AIEEE 2011 Chemistry Question Paper with Answer and Solution

(B) The heat supplied to the water is given by the formula $\Delta Q = mC\Delta T$.
Here,mass $m = 100 \ g = 0.1 \ kg$,specific heat $C = 4184 \ J/kg\cdot K$,and temperature change $\Delta T = 50^{\circ}C - 30^{\circ}C = 20 \ K$.
Substituting the values: $\Delta Q = 0.1 \times 4184 \times 20 = 8368 \ J = 8.368 \ kJ \approx 8.4 \ kJ$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Since the expansion of water is neglected,the work done $\Delta W = 0$.
Therefore,the change in internal energy $\Delta U = \Delta Q = 8.4 \ kJ$.

(A) Let the point on the curve $x = y^2$ be $P(a^2, a)$.
The equation of the line is $x - y + 1 = 0$.
The perpendicular distance $D$ from point $P(a^2, a)$ to the line $x - y + 1 = 0$ is given by the formula $D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.
Substituting the values,we get $D = \frac{|a^2 - a + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{a^2 - a + 1}{\sqrt{2}}$ (since $a^2 - a + 1 > 0$ for all real $a$).
To find the minimum distance,we minimize the expression $f(a) = a^2 - a + 1$.
Completing the square,$f(a) = (a - \frac{1}{2})^2 + \frac{3}{4}$.
The minimum value of $f(a)$ occurs at $a = \frac{1}{2}$,which is $\frac{3}{4}$.
Therefore,the minimum distance $D_{\min} = \frac{3/4}{\sqrt{2}} = \frac{3}{4\sqrt{2}} = \frac{3\sqrt{2}}{8}$.

(D) Given $\vec{a} = \frac{1}{\sqrt{10}}(3\hat{i} + \hat{k})$ and $\vec{b} = \frac{1}{7}(2\hat{i} + 3\hat{j} - 6\hat{k})$.
First,note that $|\vec{a}|^2 = \frac{1}{10}(3^2 + 1^2) = 1$ and $|\vec{b}|^2 = \frac{1}{49}(2^2 + 3^2 + (-6)^2) = \frac{4+9+36}{49} = 1$.
Also,$\vec{a} \cdot \vec{b} = \frac{1}{7\sqrt{10}}(3(2) + 0(3) + 1(-6)) = 0$.
Now,expand the expression: $(2\vec{a} - \vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} + 2\vec{b})]$.
Using the vector triple product formula $\vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w}$,we have:
$(\vec{a} \times \vec{b}) \times (\vec{a} + 2\vec{b}) = (\vec{a} \times \vec{b}) \times \vec{a} + 2((\vec{a} \times \vec{b}) \times \vec{b})$.
Using $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$,we get:
$(\vec{a} \times \vec{b}) \times \vec{a} = -(\vec{a} \times (\vec{a} \times \vec{b})) = -((\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}) = - (0\vec{a} - 1\vec{b}) = \vec{b}$.
$2((\vec{a} \times \vec{b}) \times \vec{b}) = 2((\vec{a} \cdot \vec{b})\vec{b} - (\vec{b} \cdot \vec{b})\vec{a}) = 2(0\vec{b} - 1\vec{a}) = -2\vec{a}$.
So,the expression becomes $(2\vec{a} - \vec{b}) \cdot (\vec{b} - 2\vec{a})$.
$= -(2\vec{a} - \vec{b}) \cdot (2\vec{a} - \vec{b}) = -|2\vec{a} - \vec{b}|^2$.
$= -(4|\vec{a}|^2 + |\vec{b}|^2 - 4\vec{a} \cdot \vec{b}) = -(4(1) + 1 - 4(0)) = -5$.

(B) To determine the hybridization,we use the formula: $\text{Hybridization} = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$1$. For $NO_2^+$: $\text{Hybridization} = \frac{1}{2} [5 + 0 - 1 + 0] = 2$,which corresponds to $sp$ hybridization.
$2$. For $NO_3^-$: $\text{Hybridization} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$,which corresponds to $sp^2$ hybridization.
$3$. For $NH_4^+$: $\text{Hybridization} = \frac{1}{2} [5 + 4 - 1 + 0] = 4$,which corresponds to $sp^3$ hybridization.
Thus,the correct order is $sp$,$sp^2$,and $sp^3$ respectively.

(C) According to the law of conservation of energy,the energy of the absorbed photon is equal to the sum of the energies of the two emitted photons.
$E_{absorbed} = E_{1} + E_{2}$
Since $E = \frac{hc}{\lambda}$,we have:
$\frac{hc}{\lambda} = \frac{hc}{\lambda_{1}} + \frac{hc}{\lambda_{2}}$
$\frac{1}{\lambda} = \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}}$
Given $\lambda = 355 \ nm$ and $\lambda_{1} = 680 \ nm$:
$\frac{1}{355} = \frac{1}{680} + \frac{1}{\lambda_{2}}$
$\frac{1}{\lambda_{2}} = \frac{1}{355} - \frac{1}{680} = \frac{680 - 355}{355 \times 680} = \frac{325}{241400}$
$\lambda_{2} = \frac{241400}{325} \approx 742.77 \ nm \approx 743 \ nm$

(A) $I$. In a period,from left to right,the metallic character decreases,so the basic nature of oxides decreases.
$II$. In a group,from top to bottom,the metallic character increases,so the basic nature of oxides increases.
$Na, Mg$,and $Al$ are in the same period ($3^{rd}$ period). The order of basic nature is $Na_2O > MgO > Al_2O_3$.
$Na$ and $K$ are in the same group ($1^{st}$ group). Since $K$ is below $Na$,the basic nature of $K_2O > Na_2O$.
Combining these,the correct order of increasing basic nature is $Al_2O_3 < MgO < Na_2O < K_2O$.

(C) According to Fajan's rule,the covalent character of an ionic bond increases with an increase in the polarizing power of the cation.
Polarizing power is directly proportional to the charge density of the cation,which depends on the charge and size of the cation.
Comparing the charges on the cations: $Fe^{2+}$,$Sn^{2+}$,$Al^{3+}$,and $Mg^{2+}$.
$Al^{3+}$ has the highest positive charge $(+3)$ among the given cations.
Therefore,$AlCl_3$ exhibits the maximum covalent character.

(D) In $IF_7$,the central iodine atom $(I)$ undergoes $sp^3d^3$ hybridization.
Due to this hybridization,the molecule adopts a pentagonal bipyramidal geometry to minimize electron pair repulsions.

(D) The value of $a$ is a measure of the magnitude of the attractive forces between the molecules of the gas. Greater the value of $a$,larger is the attractive inter-molecular force between the gas molecules.
The value of $b$ is related to the effective size of the gas molecules. It is also termed as excluded volume.
The gases with higher value of $a$ and lower value of $b$ are more liquefiable. Since $Cl_2$ is more easily liquefied than $C_2H_6$,$a$ for $Cl_2$ must be greater than $a$ for $C_2H_6$,and $b$ for $Cl_2$ must be less than $b$ for $C_2H_6$.

(A) The entropy change for an isothermal reversible process is given by the formula: $\Delta S = n R \ln \frac{V_2}{V_1}$
Substituting the given values: $n = 2 \, \text{mol}$,$R = 8.314 \, J \, K^{-1} \, \text{mol}^{-1}$,$V_1 = 10 \, dm^3$,$V_2 = 100 \, dm^3$.
$\Delta S = 2 \times 8.314 \times 2.303 \log \left( \frac{100}{10} \right)$
$\Delta S = 2 \times 8.314 \times 2.303 \times \log(10)$
$\Delta S = 2 \times 8.314 \times 2.303 \times 1 = 38.29 \, J \, K^{-1} \approx 38.3 \, J \, K^{-1}$

(A) The chemical equation for the reaction is: $CO_2(g) + C(s) \rightleftharpoons 2CO(g)$
Initially,the pressure of $CO_2$ is $0.5 \ atm$ and $CO$ is $0 \ atm$.
Let $x$ be the decrease in pressure of $CO_2$ at equilibrium.
At equilibrium,the partial pressures are: $P_{CO_2} = (0.5 - x) \ atm$ and $P_{CO} = 2x \ atm$.
The total pressure at equilibrium is given as $0.8 \ atm$.
$P_{\text{total}} = P_{CO_2} + P_{CO} = (0.5 - x) + 2x = 0.5 + x$
$0.5 + x = 0.8 \implies x = 0.3 \ atm$.
Now,calculate the equilibrium constant $K_p$:
$K_p = \frac{(P_{CO})^2}{P_{CO_2}} = \frac{(2x)^2}{(0.5 - x)}$
$K_p = \frac{(2 \times 0.3)^2}{(0.5 - 0.3)} = \frac{(0.6)^2}{0.2} = \frac{0.36}{0.2} = 1.8 \ atm$.

(B) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
If the alkene has a terminal $CH_2$ group (i.e.,a vinyl group,$R-CH=CH_2$),the ozonolysis reaction produces formaldehyde $(HCHO)$ as one of the products.
The reaction is represented as: $R-CH=CH_2 + O_3 \rightarrow R-CHO + HCHO$.
Thus,the formation of formaldehyde confirms the presence of a terminal vinyl group.

(C) The reduction half-reaction for the hydrogen electrode is: $H^{+} + e^{-} \longrightarrow \frac{1}{2} H_{2}$.
Using the Nernst equation: $E = E^{\circ} - \frac{0.0591}{1} \log \frac{(P_{H_{2}})^{1/2}}{[H^{+}]}$.
Since $E^{\circ} = 0 \, V$ for the standard hydrogen electrode,the equation becomes $E = -0.0591 \log \frac{(P_{H_{2}})^{1/2}}{[H^{+}]}$.
For $E$ to be negative,$\log \frac{(P_{H_{2}})^{1/2}}{[H^{+}]}$ must be positive,which means $\frac{(P_{H_{2}})^{1/2}}{[H^{+}]} > 1$.
In option $C$,$P_{H_{2}} = 2 \, atm$ and $[H^{+}] = 1.0 \, M$,so $\frac{(2)^{1/2}}{1} = \sqrt{2} \approx 1.414 > 1$. Thus,$E < 0$.

(A) Boron belongs to the second period and has a valence shell configuration of $2s^2 2p^1$.
It lacks $d$-orbitals in its valence shell,which limits its maximum covalency to $4$.
Therefore,boron cannot expand its octet to form $BF_6^{3-}$ because it would require a coordination number of $6$ and the presence of $d$-orbitals to accommodate the extra electrons.

(D) According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.
Since the expansion of water is ignored,the change in volume $\Delta V = 0$,which implies that the work done $W = P \Delta V = 0$.
Therefore,the change in internal energy $\Delta U$ is equal to the heat supplied $\Delta Q$.
The formula for heat supplied is $\Delta Q = m c \Delta T$.
Given: mass $m = 100 \ g = 0.1 \ kg$,specific heat $c = 4184 \ J/kg/K$,and change in temperature $\Delta T = 50^\circ C - 30^\circ C = 20 \ K$.
Substituting the values: $\Delta U = 0.1 \ kg \times 4184 \ J/kg/K \times 20 \ K = 8368 \ J$.
Converting to $kJ$: $\Delta U = 8368 / 1000 \approx 8.4 \ kJ$.

(D) The mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Differentiating with respect to time $t$,we get $-\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0$.
Thus,the velocity of the image $v_I = \frac{dv}{dt} = -(\frac{v}{u})^2 \frac{du}{dt}$.
Since $\frac{v}{u} = \frac{f}{u-f}$,we have $v_I = -(\frac{f}{u-f})^2 v_O$.
Given $f = 20\ cm = 0.2\ m$,$u = -2.8\ m$,and $v_O = -15\ m/s$ (approaching the mirror).
Substituting the values: $v_I = -(\frac{0.2}{-2.8 - 0.2})^2 \times (-15) = -(\frac{0.2}{-3.0})^2 \times (-15) = -(\frac{1}{15})^2 \times (-15) = \frac{1}{15}\ m/s$.

(B) The kinetic energy of the vessel is converted into the internal energy of the gas when it is suddenly brought to rest.
Let $m$ be the mass of the gas.
The kinetic energy of the gas is $K.E. = \frac{1}{2} m v^2$.
The change in internal energy of an ideal gas is given by $\Delta U = n C_V \Delta T$.
Here,$n = \frac{m}{M}$ and $C_V = \frac{R}{\gamma - 1}$.
Equating the kinetic energy to the change in internal energy: $\frac{1}{2} m v^2 = \frac{m}{M} \cdot \frac{R}{\gamma - 1} \cdot \Delta T$.
Solving for $\Delta T$: $\Delta T = \frac{(\gamma - 1) M v^2}{2R}$.
Thus,the temperature increases by $\frac{(\gamma - 1) M v^2}{2R} \text{ K}$.

(A) For the hanging mass $m$,the equation of motion is:
$m g - T = m a$ --- $(1)$
For the pulley,the torque equation is:
$\tau = I \alpha$
$T R = (\frac{1}{2} m R^2) \alpha$
Since the string does not slip,$a = R \alpha$,so $\alpha = \frac{a}{R}$.
$T R = \frac{1}{2} m R^2 (\frac{a}{R})$
$T = \frac{m a}{2}$ --- $(2)$
Substituting equation $(2)$ into equation $(1)$:
$m g - \frac{m a}{2} = m a$
$m g = m a + \frac{m a}{2}$
$m g = \frac{3 m a}{2}$
$a = \frac{2}{3} g$

(A) Consider an infinitely long wire with a semicircular cross-section of radius $R$. The total current $I$ is distributed over the semicircular arc of length $\pi R$.
The linear current density is $\lambda = \frac{I}{\pi R}$.
Consider a small element of arc length $d\ell = R d\theta$ at an angle $\theta$ from the axis of symmetry. The current in this element is $di = \lambda d\ell = \frac{I}{\pi R} (R d\theta) = \frac{I}{\pi} d\theta$.
This element acts as an infinitely long wire carrying current $di$. The magnetic field $dB$ at the center $O$ due to this element is $dB = \frac{\mu_0 di}{2\pi R} = \frac{\mu_0}{2\pi R} \left( \frac{I}{\pi} d\theta \right) = \frac{\mu_0 I}{2\pi^2 R} d\theta$.
By symmetry,the components of the magnetic field perpendicular to the axis of symmetry cancel out,while the components parallel to the axis add up. The component of $dB$ along the axis is $dB_x = dB \cos \theta$.
Integrating from $\theta = -\pi/2$ to $\pi/2$,or $2 \int_{0}^{\pi/2} dB \cos \theta$:
$B = 2 \int_{0}^{\pi/2} \frac{\mu_0 I}{2\pi^2 R} \cos \theta d\theta = \frac{\mu_0 I}{\pi^2 R} [\sin \theta]_{0}^{\pi/2} = \frac{\mu_0 I}{\pi^2 R} (1 - 0) = \frac{\mu_0 I}{\pi^2 R}$.

(D) The torque applied is $\tau = rF = 2(20t - 5t^2) = 40t - 10t^2\, N\cdot m$.
Using $\tau = I\alpha$,we have $40t - 10t^2 = 10\alpha$,so $\alpha = 4t - t^2\, rad/s^2$.
Integrating $\alpha$ with respect to time to find angular velocity $\omega$ (assuming $\omega_0 = 0$): $\omega = \int (4t - t^2) dt = 2t^2 - \frac{t^3}{3}$.
The direction of motion reverses when $\omega = 0$,so $2t^2 - \frac{t^3}{3} = 0$,which gives $t^2(2 - \frac{t}{3}) = 0$. Thus,$t = 6\, s$.
To find the angular displacement $\theta$,we integrate $\omega$: $\theta = \int_0^6 (2t^2 - \frac{t^3}{3}) dt = [\frac{2t^3}{3} - \frac{t^4}{12}]_0^6$.
$\theta = \frac{2(216)}{3} - \frac{1296}{12} = 144 - 108 = 36\, rad$.
The number of rotations is $N = \frac{\theta}{2\pi} = \frac{36}{2\pi} \approx \frac{36}{6.28} \approx 5.73$.
Since $5.73$ is between $3$ and $6$,the correct option is $D$.

(B) The kinetic energy of the gas is converted into internal energy of the gas when the vessel is suddenly brought to rest.
Let $m$ be the mass of the gas. The kinetic energy is $K.E. = \frac{1}{2} m v^2$.
The change in internal energy is given by $\Delta U = n C_v \Delta T$,where $n = \frac{m}{M}$ is the number of moles and $C_v = \frac{R}{\gamma - 1}$.
Equating the kinetic energy to the change in internal energy:
$\frac{1}{2} m v^2 = \frac{m}{M} \left( \frac{R}{\gamma - 1} \right) \Delta T$
Canceling $m$ from both sides:
$\frac{1}{2} v^2 = \frac{R}{M(\gamma - 1)} \Delta T$
Solving for $\Delta T$:
$\Delta T = \frac{(\gamma - 1) M v^2}{2R}$

(D) Given the deceleration equation: $\frac{dv}{dt} = -2.5\sqrt{v}$.
Rearranging the terms to integrate: $\frac{dv}{\sqrt{v}} = -2.5\, dt$.
Integrating both sides with limits from initial speed $v = 6.25\, m/s$ to final speed $v = 0\, m/s$ at time $t$:
$\int_{6.25}^{0} v^{-1/2} dv = \int_{0}^{t} -2.5\, dt$.
Evaluating the integral: $[2\sqrt{v}]_{6.25}^{0} = -2.5[t]_{0}^{t}$.
Substituting the limits: $2(\sqrt{0} - \sqrt{6.25}) = -2.5(t - 0)$.
$2(0 - 2.5) = -2.5t$.
$-5 = -2.5t$.
$t = \frac{5}{2.5} = 2\, s$.

(C) The internal energy of an ideal gas is given by $U = \frac{f}{2} n R T$,where $f$ is the degrees of freedom,$n$ is the number of moles,and $T$ is the absolute temperature.
Since the gases are mixed and there is no loss of energy,the total internal energy of the mixture is the sum of the internal energies of the individual gases.
Total internal energy $U_{total} = U_1 + U_2 + U_3$.
Using $n = \frac{N}{N_A}$ (where $N$ is the number of molecules and $N_A$ is Avogadro's number),the energy is $U = \frac{f}{2} \frac{N}{N_A} R T = \frac{f}{2} N k_B T$.
Assuming the gases have the same degrees of freedom $f$,the total energy is $\frac{f}{2} k_B (n_1 T_1 + n_2 T_2 + n_3 T_3) = \frac{f}{2} k_B (n_1 + n_2 + n_3) T_{mix}$.
Canceling the common terms $\frac{f}{2} k_B$,we get $n_1 T_1 + n_2 T_2 + n_3 T_3 = (n_1 + n_2 + n_3) T_{mix}$.
Therefore,the final temperature of the mixture is $T_{mix} = \frac{n_1 T_1 + n_2 T_2 + n_3 T_3}{n_1 + n_2 + n_3}$.

(D) According to the law of conservation of linear momentum,the total momentum of the system must remain constant.
Since the initial momentum of the system is approximately zero (as the neutron is moving slowly),the final momentum of the two resulting nuclei must also be zero.
Let $p_1$ be the momentum of the nucleus with mass $m_1$ and $p_2$ be the momentum of the nucleus with mass $5m_1$.
Therefore,$p_1 + p_2 = 0$,which implies $p_1 = -p_2$,or $|p_1| = |p_2|$.
The de-Broglie wavelength $\lambda$ is given by the relation $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the magnitude of the momentum.
Since both nuclei have the same magnitude of momentum $(|p_1| = |p_2|)$,their de-Broglie wavelengths must also be equal.
Thus,the de-Broglie wavelength of the second nucleus is also $\lambda$.

(A) Consider a cross-section of the wire as a semicircular arc of radius $R$. The total current $I$ is distributed over the arc length $\pi R$.
The current per unit arc length is $\lambda = \frac{I}{\pi R}$.
Consider a small element of arc length $dl = R d\theta$ at an angle $\theta$ from the axis. The current in this element is $di = \lambda dl = \frac{I}{\pi R} (R d\theta) = \frac{I}{\pi} d\theta$.
The magnetic field $dB$ at the center $O$ due to this infinite wire element is given by the formula for an infinite wire: $dB = \frac{\mu_0 di}{2\pi R} = \frac{\mu_0}{2\pi R} \left( \frac{I}{\pi} d\theta \right) = \frac{\mu_0 I}{2\pi^2 R} d\theta$.
Due to symmetry,the components of the magnetic field perpendicular to the axis cancel out. The net magnetic field is the integral of the components along the axis: $B = \int_{-\pi/2}^{\pi/2} dB \cos\theta = 2 \int_{0}^{\pi/2} \frac{\mu_0 I}{2\pi^2 R} \cos\theta d\theta$.
$B = \frac{\mu_0 I}{\pi^2 R} \int_{0}^{\pi/2} \cos\theta d\theta = \frac{\mu_0 I}{\pi^2 R} [\sin\theta]_{0}^{\pi/2} = \frac{\mu_0 I}{\pi^2 R}$.

(A) The basic nature of oxides depends on the metallic character of the element.
Metallic character increases down a group and decreases across a period from left to right.
$Al$ is in period $3$,group $13$ (amphoteric oxide).
$Mg$ is in period $3$,group $2$ (basic oxide).
$Na$ is in period $3$,group $1$ (strongly basic oxide).
$K$ is in period $4$,group $1$ (more strongly basic than $Na_2O$ due to larger size).
Thus,the order of increasing basic nature is: $Al_2O_3 < MgO < Na_2O < K_2O$.

(B) Let the point where the gravitational field is zero be at a distance $x$ from mass $m$. The gravitational field due to both masses must be equal in magnitude and opposite in direction.
$\frac{Gm}{x^2} = \frac{G(4m)}{(r-x)^2}$
Taking the square root on both sides:
$\frac{1}{x} = \frac{2}{r-x} \Rightarrow r-x = 2x \Rightarrow 3x = r \Rightarrow x = \frac{r}{3}$
The distance from mass $4m$ is $r-x = r - \frac{r}{3} = \frac{2r}{3}$.
The gravitational potential $V$ at this point is the sum of potentials due to both masses:
$V = -\frac{Gm}{x} - \frac{G(4m)}{r-x}$
Substituting the values of $x$ and $r-x$:
$V = -\frac{Gm}{r/3} - \frac{4Gm}{2r/3} = -\frac{3Gm}{r} - \frac{6Gm}{r} = -\frac{9Gm}{r}$

(C) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$.
At time $t_1$,the amount decayed is $\frac{1}{3} N_0$,so the remaining amount is $N_1 = N_0 - \frac{1}{3} N_0 = \frac{2}{3} N_0$.
At time $t_2$,the amount decayed is $\frac{2}{3} N_0$,so the remaining amount is $N_2 = N_0 - \frac{2}{3} N_0 = \frac{1}{3} N_0$.
We observe that $N_2 = \frac{1}{2} N_1$.
Since the amount of substance reduces to half of its value in one half-life period $(T_{1/2})$,the time interval required for the substance to decay from $N_1$ to $N_2$ is exactly one half-life.
Therefore,$t_2 - t_1 = T_{1/2} = 20 \, min$.

(B) Given:
Length of the wire,$\ell = 20\, m$
Speed of the wire,$v = 5.0\, m/s$
Horizontal component of Earth's magnetic field,$B_H = 0.30 \times 10^{-4}\, Wb/m^2$
The motional $e.m.f.$ induced in a conductor moving in a magnetic field is given by the formula:
$e = B_H v \ell$
Substituting the given values:
$e = (0.30 \times 10^{-4}) \times 5.0 \times 20$
$e = 0.30 \times 10^{-4} \times 100$
$e = 0.30 \times 10^{-2}\, V$
$e = 3 \times 10^{-3}\, V$
Since $1\, mV = 10^{-3}\, V$,the induced $e.m.f.$ is:
$e = 3\, mV$

(B) soap bubble has two surfaces (inner and outer). The work done $W$ in changing the radius of a soap bubble from $r_1$ to $r_2$ is given by:
$W = 2 \times T \times \Delta A = 2 \times T \times 4\pi(r_2^2 - r_1^2) = 8\pi T(r_2^2 - r_1^2)$
Given: $T = 0.03\, Nm^{-1}$,$r_1 = 3\, cm = 0.03\, m$,$r_2 = 5\, cm = 0.05\, m$.
Substituting the values:
$W = 8 \times \pi \times 0.03 \times ((0.05)^2 - (0.03)^2)$
$W = 8 \times \pi \times 0.03 \times (0.0025 - 0.0009)$
$W = 8 \times \pi \times 0.03 \times 0.0016$
$W = 0.24 \times \pi \times 0.0016 = 0.000384\pi \,J$
$W = 0.384\pi \,mJ \approx 0.4\pi \,mJ$.

(A) The lanthanoids exhibit a common oxidation state of $+3$.
Not all members exhibit a $+4$ oxidation state; only a few,such as $Ce^{+4}$,$Pr^{+4}$,$Nd^{+4}$,$Tb^{+4}$,and $Dy^{+4}$,show this state due to stable configurations.
Therefore,the statement that all members form compounds in the $+4$ state is incorrect.

(B) Let $I_0$ be the intensity of each individual wave. The resultant intensity is given by $I = 2I_0 + 2I_0 \cos(\Delta \phi) = 4I_0 \cos^2(\frac{\Delta \phi}{2})$.
At point $P$,the path difference $\Delta x = 0$,so the phase difference $\Delta \phi = 0$. The intensity $I_P = 4I_0 \cos^2(0) = 4I_0$.
At point $Q$,the path difference $\Delta x = \frac{\lambda}{4}$,so the phase difference $\Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$. The intensity $I_Q = 4I_0 \cos^2(\frac{\pi}{4}) = 4I_0 \cdot (\frac{1}{\sqrt{2}})^2 = 2I_0$.
The ratio of intensities is $\frac{I_P}{I_Q} = \frac{4I_0}{2I_0} = 2:1$.

(B) soap bubble has two surfaces (inner and outer). The work done $W$ in increasing the radius from $r_1$ to $r_2$ is given by $W = 2 \times [4\pi T(r_2^2 - r_1^2)] = 8\pi T(r_2^2 - r_1^2)$.
Given: $T = 0.03\, Nm^{-1}$,$r_1 = 3\, cm = 0.03\, m$,$r_2 = 5\, cm = 0.05\, m$.
Substituting the values:
$W = 8 \times \pi \times 0.03 \times ((0.05)^2 - (0.03)^2)$
$W = 8 \times \pi \times 0.03 \times (0.0025 - 0.0009)$
$W = 8 \times \pi \times 0.03 \times 0.0016$
$W = 0.000384\pi \,J = 0.384\pi \,mJ$.
Rounding to the nearest value,$W \approx 0.4\pi \,mJ$.

(C) Since the vessel is thermally insulated,no heat is exchanged with the surroundings.
The kinetic energy of the gas is converted into internal energy of the gas when it is suddenly brought to rest.
Loss in kinetic energy = Gain in internal energy
$\frac{1}{2} m v^2 = n C_V \Delta T$
Here,$n = \frac{m}{M}$ is the number of moles and $C_V = \frac{R}{\gamma - 1}$ is the molar specific heat at constant volume.
Substituting these values:
$\frac{1}{2} m v^2 = \left( \frac{m}{M} \right) \left( \frac{R}{\gamma - 1} \right) \Delta T$
Canceling $m$ from both sides:
$\frac{1}{2} v^2 = \frac{R \Delta T}{M(\gamma - 1)}$
Solving for $\Delta T$:
$\Delta T = \frac{M v^2 (\gamma - 1)}{2R} \ K$

(A) The electric field $E$ is related to the potential $\phi$ by the relation $E = -\frac{d\phi}{dr}$.
Given $\phi = ar^2 + b$,we have $E = -\frac{d}{dr}(ar^2 + b) = -2ar$.
According to Gauss's Law in differential form,the charge density $\rho$ is given by $\rho = \varepsilon_0 (\nabla \cdot E)$.
In spherical coordinates,for a radial field $E_r$,the divergence is $\nabla \cdot E = \frac{1}{r^2} \frac{d}{dr}(r^2 E_r)$.
Substituting $E_r = -2ar$,we get $\nabla \cdot E = \frac{1}{r^2} \frac{d}{dr}(r^2 (-2ar)) = \frac{1}{r^2} \frac{d}{dr}(-2ar^3) = \frac{1}{r^2} (-6ar^2) = -6a$.
Therefore,the charge density is $\rho = \varepsilon_0 (-6a) = -6a\varepsilon_0$.

(D) The central iodine atom $(I)$ in $IF_{7}$ has $7$ valence electrons. It forms $7$ bonds with $7$ fluorine atoms.
The steric number is calculated as:
$\text{Steric Number} = \frac{1}{2} \times (V + M - C + A) = \frac{1}{2} \times (7 + 7) = 7$.
$A$ steric number of $7$ corresponds to $sp^{3}d^{3}$ hybridization,which results in a pentagonal bipyramidal geometry.
In this structure,five fluorine atoms lie in a pentagonal plane with $I-F$ bond angles of $72^{\circ}$,and two fluorine atoms are positioned above and below the plane at $90^{\circ}$ to the equatorial plane.

(B) Let the point $P$ be at a distance $x$ from mass $m$. At this point,the gravitational field is zero.
Equating the gravitational field intensities due to both masses:
$\frac{Gm}{x^2} = \frac{G(4m)}{(r-x)^2}$
Taking the square root on both sides:
$\frac{1}{x} = \frac{2}{r-x}$
$r - x = 2x \Rightarrow 3x = r \Rightarrow x = \frac{r}{3}$
The distance of point $P$ from mass $4m$ is $r - x = r - \frac{r}{3} = \frac{2r}{3}$.
The gravitational potential $V$ at point $P$ is the sum of potentials due to both masses:
$V = -\frac{Gm}{x} - \frac{G(4m)}{r-x}$
Substituting $x = \frac{r}{3}$ and $r-x = \frac{2r}{3}$:
$V = -\frac{Gm}{r/3} - \frac{4Gm}{2r/3}$
$V = -\frac{3Gm}{r} - \frac{12Gm}{2r}$
$V = -\frac{3Gm}{r} - \frac{6Gm}{r} = -\frac{9Gm}{r}$

(C) Tautomerism is a special type of functional isomerism in which a compound exists in two or more interconvertible forms that differ in the relative position of at least one atomic nucleus,generally a hydrogen atom.
For a carbonyl compound to exhibit tautomerism,it must possess at least one $\alpha$-hydrogen atom.
$2-$Pentanone $(CH_3COCH_2CH_2CH_3)$ contains $\alpha$-hydrogen atoms on the carbon adjacent to the carbonyl group,and therefore it exhibits keto-enol tautomerism.

(D) In a face centred cubic $(FCC)$ lattice:
Number of atoms of $A$ at corners $= 8 \times \frac{1}{8} = 1$.
Total number of face centres in a cube is $6$. Since one atom of $B$ is missing,the number of atoms of $B$ present $= 6 - 1 = 5$.
Number of atoms of $B$ at face centres $= 5 \times \frac{1}{2} = 2.5$.
The ratio of $A : B = 1 : 2.5 = 2 : 5$.
Therefore,the formula of the compound is $A_2B_5$.

(C) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Given $m = 5.2 \ mol/kg$,this means $5.2$ moles of $CH_3OH$ are dissolved in $1000 \ g$ of water $(H_2O)$.
Moles of water $(n_{H_2O})$ $= \frac{1000 \ g}{18 \ g/mol} \approx 55.56 \ mol$.
Mole fraction of methyl alcohol $(X_{CH_3OH})$ $= \frac{n_{CH_3OH}}{n_{CH_3OH} + n_{H_2O}}$.
$X_{CH_3OH} = \frac{5.2}{5.2 + 55.56} = \frac{5.2}{60.76} \approx 0.0856$.
Rounding to two significant figures,we get $0.086$.

(A) Given:
$K_f = 1.86 \ K \ kg \ mol^{-1}$
Mass of solvent $(W_A)$ $= 4 \ kg = 4000 \ g$
Freezing point depression $\Delta T_f = 0 - (-6) = 6 \ K$
Molar mass of solute $(M_B)$ $= 62 \ g \ mol^{-1}$
The formula for freezing point depression is $\Delta T_f = K_f \times m$,where $m$ is molality.
$\Delta T_f = K_f \times \frac{w \times 1000}{M_B \times W_A \text{ (in g)}}$
$6 = 1.86 \times \frac{w \times 1000}{62 \times 4000}$
$6 = 1.86 \times \frac{w}{62 \times 4}$
$w = \frac{6 \times 62 \times 4}{1.86}$
$w = \frac{1488}{1.86} = 800 \ g$

(A) For the dissociation of a weak electrolyte $A_xB_y$:
$A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$
At $t=0$,the moles are $1, 0, 0$ respectively.
At equilibrium,the moles are $(1 - \alpha), x\alpha, y\alpha$.
The total number of moles at equilibrium is:
$n_{total} = 1 - \alpha + x\alpha + y\alpha = 1 + \alpha(x + y - 1)$
The van't Hoff factor $(i)$ is defined as the ratio of observed moles to initial moles:
$i = \frac{1 + \alpha(x + y - 1)}{1} = 1 + \alpha(x + y - 1)$
Rearranging for $\alpha$:
$i - 1 = \alpha(x + y - 1)$
$\alpha = \frac{i - 1}{x + y - 1}$

(C) The temperature coefficient is $2$ for a $10\,^{\circ}C$ rise in temperature.
For a rise of $50\,^{\circ}C,$ the number of $10\,^{\circ}C$ intervals is $n = \frac{50}{10} = 5$.
The increase in the rate of reaction is given by $2^n = 2^5$.
$2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$.
Therefore,the rate of the reaction increases by $32$ times.

(A) The stability of hydrides decreases from $NH_3$ to $BiH_3$ as we move down the group $15$. This is because the size of the central atom increases,leading to a weaker $M-H$ bond due to poor orbital overlap.
Therefore,the statement in option $A$ is incorrect.
Additionally,nitrogen cannot form $d\pi - p\pi$ bonds because it lacks $d$-orbitals in its valence shell.
The single $N-N$ bond is weaker than the $P-P$ bond due to high inter-electronic repulsion of the non-bonding electrons in the small $N$ atoms.
$N_2O_4$ exhibits resonance structures.

(D) The oxidation state of sulphur ranges from $-2$ to $+6$ in its various compounds. For example,in $H_2S$,the oxidation state of sulphur is $-2$. Therefore,the statement that the oxidation state of sulphur is never less than $+4$ is incorrect.

(C) In the complex $[Cr(NH_3)_6]Cl_3$,the oxidation state of $Cr$ is $+3$. The electronic configuration of $Cr^{3+}$ is $[Ar]3d^3$.
Since $NH_3$ is a ligand,it forms an octahedral complex using $d^2sp^3$ hybridization.
Because it uses $(n-1)d$ orbitals ($3d$ orbitals),it is classified as an inner orbital complex,not an outer orbital complex.
Therefore,the statement that it is an outer orbital complex is wrong.

(D) Lanthanoids exhibit a common $+3$ oxidation state.
The $4f$ electrons are deeply buried and shielded by $5s$,$5p$,and $5d$ orbitals,making them less available for bonding compared to $d$-block elements.
While some lanthanoids like $Ce$ $(+4)$ and $Tb$ $(+4)$ show $+4$ oxidation states due to stable electronic configurations (like $f^0$ or $f^7$),it is not a property of all members of the series.
Therefore,the statement that all members form compounds in the $+4$ state is incorrect.

(C) In $[NiCl_{4}]^{2-}$,the oxidation state of $Ni$ is $+2$.
$Ni^{2+}$ has the electronic configuration $[Ar] 3d^{8}$.
Since $Cl^{-}$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.
Thus,there are $n = 2$ unpaired electrons.
The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ B.M.$
$\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ B.M.$

(D) The atomic number of $Gd$ (Gadolinium) is $64$.
Its electronic configuration is based on the Aufbau principle,but it is modified to achieve extra stability.
The $4f$ subshell is half-filled $(4f^7)$,which provides extra stability to the atom.
Therefore,the correct configuration is $[Xe] 4f^7 5d^1 6s^2$.

(D) When phenol is treated with a mixture of $KBr$ and $KBrO_3$ in an acidic medium,$Br_2$ is generated in situ.
$5KBr + KBrO_3 + 6H^+ \rightarrow 3Br_2 + 6K^+ + 3H_2O$
Due to the highly activating nature of the $-OH$ group,phenol undergoes electrophilic aromatic substitution with bromine water to form a white precipitate of $2, 4, 6-$ tribromophenol.

(D) Phenol reacts with neutral $FeCl_3$ to form a violet-colored complex.
Benzoic acid reacts with neutral $FeCl_3$ to form a buff-colored (pale dull yellow) precipitate of ferric benzoate.
Therefore,neutral $FeCl_3$ can be used to distinguish between them.

(A) Cannizzaro reaction is a disproportionation reaction of aldehydes lacking $\alpha$-hydrogen atoms in the presence of a strong base.
Trichloroacetaldehyde $(CCl_3CHO)$ does not have $\alpha$-hydrogen atoms.
In the presence of $NaOH$,it undergoes self-oxidation and reduction to form sodium trichloroacetate $(CCl_3COO^-)$ and $2, 2, 2-$trichloroethanol $(CCl_3CH_2OH)$.

(C) The acidity of carboxylic acids is increased by the presence of electron-withdrawing groups $(EWG)$ due to the inductive effect ($-I$ effect).
These groups stabilize the carboxylate anion by dispersing the negative charge.
The strength of the $-I$ effect depends on the distance of the substituent from the carboxylic acid group.
In $CH_3CH_2CH(Cl)CO_2H$,the chlorine atom is at the $\alpha$-position (closest to the $-COOH$ group),which exerts a strong $-I$ effect compared to the $\gamma$-position in $ClCH_2CH_2CH_2COOH$.
Therefore,$CH_3CH_2CH(Cl)CO_2H$ is the strongest acid among the given options.

(D) The reaction between sodium ethoxide $(C_2H_5ONa)$ and ethanoyl chloride $(CH_3COCl)$ is a nucleophilic acyl substitution reaction.
In this reaction,the ethoxide ion $(C_2H_5O^-)$ acts as a nucleophile and attacks the carbonyl carbon of the ethanoyl chloride,displacing the chloride ion $(Cl^-)$.
The chemical equation is: $CH_3COCl + C_2H_5ONa \rightarrow CH_3COOC_2H_5 + NaCl$.
The product formed is ethyl ethanoate $(CH_3COOC_2H_5)$.

(D) The Silver Mirror test (Tollens' test) is given by aldehydes because they can be easily oxidized to carboxylic acids.
Both formaldehyde $(HCHO)$ and acetaldehyde $(CH_3CHO)$ are aldehydes and thus give a positive Silver Mirror test.
$HCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow HCOO^- + 2Ag \downarrow + 4NH_3 + 2H_2O$
$CH_3CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3COO^- + 2Ag \downarrow + 4NH_3 + 2H_2O$
Ketones like acetone do not give this test.

(B) The sugar molecule found in $RNA$ is $D$-ribose,while the sugar in $DNA$ is $D$-$2$-deoxyribose.
$D$-$2$-deoxyribose differs from ribose only in the substitution of a hydrogen atom for an $-OH$ group at the $2$-position,as shown in the figure.

(B) For an $fcc$ (face-centered cubic) unit cell, the relationship between the edge length $a$ and the atomic radius $r$ is given by:
$r = \frac{\sqrt{2} a}{4} = \frac{a}{2\sqrt{2}}$
Given the edge length $a = 361 \ pm$:
$r = \frac{361}{2 \times 1.414}$
$r = \frac{361}{2.828}$
$r \approx 127.65 \ pm$
Rounding to the nearest whole number, we get $r = 128 \ pm$.

(C) The reduction reaction for a hydrogen half-cell is: $2H^{+} + 2e^{-} \rightarrow H_{2}$
Here,$n = 2$ and the reaction quotient $Q = \frac{p(H_{2})}{[H^{+}]^{2}}$.
Using the Nernst equation: $E_{H^{+}/H_{2}} = E^{0}_{H^{+}/H_{2}} - \frac{0.059}{n} \log Q$.
Since $E^{0}_{H^{+}/H_{2}} = 0 \ V$,we have $E_{H^{+}/H_{2}} = -\frac{0.059}{2} \log Q$.
For $E_{H^{+}/H_{2}}$ to be negative,$\log Q$ must be positive,which means $Q > 1$.
Evaluating the options:
$(A)$ $Q = \frac{1}{1^{2}} = 1$ (Not negative)
$(B)$ $Q = \frac{1}{2^{2}} = 0.25 < 1$ (Positive potential)
$(C)$ $Q = \frac{2}{1^{2}} = 2 > 1$ (Negative potential)
$(D)$ $Q = \frac{2}{2^{2}} = 0.5 < 1$ (Positive potential)
Therefore,the correct option is $(C)$.