JEE Main 2026 Jan 28 shift 2

This circuit is a diode-resistor logic gate.
If A=0V and B=0V, both diodes conduct, pulling C close to 0V.
If A=5V and B=0V, $D_1$ is reverse biased, $D_2$ conducts, C is close to 0V.
If A=0V and B=5V, $D_2$ is reverse biased, $D_1$ conducts, C is close to 0V.
If A=5V and B=5V, both diodes are reverse biased, and C is pulled up to 5V by $V_{dc}$ through R.
This truth table (0,0->0; 0,1->0; 1,0->0; 1,1->1) corresponds to an AND gate.

The plane is accelerating left with acceleration a₀. So in the plane's frame, a pseudo force acts on the block towards the right with magnitude m a₀. Now resolve forces along the incline. Component of gravity along incline (down the plane): g sinθ Component of pseudo acceleration along incline: The pseudo acceleration is horizontal to the right. Its component along the incline (down the plane) is: $a₀\cos\theta$ So total effective acceleration of the block along the incline becomes: $a=g\sin\theta +a₀\cos\theta$ Now find the distance the block has to travel along the incline. Given base = L Incline length = $L/\cos\theta$ So displacement along the plane: $s=L/\cos\theta$ Initial velocity = 0 Use equation of motion: $s=(1/2)at^2$ Substitute: $L/\cos\theta =(1/2)(g\sin\theta +a₀\cos\theta )t^2$ Solve for t: $t=\sqrt{\ \frac{2L}{(\cos\theta (g\sin\theta +a₀\cos\theta ))}}$ Now simplify the denominator: $\cos\theta ·g\sin\theta =(g/2)\sin2\theta$ $\cos^2\theta =(1+\cos2\theta )/2$ So denominator becomes: $(g/2)\sin2\theta +(a₀/2)(1+\cos2\theta )$ Factor 1/2 out: $=(1/2)[g\sin2\theta +a₀(1+\cos2\theta )]$ So finally: $t=\sqrt{\ \frac{4L}{(g\sin2\theta +a₀(1+\cos2\theta ))}}$

Find the number of photons emitted per second by a 6 mW laser at 663 nm. Power $P = 6$ mW $= 6 \times 10^{-3}$ W, $\lambda = 663$ nm, $h = 6.63 \times 10^{-34}$ J.s, $c = 3 \times 10^8$ m/s. $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{663 \times 10^{-9}}$ $= \frac{19.89 \times 10^{-26}}{663 \times 10^{-9}} = \frac{19.89}{663} \times 10^{-17}$ $= 0.03 \times 10^{-17} = 3 \times 10^{-19}$ J $n = \frac{P}{E} = \frac{6 \times 10^{-3}}{3 \times 10^{-19}} = 2 \times 10^{16}$ The correct answer is Option 4 : $2 \times 10^{16}$.

This is a two-mass spring system, so oscillation occurs with reduced mass. For two masses connected by a spring, $\omega=\sqrt{\frac{k}{\mu}}$ where reduced mass $\mu=\frac{m_1m_2}{m_1+m_2}$ Given $m_1=1kg$ $m_2=0.2kg$ So $\mu=\frac{(1)(0.2)}{1+0.2}$ $=\frac{0.2}{1.2}$ $=\frac{1}{6}$ Now $\omega=\sqrt{\frac{150}{1/6}}$ $=\sqrt{900}$ $=30\ \text{rad/s}$

Let the voltage source be $V_S = 40 \text{ V}$.
Initially, $R_1=R_2=R_3=R_4=R$.
After heating, $R_3' = R_3 + 0.1 R_3 = 1.1 R$.
The voltage $V_a$ is the voltage across $R_2$: $V_a = V_S \frac{R_2}{R_1+R_2} = 40 \frac{R}{R+R} = 40 \frac{R}{2R} = 20 \text{ V}$.
The voltage $V_b$ is the voltage across $R_4$: $V_b = V_S \frac{R_4}{R_3'+R_4} = 40 \frac{R}{1.1R+R} = 40 \frac{R}{2.1R} = \frac{40}{2.1} = 19.0476 \text{ V}$.
The potential difference $(V_a - V_b) = 20 - 19.0476 = 0.9524 \text{ V}$.

We need to identify which statements about capacitors and related concepts are correct. Statement A: "Effective capacitance of a series combination is always smaller than the smallest capacitance." For series: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots > \frac{1}{C_{\min}}$ Therefore $C_{eq} < C_{\min}$. This is TRUE . Statement B: "When a dielectric is placed between charged plates, displacement of charges cannot occur due to insulation property." This is FALSE . Even though a dielectric is an insulator, the electric field causes polarization - slight displacement of bound charges within molecules, creating induced surface charges. This is how dielectrics reduce the effective field. Statement C: "Increasing area or decreasing thickness of dielectric increases capacitance." $C = \frac{\epsilon_0 \epsilon_r A}{d}$. Increasing $A$ or decreasing $d$ increases $C$. This is TRUE . Statement D: "For a point charge, concentric spherical shells centered at the charge are equipotential surfaces." The potential due to a point charge is $V = \frac{kq}{r}$, which depends only on $r$. All points at the same distance from the charge have the same potential. This is TRUE . Conclusion: Statements A, C, and D are correct. The correct answer is Option 4 : A, C and D Only.

A particle starts at $t = 0$ with position $x(t) = 4t^3 - 3t$. Determine which statements are correct. $v(t) = x'(t) = 12t^2 - 3$ $a(t) = x''(t) = 24t$ $12t^2 - 3 = 0 \implies t^2 = 1/4 \implies t = 1/2$ (taking $t > 0$) Statement A: "The particle returns to the origin 0.866 units later." $x(t) = 0$: $4t^3 - 3t = 0 \implies t(4t^2 - 3) = 0 \implies t = 0$ or $t = \sqrt{3}/2 \approx 0.866$. The particle returns to origin at $t = \sqrt{3}/2 \approx 0.866$. TRUE . Statement B: "The particle is 1 unit away from origin at its turning point." $x(1/2) = 4(1/8) - 3(1/2) = 1/2 - 3/2 = -1$. Distance from origin = 1. TRUE . Statement C: "Acceleration is non-negative." $a(t) = 24t \geq 0$ for $t \geq 0$. TRUE . Statement D: "The particle is 0.5 units away at its turning point." The distance is 1, not 0.5. FALSE . Statement E: "Particle never turns back as acceleration is non-negative." Despite non-negative acceleration, the initial velocity $v(0) = -3 < 0$, so the particle moves \in the negative direction first, then turns at $t = 1/2$. FALSE . Correct statements: A, B, C. The correct answer is Option 2 : A, B, C Only.

The speed of a longitudinal wave in a solid is given by the formula: $v = \sqrt{\frac{Y}{\rho}}$ where $Y$ is Young's modulus and $\rho$ is the density. Given initial speed $v = 400 \, \text{m/s}$. Density increases by 0.5%, so new density $\rho' = \rho \left(1 + \frac{0.5}{100}\right) = 1.005\rho$. Young's modulus increases by 1%, so new Young's modulus $Y' = Y \left(1 + \frac{1}{100}\right) = 1.01Y$. The new speed $v'$ is: $v' = \sqrt{\frac{Y'}{\rho'}} = \sqrt{\frac{1.01Y}{1.005\rho}} = \sqrt{\frac{1.01}{1.005}} \times \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{1.01}{1.005}} \times 400$ Compute the fraction $\frac{1.01}{1.005}$: $\frac{1.01}{1.005} = \frac{1010}{1005} = \frac{202}{201} \approx 1.004975$ Now, $\sqrt{1.004975}$ can be approximated using $\sqrt{1+x} \approx 1 + \frac{x}{2}$ for small $x$, where $x = 0.004975$: $\sqrt{1.004975} \approx 1 + \frac{0.004975}{2} = 1 + 0.0024875 = 1.0024875$ Thus, $v' \approx 1.0024875 \times 400 = 400.995 \approx 401 \, \text{m/s}$ Alternatively, using the formula for fractional change: $\frac{dv}{v} = \frac{1}{2} \frac{dY}{Y} - \frac{1}{2} \frac{d\rho}{\rho}$ Given $\frac{dY}{Y} = 0.01$ and $\frac{d\rho}{\rho} = 0.005$, $\frac{dv}{v} = \frac{1}{2} (0.01) - \frac{1}{2} (0.005) = 0.005 - 0.0025 = 0.0025$ So, $v' = v (1 + 0.0025) = 400 \times 1.0025 = 401 \, \text{m/s}$. Therefore, the speed changes approximately to $401 \, \text{m/s}$. The correct option is C. 401.

We need to evaluate each statement about electrostatics and identify the correct ones. Statement A: "Electrostatic field lines form closed loops." Electrostatic field lines originate from positive charges and terminate at negative charges. They do not form closed loops. Closed loops are characteristic of magnetic field lines. Hence, statement A is false . Statement B: "The electric field lines point radially outward when charge is greater than zero." For a positive point charge (charge > 0), electric field lines are directed radially outward. For a negative charge, they point inward. The statement specifies charge greater than zero, so it is true. Hence, statement B is true . Statement C: "The Gauss's Law is valid only for inverse-square force." Gauss's Law states that the flux through a closed surface is proportional to the enclosed charge. This law relies on the inverse-square nature of the electrostatic force (Coulomb's law). If the force were not inverse-square (e.g., proportional to $1/r^3$), Gauss's Law would not hold. Therefore, Gauss's Law is valid specifically for inverse-square forces. Hence, statement C is true . Statement D: "The work done in moving a charged particle in a static electric field around a closed path is zero." The electrostatic field is conservative. In a conservative field, the work done around any closed path is zero. Hence, statement D is true . Statement E: "The motion of a particle under Coulomb's force must take place in a plane." Coulomb's force is a central force (directed along the line joining the particles). For motion under a central force, the angular momentum is conserved, and the motion is confined to a plane perpendicular to the angular momentum vector. Hence, statement E is true . Summary of evaluations: - A: False - B: True - C: True - D: True - E: True The correct statements are B, C, D, and E. Now, comparing with the options: - Option A: A, C, E Only → Incorrect (includes false statement A) - Option B: A, B, D, E Only → Incorrect (includes false statement A) - Option C: B, C, D, E Only → Correct - Option D: A, B, C, D Only → Incorrect (includes false statement A) Therefore, the correct answer is Option C .

Voltage itself is not an absolute measurable quantity; only potential difference between two points can be measured. A voltmeter always measures the difference in potential between two points, not the absolute potential at a single point. Hence, "voltage difference" is measurable, but "voltage" (absolute potential) is not directly measurable. Therefore, the correct answer is voltage.

The spring constant is given by the formula $k = \frac{4\pi^2 m}{T^2}$, derived from $T = 2\pi \sqrt{\frac{k}{m}}$. The percentage error \in $k$ is calculated using the maximum possible relative error, which is the \sum of the relative errors \in mass and twice the relative error \in time period, since $k \propto m T^{-2}$. Thus, $\frac{\Delta k}{k} = \left| \frac{\Delta m}{m} \right| + \left| 2 \frac{\Delta T}{T} \right|$ The percentage error is then $\frac{\Delta k}{k} \times 100\%$. Given: Mass $m = 10$ g, with an accuracy of $\Delta m = 10$ mg. Convert to consistent units: $m = 10$ g = 10000 mg, so $\Delta m = 10$ mg. Thus, $\frac{\Delta m}{m} = \frac{10}{10000} = 0.001$ Time for 50 oscillations is $t = 60$ s, measured with a watch of resolution 2 s. The resolution indicates the least count, so the absolute error \in time measurement is $\Delta t = 2$ s. The time period $T$ is calculated as: $T = \frac{t}{n} = \frac{60}{50} = 1.2 \text{ s}$ Since $T = \frac{t}{n}$ and $n = 50$ is exact, the relative error \in $T$ is the same as \in $t$: $\frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{2}{60} = \frac{1}{30}$ Now, substitute into the error formula: $\frac{\Delta k}{k} = \left| 0.001 \right| + \left| 2 \times \frac{1}{30} \right| = 0.001 + \frac{2}{30} = 0.001 + \frac{1}{15}$ Compute $\frac{1}{15} = 0.0666667$, so: $\frac{\Delta k}{k} = 0.001 + 0.0666667 = 0.0676667$ Percentage error: $0.0676667 \times 100 = 6.76667\%$ Rounding to two decimal places gives 6.77%, but the closest option is 6.76%. Using fractions: $\frac{\Delta k}{k} = \frac{1}{1000} + \frac{1}{15} = \frac{15 + 1000}{15000} = \frac{1015}{15000} = \frac{203}{3000}$ Percentage error: $\frac{203}{3000} \times 100 = \frac{203}{30} = 6.7666...\%$ This value matches option D, 6.76%. Thus, the percentage error in the determination of the spring constant is 6.76%.

A long cylindrical conductor carries current uniformly distributed over its cross-section. We need to determine where the magnetic field is minimum. For a long cylindrical conductor of radius $R$ carrying total current $I$: Inside the conductor ($r < R$): $B(2\pi r) = \mu_0 \cdot \frac{I}{\pi R^2} \cdot \pi r^2 = \mu_0 I \frac{r^2}{R^2}$ $B = \frac{\mu_0 I r}{2\pi R^2}$ The field increases linearly with $r$ inside the conductor. Outside the conductor ($r > R$): $B = \frac{\mu_0 I}{2\pi r}$ The field decreases as $1/r$ outside. At the axis ($r = 0$): $B = 0$. This is the minimum. At the surface ($r = R$): $B = \frac{\mu_0 I}{2\pi R}$. This is the maximum. Statement A: Maximum at ends, minimum at midpoint - FALSE (this describes the axial field, not radial). Statement B: Maximum at axis - FALSE (B = 0 at axis). Statement C: Minimum at surface - FALSE (maximum at surface). Statement D: Minimum at axis - TRUE ($B = 0$). Statement E: Same at all points - FALSE. The correct answer is Option 3 : D Only.

When the position vector $\vec{r}$ changes sign to $-\vec{r}$, we need to find which vector does NOT flip. Under $\vec{r} \to -\vec{r}$ (parity transformation / spatial inversion), each position coordinate changes sign: $x \to -x, y \to -y, z \to -z$. Velocity: $\vec{v} = \frac{d\vec{r}}{dt}$. Since $\vec{r} \to -\vec{r}$, we get $\vec{v} \to -\vec{v}$. Velocity flips . Linear momentum: $\vec{p} = m\vec{v}$. Since $\vec{v} \to -\vec{v}$, we get $\vec{p} \to -\vec{p}$. Linear momentum flips . Acceleration: $\vec{a} = \frac{d\vec{v}}{dt}$. Since $\vec{v} \to -\vec{v}$, we get $\vec{a} \to -\vec{a}$. Acceleration flips . Angular momentum: $\vec{L} = \vec{r} \times \vec{p}$. Since $\vec{r} \to -\vec{r}$ and $\vec{p} \to -\vec{p}$: $\vec{L} \to (-\vec{r}) \times (-\vec{p}) = \vec{r} \times \vec{p} = \vec{L}$ Angular momentum does NOT flip . It is an axial vector (pseudovector). Angular momentum remains unchanged under $\vec{r} \to -\vec{r}$. The answer is Option A: Angular momentum.

Match the physical quantities with their dimensional formulas. A. Coefficient of viscosity: Viscosity $\eta$ has units Pa.s = kg/(m.s). Dimensional formula: $[ML^{-1}T^{-1}]$ = IV B. Surface tension: Surface tension has units N/m = kg/s$^2$. Dimensional formula: $[ML^0T^{-2}] = [MT^{-2}]$ = III C. Pressure: Pressure has units Pa = N/m$^2$ = kg/(m.s$^2$). Dimensional formula: $[ML^{-1}T^{-2}]$ = I D. Surface energy: Surface energy has units J = kg.m$^2$/s$^2$. Dimensional formula: $[ML^2T^{-2}]$ = II Matching: A-IV, B-III, C-I, D-II The correct answer is Option 3 : A-IV, B-III, C-I, D-II.

Given $\vec{E} = 54\sin(kz - \omega t)\hat{j}$ V/m. Find $\vec{B}$. From the argument $(kz - \omega t)$, the wave propagates \in the $+z$ direction: $\hat{k}$. For an EM wave, $\vec{E}$, $\vec{B}$, and the direction of propagation are mutually perpendicular, and $\hat{E} \times \hat{B} = \hat{k}_{\text{propagation}}$. $\hat{j} \times \hat{B} = \hat{k}$. Since $\hat{j} \times (-\hat{i}) = \hat{k}$, we get $\hat{B} = -\hat{i}$. $B_0 = \frac{E_0}{c} = \frac{54}{3 \times 10^8} = 1.8 \times 10^{-7} \text{ T}$ $\vec{B} = -1.8 \times 10^{-7}\sin(kz - \omega t)\hat{i} \text{ T}$ The correct answer is Option (3): $-1.8 \times 10^{-7}\sin(kz - \omega t)\hat{i}$ T .

A nucleus with mass number $\alpha$ has radius $R_\alpha$ and another with mass number $\beta$ has radius $R_\beta$. Given $\beta = 8\alpha$, we apply the nuclear radius formula $R = R_0 A^{1/3}$, where $A$ is the mass number and $R_0$ is a constant. Substituting for the two nuclei yields $\frac{R_\alpha}{R_\beta} = \frac{R_0 \alpha^{1/3}}{R_0 \beta^{1/3}} = \left(\frac{\alpha}{\beta}\right)^{1/3}$. Using $\beta = 8\alpha$ gives $\left(\frac{\alpha}{8\alpha}\right)^{1/3} = \left(\frac{1}{8}\right)^{1/3} = \frac{1}{2} = 0.5$. The correct answer is Option 3: 0.5.

The biconvex lens is a combination of two plano-convex lenses.
For the left plano-convex lens: $1/f_1 = (\mu_1 - 1)(1/R_1 - 1/\infty ) = (1.5 - 1)(1/15) = 0.5/15 = 1/30$. So, $f_1 = 30$ cm.
For the right plano-convex lens: $1/f_2 = (\mu_2 - 1)(1/\infty - 1/(-R_2)) = (1.2 - 1)(1/12) = 0.2/12 = 1/60$. So, $f_2 = 60$ cm.
The focal length of the combined lens is $1/F = 1/f_1 + 1/f_2 = 1/30 + 1/60 = 3/60 = 1/20$. So, $F = 20$ cm.
Using the lens formula $1/v - 1/u = 1/F$, we have $1/v - 1/(-30) = 1/20$. $1/v = 1/20 - 1/30 = 1/60$. So, $v = 60$ cm.
The magnification is $m = v/u = 60/(-30) = -2$.

Readings obtained: 1.24 mm, 1.25 mm, 1.23 mm, 1.21 mm. Find the expected least count of the instrument. All readings are given to the second decimal place (hundredths of a mm), i.e., 1.21, 1.23, 1.24, 1.25 mm. The least count of a measuring instrument is the smallest division it can measure. Since the readings vary in the hundredths place (0.01 mm), the instrument can measure up to 0.01 mm precision. The differences between consecutive readings are multiples of 0.01 mm, confirming the least count is 0.01 mm. The correct answer is Option 1: 0.01 mm.

Find the mean free path of a molecule with diameter $d = 5 \times 10^{-10}$ m at temperature $T = 41°C = 314$ K and pressure $P = 1.38 \times 10^5$ Pa. We apply the mean free path formula $\lambda = \frac{k_B T}{\sqrt{2}\pi d^2 P}$. Substituting the values gives $\lambda = \frac{1.38 \times 10^{-23} \times 314}{\sqrt{2} \times \pi \times (5 \times 10^{-10})^2 \times 1.38 \times 10^5}$. Calculating the numerator yields $1.38 \times 10^{-23} \times 314 = 433.32 \times 10^{-23} = 4.3332 \times 10^{-21}$. For the denominator, we first compute $d^2 = 25 \times 10^{-20}$, so $\sqrt{2} \times \pi \times 25 \times 10^{-20} \times 1.38 \times 10^5 = \sqrt{2} \times \pi \times 25 \times 1.38 \times 10^{-15}$ $= \sqrt{2} \times 3.14159 \times 34.5 \times 10^{-15} = 1.4142 \times 108.34 \times 10^{-15} = 153.18 \times 10^{-15} = 1.5318 \times 10^{-13}$. Finally, $\lambda = \frac{4.3332 \times 10^{-21}}{1.5318 \times 10^{-13}} = 2.828 \times 10^{-8} = 2\sqrt{2} \times 10^{-8}$ m. Hence, the mean free path is $2\sqrt{2} \times 10^{-8}$ m, corresponding to option 3.

For a prism, the angle of minimum deviation $\delta_{\text{min}}$ is related to the refracting angle $A$ and the refractive index $n$ by the formula: $n = \frac{\sin\left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ Given that $\delta_{\text{min}} = A$, substitute this into the formula: $n = \frac{\sin\left(\frac{A + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{2A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin A}{\sin\left(\frac{A}{2}\right)}$ Using the double-angle identity $\sin A = 2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right)$: $n = \frac{2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = 2 \cos\left(\frac{A}{2}\right)$ Thus, $n = 2 \cos\left(\frac{A}{2}\right)$. For a prism, the refracting angle $A$ is typically between $0^\circ$ and $90^\circ$ (acute). As $A$ increases from $0^\circ$ to $90^\circ$, $\frac{A}{2}$ increases from $0^\circ$ to $45^\circ$. The cosine function decreases from $\cos 0^\circ = 1$ to $\cos 45^\circ = \frac{\sqrt{2}}{2}$. Therefore: When $A \to 0^\circ$, $n \to 2 \times 1 = 2$. When $A \to 90^\circ$, $n \to 2 \times \frac{\sqrt{2}}{2} = \sqrt{2}$. However, $A = 0^\circ$ is not possible for a physical prism, so $n < 2$. Also, $A = 90^\circ$ is not typical for prisms, and the ray emergence condition requires $A < 90^\circ$ to avoid grazing emergence, so $n > \sqrt{2}$. Thus, $n$ lies \in the open interval $(\sqrt{2}, 2)$. Now, comparing with the options: A. $n \geq 2$ \to Incorrect, as $n < 2$. B. $\sqrt{2} < n < 2\sqrt{2}$ \to Incorrect, as $2\sqrt{2} \approx 2.828$ is larger than 2, and $n < 2$. C. $1 < n < 2$ \to Incorrect, as it includes values less than $\sqrt{2} \approx 1.414$. D. $\sqrt{2} < n < 2$ \to Correct, as it matches the interval $\sqrt{2} < n < 2$. Therefore, the refractive index $n$ satisfies $\sqrt{2} < n < 2$.

Two tuning forks A and B produce 8 beats in 2 seconds, so the beat frequency is 4 Hz. The frequency of B is 380 Hz. To determine possible frequencies of A, note that the beat frequency is $|f_A - f_B| = 4$ Hz. Thus $f_A = 380 + 4 = 384$ Hz or $f_A = 380 - 4 = 376$ Hz. Using the wax-loading condition, when fork A is loaded with wax its frequency decreases (wax increases mass, lowering the frequency). After loading, the beat frequency becomes $\dfrac{4}{2} = 2$ beats/s. Let $f_A'$ be the new frequency of A after loading. Then $f_A' < f_A$ and $|f_A' - 380| = 2$. For the case $f_A = 384$ Hz, after loading $f_A'$ decreases. Solving $|f_A' - 380| = 2$ gives $f_A' = 382$ or $f_A' = 378$. Since both are less than 384 and the beat frequency decreased from 4 to 2, the result $f_A' = 382$ Hz (getting closer to 380) is consistent. For the case $f_A = 376$ Hz, after loading $f_A'$ would decrease further below 376. The condition $|f_A' - 380| = 2$ again gives $f_A' = 378$ or $f_A' = 382$, but neither is less than 376, so this case is invalid. Therefore, the original frequency of tuning fork A is $\boxed{384}$ Hz.

In a cyclic process on a P-V diagram, net work done equals area enclosed by the loop. From graph: $A=(2m^3,100Pa)$ $B=(5m^3,300Pa)$ $C=(5m^3,100Pa)$ The path ABCA forms a triangle. Work done = area of triangle $W=\frac{1}{2}\times \text{base}\times \text{height}$ Base (change \in volume): $5-2=3\ \text{m}^3$ Height (change \in pressure): $300-100=200\ \text{Pa}$ So, $W=\frac{1}{2}(3)(200)$ $W=300\ \text{J}$

The stored magnetic energy $U = \frac{1}{2}LI^2$. Given $U = 16$ J and $I_{peak} = I_{rms}\sqrt{2} = 2\sqrt{2}$ A.
$16 = \frac{1}{2}L(2\sqrt{2})^2 \Rightarrow 16 = \frac{1}{2}L(8) \Rightarrow L = 4$ H.

The dissipated power $P = I_{rms}^2R$. Given $P = 32$ W and $I_{rms} = 2$ A.
$32 = (2)^2R \Rightarrow 32 = 4R \Rightarrow R = 8 \, \Omega$.

Inductive reactance $X_L = 2\pi f L = 2\pi (50)(4) = 400\pi = 400(3.14) = 1256 \, \Omega$.

The ratio of inductive reactance to its resistance is $\frac{X_L}{R} = \frac{1256}{8} = 157$.

The target answer 314 is incorrect.

Wavelengths: $\lambda_1 = 650$ nm and $\lambda_2 = 550$ nm, slit separation: $d = 2$ mm $= 2 \times 10^{-3}$ m, and screen distance: $D = 1.2$ m. The bright fringe positions for wavelength $\lambda$ \in Young's double slit experiment are given by $y_n = \frac{n \lambda D}{d}$. For bright fringes of both wavelengths to coincide at the same point, we require $n_1 \lambda_1 = n_2 \lambda_2$. Since $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{550}{650} = \frac{11}{13}$, the smallest positive integers satisfying this are $n_1 = 11$ and $n_2 = 13$. The distance of their least coincidence from the central maximum is $y = \frac{n_1 \lambda_1 D}{d} = \frac{11 \times 650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}}$. $y = \frac{11 \times 650 \times 1.2}{2} \times 10^{-9+3}$ $y = \frac{8580}{2} \times 10^{-6} = 4290 \times 10^{-6} = 429 \times 10^{-5} \text{ m}$ The answer is 429.

A flywheel (mass $M = 3$ kg, radius $R = 5$ m) has a string wound around it connected to a hanging mass ($m = 3$ kg). The mass descends by $h = 3$ m. We need the kinetic energy of the wheel. The flywheel is a solid disc, so its moment of inertia is $I = \dfrac{1}{2}MR^2$. Since the string is inextensible, the linear velocity of the mass equals the tangential velocity at the rim, giving $v = R\omega$. Using conservation of energy and noting that the initial state is at rest while the final state has the mass descended by $h = 3$ m, we write: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ Substituting $\omega = \dfrac{v}{R}$ and $I = \dfrac{1}{2}MR^2$ into the energy equation yields: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}\cdot \frac{1}{2}MR^2 \cdot \frac{v^2}{R^2} = \frac{1}{2}mv^2 + \frac{1}{4}Mv^2$ Combining terms gives $mgh = \frac{v^2}{2}\left(m + \frac{M}{2}\right) = \frac{v^2}{2} \cdot \frac{2m + M}{2}$ Solving for $v^2$ results \in $v^2 = \frac{4mgh}{2m + M} = \frac{4 \times 3 \times 10 \times 3}{6 + 3} = \frac{360}{9} = 40 \text{ m}^2/\text{s}^2$ Now, the kinetic energy of the wheel is $KE_{\text{wheel}} = \frac{1}{2}I\omega^2 = \frac{1}{4}Mv^2 = \frac{1}{4} \times 3 \times 40 = 30 \text{ J}$ From the above, the answer is $\boxed{30}$ J.

The effect of the nitro group ($-NO_2$) on the benzene ring arises from its strong electron-withdrawing nature, which is due to both its $-I$ (inductive) and $-M$ (mesomeric/resonance) effects. It pulls electron density away from the benzene ring. Electrophilic aromatic substitution (EAS) requires electron density on the ring to attack the electrophile. Since $-NO_2$ withdraws electron density, it deactivates the ring towards electrophilic substitution. Statement B is correct. Nucleophilic aromatic substitution (NAS) is favoured when the ring is electron-deficient, because the nucleophile attacks a carbon bearing a partial positive charge. Since $-NO_2$ makes the ring electron-poor, it activates the ring towards nucleophilic substitution. Statement C is correct. Statement A (activates ring towards EAS) is incorrect because the nitro group instead deactivates the ring. Similarly, Statement D (deactivates ring towards NAS) is incorrect because it activates the ring towards nucleophilic substitution. Therefore, the correct statements are B and C, which corresponds to Option 2.

The energy of a photon is given by $E = h\nu$, where $h$ is Planck's constant and $\nu$ is the frequency. Since $h$ is constant, the energy is proportional to the frequency. Therefore, we need to find the frequencies of photons A, B, and C to compare their energies. We have photon A with wavelength $\lambda_A = 400 \text{ nm}$, photon B with frequency $\nu_B = 10^{16} \text{ s}^{-1}$, and photon C with wave number $\bar{\nu}_C = 10^4 \text{ cm}^{-1}$. First, we convert all quantities to SI units. For photon A, $\lambda_A = 400 \text{ nm} = 400 \times 10^{-9} \text{ m} = 4 \times 10^{-7} \text{ m}$. For photon C, since $1 \text{ cm} = 0.01 \text{ m}$, we have $\bar{\nu}_C = 10^4 \text{ cm}^{-1} = 10^4 \times 100 \text{ m}^{-1} = 10^6 \text{ m}^{-1}$. Next, using the speed of light $c = 3 \times 10^8 \text{ m/s}$, the frequency relates to wavelength and wave number by $\nu = \frac{c}{\lambda}$ and $\nu = c\,\bar{\nu}$. For photon A, $\nu_A = \frac{c}{\lambda_A} = \frac{3 \times 10^8}{4 \times 10^{-7}} = \frac{3}{4} \times 10^{15} = 7.5 \times 10^{14} \text{ Hz}$. Photon B has $\nu_B = 10^{16} \text{ Hz}$ directly. For photon C, $\nu_C = c \times \bar{\nu}_C = (3 \times 10^8) \times (10^6) = 3 \times 10^{14} \text{ Hz}$. The frequencies are $\nu_A = 7.5 \times 10^{14} \text{ Hz}$, $\nu_B = 10^{16} \text{ Hz} = 100 \times 10^{14} \text{ Hz}$, and $\nu_C = 3 \times 10^{14} \text{ Hz}$. Comparing the numerical coefficients gives $100 > 7.5 > 3$, so $\nu_B > \nu_A > \nu_C$. Since energy $E \propto \nu$, the order of energy is the same as the order of frequency: $E_B > E_A > E_C$, giving B > A > C, which corresponds to option A.

Empirical formula of X: C=61.01/12 = 5.08; H=15.25/1 = 15.25; N=23.74/14 = 1.69. Divide by smallest (1.69): C=3, H=9, N=1. So, the empirical formula is $C_3H_9N$.
Reaction with $HNO_2/H_2O$ produces 'Y' (an alcohol) and removes nitrogen.
Compound 'Y' upon oxidation produces 'Z' which gives a positive iodoform test. This implies 'Y' is a secondary alcohol and 'Z' is a methyl ketone ($CH_3COR$).
For $C_3H_9N$, the only primary amine (Option A) is $CH_3CH_2CH_2NH_2$, which forms primary alcohol $CH_3CH_2CH_2OH$ (1-propanol), which oxidizes to propanal, not a methyl ketone.
The only secondary amine (Option B) is 2-propylamine $(CH_3)_2CHNH_2$. This reacts with $HNO_2$ to give isopropyl alcohol $(CH_3)_2CHOH$ (2-propanol, Y).
Oxidation of 2-propanol (Y) gives acetone $(CH_3)_2CO$ (Z), which is a methyl ketone and gives a positive iodoform test.

The target answer is incorrect.
Option A: Benzylic bromination of toluene with $Br_2$ in the presence of $h\nu$ (light) is correct.
Option B: Sandmeyer reaction for converting diazonium salt to aryl bromide is correct.
Option C: Electrophilic aromatic substitution of toluene with $Br_2$ and $FeBr_3$ (Lewis acid) in the dark yields ortho and para bromotoluenes due to the activating and ortho/para directing effect of the methyl group. This is correct.
Option D: Allylic bromination of methylcyclohexene with $Br_2$ in the presence of $h\nu$ (light) would occur at the allylic position. However, $Br_2$ under $h\nu$ also causes addition across the double bond. For selective allylic bromination, N-bromosuccinimide (NBS) is preferred. Therefore, this reaction is not correctly represented as written, as a mixture of products or predominantly addition across the double bond would occur.