,$Pr$,$Pm$,$Sm$),the atomic numbers are $Ce(58)$,$Pr(59)$,$Pm(61)$,and $Sm(62)$.
Since atomic size decreases as atomic number increases,$Sm$ has the smallest atomic size among the options provided.
The order is $Ce > Pr > Pm > Sm$.

(C) In the lanthanoid series,as the atomic number increases from $Ce$ $(Z=58)$ to $Lu$ $(Z=71)$,the ionic radii of the trivalent ions $(Ln^{3+})$ decrease steadily. This phenomenon is known as lanthanoid contraction.
Therefore,the decreasing order of ionic radii for the given ions is: $Ce^{3+} > Pm^{3+} > Sm^{3+} > Gd^{3+}$.

(A) The general oxidation state for lanthanoids is $+3$.
$Ce$ (Cerium) has the electronic configuration $[Xe] 4f^1 5d^1 6s^2$.
Due to the stability of the empty $f$-orbital after losing four electrons,$Ce$ exhibits a $+4$ oxidation state in addition to $+3$.

(B) The correct answer is $B$.
Lanthanoids generally exhibit a $+3$ oxidation state.
However,some elements show $+2$ or $+4$ oxidation states due to the stability of $f^0$,$f^7$,or $f^{14}$ configurations.
Gadolinium $(Gd)$ has the electronic configuration $[Xe] \ 4f^7 \ 5d^1 \ 6s^2$.
Upon losing three electrons,it forms $Gd^{3+}$ with the configuration $[Xe] \ 4f^7$,which is a half-filled stable $f$-subshell configuration.
Due to this extra stability,$Gd$ predominantly exhibits only the $+3$ oxidation state.

(A) The correct option is $A$.
$Lr$ $(Z=103)$ has the electronic configuration $[Rn] 5f^{14} 6d^{1} 7s^{2}$.
Due to the completely filled $5f$ orbital,$Lr$ exhibits only the $+3$ oxidation state.

(C) The first inner transition series is known as the $Lanthanoid$ series,which includes elements from atomic number $58$ $(Ce)$ to $71$ $(Lu)$.
$Pr$ $(Praseodymium)$ has an atomic number of $59$,which falls within the range of the $Lanthanoid$ series.
$Bk$ $(Berkelium)$,$Pu$ $(Plutonium)$,and $Fm$ $(Fermium)$ belong to the second inner transition series,known as the $Actinoid$ series.
Therefore,the correct option is $C$.

(A) The actinoids exhibit a range of oxidation states due to the comparable energies of $5f$,$6d$,and $7s$ orbitals.
While the most common oxidation state is $+3$,the elements like $Np$ and $Pu$ exhibit oxidation states up to $+7$.
Therefore,the highest oxidation state exhibited by actinoids is $+7$.

(D) Promethium $(Pm)$ is the only radioactive lanthanide.

(A) In an electrolytic cell,the anode is the positive electrode where oxidation takes place,and the cathode is the negative electrode where reduction takes place. Therefore,oxidation occurs at the positive electrode.

(C) The standard cell potential is calculated using the formula: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$.
In the given cell $Ni|Ni^{2+} || Cu^{2+}| Cu$,$Cu$ acts as the cathode and $Ni$ acts as the anode.
$E^{\circ}_{cell} = E^{\circ}_{Cu^{2+}/Cu} - E^{\circ}_{Ni^{2+}/Ni}$.
Substituting the given values: $E^{\circ}_{cell} = 0.337 \ V - (-0.236 \ V)$.
$E^{\circ}_{cell} = 0.337 \ V + 0.236 \ V = 0.573 \ V$.

(B) During the electrolysis of aqueous $NaCl$ solution,the following reactions occur:
At the cathode: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$
At the anode: $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$
Thus,$Cl_2$ gas is released at the anode.

(B) The reduction reaction is: $Al^{3+} + 3e^{-} \longrightarrow Al$.
According to the reaction,$1 \ \text{mole}$ of $Al^{3+}$ requires $3 \ \text{moles}$ of electrons for reduction to $Al$.
We know that $1 \ \text{mole}$ of electrons carries a charge of $1 \ F = 96500 \ C$.
Therefore,the charge required for $3 \ \text{moles}$ of electrons is $3 \times 96500 \ C = 289500 \ C$.
This can be expressed in scientific notation as $2.895 \times 10^{5} \ C$.

(D) The total charge $Q$ passed through the cell is calculated using the formula $Q = I \times t$.
Given $I = 5 \ A$ and $t = 200 \ s$,we have $Q = 5 \ A \times 200 \ s = 1000 \ C$.
The charge of a single electron is approximately $1.602 \times 10^{-19} \ C$.
The number of electrons $n$ is given by $n = \frac{Q}{e} = \frac{1000 \ C}{1.602 \times 10^{-19} \ C} \approx 6.24 \times 10^{21}$ electrons.

(A) The cell reaction is: $2 Ag_{(aq)}^{+} + Cd_{(s)} \longrightarrow 2 Ag_{(s)} + Cd_{(aq)}^{2+}$
Here,the number of electrons transferred,$n = 2$.
The standard free energy change is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$
Substituting the values: $\Delta G^{\circ} = -2 \times 96500 \ C/mol \times 1.20 \ V$
$\Delta G^{\circ} = -231600 \ J/mol = -231.6 \ kJ/mol$

(B) The standard cell potential is given by the formula: $E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0}$.
Here,the copper electrode acts as the anode and the silver electrode acts as the cathode.
Given: $E_{cell}^{0} = 0.463 \ V$ and $E_{Cu}^{0} = 0.337 \ V$.
Substituting the values: $0.463 \ V = E_{Ag}^{0} - 0.337 \ V$.
Therefore,$E_{Ag}^{0} = 0.463 \ V + 0.337 \ V = 0.800 \ V$.

(B) $1 \ F$ is the charge of $1 \ \text{mole}$ of electrons.
$1 \ \text{mole}$ of electrons = $6.022 \times 10^{23} \ \text{electrons}$.
The number of electrons involved for $0.40 \ F$ is calculated as:
$\text{Number of electrons} = 0.40 \times 6.022 \times 10^{23} = 2.4088 \times 10^{23}$.

(D) The half-reaction at the cathode is:
$Cu^{2+}_{(aq)} + 2e^- \longrightarrow Cu_{(s)}$
Given: Current $I = 1.5 \ A$,Time $t = 10 \ min = 600 \ s$.
According to Faraday's law of electrolysis,the mass of substance deposited is given by:
$m = \frac{I \times t \times M}{n \times F}$
Where $M = 63.7 \ g/mol$,$n = 2$ (electrons involved),and $F = 96500 \ C/mol$.
$m = \frac{1.5 \times 600 \times 63.7}{2 \times 96500} = \frac{57330}{193000} \approx 0.297 \ g$.
Therefore,the correct option is $D$.

(B) The electrochemical equivalent $(Z)$ is defined as the mass of a substance deposited or liberated at an electrode per unit charge $(Q)$ passed through the electrolyte.
According to Faraday's law of electrolysis,$m = Z \times Q$,where $m$ is the mass in $Kg$ and $Q$ is the charge in $Coulombs$ $(C)$.
Therefore,$Z = \frac{m}{Q}$.
The $SI$ unit of mass is $Kg$ and the $SI$ unit of charge is $C$.
Thus,the $SI$ unit of electrochemical equivalent is $Kg \ C^{-1}$.

(A) The reduction reaction at the cathode is: $Mg^{2+} + 2e^{-} \longrightarrow Mg$.
According to the stoichiometry,$1 \ mol$ of $Mg$ $(24 \ g)$ requires $2 \ F$ of electricity.
Therefore,the number of faradays required for $4.8 \ g$ of $Mg$ is calculated as:
$\text{Faradays} = \frac{4.8 \ g}{24 \ g/mol} \times 2 \ F/mol = 0.2 \ mol \times 2 \ F/mol = 0.4 \ F$.

(D) $1 \ Faraday$ is defined as the total charge carried by $1 \ mole$ of electrons.
Since $1 \ mole$ of any substance contains $6.022 \times 10^{23}$ particles,$1 \ Faraday$ of electricity corresponds to $6.022 \times 10^{23}$ electrons.

(A) Given: $E^{\circ}_{\text{cell}} = 1.049 \ V$,$n = 2$.
The relationship between $E^{\circ}_{\text{cell}}$ and the equilibrium constant $K$ is given by the Nernst equation at $298 \ K$:
$E^{\circ}_{\text{cell}} = \frac{0.0592}{n} \log_{10} K$
Rearranging for $\log_{10} K$:
$\log_{10} K = \frac{E^{\circ}_{\text{cell}} \times n}{0.0592} = \frac{1.049 \times 2}{0.0592}$
$\log_{10} K = \frac{2.098}{0.0592} \approx 35.439$
Taking the antilog:
$K = 10^{35.439} = 2.75 \times 10^{35}$

(C) metal-metal ion electrode consists of a metal rod dipped in a solution containing its own ions.
In the option $Zn^{2+} (aq.) \mid Zn (s)$,a zinc metal rod is in contact with $Zn^{2+}$ ions in the solution.
Therefore,it represents a metal-metal ion electrode.

(B) metal-sparingly soluble salt electrode consists of a metal in contact with its sparingly soluble salt and a solution containing the anion of that salt.
In the electrode $Cl^{-} (aq) | AgCl_{(s)} | Ag_{(s)}$,the metal $Ag$ is in contact with its sparingly soluble salt $AgCl$ and the anion $Cl^{-}$.
The half-cell reaction is: $AgCl_{(s)} + e^{-} \rightarrow Ag_{(s)} + Cl^{-} (aq)$.
This is a classic example of a metal-sparingly soluble salt electrode,often used as a reference electrode.

(C) In a lead accumulator (lead-acid battery),the cathode is made of lead dioxide $(PbO_{2_{(s)}})$.
During the discharging process,the reaction at the cathode is: $PbO_{2\text{(s)}} + 4H^{+}{_{\text{(aq)}}} + SO_4^{2-}{_{\text{(aq)}}} + 2e^{-} \rightarrow PbSO_{4\text{(s)}} + 2H_2O_{\text{(l)}}$.
In this reaction,the oxidation state of $Pb$ in $PbO_{2}$ changes from $+4$ to $+2$ in $PbSO_{4}$.
Since the oxidation state decreases,$PbO_{2_{(s)}}$ is reduced to $Pb^{2+}$ ions (which then form $PbSO_{4_{(s)}}$).

(A) Given: Conductivity,$\kappa = 1.06 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$ and Concentration,$C = 0.1 \ M$.
The formula for molar conductivity is $\Lambda_m = \frac{\kappa \times 1000}{C}$.
Substituting the values: $\Lambda_m = \frac{1.06 \times 10^{-2} \times 1000}{0.1} = \frac{10.6}{0.1} = 106 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.
Thus,$\Lambda_m = 1.06 \times 10^{2} \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.

(B) Conductivity $(\kappa)$ is the reciprocal of resistivity $(\rho)$.
$\kappa = \frac{1}{\rho}$
Since resistivity $\rho = R \times \frac{A}{l}$,where $R$ is resistance in $\Omega$,$A$ is area in $cm^2$,and $l$ is length in $cm$.
$\kappa = \frac{1}{R} \times \frac{l}{A} = \Omega^{-1} \times \frac{cm}{cm^2} = \Omega^{-1} \ cm^{-1}$.
Thus,the common unit of conductivity in the $C.G.S.$ system is $\Omega^{-1} \ cm^{-1}$ or $S \ cm^{-1}$.

(B) The formula for conductivity $(k)$ is given by: $k = \frac{\text{Cell constant}}{R}$
Given that the cell constant is $0.5 \ cm^{-1}$ and the resistance $(R)$ is $375 \ \Omega$.
Substituting these values into the formula:
$k = \frac{0.5 \ cm^{-1}}{375 \ \Omega} = 1.333 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$

(A) Given: Resistance $(R) = 484 \ \Omega$,Conductivity $(\kappa) = 0.00141 \ \Omega^{-1} \ cm^{-1}$.
Formula for cell constant $(G^*)$ is: $G^* = \kappa \times R$.
Substituting the values: $G^* = 0.00141 \ \Omega^{-1} \ cm^{-1} \times 484 \ \Omega$.
$G^* = 0.68244 \ cm^{-1} \approx 0.682 \ cm^{-1}$.

(C) According to Kohlrausch's Law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions.
$\wedge_{m}^{0}(CaCl_{2}) = \lambda_{Ca^{2+}}^{0} + 2 \lambda_{Cl^{-}}^{0}$
Given,$\lambda_{Ca^{2+}}^{0} = 119 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$ and $\lambda_{Cl^{-}}^{0} = 71 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.
Substituting the values:
$\wedge_{m}^{0}(CaCl_{2}) = 119 + 2(71)$
$\wedge_{m}^{0}(CaCl_{2}) = 119 + 142 = 261.0 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.