27 Jan 2024 Shift 1

We need to check the dimensional correctness of both statements. Statement (I): Planck's constant and angular momentum have the same dimensions. Planck's constant $h$ has the relation $E = h\nu$, so: $[h] = \frac{[E]}{[\nu]} = \frac{ML^2T^{-2}}{T^{-1}} = ML^2T^{-1}$ Angular momentum $L = mvr$, so: $[L] = M \cdot LT^{-1} \cdot L = ML^2T^{-1}$ Both have the same dimensions $ML^2T^{-1}$. Statement (I) is true . Statement (II): Linear momentum and moment of force have the same dimensions. Linear momentum $p = mv$: $[p] = MLT^{-1}$ Moment of force (torque) $\tau = r \times F$: $[\tau] = L \cdot MLT^{-2} = ML^2T^{-2}$ The dimensions are different: $MLT^{-1} \neq ML^2T^{-2}$. Statement (II) is false . Therefore, Statement I is true but Statement II is false. The correct answer is Option A .

The position is given by $\vec{S} = 2t^2 \hat{j} + 5\hat{k}$. Velocity is the time derivative of position: $\vec{v} = \frac{d\vec{S}}{dt} = 4t\hat{j} + 0\hat{k} = 4t\hat{j}$ At $t = 1$ s: $\vec{v} = 4(1)\hat{j} = 4\hat{j}$ m/s The magnitude is $4$ m/s and the direction is along the y-direction. The answer is $4 \text{ m s}^{-1}$ in y-direction, which corresponds to Option (4).

For banking of tracks, the formula for the height of the outer rail is: $\tan\theta = \frac{v^2}{Rg}$ For small angles, $\tan\theta \approx \sin\theta \approx \frac{h}{l}$, where $h$ is the height difference and $l$ is the distance between rails. $\frac{h}{l} = \frac{v^2}{Rg}$ $h = \frac{v^2 l}{Rg} = \frac{(12)^2 \times 1.5}{400 \times 10} = \frac{144 \times 1.5}{4000} = \frac{216}{4000} = 0.054 \text{ m} = 5.4 \text{ cm}$ The answer is $5.4$ cm, which corresponds to Option (2).

The relationship between kinetic energy and momentum is: $KE = \frac{p^2}{2m}$, so $p = \sqrt{2mKE}$. Since both bodies have equal kinetic energy: $\frac{p_1}{p_2} = \frac{\sqrt{2m_1 KE}}{\sqrt{2m_2 KE}} = \sqrt{\frac{m_1}{m_2}} = \sqrt{\frac{4}{25}} = \frac{2}{5}$ The ratio of magnitudes of their linear momenta is $2 : 5$, which corresponds to Option (3).

By conservation of momentum (assuming the extra mass is added with zero velocity in the horizontal direction): $m_1 v_1 = (m_1 + m_2)v_f$ $1000 \times 6 = (1000 + 200) \times v_f$ $6000 = 1200 \times v_f$ $v_f = 5 \text{ m s}^{-1}$ The answer is $5$ m/s, which corresponds to Option (4).

The acceleration due to gravity on the surface of Earth is: $g = \frac{GM}{R^2}$ If the diameter is halved, the radius becomes $R' = R/2$. $g' = \frac{GM}{(R/2)^2} = \frac{GM}{R^2/4} = \frac{4GM}{R^2} = 4g$ The answer is $4g$, which corresponds to Option (4).

Viscosity of gases is not greater than that of liquids; in fact, liquids generally exhibit much higher viscosities than gases. For example, the viscosity of water at 20 °C is approximately $1 \times 10^{-3}$ Pa·s, while the viscosity of air at 20 °C is approximately $1.8 \times 10^{-5}$ Pa·s, which is about fifty times smaller. This difference arises because in liquids, viscosity is governed by strong intermolecular cohesive forces, whereas in gases it results from momentum transfer between molecular layers; the cohesive forces in liquids are far stronger, leading to higher viscosity. Insoluble impurities such as dust particles, powders, or oils that spread on the surface of a liquid reduce the cohesive forces between the liquid molecules at the interface, thereby lowering its surface tension. For instance, camphor particles on water demonstrate this effect by decreasing the water's surface tension. The correct answer is Option (2): Statement I is incorrect but Statement II is correct.

For heating at constant volume, the change in internal energy equals the heat added: $\Delta U = m c_v \Delta T$ Given: $m = 0.08$ kg, $c_v = 0.17$ kcal kg$^{-1}$ °C$^{-1}$, $\Delta T = 5°C$. $\Delta U = 0.08 \times 0.17 \times 5 = 0.068 \text{ kcal}$ Converting to joules ($1 \text{ cal} = 4.18 \text{ J}$, so $1 \text{ kcal} = 4180 \text{ J}$): $\Delta U = 0.068 \times 4180 = 284.24 \text{ J} \approx 284 \text{ J}$ The answer is approximately $284$ J, which corresponds to Option (3).

The average kinetic energy of a monatomic molecule is: $KE = \frac{3}{2}k_BT$ Given $KE = 0.414$ eV $= 0.414 \times 1.6 \times 10^{-19} = 6.624 \times 10^{-20}$ J. $T = \frac{2 \times KE}{3k_B} = \frac{2 \times 6.624 \times 10^{-20}}{3 \times 1.38 \times 10^{-23}} = \frac{13.248 \times 10^{-20}}{4.14 \times 10^{-23}} = \frac{13.248}{0.0414} \approx 3200 \text{ K}$ The answer is $3200$ K, which corresponds to Option (2).

A point charge $q = 10^{-6}\mu C = 10^{-12}$ C is placed at the origin, and we consider the points P $(\sqrt{3},\sqrt{3})$ and Q $(\sqrt{6},0)$. The distance of point P from the origin is $r_P = \sqrt{3+3} = \sqrt{6}$ m, while the distance of point Q is $r_Q = \sqrt{6}$ m. The electric potential due to a point charge is given by $V = \frac{kq}{r}$. Since $r_P = r_Q = \sqrt{6}$, it follows that $V_P = V_Q = \frac{kq}{\sqrt{6}}$ Consequently, the potential difference between P and Q is $V_P - V_Q = 0$ The correct answer is Option 3 : 0 V.

A wire of resistance $R$ and length $L$ is cut into 5 equal parts. Each part has length $L/5$ and resistance $R/5$. When 5 resistors of resistance $R/5$ each are joined \in parallel: $\frac{1}{R_{eq}} = \frac{5}{R/5} = \frac{25}{R}$ $R_{eq} = \frac{R}{25}$ The answer is $\frac{R}{25}$, which corresponds to Option (1).

In a meter bridge, at balance: $\frac{R_1}{R_2} = \frac{l}{100 - l}$ Given: $R_1 = 4.5 \; \Omega$ (left gap), balance length $l = 60$ cm. $\frac{4.5}{R_2} = \frac{60}{40} = \frac{3}{2}$ $R_2 = \frac{4.5 \times 2}{3} = 3 \; \Omega$ The wire \in the right gap has length $l_w = 10$ cm $= 0.1$ m and radius $r = \sqrt{7} \times 10^{-4}$ m. Using $R_2 = \frac{\rho l_w}{\pi r^2}$: $3 = \frac{\rho \times 0.1}{\pi \times 7 \times 10^{-8}}$ $\rho = \frac{3 \times \pi \times 7 \times 10^{-8}}{0.1} = \frac{21\pi \times 10^{-8}}{0.1} = 21\pi \times 10^{-7}$ $\rho = 21 \times 3.14159 \times 10^{-7} \approx 65.97 \times 10^{-7} \approx 66 \times 10^{-7} \; \Omega \text{m}$ So $R = 66$. The answer is $66$, which corresponds to Option (3).

A proton moves with constant velocity, meaning the net force on it is zero. (A) $E = 0, B = 0$: No force, proton moves at constant velocity. Possible. (B) $E = 0, B \neq 0$: If the proton moves parallel to $\vec{B}$, the magnetic force $q\vec{v} \times \vec{B} = 0$. Possible. (C) $E \neq 0, B = 0$: The electric force $q\vec{E} \neq 0$, so the proton would accelerate. Not possible. (D) $E \neq 0, B \neq 0$: If $q\vec{E} + q\vec{v} \times \vec{B} = 0$, the electric and magnetic forces can cancel. Possible. The correct options are (A), (B) and (D) only, which corresponds to Option (3).

A rectangular loop of length 2.5 m and width 2 m is placed at an angle of 60° with a magnetic field of 4 T and is removed from the field in 10 s. The magnetic flux through the loop initially is given by $\Phi = BA\cos\theta = 4 \times (2.5 \times 2) \times \cos 60° = 4 \times 5 \times 0.5 = 10$ Wb. According to Faraday's law, the average induced emf is $\mathcal{E} = -\frac{\Delta\Phi}{\Delta t} = -\frac{0 - 10}{10} = \frac{10}{10} = +1$ V. The correct answer is Option C: +1 V.

The intensity of an electromagnetic wave is: $I = \frac{1}{2}\epsilon_0 c E_0^2$ where $E_0 = 200$ V/m is the amplitude of the electric field and $c = 3 \times 10^8$ m/s. $I = \frac{1}{2} \times 8.85 \times 10^{-12} \times 3 \times 10^8 \times (200)^2$ $= \frac{1}{2} \times 8.85 \times 10^{-12} \times 3 \times 10^8 \times 4 \times 10^4$ $= \frac{1}{2} \times 8.85 \times 3 \times 4 \times 10^{-12+8+4}$ $= \frac{1}{2} \times 106.2 \times 10^0$ $= 53.1 \text{ W m}^{-2}$ The answer is $53.1 \text{ W m}^{-2}$, which corresponds to Option (2).

The refractive index ($n$) of a prism at minimum deviation is given by the formula $n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$, where $A$ is the prism angle and $\delta_m$ is the angle of minimum deviation. We are given that the refractive index is $n = \cot\left(\frac{A}{2}\right)$. Equating the two expressions for $n$ gives $\frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \cot\left(\frac{A}{2}\right) = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$. This simplifies to $\sin\left(\frac{A + \delta_m}{2}\right) = \cos\left(\frac{A}{2}\right)$. Using the identity $\cos(\theta) = \sin\left(\frac{\pi}{2} - \theta\right)$, we can write $\sin\left(\frac{A + \delta_m}{2}\right) = \sin\left(\frac{\pi}{2} - \frac{A}{2}\right)$. Therefore, $\frac{A + \delta_m}{2} = \frac{\pi}{2} - \frac{A}{2}$, which gives $A + \delta_m = \pi - A$. Solving for $\delta_m$, we find the angle of minimum deviation to be $\delta_m = \pi - 2A$.

The photoelectric current ($I$) is directly proportional to the total number of photons striking the photoelectric cell per second. This, in turn, is proportional to the total light energy (or power) incident on the cell.

The lens collects light from the source and focuses it onto the cell. The total amount of light energy collected by the lens is determined by its light-gathering area, which depends on its diameter $D$. The area of the lens is given by $A = \pi (D/2)^2$.

Since both lenses have the same diameter, their areas are identical. Consequently, they collect the same amount of light energy from the extended source and focus it onto the photoelectric cell.

The focal length affects the size and intensity (power per unit area) of the image, but not the total energy collected. As the total incident energy on the cell remains the same, the photoelectric current does not change. Therefore, the new current is also $I$.

In Bohr's model, the radius of the $n$th orbit is proportional to $n^2$: $r_n \propto n^2$ For the third orbit: $r_3 = R \propto 3^2 = 9$ For the fourth orbit: $r_4 \propto 4^2 = 16$ $\frac{r_4}{r_3} = \frac{16}{9}$ $r_4 = \frac{16}{9}R$ The answer is $\frac{16}{9}R$, which corresponds to Option (2).

A diode is considered reverse-biased when the voltage at its anode (p-side, the base of the triangle) is lower than the voltage at its cathode (n-side, the line). Let the anode voltage be $V_A$ and the cathode voltage be $V_K$. The condition for reverse bias is $V_A < V_K$.

Let's examine circuit D:
The anode is connected to a voltage source of $+2V$, so we have $V_A = +2V$.
The cathode is connected through a resistor to a voltage source of $+4V$.
Comparing the potentials at the two terminals of the diode, we have:
$V_A = +2V$
$V_K \text{ is connected to } +4V$
Since $V_A < V_K$ ($+2V < +4V$), the diode in circuit D is reverse-biased. In circuits A, B, and C, the anode potential is higher than the cathode potential, so they are forward-biased.

A) Radius of curvature of concave surface A spherometer is specifically designed to measure radius of curvature (both concave and convex). So this can be measured . B) Specific rotation of liquids This is an optical property measured using a polarimeter , not a spherometer. Spherometer only measures mechanical dimensions (height, curvature) . So this cannot be measured . C) Thickness of thin plates Yes, a spherometer can measure very small thickness by comparing readings. So this can be measured . D) Radius of curvature of convex surface Again, one of the main uses of a spherometer. So this can be measured . Final conclusion: Only option B is not measurable using a spherometer. Answer: Specific rotation of liquids

Initial velocity: $u_x = 5$ m/s, $u_y = 0$. Acceleration: $a_x = 3$ m/s$^2$, $a_y = 2$ m/s$^2$. The x-coordinate: $x = u_x t + \frac{1}{2}a_x t^2 = 5t + \frac{3}{2}t^2$ Setting $x = 84$: $5t + \frac{3}{2}t^2 = 84$ $3t^2 + 10t - 168 = 0$ Using the quadratic formula: $t = \frac{-10 + \sqrt{100 + 2016}}{6} = \frac{-10 + \sqrt{2116}}{6} = \frac{-10 + 46}{6} = \frac{36}{6} = 6$ s Velocity components at $t = 6$ s: $v_x = u_x + a_x t = 5 + 3(6) = 23$ m/s $v_y = u_y + a_y t = 0 + 2(6) = 12$ m/s Speed: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{529 + 144} = \sqrt{673}$ m/s So $\alpha = 673$. The answer is $\boxed{673}$.

Four particles each of mass 1 kg at corners of a square with side 2 m. The axis passes through one vertex, perpendicular to the plane. Taking the vertex through which the axis passes as origin, the distances of the four particles from the axis are: - Particle at the vertex: $r_1 = 0$ - Two adjacent particles: $r_2 = r_3 = 2$ m (side length) - Diagonally opposite particle: $r_4 = 2\sqrt{2}$ m (diagonal) $I = m(r_1^2 + r_2^2 + r_3^2 + r_4^2) = 1(0 + 4 + 4 + 8) = 16 \text{ kg m}^2$ The answer is $\boxed{16}$ kg m$^2$.

The fractional compression is given by: $\frac{\Delta V}{V} = \frac{P}{B}$ where $P$ is the pressure at the bottom and $B$ is the bulk modulus. Pressure at the bottom: $P = \rho g h = 1000 \times 10 \times 4000 = 4 \times 10^7$ N/m$^2$ $\frac{\Delta V}{V} = \frac{4 \times 10^7}{2 \times 10^9} = 2 \times 10^{-2}$ So $\alpha = 2$. The answer is $\boxed{2}$.

In SHM, the velocity at displacement $x$ from mean position is: $v = \omega\sqrt{A^2 - x^2}$ At mean position ($x = 0$): $v_{max} = \omega A = 10$ cm/s, with $A = 4$ cm. So $\omega = \frac{10}{4} = 2.5$ rad/s. When $v = 5$ cm/s: $5 = 2.5\sqrt{16 - x^2}$ $2 = \sqrt{16 - x^2}$ $4 = 16 - x^2$ $x^2 = 12$ $x = \sqrt{12}$ cm So $\alpha = 12$. The answer is $\boxed{12}$.

The radius of the ring is $R = 30\text{ cm}=0.30\text{ m}.$ The charge uniformly spread over the ring is $Q = 2\pi\text{ C},$ and the point charge kept at the centre is $q = 30\text{ pC}=30\times10^{-12}\text{ C}.$ Choose a small element of the ring that subtends an angle $d\theta$ at the centre. Length of the element $dl = R\,d\theta,$ charge on the element $dq = \frac{Q}{2\pi R}\,dl \;=\; \frac{Q}{2\pi}\,d\theta.$ The electrostatic force on this element due to the central charge $q$ is radially outward and equals $dF = \frac{1}{4\pi\varepsilon_0}\,\frac{q\,dq}{R^2} = k\,\frac{q}{R^2}\,\frac{Q}{2\pi}\,d\theta,$ where $k=\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\text{N\,m}^2\text{/C}^2.$ The element is held \in equilibrium by the tension $T$ \in the ring. The two tension forces at its ends are tangential; their radial components add up to an inward force $2T\sin\!\left(\frac{d\theta}{2}\right)\;\approx\;T\,d\theta$ (because $d\theta$ is infinitesimal). Equating the outward electrostatic force and the inward force due to tension: $T\,d\theta = k\,\frac{qQ}{2\pi R^{2}}\,d\theta \;\;\Longrightarrow\;\; T = k\,\frac{qQ}{2\pi R^{2}}.$ In the given data Q = 2\pi\,$C, so the factors 2\pi$$$$ cancel, giving a very simple result: T = k,\frac{q}{R^{2}}.$Substituting the numerical values:$T = 9$\times$10^{9},\frac{30\times10^{-12}}{(0.30)^{2}} = 9$\times$10^{9}$\times$30$\times$10^{-12}\times $$$$\frac{1}{0.09}$$$$ \phantom{T} = 9$\times$30$\times$10^{-3}\times $$$$\frac{1}{0.09} = 270$\times$10^{-3}\times $$$$\frac{1}{0.09} = 0.27\times $$$$\frac{1}{0.09} = 3;$\text{N}$.$Hence the tension \in the metallic ring is$\mathbf{3;N}.$$

step 1: steady state capacitor is open → no current through AB So the diamond reduces to two parallel branches between top and bottom nodes • right branch: 6Ω + 4Ω = 10Ω • left branch: 1Ω + 2Ω = 3Ω step 2: equivalent of parallel $R_{eq}=10∥3=\frac{30}{13}\Omega$ Total current from 10 V source: $I=\frac{10}{30/13}=\frac{130}{30}=\frac{13}{3}\text{ A}$ step 3: current division current \in left branch (1\Omega + 2\Omega ): $I_L=I\cdot\frac{10}{10+3}=\frac{13}{3}\cdot\frac{10}{13}=\frac{10}{3}\text{ A}$ current \in right branch: IR=133−103=1 AI_R = \frac{13}{3} - \frac{10}{3} = 1 \text{ A}IR​=313​−310​=1 A step 4: find potentials take bottom node = 0 V, top = 10 V Point A (on left branch): drop across 1\Omega : $V_A=10-\frac{10}{3}=\frac{20}{3}$ Point B (on right branch): drop across 6\Omega : $V_B=10-6(1)=4$ step 5: capacitor voltage $V_{AB}=\frac{20}{3}-4=\frac{8}{3}$ step 6: charge $Q=CV=150\times \frac{8}{3}=400\mu C$

The magnetic field from a long, straight wire is given by $B = \frac{\mu_0 I}{2\pi r}$. Point P is midway between the wires, so the distance from each wire is $r = \frac{5.0 \text{ cm}}{2} = 2.5 \text{ cm} = 0.025 \text{ m}$. By the right-hand rule, the magnetic fields from the left wire (current up) and the right wire (current down) both point into the page at point P. The total magnetic field is therefore the sum of the magnitudes of the individual fields.

$B_P = B_1 + B_2 = 2 \times \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{\pi r}$

Substituting the given values:
$B_P = \frac{(4\pi \times 10^{-7} \text{ T m A}^{-1}) \times (10 \text{ A})}{\pi \times (0.025 \text{ m})}$
$B_P = \frac{4 \times 10^{-6}}{0.025} \text{ T} = 1.6 \times 10^{-4} \text{ T}$

Converting the result to microteslas ($1 \text{ T} = 10^6 \mu\text{T}$):
$B_P = 1.6 \times 10^{-4} \times 10^6 \mu\text{T} = 160 \mu\text{T}$

The mutual inductance is $M = 0.002$ H. The current \in the first coil is $i = i_0 \sin\omega t$ with $i_0 = 5$ A and $\omega = 50\pi$ rad/s. The induced EMF \in the second coil: $\varepsilon = -M\frac{di}{dt} = -M i_0 \omega \cos\omega t$ Maximum EMF: $\varepsilon_{max} = M i_0 \omega = 0.002 \times 5 \times 50\pi = 0.5\pi$ V $= \frac{\pi}{2}$ V So $\frac{\pi}{\alpha} = \frac{\pi}{2}$, giving $\alpha = 2$. The answer is $\boxed{2}$.

For near normal vision, the total apparent depth is the sum of the apparent depths contributed by each liquid layer. The apparent depth for each layer is its real depth divided by its refractive index.

The total apparent depth ($D_{app}$) is calculated as:
$D_{app} = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2}$
Here, the real depth of the bottom liquid is $h_1 = 6$ cm with refractive index $\mu_1 = 8/5$, and the real depth of the top liquid is $h_2 = 6$ cm with refractive index $\mu_2 = 3/2$.

Substituting these values into the formula:
$D_{app} = \frac{6}{8/5} + \frac{6}{3/2} = 6 \times \frac{5}{8} + 6 \times \frac{2}{3}$
$D_{app} = \frac{30}{8} + 4 = \frac{15}{4} + \frac{16}{4} = \frac{31}{4} \text{ cm}$
The problem states the apparent depth is $\frac{\alpha}{4}$ cm. By comparing this with our calculated value, we get:
$\frac{\alpha}{4} = \frac{31}{4} \implies \alpha = 31$

[CORRECT_OPTION: 31]

In nuclear fission: - Parent nucleus: $A = 236$, binding energy per nucleon = 7.6 MeV - Two daughter nuclei: each $A = 118$, binding energy per nucleon = 8.6 MeV Total binding energy before = $236 \times 7.6 = 1793.6$ MeV Total binding energy after = $2 \times 118 \times 8.6 = 236 \times 8.6 = 2029.6$ MeV Energy released = Final BE - Initial BE = $2029.6 - 1793.6 = 236$ MeV The answer is $\boxed{236}$ MeV.

Neodymium (Nd) has atomic number 60. It is a lanthanide element. The noble gas core is Xenon (Xe, Z=54). After Xe, the electron filling order is: 6s, then 4f. $60 - 54 = 6$ electrons to fill after Xe core. First, 6s is filled: 2 electrons \to $6s^2$ (4 remaining) Then, 4f is filled: 4 electrons \to $4f^4$ Electronic configuration: $[Xe]\,4f^4\,6s^2$. The correct answer is Option 1 : $[Xe]\,4f^4\,6s^2$.

The magnetic moment is given by $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. Higher $n$ gives higher magnetic moment. Counting unpaired electrons for each configuration: - $[Ar] \; 3d^7$: $\uparrow\downarrow \; \uparrow\downarrow \; \uparrow \; \uparrow \; \uparrow$ = 3 unpaired - $[Ar] \; 3d^8$: $\uparrow\downarrow \; \uparrow\downarrow \; \uparrow\downarrow \; \uparrow \; \uparrow$ = 2 unpaired - $[Ar] \; 3d^3$: $\uparrow \; \uparrow \; \uparrow \; \_ \; \_$ = 3 unpaired - $[Ar] \; 3d^6$: $\uparrow\downarrow \; \uparrow \; \uparrow \; \uparrow \; \uparrow$ = 4 unpaired $[Ar] \; 3d^6$ has the maximum number of unpaired electrons (4), giving the highest magnetic moment: $\mu = \sqrt{4(4+2)} = \sqrt{24} = 2\sqrt{6}$ BM The answer corresponds to Option (4).

Let us analyze each molecule for polarity: - $CCl_4$: Tetrahedral, symmetric, dipole moments cancel. Non-polar . - $CO_2$: Linear, symmetric, dipole moments cancel. Non-polar . - $CH_2=CH_2$: Planar, symmetric. Non-polar . - $CHCl_3$: Tetrahedral but asymmetric (one H, three Cl). Net dipole moment exists. Polar . The answer is $CHCl_3$, which corresponds to Option (4).

The strength of a Brønsted base is determined by the availability of its lone pair of electrons to accept a proton ($H^+$).

1. In aniline (A) and diphenylamine (B), the nitrogen lone pair is delocalized into the benzene ring(s) through resonance, which decreases its availability for protonation. Diphenylamine (B) is weaker than aniline (A) as the lone pair is delocalized over two rings.

2. In pyrrole (C), the nitrogen lone pair is an integral part of the aromatic $\pi$ system (6$\pi$ electrons). Protonating the nitrogen would destroy the aromaticity, making it an extremely weak base.

3. In pyrrolidine (D), the nitrogen is $sp^3$ hybridized, and its lone pair is localized in an $sp^3$ orbital. There is no resonance or aromaticity to delocalize this lone pair.

4. Since the lone pair in pyrrolidine (D) is highly localized and not involved in resonance, it is the most available for protonation, making it the strongest base among the given options.

Step 1: Analyze Statement (I)
Ammonium carbonate, $(NH_4)_2CO_3$, is a salt formed from a weak base, ammonium hydroxide ($NH_4OH$), and a weak acid, carbonic acid ($H_2CO_3$). To determine the nature of its aqueous solution, we compare the extent of hydrolysis of the cation ($NH_4^+$) and the anion ($CO_3^{2-}$).

The hydrolysis reactions are:
$NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+$
$CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$

The respective hydrolysis constants are related to the dissociation constants of the parent base and acid:
$K_h(\text{cation}) = K_a(NH_4^+) = \frac{K_w}{K_b(NH_4OH)} = \frac{10^{-14}}{1.8 \times 10^{-5}} \approx 5.6 \times 10^{-10}$
$K_h(\text{anion}) = K_b(CO_3^{2-}) = \frac{K_w}{K_{a2}(H_2CO_3)} = \frac{10^{-14}}{5.6 \times 10^{-11}} \approx 1.8 \times 10^{-4}$

Since $K_h(\text{anion}) > K_h(\text{cation})$, the production of $OH^-$ ions from carbonate hydrolysis is greater than the production of $H^+$ ions from ammonium hydrolysis. Therefore, the aqueous solution of ammonium carbonate is basic. Statement (I) is correct.

Step 2: Analyze Statement (II)
This statement correctly describes the principle of salt hydrolysis for salts of a weak acid and a weak base. The overall pH of the solution is determined by the relative strengths of the weak acid and weak base from which the salt is formed. This is quantitatively expressed by comparing their dissociation constants, $K_a$ and $K_b$. The pH is given by the formula $pH = 7 + \frac{1}{2}(pK_a - pK_b)$. If $K_b > K_a$ ($pK_b < pK_a$), the solution is basic. If $K_a > K_b$ ($pK_a < pK_b$), the solution is acidic. Thus, Statement (II) is correct.

Step 3: Conclude
Both Statement (I) and Statement (II) are correct. Statement (I) is a specific example of the general principle described in Statement (II).

Assertion (A): Melting point of Boron (2453 K) is unusually high in group 13 elements. This is true - Boron has the highest melting point among group 13 elements. Reason (R): Solid Boron has very strong crystalline lattice. This is true - Boron exists in an icosahedral crystalline structure with very strong covalent bonds, giving it a very high melting point. The reason correctly explains the assertion - the strong crystalline lattice of boron is why it has an unusually high melting point. Both (A) and (R) are correct and (R) is the correct explanation of (A). The answer corresponds to Option (2).