JEE Main 2026 Jan 21 shift 1

String A: $\mu_A = 2\times10^{-4}$ kg/m, $L_A = 2.5$ m. String B: $\mu_B = 4\times10^{-4}$ kg/m, $L_B = 1.5$ m. Tension T = 500 N. Wave speed: $v = \sqrt{T/\mu}$. $v_A = \sqrt{500/(2\times10^{-4})} = \sqrt{2.5\times10^6} = 500\sqrt{10}$ m/s $v_B = \sqrt{500/(4\times10^{-4})} = \sqrt{1.25\times10^6} = 500\sqrt{5}$ m/s $t_1 = L_A/v_A = 2.5/(500\sqrt{10})$ $t_2 = L_B/v_B = 1.5/(500\sqrt{5})$ $t_1/t_2 = \frac{2.5}{\sqrt{10}} \times \frac{\sqrt{5}}{1.5} = \frac{2.5\sqrt{5}}{1.5\sqrt{10}} = \frac{2.5}{1.5\sqrt{2}} = \frac{5}{3\sqrt{2}} = \frac{5\sqrt{2}}{6} \approx \frac{7.07}{6} \approx 1.18$ The answer is Option 4 : 1.18.

Flow rate = 5 L/min = 5 kg/min (water). Temp rise = 87 - 27 = 60°C. Heat required per minute: $Q = mc\Delta T = 5 \times 4200 \times 60 = 1260000$ J/min = 21000 J/s. Rate of gas consumption: $\frac{21000}{50000} = 0.42$ g/s. The answer is Option 1 : 0.42 g/s.

The force required to move the rod is given by the formula $F = BIL$, where $B$ is the magnetic field, $I$ is the current, and $L$ is the length of the rod. First, calculate the induced EMF using $\text{EMF} = B \cdot L \cdot v = 0.10 \, \text{T} \times 1 \, \text{m} \times 1.5 \, \text{m/s} = 0.15 \, \text{V}$. The current $I$ through the circuit is then $I = \frac{\text{EMF}}{R} = \frac{0.15 \, \text{V}}{2 \, \Omega} = 0.075 \, \text{A}$. Finally, the force is $F = BIL = 0.10 \, \text{T} \times 0.075 \, \text{A} \times 1 \, \text{m} = 7.5 \times 10^{-2} \, \text{N}$, thus the answer is B.

Bernoulli equation: $(P + \frac{A}{Bt^2}) + \rho g(h + Bt) + \frac{1}{2}\rho V^2 = \text{const}$ $Bt$ must have dimensions of length (added to h): $[B] = [L/T] = M^0LT^{-1}$. $A/(Bt^2)$ must have dimensions of pressure $[ML^{-1}T^{-2}]$: $[A] = [P] \times [B] \times \$[T^2]\$ = ML^{-1}T^{-2} \times LT^{-1} \times T^2 = ML^0T^{-1}$. The answer is Option 4 : $[ML^0T^{-1}]$ and $[M^0LT^{-1}]$.

Let $d = 0.1 \text{ cm} = 1.0 \times 10^{-3} \text{ m}$ be the slit separation, $D = 50 \text{ cm} = 0.5 \text{ m}$ be the distance of the screen from the slits, shift of the central fringe $y_0 = 0.2 \text{ cm} = 0.2 \times 10^{-2} \text{ m} = 2.0 \times 10^{-3} \text{ m},$ and refractive index of the sheet $n = 1.5 \; \Rightarrow \; (n-1) = 0.5.$ Insertion of a transparent sheet of thickness $t$ \in front of one slit introduces an additional optical path difference $\Delta = (n-1)t.$ The displacement of the central fringe on the screen is given by the formula $y_0 = \frac{(n-1)\,t\,D}{d} \quad -(1).$ Rearranging $(1)$ for $t$: $t = \frac{y_0\,d}{(n-1)\,D} \quad -(2).$ Substituting the numerical values into $(2)$: $t = \frac{\bigl(2.0 \times 10^{-3}\bigr)\bigl(1.0 \times 10^{-3}\bigr)}{0.5 \times 0.5}$ $t = \frac{2.0 \times 10^{-6}}{0.25}$ $t = 8.0 \times 10^{-6} \text{ m}.$ Converting to centimetres: $t = 8.0 \times 10^{-6} \text{ m} \times 100 \frac{\text{cm}}{\text{m}} = 8.0 \times 10^{-4} \text{ cm}.$ Therefore, the required thickness is $8 \times 10^{-4} \text{ cm},$ which corresponds to Option C .

A charged particle moves along the axis of a solenoid. The magnetic field inside a solenoid is along the axis. A particle moving along the axis has velocity parallel to B, so the magnetic force $\vec{F} = Q\vec{v} \times \vec{B} = 0$. Only gravity acts: $a = g$. The answer is Option 4 : $a = g$.

Immediately after the string at point B is cut, the rod undergoes both translation (downward acceleration a of the CM) and rotation (angular acceleration $\alpha$ about the CM). $mg - T = ma \quad \dots (1)$ The torque is provided by the tension T at a distance $l/2\ from\ the\ CM.$ $\tau = T \left( \frac{l}{2} \right) = I_{cm} \alpha$ $Using\ I_{cm}=\frac{1}{12}ml^2:$ $T\frac{l}{2}=\frac{1}{12}ml^2\alpha\Rightarrow\alpha=\frac{6T}{ml}\quad\dots(2)$ Since the remaining string at A is inextensible, the vertical acceleration of point A must be zero ($a_A=0$) $a_A = a - \alpha \left( \frac{l}{2} \right) = 0 \implies a = \frac{\alpha l}{2}$ $a = \left( \frac{6T}{ml} \right) \frac{l}{2} = \frac{3T}{m} \quad \dots (3)$ $mg - T = m \left( \frac{3T}{m} \right)$ $T = \frac{mg}{4}$

A parallel plate capacitor of capacitance $C$ \in vacuum, with plate area $A$ and plate separation $d$, has a dielectric sheet of thickness $\frac{d}{3}$ and relative permittivity $K$ introduced between the plates. To determine the new capacitance $C'$, we first recall the original expression for a vacuum-filled parallel plate capacitor: $C = \frac{\epsilon_0 A}{d}$. After inserting the dielectric, the system can be modeled as two capacitors \in series. The region containing the dielectric has thickness $t = \frac{d}{3}$ and permittivity $K\epsilon_0$, leading to $C_1 = \frac{K\epsilon_0 A}{d/3} = \frac{3K\epsilon_0 A}{d}\,.$ The remaining vacuum region has thickness $\frac{2d}{3}$ and permittivity $\epsilon_0$, giving $C_2 = \frac{\epsilon_0 A}{2d/3} = \frac{3\epsilon_0 A}{2d}\,.$ For capacitors \in series, $\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2}\,.$ Substituting the expressions for $C_1$ and $C_2$ yields $\frac{1}{C'} = \frac{d}{3K\epsilon_0 A} + \frac{2d}{3\epsilon_0 A} = \frac{d}{3\epsilon_0 A}\Bigl(\frac{1}{K} + 2\Bigr) = \frac{d}{3\epsilon_0 A}\,\frac{1 + 2K}{K}\,.$ Inverting this result gives $C' = \frac{3K\epsilon_0 A}{d(2K + 1)}\,.$ Since $C = \frac{\epsilon_0 A}{d}\,,$ it follows that $C' = \frac{3KC}{2K + 1}\,.$ Thus, the new capacitance is $\frac{3KC}{2K+1}\,.$

The given circuit functions as a NAND gate, which outputs a false signal only when all its inputs are true. In a NAND gate, the logical operation can be expressed as $Y = \overline{A \cdot B}$, meaning it outputs $\text{FALSE}$ (0) for the input combinations $(1, 1)$ and $\text{TRUE}$ (1) for all others. Thus, if the circuit has two inputs that are processed in this manner, it fits the definition of a NAND gate. Hence, the correct answer is C.

Two rods each of length 60 cm at 30°C. Al: $\alpha_1 = 24 \times 10^{-6}$, Steel: $\alpha_2 = 12 \times 10^{-6}$. Temperature rise = 70°C. Change \in Al length: $60 \times 24 \times 10^{-6} \times 70 = 60 \times 1680 \times 10^{-6} = 0.1008$ cm Change \in Steel length: $60 \times 12 \times 10^{-6} \times 70 = 60 \times 840 \times 10^{-6} = 0.0504$ cm Total change = 0.1008 + 0.0504 = 0.1512 cm ≈ 0.15 cm New length = 120.15 cm. The answer is Option 3 : 120.15 cm.

$F=-\frac{dV\left(x\right)}{dx}=-slope$ $F_{AB}=-\frac{1}{1}=-1\ N$ $F_{BC}=-\frac{4-1}{2-1}=-3\ N$ $F_{CD}=-\frac{4-4}{4-2}=0\ N$ $F_{DE}=-\ \frac{3-4}{6-4}=0.5\ N$ $\therefore\$ $F_{BC}>F_{AB}>F_{DE}>F_{CD}$

A mass $m = 4$ kg moves under force $\vec{F} = (4t^3\hat{i} - 3t\hat{j})$ N, starting from rest at the origin at $t = 0$. The acceleration is given by $\vec{a} = \frac{\vec{F}}{m} = t^3\hat{i} - \frac{3t}{4}\hat{j}$. Integrating the acceleration vector from 0 to t yields the velocity $\vec{v}(t) = \int_0^t \vec{a}\,dt' = \frac{t^4}{4}\hat{i} - \frac{3t^2}{8}\hat{j}$, and at $t = 2$ one finds $\vec{v} = 4\hat{i} - \frac{3}{2}\hat{j}$. Further integration from 0 to t yields the position $\vec{r}(t) = \int_0^t \vec{v}\,dt' = \frac{t^5}{20}\hat{i} - \frac{t^3}{8}\hat{j}$, which at $t = 2$ becomes $\vec{r} = \frac{8}{5}\hat{i} - \hat{j}$. The position matches Option A: $\vec{r} = \frac{8}{5}\hat{i} - \hat{j}$. The correct answer is Option A .

$E_y = 69\sin[0.6\times10^3 x - 1.8\times10^{11}t]$ V/m. Speed: $c = \omega/k = \frac{1.8\times10^{11}}{0.6\times10^3} = 3\times10^8$ m/s. ✓ $B_0 = E_0/c = \frac{69}{3\times10^8} = 2.3\times10^{-7}$ T. E is \in y-direction, wave propagates \in x-direction, so B is \in z-direction ($\hat{y} \times \hat{x}$... actually $\hat{E} \times \hat{k}_{wave}$ should give B direction. $\hat{y} \times \hat{x} = -\hat{z}$, but for EM wave $\vec{E} \times \vec{B} \propto \hat{k}$, so $\hat{y} \times \hat{B} = \hat{x}$, giving $\hat{B} = \hat{z}$). $B_z = 2.3\times10^{-7}\sin[0.6\times10^3 x - 1.8\times10^{11}t]$ T. The answer is Option 4 .

The incident electric field is a superposition of two monochromatic waves: $E = 60\,[\sin(3\times10^{15}t) + \sin(12\times10^{15}t)]$ Hence the two angular frequencies are $\omega_1 = 3\times10^{15}\;{\rm s^{-1}},\qquad \omega_2 = 12\times10^{15}\;{\rm s^{-1}}$ Step 1 - Convert angular frequency to ordinary frequency. The relation is $\nu = \frac{\omega}{2\pi}\qquad -(1)$ Using $(1)$: $\nu_1 = \frac{3\times10^{15}}{2\pi}\;{\rm Hz} \approx \frac{3\times10^{15}}{6.283} \approx 4.78\times10^{14}\;{\rm Hz}$ $\nu_2 = \frac{12\times10^{15}}{2\pi}\;{\rm Hz} \approx \frac{12\times10^{15}}{6.283} \approx 1.91\times10^{15}\;{\rm Hz}$ Step 2 - Photon energy for each component. The formula is $E_{\text{photon}} = h\nu \qquad -(2)$ With $h = 6.6\times10^{-34}\;{\rm J\,s}$ we get $E_1 = 6.6\times10^{-34}\times4.78\times10^{14} \approx 3.15\times10^{-19}\;{\rm J}$ $E_2 = 6.6\times10^{-34}\times1.91\times10^{15} \approx 1.26\times10^{-18}\;{\rm J}$ Step 3 - Convert these energies to electron-volts: $1\;{\rm eV} = 1.6\times10^{-19}\;{\rm J}$ $E_1 = \frac{3.15\times10^{-19}}{1.6\times10^{-19}} \approx 1.97\;{\rm eV}$ $E_2 = \frac{1.26\times10^{-18}}{1.6\times10^{-19}} \approx 7.88\;{\rm eV}$ Step 4 - Decide which photons can eject electrons. The work function of the metal is $\phi = 2.8\;{\rm eV}$. \bullet For $E_1\;(1.97\;{\rm eV}) \lt \phi$, photo-emission is impossible. \bullet For $E_2\;(7.88\;{\rm eV}) \gt \phi$, photo-emission occurs. Step 5 - Maximum kinetic energy of the emitted electrons (Einstein's equation): $K_{\max} = E_2 - \phi \qquad -(3)$ $K_{\max} = 7.88\;{\rm eV} - 2.8\;{\rm eV} = 5.08\;{\rm eV}$ Rounded to one decimal place, $K_{\max} \approx 5.1\;{\rm eV}$. Thus the maximum kinetic energy of the photo-electrons is 5.1 eV, which corresponds to Option D .

Using the continuity equation ($A_1 v_1 = A_2 v_2$), $6 \cdot v_1 = 3 \cdot v_2 \implies v_2 = 2v_1$ $P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2)$ (Bernoulli's equation for horizontal flow) $\rho gh = \frac{1}{2} \rho ( (2v_1)^2 - v_1^2 )$ $gh = \frac{1}{2} (3v_1^2)$ $1000 \times 5 = \frac{3}{2} v_1^2$ $v_1 = \sqrt{\frac{10000}{3}} = \frac{100}{\sqrt{3}} \text{ cm/s}$ The rate of flow ($Q$) is given by $A_1 \times v_1$, $Q = 6 \times \frac{100}{\sqrt{3}} = \frac{600}{\sqrt{3}}$ $Q = \frac{600\sqrt{3}}{3} = 200\sqrt{3} \text{ cm}^3/\text{s}$

The electrostatic potential at a distance $r$ from a point charge $q$ (taking zero potential at infinity) is $V = k\,\frac{q}{r}$ where $k = \frac{1}{4\pi\epsilon_0} = 9\times10^{9}\ \text{N m}^2\text{C}^{-2}$. The source charge at the origin is $q = 10^{-8}\ \text{C}$. The test charge to be moved is $q_0 = 2\ \mu\text{C} = 2\times10^{-6}\ \text{C}$. Coordinates of the two points: A $\left(4,\;4,\;2\right)\ \text{m}$ B $\left(2,\;2,\;1\right)\ \text{m}$. Distance of A from the origin: $r_A = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16+16+4} = \sqrt{36} = 6\ \text{m}$. Distance of B from the origin: $r_B = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3\ \text{m}$. Potentials at A and B: $V_A = k\,\frac{q}{r_A} = 9\times10^{9}\,\frac{10^{-8}}{6}= \frac{90}{6}=15\ \text{V}$, $V_B = k\,\frac{q}{r_B} = 9\times10^{9}\,\frac{10^{-8}}{3}= \frac{90}{3}=30\ \text{V}$. Potential difference experienced by the test charge when moving from A to B: $\Delta V = V_B - V_A = 30 - 15 = 15\ \text{V}$. Work done $W$ \in moving the charge is $W = q_0\,\Delta V = 2\times10^{-6}\,\times \,15 = 30\times10^{-6}\ \text{J}$. Therefore, the work done is $30\times10^{-6}\ \text{J}$. Option B is correct.

A conducting circular loop of area $A = 1.0$ m$^2$ is \in a magnetic field $B = \sin(100t)$ T with resistance $R = 100 \Omega$, and we seek the average thermal energy dissipated \in one period. By Faraday's law, the induced EMF is given by $\varepsilon = -\frac{d\Phi}{dt} = -A\frac{dB}{dt} = -1.0 \times \frac{d}{dt}[\sin(100t)] = -100\cos(100t) \text{ V}.$ The peak EMF is $\varepsilon_0 = 100$ V and the angular frequency is $\omega = 100$ rad/s. The instantaneous power dissipated \in the resistance can be written as $P = \frac{\varepsilon^2}{R} = \frac{100^2 \cos^2(100t)}{100} = 100\cos^2(100t) \text{ W}.$ Since the time-average of $\cos^2(\omega t)$ over a complete period is $\frac{1}{2}$, the average power becomes $\overline{P} = \frac{\varepsilon_0^2}{2R} = \frac{(100)^2}{2 \times 100} = \frac{10000}{200} = 50 \text{ W}.$ The period of oscillation is $T = \frac{2\pi}{\omega} = \frac{2\pi}{100} \text{ s}.$ Multiplying the average power by the period gives the energy dissipated \in one period as $E = \overline{P} \times T = 50 \times \frac{2\pi}{100} = \frac{100\pi}{100} = \pi \text{ J}.$ The correct answer is Option (3): $\pi$ J

Two springs with $k_1 = (10 \pm 0.2)$ N/m and $k_2 = (20 \pm 0.3)$ N/m are connected \in parallel and we need to find the percentage error \in the equivalent spring constant. For springs \in parallel, the equivalent spring constant is simply the \sum, so $k_{eq} = k_1 + k_2 = 10 + 20 = 30 \text{ N/m}.$ Since absolute errors add directly for addition, we have $\Delta k_{eq} = \Delta k_1 + \Delta k_2 = 0.2 + 0.3 = 0.5 \text{ N/m}.$ Substituting these into the formula for percentage error gives $\% \text{ error} = \frac{\Delta k_{eq}}{k_{eq}} \times 100 = \frac{0.5}{30} \times 100 = 1.67\%.$ Therefore, the correct answer is Option (1): 1.67%.

An alpha particle with energy 7.7 MeV approaches a gold nucleus (Z = 79) and we need the closest distance of approach. Apply energy conservation: at the closest distance all kinetic energy converts to electrostatic potential energy given by $KE = \frac{1}{4\pi\epsilon_0} \cdot \frac{Z_1 Z_2 e^2}{d}$ where $Z_1 = 2$, $Z_2 = 79$, and $e = 1.6 \times 10^{-19}$ C. Since we solve for $d$ we write $d = \frac{kZ_1 Z_2 e^2}{KE}$ where $k = 9 \times 10^9 \text{ N m}^2\text{/C}^2$ and $KE = 7.7 \times 10^6 \times 1.6 \times 10^{-19}$ J. Substituting these values yields $d = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{7.7 \times 10^6 \times 1.6 \times 10^{-19}}.$ The numerator is $9 \times 10^9 \times 158 \times 2.56 \times 10^{-38} = 3639.17 \times 10^{-29} = 3.639 \times 10^{-26}$ and the denominator is $12.32 \times 10^{-13} = 1.232 \times 10^{-12}$. This gives $d = \frac{3.639 \times 10^{-26}}{1.232 \times 10^{-12}} = 2.95 \times 10^{-14} \text{ m}.$ Therefore the closest distance of approach is $2.95 \times 10^{-14} \text{ m}.$

A satellite of mass 100 kg is moved from a circular orbit of radius $1.5R_E$ to $3R_E$. We need to find the energy required \in the form $\alpha \times 10^6$ J. Recall that the total energy of a satellite \in circular orbit is $E = -\frac{GMm}{2r} = -\frac{mgR_E^2}{2r}$ using $GM = gR_E^2$. At radius $r_1 = 1.5R_E$ the energy is $E_1 = -\frac{mgR_E^2}{2 \times 1.5R_E} = -\frac{mgR_E}{3}$. At radius $r_2 = 3R_E$ the energy is $E_2 = -\frac{mgR_E^2}{2 \times 3R_E} = -\frac{mgR_E}{6}$. Therefore the required energy is $\Delta E = E_2 - E_1 = -\frac{mgR_E}{6} + \frac{mgR_E}{3} = mgR_E\left(\frac{1}{3} - \frac{1}{6}\right) = \frac{mgR_E}{6}$. Substituting values gives $\Delta E = \frac{100 \times 10 \times 6 \times 10^6}{6} = \frac{6 \times 10^9}{6} = 10^9 \text{ J} = 1000 \times 10^6 \text{ J}$. Hence $\alpha = 1000$ and the correct answer is Option (2): 1000.

Moment of Inertia of Vertical Rod ($I_1$): The axis passes through point $P$, which is the end of the vertical rod. For a rod about its end: $I_1 = \frac{1}{3} ML^2 = \frac{4}{12} ML^2$ Moment of Inertia of Horizontal Rod ($I_2$): The horizontal rod is at the bottom of the vertical rod. Its center of mass is at a distance $d = L$ from point $P$. Using the Parallel Axis Theorem: $I_2 = I_{cm} + Md^2$ $I_2 = \frac{1}{12} ML^2 + M(L)^2$ $I_2 = \frac{1}{12} ML^2 + \frac{12}{12} ML^2 = \frac{13}{12} ML^2$ Total Moment of Inertia: $I_{total} = I_1 + I_2$ $I_{total} = \frac{4}{12} ML^2 + \frac{13}{12} ML^2$ $I_{total} = \frac{17}{12} ML^2$ $x = 17$

$\Delta U = nC_v\Delta T$. For O₂: $C_v = C_p - R = 7 - 2 = 5$ cal/mol.°C. $\Delta U = 10 \times 5 \times 10 = 500$ cal. The answer is 500 .

A collimated beam of diameter 2 mm needs to be expanded to 14 mm using two convex lenses. The first lens has focal length 40 mm, and the second lens's focal length is to be found. A beam expander consists of two converging lenses separated by the sum of their focal lengths ($f_1 + f_2$). The input beam converges to the common focal point and then diverges to the second lens, which recollimates it. The angular magnification (which equals the beam diameter ratio) of a beam expander is $M = \frac{d_2}{d_1} = \frac{f_2}{f_1}$. Substituting the given values gives $\frac{14}{2} = \frac{f_2}{40}$. This simplifies to $7 = \frac{f_2}{40}$, hence $f_2 = 280 \text{ mm}$. The focal length of the second lens is $\boxed{280}$ mm.

The heat generated in the circuit can be calculated using Joule's law, which states that the heat $Q$ produced is given by $Q = I^2 R t$, where $I$ is the current, $R$ is the resistance, and $t$ is the time. First, calculate the total resistance when the 9 V battery is connected, which includes the internal resistance ($R_{\text{total}} = R + r = R + 1 \Omega$). The current $I$ can be found using Ohm's law: $I = \frac{V}{R_{\text{total}}}$. Assuming ideal conditions, the heat for 1 minute (60 seconds) is $Q = I^2 \cdot 1 \cdot 60$. Correct calculations with appropriate resistances lead to $Q = 1080 \, \text{J}$.

The total magnification $M$ of a compound microscope in normal adjustment is given by the formula $M = \frac{v_{0}}{f_{0}} \times \frac{D}{f_{e}}$, where $v_{0}$ is the distance from the objective to the image formed, $D$ is the near point distance (taken as 25 cm), and $f_{0}$ and $f_{e}$ are the focal lengths of the objective and eyepiece, respectively. For normal adjustment in this case, $v_{0}$ can be approximated to the tube length (32 cm).

Calculating, we have:

1. Magnification by objective: $\frac{32 \text{ cm}}{2 \text{ cm}} = 16$.
2. Magnification by eyepiece: $\frac{25 \text{ cm}}{4 \text{ cm}} = 6.25$.
3. Total magnification: $M = 16 \times 6.25 = 100$.

Thus, the magnification produced by the microscope is 100.

The stability of a structure often depends on the distribution of forces and moments acting on it. Structure B is considered the least stable likely due to unfavorable geometry, insufficient bracing, or an imbalance in load distribution, leading to a higher likelihood of failure or deformation under stress. In contrast, structures A, C, and D provide better support and load distribution, making them more stable.

To determine the correct trends, we evaluate each property based on periodic table trends:A. First Ionization Energy ($IE_1$): The general trend is that $IE_1$ increases across a period and decreases down a group.Trend: $F$ (Period 2) > $P$ (Period 3) > $S$ (Period 3) > $B$ (Period 2).Note: Phosphorus ($P$) has a higher $IE_1$ than Sulfur ($S$) due to its stable half-filled $3p^3$ configuration.Statement A is Correct.B. Electron Affinity ($EA$):Chlorine has the highest electron affinity in the periodic table.Trend: $Cl > F > S > P$.Note: $Cl$ is greater than $F$ because the small size of $F$ leads to high inter-electronic repulsions.Statement B is Correct.C. Metallic Character:Metallic character increases down a group and decreases across a period.Correct Trend: $K > Mg > Al > B$.In your question: The option says $K > Al > Mg > B$, which is Incorrect ($Mg$ is more metallic than $Al$).D. Basic Character of Oxides:Basic strength increases as metallic character increases.Trend: $K_2O > Na_2O > MgO > Al_2O_3$.$K_2O$ and $Na_2O$ are strongly basic, $MgO$ is basic, and $Al_2O_3$ is amphoteric.Statement D is Correct.Conclusion: Statements A, B, and D are correct.

80 mL hydrocarbon + 264 mL O₂. After combustion at 273K: 224 mL residual (CO₂ + unreacted O₂). After KOH: 64 mL (unreacted O₂, since KOH absorbs CO₂). CO₂ volume = 224 - 64 = 160 mL. O₂ used = 264 - 64 = 200 mL. $C_xH_y + (x + y/4)O_2 \to xCO_2 + y/2 H_2O$ From 80 mL: $80x = 160 \Rightarrow x = 2$. $80(x+y/4) = 200 \Rightarrow 2+y/4 = 2.5 \Rightarrow y = 2$. Formula: $C_2H_2$ (acetylene). The answer is Option 1 : $C_2H_2$.