6 Apr 2024 Shift 1

Find percentage error in spring constant $k$ given 2% error \in time and 1% error \in mass. Relation between k, m, and T. $T = 2\pi\sqrt{\frac{m}{k}}$, so $k = \frac{4\pi^2 m}{T^2}$. Error propagation. $\frac{\Delta k}{k} = \frac{\Delta m}{m} + 2\frac{\Delta T}{T} = 1\% + 2(2\%) = 1\% + 4\% = 5\%$ Note: we take absolute values of errors for maximum percentage error regardless of sign. The correct answer is Option A : 5%.

Torque: Torque is given by force × distance. $\tau=r\times F$ Dimensions of force: $[F]=[MLT^{-2}]$ So, $[\tau]=[L]\cdot\$[MLT^{-2}]\$=[ML^2T^{-2}]$ So, Torque \to (IV) Magnetic field: From Lorentz force, $F=qvB$ $B=\frac{F}{qv}$ Now substitute dimensions: $[q]=[AT]$ $\quad[v]=[LT^{-1}]$ $[B]=\frac{MLT^{-2}}{(AT)(LT^{-1})}=[MT^{-2}A^{-1}]$ So, Magnetic field \to (III) Magnetic moment: Magnetic moment is given by: $\mu =IA$ $[\mu]=[A]\cdot\$[L^2]\$=[L^2A]$ So, Magnetic moment \to (II) Permeability of free space: Using relation: $B=\mu_0\frac{I}{L}$ $\mu_0=\frac{BL}{I}$ Substitute dimensions: $[B]=[MT^{-2}A^{-1}]$ $[\mu_0]=\frac{(MT^{-2}A^{-1})\cdot L}{A}=[MLT^{-2}A^{-2}]$ So, Permeability → (I) Final matching: (A)−(IV), (B)−(III), (C)−(II), (D)−(I)

A train starts from rest, accelerates uniformly to 80 km/h in time $t$, then moves at constant speed for time $3t$. We need the average speed. Calculate the distance during acceleration. During uniform acceleration from 0 to 80 km/h over time $t$, the average speed is $\frac{0 + 80}{2} = 40$ km/h. $d_1 = \text{average speed} \times \text{time} = 40t \text{ (\in km, with } t \text{ \in hours)}$ Calculate the distance during constant speed. $d_2 = 80 \times 3t = 240t \text{ km}$ Calculate the average speed. $\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{d_1 + d_2}{t + 3t} = \frac{40t + 240t}{4t} = \frac{280t}{4t} = 70 \text{ km/h}$ The correct answer is Option (4): 70 .

$a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{g}{\sqrt{2}}$. $\frac{m_2 - m_1}{m_1 + m_2} = \frac{1}{\sqrt{2}}$. Let $r = m_1/m_2$: $\frac{1-r}{1+r} = \frac{1}{\sqrt{2}}$. $\sqrt{2}(1-r) = 1+r \Rightarrow \sqrt{2} - \sqrt{2}r = 1 + r \Rightarrow r(1+\sqrt{2}) = \sqrt{2}-1 \Rightarrow r = \frac{\sqrt{2}-1}{\sqrt{2}+1}$. The correct answer is Option (2): $\frac{\sqrt{2}-1}{\sqrt{2}+1}$.

$KE_i = \frac{1}{2}(0.05)(100^2) = 250$ J. $KE_f = \frac{1}{2}(0.05)(40^2) = 40$ J. Loss = $250 - 40 = 210$ J. Percentage loss = $\frac{210}{250} \times 100 = 84\%$. The correct answer is Option (1): 84%.

All have same momentum $p$. $KE = \frac{p^2}{2m}$. Smallest mass has maximum KE. Particle A has mass $m/2$ (smallest). The correct answer is Option (2): A.

To just reach infinity with zero final speed, total mechanical energy at infinity should be zero. At Earth's surface: Potential energy is $U=-\frac{GMm}{R_E}$ Required kinetic energy = escape energy so that $K+U=0$ Thus $K=\frac{GMm}{R_E}$ Now use $g=\frac{GM}{R_E^2}$ So $GM=gR_E^2$ Substitute: $K=\frac{(gR_E^2)m}{R_E}$ $K=mgR_E$

At terminal velocity: Weight = Buoyant force + Viscous force. $mg = \frac{m\rho_0}{\rho}g + F_v$ (buoyant force = $V\rho_0 g = \frac{m}{\rho}\rho_0 g$). $F_v = mg - \frac{m\rho_0 g}{\rho} = mg\left(1 - \frac{\rho_0}{\rho}\right)$. The correct answer is Option (4): $mg\left(1 - \frac{\rho_0}{\rho}\right)$.

We need to find the ratio of r.m.s. speeds of helium to oxygen at the same temperature. Recall the r.m.s. speed formula. $v_{rms} = \sqrt{\frac{3RT}{M}}$ where $R$ is the gas constant, $T$ is absolute temperature, and $M$ is the molar mass. Calculate the ratio. At the same temperature, $R$ and $T$ are the same for both gases: $\frac{v_{He}}{v_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{He}}}$ Molar masses: $M_{He} = 4$ g/mol, $M_{O_2} = 32$ g/mol $\frac{v_{He}}{v_{O_2}} = \sqrt{\frac{32}{4}} = \sqrt{8} = 2\sqrt{2}$ The correct answer is Option (2): $2\sqrt{2}$ .

For a real gas, the relation between the molar specific heats is $C_P - C_V \;=\; T\Bigl(\frac{\partial P}{\partial T}\Bigr)_V \Bigl(\frac{\partial V}{\partial T}\Bigr)_P$ This identity is obtained from the first & second laws together with exact-differential criteria, and it is valid for one mole of any substance. The gas \in the question obeys the equation of state $P\,V^2 \;=\; R\,T \;\;-(1)$ Step 1: Find $\Bigl(\dfrac{\partial P}{\partial T}\Bigr)_V$. From $(1)$: $P \;=\; \dfrac{R\,T}{V^{2}}$. Keeping $V$ constant and differentiating with respect to $T$: $\Bigl(\frac{\partial P}{\partial T}\Bigr)_V = \frac{R}{V^{2}} \;\;-(2)$ Step 2: Find $\Bigl(\dfrac{\partial V}{\partial T}\Bigr)_P$. Rewrite $(1)$ to express $V$ \in terms of $T$ at fixed $P$: $V^{2} = \frac{R\,T}{P} \quad\Longrightarrow\quad 2V\,dV = \frac{R}{P}\,dT$ so $\Bigl(\frac{\partial V}{\partial T}\Bigr)_P = \frac{R}{2\,P\,V} \;\;-(3)$ Step 3: Compute $C_P - C_V$. Substitute $(2)$ and $(3)$ into the heat-capacity relation:

$

$ Step 4: Eliminate $P$ using the equation of state. From $(1)$: $P = \dfrac{R\,T}{V^{2}}$. Insert this \in $(4)$:

$

\begin{aligned} C_P - C_V &= \frac{T,R^{2}}{2,(\tfrac{R,T}{V^{2}}),V^{3}}$$6pt] &= \frac{T,R^{2}}{2,R,T,V}$$4pt] &= \frac{R}{2,V} \end{aligned}

$ Therefore $C_P = C_V + \frac{R}{2\,V}$ Among the given options this matches Option C. Answer: Option C

We need to find the electric field at any point on the surface of a thin spherical shell with uniform surface charge density $\sigma$ and radius $R$. The total charge on the shell is $Q = \sigma \cdot 4\pi R^2$. By Gauss's law, the electric field just outside the shell (at distance $R$ from center) is: $E_{\text{outside}} = \frac{Q}{4\pi\epsilon_0 R^2} = \frac{\sigma \cdot 4\pi R^2}{4\pi\epsilon_0 R^2} = \frac{\sigma}{\epsilon_0}$ The electric field just inside the shell is $E_{\text{inside}} = 0$ (no enclosed charge for a Gaussian surface inside). For a conducting shell or a charged surface, the electric field AT the surface is taken as the value just outside the surface, which is $\frac{\sigma}{\epsilon_0}$. This is the standard convention used \in JEE for the field "on the surface" of a spherical shell. The correct answer is Option B: $\sigma/\epsilon_0$.

given that potential difference between B and D is zero, the bridge is balanced so we use: left top / left bottom = right top / right bottom left top: 24 ∥ (12 + 12) = 24 ∥ 24 = 12 left bottom: 12 ∥ 12 = 6, then in series with x → 6 + x right top: 1 ∥ 1 = 0.5 right bottom: 0.5 so, 12 / (6 + x) = 0.5 / 0.5 = 1 12 = 6 + x x = 6 Ω

The magnetic field produced by a small current-carrying element is described by the Biot-Savart law. The magnitude of the magnetic field $dB$ is given by:
$dB = \frac{\mu_0}{4\pi} \frac{I (dl) \sin\theta}{r^2}$
Here, the current element is $\Delta \vec{l} = \Delta x \hat{i}$ and the point P is on the y-axis, so the position vector is $\vec{r} = y \hat{j}$. The angle $\theta$ between $\Delta \vec{l}$ and $\vec{r}$ is $90^\circ$, so $\sin(90^\circ) = 1$.
We are given $I = 10 \text{ A}$, $\Delta l = \Delta x = 1 \text{ cm} = 10^{-2} \text{ m}$, and $r = 0.5 \text{ m}$. We also know that $\frac{\mu_0}{4\pi} = 10^{-7} \text{ T}\cdot\text{m/A}$.
Substituting these values into the formula for the magnetic field $B$:
$B = (10^{-7} \text{ T}\cdot\text{m/A}) \frac{(10 \text{ A})(10^{-2} \text{ m})}{(0.5 \text{ m})^2}$
$B = 10^{-7} \frac{0.1}{0.25} \text{ T} = 10^{-7} \times 0.4 \text{ T} = 4 \times 10^{-8} \text{ T}$
The direction of the field, given by the cross product $\hat{i} \times \hat{j} = \hat{k}$, is along the positive z-axis.

We need to evaluate two statements about LCR circuits and resistive circuits. Analysis of Statement I: "In an LCR series circuit, current is maximum at resonance." At resonance in a series LCR circuit, the inductive reactance equals the capacitive reactance: $X_L = X_C$. The impedance becomes: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + 0} = R$. Since impedance is minimum (equal to R alone), the current $I = V/Z = V/R$ is maximum. Statement I is TRUE. Analysis of Statement II: "Current \in a purely resistive circuit can never be less than that \in a series LCR circuit when connected to same voltage source." For a purely resistive circuit: $I_R = V/R$ For a series LCR circuit: $I_{LCR} = V/Z = V/\sqrt{R^2 + (X_L - X_C)^2}$ Since $Z = \sqrt{R^2 + (X_L - X_C)^2} \geq R$ for all values of $X_L$ and $X_C$, we always have $I_R \geq I_{LCR}$. The equality holds only at resonance when $X_L = X_C$. Statement II is TRUE. The correct answer is Option (3): Both Statement I and Statement II are true .

$v = \frac{c}{\sqrt{\mu_r \epsilon_r}}$. Given $v = 1.5 \times 10^8$ m/s, $\mu_r = 2$. $\frac{c}{v} = \sqrt{\mu_r \epsilon_r} \Rightarrow \left(\frac{3 \times 10^8}{1.5 \times 10^8}\right)^2 = \mu_r \epsilon_r \Rightarrow 4 = 2\epsilon_r \Rightarrow \epsilon_r = 2$. The correct answer is Option (1): 2.

In a photoelectric experiment, light of energy 2.48 eV is incident on a photosensitive material, and the stopping potential is 0.5 V. We need to find the work function. Recall Einstein's photoelectric equation. $E_{photon} = \phi + KE_{max}$ where $\phi$ is the work function and $KE_{max} = eV_0$ (stopping potential \times electron charge). Express \in electron volts. Since the stopping potential is 0.5 V, the maximum kinetic energy is $eV_0 = 0.5$ eV. Solve for the work function. $\phi = E_{photon} - KE_{max} = 2.48 - 0.5 = 1.98 \text{ eV}$ The correct answer is Option (3): 1.98 eV .

We need to identify which phenomenon cannot be explained by the wave nature of light. A. Reflection: Explained by wave theory (Huygens' principle). Each point on a wavefront acts as a source of secondary wavelets, and the reflected wavefront follows the law of reflection. B. Diffraction: Explained by wave theory. Diffraction is the bending of waves around obstacles, a characteristic wave phenomenon described by Huygens-Fresnel principle. C. Photoelectric effect: This cannot be explained by the wave nature of light. The wave theory predicts that any frequency of light should eject electrons if given enough time and intensity, but experimentally, there is a threshold frequency below which no electrons are emitted regardless of intensity. Einstein explained this using the particle (photon) nature of light: each photon carries energy $h\nu$, and only photons with energy greater than the work function can eject electrons. D. Interference: Explained by wave theory. Constructive and destructive interference patterns arise from the superposition of waves. E. Polarization: Explained by wave theory. Polarization demonstrates that light is a transverse wave. The correct answer is Option (3): C only .

We need to find the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series for hydrogen. Recall the Rydberg formula: $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ For the Balmer series with $n_1 = 2$, the shortest wavelength (highest energy) occurs as $n_2 \to \infty$. Thus: $\frac{1}{\lambda_B} = R\left(\frac{1}{4} - 0\right) = \frac{R}{4} \implies \lambda_B = \frac{4}{R}$ For the Lyman series with $n_1 = 1$, the shortest wavelength occurs as $n_2 \to \infty$. Hence: $\frac{1}{\lambda_L} = R\left(1 - 0\right) = R \implies \lambda_L = \frac{1}{R}$ Therefore the ratio is given by: $\frac{\lambda_B}{\lambda_L} = \frac{4/R}{1/R} = \frac{4}{1} = 4:1$ The correct answer is Option (1): 4:1.

The given logic circuit has inputs A and B and output Y. The inputs A and B are connected to an AND gate, and input A is also connected to a NOT gate. The outputs of these two gates are then fed into an OR gate.

The Boolean expression for the output Y can be written as:
$Y = \bar{A} + (A \cdot B)$
Using the distributive law of Boolean algebra, $X + (Y \cdot Z) = (X+Y) \cdot (X+Z)$, we simplify the expression:
$Y = (\bar{A} + A) \cdot (\bar{A} + B)$
Since $\bar{A} + A = 1$, the expression simplifies to $Y = 1 \cdot (\bar{A} + B)$, or simply:
$Y = \bar{A} + B$
We can now construct the truth table for $Y = \bar{A} + B$:

• If A=0, B=0, then $Y = \bar{0} + 0 = 1 + 0 = 1$.
• If A=0, B=1, then $Y = \bar{0} + 1 = 1 + 1 = 1$.
• If A=1, B=0, then $Y = \bar{1} + 0 = 0 + 0 = 0$.
• If A=1, B=1, then $Y = \bar{1} + 1 = 0 + 1 = 1$.
  This set of outputs (1, 1, 0, 1) matches the truth table in option A.

We need to find the diameter of a wire measured using a screw gauge. Recall that the screw gauge reading is given by: $\text{Reading} = \text{MSR} + \text{CSR} \times \text{Least Count}$ where MSR = Main Scale Reading, CSR = Circular Scale Reading, and Least Count = Pitch / Number of divisions on circular scale. Since the pitch is 1 mm and the circular scale has 100 divisions, the least count is: $LC = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{1 \text{ mm}}{100} = 0.01 \text{ mm}$. Substituting \in the formula gives the diameter: $d = 1 + 42 \times 0.01 = 1 + 0.42 = 1.42 \text{ mm}$. Expressing 1.42 mm \in fractional form: $d = 1.42 = \frac{142}{100} = \frac{71}{50} \text{ mm}$. Therefore x = 71 . The correct answer is Option (3): 71 .

$\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$ (scalar triple product). $\vec{B} \times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix} = (12+3)\hat{i} - (-3+24)\hat{j} + (1+32)\hat{k} = 15\hat{i} - 21\hat{j} + 33\hat{k}$. $\vec{A} \cdot (\vec{B} \times \vec{C}) = -15x + (-6)(-21) + (-2)(33) = -15x + 126 - 66 = -15x + 60 = 0$. $x = 4$. The answer is 4.

Earth's angular momentum is conserved: $I\omega = const$. $I = \frac{2}{5}MR^2$. $I_1\omega_1 = I_2\omega_2 \Rightarrow R^2 \omega_1 = (3R/4)^2 \omega_2 \Rightarrow \omega_2 = \frac{16}{9}\omega_1$. $T_2 = \frac{2\pi}{\omega_2} = \frac{9}{16} T_1 = \frac{9}{16} \times 24 = 13.5$ hours = 13 hours 30 minutes. The answer is 13.

1000 small drops coalesce to 1 big drop. Volume conserved: $1000 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow R = 10r$. Surface energy: $E = \sigma \times 4\pi r^2 \times n$ for droplets, $E_{big} = \sigma \times 4\pi R^2$. Ratio = $\frac{1000 \times r^2}{R^2} = \frac{1000r^2}{100r^2} = 10 = \frac{10}{1}$. So $x = 1$. The answer is 1.

$v_{max} = A\omega = A \times \frac{2\pi}{T} = 0.06 \times \frac{2\pi}{3.14} = 0.06 \times 2 = 0.12$ m/s = 12 cm/s. The answer is 12.

The electric field ($E$) due to an infinitely long thin charged sheet with surface charge density $\sigma$ is $E = \frac{\sigma}{2\epsilon_0}$. Field points away from a positive charge ($+\sigma$) Field points towards a negative charge ($-\sigma$) Sheet 1 (at $x = -a$, charge $-\sigma$): Point $P$ is to the right of this sheet. Since it is negatively charged, the field points left. $\vec{E}_1 = -\frac{\sigma}{2\epsilon_0} \hat{i}$ Sheet 2 (at $x = a$, charge $-2\sigma$): Point $P$ is to the right of this sheet. Since it is negatively charged, the field points left. $\vec{E}_2 = -\frac{2\sigma}{2\epsilon_0} \hat{i} = -\frac{\sigma}{\epsilon_0} \hat{i}$ Sheet 3 (at $x = 3a$, charge $\sigma$): Point $P$ is to the left of this sheet. Since it is positively charged, the field points left (away from the sheet). $\vec{E}_3 = -\frac{\sigma}{2\epsilon_0} \hat{i}$ $\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3$ $\vec{E}_{net} = \left( -\frac{\sigma}{2\epsilon_0} - \frac{\sigma}{\epsilon_0} - \frac{\sigma}{2\epsilon_0} \right) \hat{i}$ $\vec{E}_{net} = \left( -\frac{2\sigma}{2\epsilon_0} - \frac{\sigma}{\epsilon_0} \right) \hat{i} = -\frac{2\sigma}{\epsilon_0} \hat{i}$ $|\vec{E}_{net}| = \frac{2\sigma}{\epsilon_0}$ $x = 2$

Volume conserved: $\pi r^2 l = \pi (r/2)^2 l' \Rightarrow l' = 4l$. $R' = \rho \frac{l'}{A'} = \rho \frac{4l}{\pi(r/2)^2} = \rho \frac{4l}{\pi r^2/4} = 16 \cdot \rho\frac{l}{\pi r^2} = 16R$. So $x = 16$. The answer is 16.

A circular coil with 200 turns, area $2.5 \times 10^{-4}$ m² carrying $100 \mu$A current is \in a uniform magnetic field of 1 T. We need the work to rotate it 90° from aligned to perpendicular orientation. The magnetic dipole moment is $M = NIA = 200 \times 100 \times 10^{-6} \times 2.5 \times 10^{-4}$ which gives $M = 200 \times 10^{-4} \times 2.5 \times 10^{-4} = 5 \times 10^{-6} \text{ A m}^2$. The potential energy of a magnetic dipole \in a field is $U = -MB\cos\theta$. Initially ($\theta = 0°$), $U_i = -MB\cos 0° = -MB$ and finally ($\theta = 90°$), $U_f = -MB\cos 90° = 0$. Therefore the work done is $W = U_f - U_i = 0 - (-MB) = MB$ which equals $W = 5 \times 10^{-6} \times 1 = 5 \times 10^{-6} \text{ J} = 5 \text{ }\mu\text{J}$. The answer is $\boxed{5}$ $\mu$J.

An inductor is connected first to DC (100 V, 5 A) and then to AC (peak 200 V, $X_L = 20\sqrt{3} \Omega$). Under DC the inductor acts as a pure resistance at zero frequency, so the resistance is $R = \frac{V_{DC}}{I_{DC}} = \frac{100}{5} = 20 \,\Omega$. Under AC the impedance is $Z = \sqrt{R^2 + X_L^2} = \sqrt{20^2 + (20\sqrt{3})^2} = \sqrt{400 + 1200} = \sqrt{1600} = 40 \,\Omega$. The RMS voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{200}{\sqrt{2}}$ V, so the RMS current is $I_{rms} = \frac{V_{rms}}{Z} = \frac{200}{\sqrt{2}\times 40} = \frac{200}{40\sqrt{2}} = \frac{5}{\sqrt{2}} \text{ A}$. Since power is dissipated only \in the resistance, $P = I_{rms}^2 \times R = \frac{25}{2} \times 20 = 250 \text{ W}$. Therefore the answer is $250$ W .

For minimum deviation: $\mu = \frac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}}$. Given $\delta_m = A$: $\sqrt{3} = \frac{\sin A}{\sin(A/2)} = \frac{2\sin(A/2)\cos(A/2)}{\sin(A/2)} = 2\cos(A/2)$. $\cos(A/2) = \frac{\sqrt{3}}{2} \Rightarrow A/2 = 30° \Rightarrow A = 60°$. The answer is 60.

$r_n = n^2 a_0$. Given $r = 8.48$ Å: $n^2 = 8.48/0.529 = 16 \Rightarrow n = 4$. $E_n = E/n^2 = E/16$. So $x = 16$. The answer is 16.

Molality = 3 m means 3 mol NaOH per 1 kg solvent. Mass of NaOH = 3 × 40 = 120 g. Total mass = 1000 + 120 = 1120 g. Volume = 1120/1.12 = 1000 mL = 1 L. Molarity = 3/1 = 3 M. The correct answer is Option (4): 3.0.

We need to identify which processes have negative electron affinity (i.e., the process is endothermic when an electron is added). Key Concept: Electron affinity (EA) is the energy change when an electron is added to a neutral gaseous atom. A negative electron affinity (in the convention where exothermic = positive EA) means the atom resists gaining an electron, i.e., energy must be supplied. Elements with negative EA (endothermic electron gain) typically have either: - Fully filled subshells (extra stability makes electron addition unfavourable) - Half-filled subshells (extra stability due to exchange energy) - Second electron additions (always endothermic due to electron-electron repulsion) A. Be $\rightarrow$ Be$^-$: Beryllium has configuration $1s^2 2s^2$ (completely filled 2s subshell). Adding an electron requires placing it \in the higher-energy 2p orbital against the nuclear charge shielded by the filled 2s. This process is endothermic (negative EA). ✓ B. N $\rightarrow$ N$^-$: Nitrogen has configuration $1s^2 2s^2 2p^3$ (half-filled 2p subshell). The half-filled 2p configuration is extra stable due to maximum exchange energy. Adding a fourth electron to the 2p shell would disrupt this stability and require pairing \in an already occupied orbital. This process is endothermic (negative EA). ✓ C. O $\rightarrow$ O$^{2-}$: This is a two-step process: O $\rightarrow$ O$^-$ $\rightarrow$ O$^{2-}$. While the first EA of oxygen is positive (exothermic), the second EA is always endothermic because adding an electron to an already negatively charged ion (O$^-$) requires overcoming strong electrostatic repulsion. The overall process of forming O$^{2-}$ is endothermic. ✓ D. Na $\rightarrow$ Na$^-$: Sodium ($1s^2 2s^2 2p^6 3s^1$) can accept an electron to complete its 3s subshell ($3s^2$). This process is exothermic (positive EA), similar to how hydrogen gains an electron. ✗ E. Al $\rightarrow$ Al$^-$: Aluminium ($1s^2 2s^2 2p^6 3s^2 3p^1$) can accept an electron into its 3p orbital. This process is exothermic (positive EA), as the 3p subshell is far from half-filled. ✗ Conclusion: Elements with negative electron affinity values are Be, N, and O (for the second electron addition). These correspond to A, B, and C. The correct answer is Option 2 : A, B and C only.

We need to identify which of the given materials is not a semiconductor. Analysing each material: Option 1: Silicon (Si) Silicon is one of the most widely used semiconductor materials. It has a band gap of approximately 1.1 eV at room temperature. It is the basis of modern electronics and integrated circuits. Silicon is a semiconductor. Option 2: Copper oxide (Cu$_2$O / CuO) Copper oxide is a p-type semiconductor. Historically, copper oxide rectifiers (based on Cu$_2$O) were among the first semiconductor devices used commercially. It has a band gap of about 1.2-2.0 eV depending on the form. Copper oxide is a semiconductor. Option 3: Germanium (Ge) Germanium is a classic semiconductor with a band gap of approximately 0.67 eV at room temperature. It was the first material used \in transistors before being largely replaced by silicon. Germanium is a semiconductor. Option 4: Graphite Graphite is an allotrope of carbon with a layered structure. In each layer, carbon atoms are sp$^2$ hybridised, and the unhybridised p-orbitals form a delocalised $\pi$-electron system extending over the entire layer. These delocalised electrons are free to move within the layers, making graphite a good electrical conductor (resistivity $\sim 3-60 \times 10^{-6}$ ohm-m). Graphite is classified as a semimetal or conductor , not a semiconductor. The correct answer is Option 4 : Graphite.

This question requires matching the type of orbital hybridization with its corresponding spatial orientation or molecular geometry. Let's analyze each pair based on Valence Bond Theory.

1. A. $sp^3$ Hybridization: One s orbital and three p orbitals combine to form four equivalent hybrid orbitals. These orbitals orient themselves to minimize repulsion, resulting in a Tetrahedral geometry with bond angles of $109.5^\circ$. Thus, A matches with III.

2. B. $dsp^2$ Hybridization: One d, one s, and two p orbitals combine to form four hybrid orbitals. These orbitals lie in a single plane, pointing towards the corners of a square, leading to a Square planar geometry. Thus, B matches with IV.

3. C. $sp^3d$ Hybridization: One s, three p, and one d orbital combine to form five hybrid orbitals. The lowest energy arrangement for these is a Trigonal bipyramidal geometry, with three orbitals in the equatorial plane and two in the axial positions. Thus, C matches with I.

4. D. $sp^3d^2$ Hybridization: One s, three p, and two d orbitals combine to form six equivalent hybrid orbitals. These point towards the vertices of an octahedron, resulting in an Octahedral geometry. Thus, D matches with II.

Combining these matches: A-III, B-IV, C-I, D-II. This corresponds to option B.

To determine the correct match, we analyze the molecular geometry of each species using VSEPR theory. The steric number (SN) is the sum of bond pairs (BP) and lone pairs (LP).

A. In $SF_4$, the central sulfur atom has 6 valence electrons. It forms 4 single bonds with fluorine and has 1 lone pair. $SN = 4(BP) + 1(LP) = 5$. The geometry is see-saw ($AX_4E_1$). So, A matches III.

B. In $BrF_3$, the central bromine atom has 7 valence electrons. It forms 3 single bonds with fluorine and has 2 lone pairs. $SN = 3(BP) + 2(LP) = 5$. The geometry is bent T-shape ($AX_3E_2$). So, B matches IV.

C. In $BrO_3^-$, the central bromine atom has 7 valence electrons. With the negative charge, we have a total of $7+1=8$ electrons to be arranged in pairs around Br after bonding. $SN = \frac{1}{2}(7 + 0 + 1) = 4$. It has 3 bond pairs and 1 lone pair ($AX_3E_1$). The geometry is pyramidal. So, C matches II.

D. In $NH_4^+$, the central nitrogen atom has 5 valence electrons. Due to the +1 charge, it has $5-1=4$ valence electrons for VSEPR. It forms 4 single bonds with hydrogen and has 0 lone pairs. $SN = 4(BP) + 0(LP) = 4$. The geometry is tetrahedral ($AX_4$). So, D matches I.

Combining these results gives the matching: A-III, B-IV, C-II, D-I.

Here is a concise, step-by-step explanation to match the molecules/species with their properties/shapes:

1. A. SO₂Cl₂ (Sulfuryl chloride): The central sulfur atom forms four bonds (two with oxygen, two with chlorine) and has zero lone pairs. The steric number is 4, which corresponds to a tetrahedral geometry according to VSEPR theory. Therefore, A matches with III.

2. B. NO (Nitric oxide): The total number of valence electrons is $5 (\text{from N}) + 6 (\text{from O}) = 11$. Since there is an odd number of total electrons, there must be an unpaired electron, making the molecule paramagnetic. Therefore, B matches with I.

3. C. NO₂⁻ (Nitrite ion): The total number of valence electrons is $5 (\text{from N}) + 2 \times 6 (\text{from O}) + 1 (\text{charge}) = 18$. With an even number of electrons, all electrons are paired, making the species diamagnetic. Therefore, C matches with II.

4. D. I₃⁻ (Triiodide ion): The central iodine atom is bonded to two other iodine atoms and has three lone pairs. The five electron pairs arrange in a trigonal bipyramidal geometry. To minimize repulsion, the three lone pairs occupy the equatorial positions, forcing the two bonded iodine atoms into axial positions, resulting in a linear shape. Therefore, D matches with IV.

Combining these findings, the correct matching is A-III, B-I, C-II, D-IV.

The reaction is $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$. The cylinder is filled with equal number of molecules of $H_2$, $I_2$, and $HI$ at $-20°C$ (253 K) and 1 atm. Since the number of moles of gaseous reactants equals the number of moles of gaseous products ($\Delta n_g = 0$), $K_p = K_c$ for this reaction. Let the number of moles of each gas be $n$. The partial pressures are all equal (since equal moles at same T, V): each gas has partial pressure $P/3$ where $P = 1$ atm. $K_p = \frac{(P_{HI})^2}{P_{H_2} \cdot P_{I_2}} = \frac{(P/3)^2}{(P/3)(P/3)} = \frac{P^2/9}{P^2/9} = 1$ So $K_p = 1 = x \times 10^{-1}$. This gives $x = 10$. The correct answer is Option B: 10.

Sulphonic acids are a class of organosulfur compounds with the general formula $R-SO_3H$, where R is an organic alkyl or aryl group. The part of the molecule responsible for its characteristic chemical properties is the functional group.

The sulphonic acid functional group consists of a central sulfur atom bonded to an organic group (R), two oxygen atoms via double bonds, and one hydroxyl ($-OH$) group. The structure is $R-S(=O)_2-OH$.

Counting the atoms in the functional group, we have one sulfur (S), three oxygen (O), and one hydrogen (H). This gives the condensed formula $-SO_3H$.

Comparing this with the given options:

• A. $-SO_4H$ is incorrect.
• C. The image provided shows a sulfinic acid group ($-SO_2H$).
• D. $-SO_2$ represents the sulfonyl group, which is only a part of the full functional group.

Thus, the correct representation for the sulphonic acid functional group is $-SO_3H$.