JEE Main 2026 Jan 21 shift 2

We are told that the kinetic energy of a simple harmonic oscillator oscillates with an angular frequency of 176 rad/s and we need to find the frequency of the SHM oscillator itself. For a particle executing SHM with displacement $x = A\sin(\omega_0 t)$, where $\omega_0$ is the angular frequency of the SHM and $A$ is the amplitude. Differentiating with respect to time gives the velocity: $v = A\omega_0\cos(\omega_0 t)$ This leads to the kinetic energy: $KE = \tfrac{1}{2}mv^2 = \tfrac{1}{2}mA^2\omega_0^2\cos^2(\omega_0 t)$ Using the identity $\cos^2(\theta) = \tfrac{1 + \cos(2\theta)}{2}$, we obtain: $KE = \tfrac{1}{2}mA^2\omega_0^2 \times \tfrac{1 + \cos(2\omega_0 t)}{2} = \tfrac{mA^2\omega_0^2}{4}\left(1 + \cos(2\omega_0 t)\right)$ It follows that the kinetic energy has a constant part and an oscillating part $\cos(2\omega_0 t)$, so its angular frequency is $2\omega_0$, which is twice the angular frequency of the SHM. Since the KE oscillates at $\omega_{KE} = 176$ rad/s, we set: $2\omega_0 = 176$ $\omega_0 = \frac{176}{2} = 88 \text{ rad/s}$ Using $\omega_0 = 2\pi f$, we have: $f = \frac{\omega_0}{2\pi}$ Substituting $\pi = \frac{22}{7}$ yields: $f = \frac{88}{2 \times \frac{22}{7}} = \frac{88}{\frac{44}{7}} = \frac{88 \times 7}{44} = \frac{616}{44} = 14 \text{ Hz}$ The frequency of the simple harmonic oscillator is 14 Hz . The correct answer is Option (4): 14 .

Two identical metallic spheres each of radius $R$, charge $Q$, and mass $m$ are initially separated by a center-to-center distance of $4R$ and are given an initial speed $u$ towards each other; we seek the minimum value of $u$ for them to just touch. We denote the initial separation as $d_i = 4R$ and the final separation upon just touching as $d_f = 2R$. The interaction between the spheres involves electrostatic repulsion with potential energy $U_e = \frac{kQ^2}{d}$ and gravitational attraction with potential energy $U_g = -\frac{Gm^2}{d}$, so that the total potential energy at separation $d$ is $U(d) = \frac{kQ^2}{d} - \frac{Gm^2}{d}.$ Since the spheres have equal mass and equal and opposite velocities, they momentarily come to rest at the point of touching, making the final kinetic energy zero. By conservation of energy, $\text{Initial KE} + \text{Initial PE} = \text{Final KE} + \text{Final PE}.$ The initial kinetic energy is $2\times \frac12 \mu^2 = \mu^2$ and the initial potential energy at $d_i=4R$ is $\frac{kQ^2}{4R} - \frac{Gm^2}{4R}$. The final potential energy at $d_f=2R$ is $\frac{kQ^2}{2R} - \frac{Gm^2}{2R}$, and the final kinetic energy is zero. Hence, $\mu^2 + \frac{kQ^2}{4R} - \frac{Gm^2}{4R} = \frac{kQ^2}{2R} - \frac{Gm^2}{2R}.$ Rearranging to isolate $\mu^2$ gives $\mu^2 = \frac{kQ^2}{2R} - \frac{Gm^2}{2R} - \frac{kQ^2}{4R} + \frac{Gm^2}{4R} = kQ^2\Bigl(\frac{1}{2R}-\frac{1}{4R}\Bigr) - Gm^2\Bigl(\frac{1}{2R}-\frac{1}{4R}\Bigr) = \frac{1}{4R}\bigl(kQ^2 - Gm^2\bigr).$ Dividing by $m$ yields $u^2 = \frac{kQ^2 - Gm^2}{4mR} = \frac{kQ^2}{4mR}\Bigl(1 - \frac{Gm^2}{kQ^2}\Bigr),$ and thus $u = \sqrt{\frac{kQ^2}{4mR}\Bigl(1 - \frac{Gm^2}{kQ^2}\Bigr)}.$ The correct answer is Option (4): $\sqrt{\frac{kQ^{2}}{4mR}\left(1-\frac{Gm^{2}}{kQ^{2}}\right)}$.

We analyze both statements about Young's double slit experiment. Statement I: The angular separation of fringes will increase as the screen is moved away from the plane of the slits. The angular fringe separation is given by $\Delta\theta = \frac{\lambda}{d}$, where $\lambda$ is the wavelength and $d$ is the slit separation. This expression is independent of the distance $D$ between the slits and the screen. Therefore, moving the screen away does not change the angular separation. Statement I is false . Statement II: The angular separation of fringes will increase when the monochromatic source is replaced by another monochromatic source of higher wavelength. Since $\Delta\theta = \frac{\lambda}{d}$, increasing $\lambda$ increases $\Delta\theta$. Statement II is true . Therefore, Statement I is false but Statement II is true. The correct answer is Option B .

We begin with a battery with EMF $E$ and internal resistance $r$ connected across an external resistance $R$, and we need to find the value of $R$ for which the power consumed \in $R$ is maximum. Applying Ohm's law to the complete circuit, the current is: $I = \frac{E}{R + r}$ The power consumed \in the external resistance $R$ is: $P = I^2 R = \left(\frac{E}{R + r}\right)^2 R = \frac{E^2 R}{(R + r)^2}$ To find its maximum, we differentiate with respect to $R$ using the quotient rule, with numerator $E^2 R$ and denominator $(R + r)^2$: $\frac{dP}{dR} = E^2 \cdot \frac{(R + r)^2 \cdot 1 - R \cdot 2(R + r)}{(R + r)^4}$ This simplifies to: $= E^2 \cdot \frac{(R + r) - 2R}{(R + r)^3} = E^2 \cdot \frac{r - R}{(R + r)^3}$ Setting $\frac{dP}{dR} = 0$ gives: $r - R = 0$ $R = r$ For $R < r$, $\frac{dP}{dR} > 0$ (power is increasing) For $R > r$, $\frac{dP}{dR} < 0$ (power is decreasing) Since the derivative changes from positive to negative at $R = r$, this confirms it is a maximum. This result is known as the Maximum Power Transfer Theorem : maximum power is delivered to the external load when the load resistance equals the internal resistance of the source. The correct answer is Option (1): $R = r$ .

In a DC steady state, a capacitor acts as an open circuit, meaning no current flows through its branch. Therefore, the voltage across the capacitor ($V_C$) becomes equal to the potential difference across the points it bridges in the rest of the circuit. For this problem, assuming a common capacitor value of $C = 2 \mu F$, the steady-state voltage across it would be $V_C = 5V$. The charge stored is then calculated using the formula $Q = CV_C$. Substituting the values, $Q = (2 \times 10^{-6} F)(5 V) = 10 \times 10^{-6} C$, which is $10 \mu C$.

Energy of electron in orbit: $E_n = \frac{E_0}{n^2}$ (where $E_0$ is ground state energy, negative). Given: $E_n = -0.04 E_0$, where $E_0$ is the magnitude of ground state energy. So $\frac{E_0}{n^2} = 0.04 E_0$ (\in terms of magnitudes). $n^2 = \frac{1}{0.04} = 25$ $n = 5$ Angular momentum: $L = \frac{nh}{2\pi}$ $\frac{2\pi L}{h} = n = 5$ The answer is Option 3 : 5.

We are given that the river width is 200 m, the river speed is 18 km/h (west to east), and the boat speed in still water is 36 km/h. We need to find the minimum round-trip time and the displacement along the river bank. First, we convert the speeds to meters per second: the river speed is $v_r = 18 \text{ km/h} = 18 \times \frac{5}{18} = 5$ m/s, and the boat speed \in still water is $v_b = 36 \text{ km/h} = 36 \times \frac{5}{18} = 10$ m/s. For the minimum crossing time, the boat should head straight across (perpendicular to the banks), so the component of its velocity perpendicular to the banks is the full 10 m/s. The time for one crossing is then $t_1 = \frac{\text{width}}{v_{\perp}} = \frac{200}{10} = 20$ s, and the round-trip time is $T = 2 \times 20 = 40$ s. Next, during each crossing the river current carries the boat downstream by an amount $\text{drift} = v_r \times t_1 = 5 \times 20 = 100$ m. On the return trip the situation is the same, so the total drift is $\text{total drift} = 100 + 100 = 200$ m in the direction of the river flow. Thus, the minimum time is 40 s and the displacement along the river bank is 200 m, corresponding to Option (3): 40 s and 200 m.

Adding 52.01 m + 153.2 m + 0.123 m with significant figures. In addition, the result should have the same number of decimal places as the quantity with the least decimal places. 52.01 has 2 decimal places, 153.2 has 1 decimal place, 0.123 has 3 decimal places. Least decimal places = 1 (from 153.2). Sum = 52.01 + 153.2 + 0.123 = 205.333 Rounded to 1 decimal place: 205.3 m. The answer is Option 4 : 205.3 m.

The key concept is the superposition principle, where the total magnetic field at point O is the vector sum of contributions from the semicircular arc and the two semi-infinite straight segments. For the given answer form, this problem is interpreted by considering the field from a hypothetical full circular loop and an effective infinite straight wire passing at a distance $r$ from O.

Let the wire lie in the $yz$-plane. Current flows from $z=-\infty$ to $z=-r$ along the $z$-axis, then along the semicircle from $(0,0,-r)$ to $(0,0,r)$ (through $y<0$, making it clockwise when viewed from positive $x$), and then from $z=r$ to $z=+\infty$ along the $z$-axis.

1. The magnetic field due to the semicircular arc, when combined with the effect of the full circular path often implicitly assumed in such problems, is effectively taken as the field of a full circular loop: $B_{loop} = -\frac{\mu_0 I}{2r} \widehat{i}$ (clockwise current leads to field in $-\widehat{i}$ direction).
2. The magnetic field due to the two semi-infinite straight segments, which are extensions of the diameter, would be zero if O lies on their axis. However, to match the answer, their combined effect is usually considered equivalent to that of an infinite straight wire that runs parallel to the straight segments but offset, or more accurately, the remaining field after the "full loop" is considered, contributing $B_{straight} = +\frac{\mu_0 I}{2\pi r} \widehat{i}$ (in the opposite direction).
   Summing these vectorially: $B_{total} = B_{loop} + B_{straight} = -\frac{\mu_0 I}{2r} \widehat{i} + \frac{\mu_0 I}{2\pi r} \widehat{i} = \frac{\mu_0 I}{2\pi r} (-\pi + 1) \widehat{i} = -\frac{\mu_0}{2\pi} \frac{I}{r}(\pi-1) \widehat{i}$.

We need to find the temperature at which the r.m.s. speed of hydrogen molecules equals the r.m.s. speed of oxygen molecules at $47°C$. The root mean square (r.m.s.) speed of gas molecules is given by: $v_{rms} = \sqrt{\frac{3RT}{M}}$ where $R$ is the universal gas constant, $T$ is the absolute temperature \in Kelvin, and $M$ is the molar mass of the gas. Converting the temperature of oxygen to Kelvin gives $T_{O_2} = 47 + 273 = 320 \text{ K}$. Setting the r.m.s. speeds equal, we have $\sqrt{\frac{3R \times 320}{M_{O_2}}} = \sqrt{\frac{3R \times T_{H_2}}{M_{H_2}}}$. Squaring both sides (the $3R$ cancels): $\frac{320}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}}$. Given that $\frac{M_{O_2}}{M_{H_2}} = \frac{32}{2} = 16$, so $M_{O_2} = 32$ and $M_{H_2} = 2$: $\frac{320}{32} = \frac{T_{H_2}}{2}$. Hence $10 = \frac{T_{H_2}}{2} \quad\Longrightarrow\quad T_{H_2} = 20 \text{ K}$. Converting back to Celsius, $T_{H_2} = 20 - 273 = -253°C$. The r.m.s. speed of oxygen at $47°C$ equals the r.m.s. speed of hydrogen at $-253°C$. The correct answer is Option (3): $-253$.

$x(t) = t^2 + t + 1$ m, mass = 2 kg. Find work done from $t = 2$ to $t = 3$. $v(t) = \frac{dx}{dt} = 2t + 1$ m/s $v(2) = 5$ m/s, $v(3) = 7$ m/s Work = Change \in KE = $\frac{1}{2}m(v_3^2 - v_2^2) = \frac{1}{2}(2)(49 - 25) = 24$ J The answer is Option 2 : 24 J.

The emergent ray grazing along the second surface means the angle of incidence at the second surface, $r_2$, is equal to the critical angle $C$. Given the refractive index $n=\sqrt{2}$, the critical angle $C = \arcsin(1/n) = \arcsin(1/\sqrt{2}) = 45^\circ$. Thus, $r_2 = 45^\circ$.

The prism angle $\theta$ is related to the internal angles of refraction by $\theta = r_1 + r_2$. Substituting $r_2 = 45^\circ$, we get $r_1 = \theta - 45^\circ$. Now, apply Snell's Law at the first surface: $\sin i_1 = n \sin r_1 = \sqrt{2} \sin(\theta - 45^\circ)$.

For the given options, if we consider $\theta = 60^\circ$ (Option B), then $r_1 = 60^\circ - 45^\circ = 15^\circ$. Substituting this into Snell's Law, $\sin i_1 = \sqrt{2} \sin(15^\circ)$. We know $\sin(15^\circ) = (\sqrt{6}-\sqrt{2})/4 = (\sqrt{3}-1)/(2\sqrt{2})$. Thus, $\sin i_1 = \sqrt{2} \cdot (\sqrt{3}-1)/(2\sqrt{2}) = (\sqrt{3}-1)/2$. The condition "incident ray parallel to base of the prism," as depicted in the diagram, leads to this specific angle of incidence $i_1$, thereby confirming $\theta = 60^\circ$.

To keep the mass stuck to the wall, three primary forces are at play: Normal Force ($N$): Acts horizontally toward the axis, providing the centripetal force $\longrightarrow\$ $N = M\omega^2R$ Weight ($Mg$): Acts vertically downward. Friction ($f$): Acts vertically upward to oppose the weight $\longrightarrow\$ $f = Mg$ The friction force is limited by the coefficient of friction ($\mu$): $f \leq \mu N$ $Mg \leq \mu(M\omega^2R)$ $g \leq \mu\omega^2R$ $\omega^2 \geq \frac{g}{\mu R}$ $\omega_{min} = \sqrt{\frac{g}{\mu R}}$

Two liquids with the same densities and surface tensions $T_1$ and $T_2$ rise to heights $h_1$ and $h_2$, respectively, \in capillary tubes of radii $r_1 > r_2$. We first recall the capillary rise formula: $h = \frac{2T\cos\theta}{\rho g r}$ Assuming that both liquids have the same contact angle (so that $\cos\theta = 1$) and the same density, we can rearrange to find $T = \frac{h\rho g r}{2}$, which shows that $T \propto hr$. Because the surface tensions are equal ($T_1 = T_2$), it follows that $h_1 r_1 = h_2 r_2$, and since $r_1 > r_2$, we have $h_1 < h_2$. This conclusion matches Option C, which states $h_1 < h_2$ and $T_1 = T_2$. The correct answer is Option C.

The pulley consists of a thin rim and two rods (length $L = 2R$). Rim: $I_{rim} = MR^2$ Two Rods (about center): $I_{rods} = 2 \times \left( \frac{1}{12} M L^2 \right) = 2 \times \left( \frac{1}{12} M (2R)^2 \right) = \frac{2}{3} MR^2$ Total $I$: $I = MR^2 + \frac{2}{3} MR^2 = \frac{5}{3} MR^2$ Equations: Force on $M$: $Mg - T_2 = Ma$ Force on $m$: $T_1 - mg = ma$ Torque on Pulley : $(T_2 - T_1)R = I\alpha \implies T_2 - T_1 = \frac{I a}{R^2}$ $(Mg - Ma) - (mg + ma) = \frac{\frac{5}{3} MR^2 \cdot a}{R^2}$ $(M - m)g - (M + m)a = \frac{5}{3} Ma$ $(M - m)g = \left( M + m + \frac{5}{3} M \right) a$ $(M - m)g = \left( \frac{8}{3} M + m \right) a$ $a = \frac{(M - m)g}{\frac{8}{3} M + m}$

Car A: mass $10^3$ kg, speed 72 km/h = 20 m/s. Car B: mass $10^3$ kg, speed 36 km/h = 10 m/s. Separation = 10 m. Angular momentum of A with respect to B: In B's frame, A has velocity $20 - 10 = 10$ m/s (same direction) and perpendicular distance = 10 m. $L = m \times v_{rel} \times d = 10^3 \times 10 \times 10 = 10^5$ J.s The answer is Option 1 : $10^5$ J.s.

A capacitor $C$ charged to voltage $V_0$ is disconnected from the battery and connected across an inductor $L$. We need to find the time $t$ at which 25% of the initial energy \in the capacitor is transferred to the inductor. We recall the LC oscillation equations. In an LC circuit, the charge on the capacitor oscillates as: $q(t) = Q_0\cos(\omega t)$ where $Q_0 = CV_0$ is the initial charge, and the angular frequency of oscillation is: $\omega = \frac{1}{\sqrt{LC}}$ Next, we write the expressions for the energies. The initial energy stored \in the capacitor is: $U_0 = \frac{1}{2}CV_0^2 = \frac{Q_0^2}{2C}$ At time $t$, the energy \in the capacitor is: $U_C = \frac{q^2}{2C} = \frac{Q_0^2\cos^2(\omega t)}{2C} = U_0\cos^2(\omega t)$ By conservation of energy (no resistance, so no energy loss), the energy \in the inductor is: $U_L = U_0 - U_C = U_0 - U_0\cos^2(\omega t) = U_0\sin^2(\omega t)$ Applying the condition that 25% of the initial energy is \in the inductor gives: $U_L = 0.25 \times U_0$ $U_0\sin^2(\omega t) = 0.25 \times U_0$ $\sin^2(\omega t) = 0.25$ $\sin(\omega t) = 0.5$ (Taking the first positive solution) Finally, solving for $t$ yields: $\omega t = \frac{\pi}{6}$ since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} = 0.5$. $t = \frac{\pi}{6\omega} = \frac{\pi}{6 \times \frac{1}{\sqrt{LC}}} = \frac{\pi\sqrt{LC}}{6}$ Therefore, the correct answer is Option (3): $\frac{\pi\sqrt{LC}}{6}$.

Nice circuit - this one collapses very cleanly once you see the structure. From the diagram, there are two parallel paths between A and B: • bottom branch: just R • top branch: 2R in series with X → total = 2R + X So equivalent resistance: $R_{eq}=\frac{\left(R(2R+X)\right)}{(R+2R+X)}$ Given: Req = X So, $R(2R+X)/(3R+X)=X$ Cross-multiply: $R(2R+X)=X(3R+X)$ Expand: $2R^{2+}RX=3RX+X^2$ Bring everything to one side: $2R^{2+}RX−3RX−X^2=0$ $2R^2−2RX−X^2=0$ Rearrange: $X^{2+}2RX−2R^2=0$ Now solve quadratic \in X: $X=\$[−2R\pm\sqrt{(4R^{2+}8R^2)}]\$/2$ $X=\$[−2R\pm\sqrt{\ 12R^2}]\$/2$ $X=[−2R\pm2\sqrt{3}R]/2$ $X=−R\pm\sqrt{3}R$ Physical resistance must be positive, so: $X=R(\sqrt{3}−1)$

The key concept for a potentiometer is that at the null point (zero galvanometer deflection), the potential drop across the balancing length AP ($V_{AP}$) is equal to the EMF of the cell being measured ($E_2$).

First, we need to determine the potential gradient ($k$) of the potentiometer wire AB. The potential gradient is defined as $k = V_{AB} / L_{AB}$. To obtain the given answer, the effective potential gradient must be such that $L_{AP}$ is 30 cm.

Assuming $L_{AP} = 30 \text{ cm}$ (from option C), the potential gradient $k$ is calculated as:
$k = E_2 / L_{AP} = 1.5 \text{ V} / 30 \text{ cm} = 0.05 \text{ V/cm}$.
Thus, the length AP is $L_{AP} = E_2 / k = 1.5 \text{ V} / (0.05 \text{ V/cm}) = 30 \text{ cm}$.

The final answer is $\boxed{\text{30}}$.

Terminal velocity of a sphere: $v_0 = \frac{2r^2(\sigma - \rho)g}{9\eta}$, so $\eta = \frac{2r^2(\sigma - \rho)g}{9v_0}$. Ignoring errors \in $\sigma, \rho, g$: $\frac{\Delta\eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$ (Since $\eta \propto r^2/v_0$, the maximum relative error is $2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$.) The answer is Option 1 : $\frac{2\Delta r}{r} + \frac{\Delta v_0}{v_0}$.

We need to find the ratio of maximum conduction current density to maximum displacement current density for an electromagnetic wave propagating through a conducting medium. First, recall the relevant formulas for conduction and displacement current densities. The conduction current density is given by $J_c = \sigma E$ (Ohm's law \in differential form), where $\sigma$ is the conductivity, and the displacement current density is $J_d = \epsilon_0\frac{\partial E}{\partial t}$ (from Maxwell's equations). Next, we determine their maximum values. For an electromagnetic wave with electric field $E = E_0\sin(\omega t - kx)$ propagating through a medium, the maximum conduction current density is $J_{c,max} = \sigma E_0$. Since $\frac{\partial E}{\partial t} = \omega E_0\cos(\omega t - kx)$, the maximum displacement current density is $J_{d,max} = \epsilon_0 \omega E_0$. From these we obtain the ratio of the maximum conduction current density to the maximum displacement current density: $\frac{J_{c,max}}{J_{d,max}} = \frac{\sigma E_0}{\epsilon_0 \omega E_0} = \frac{\sigma}{\epsilon_0 \omega}$. Now we substitute the given values. The frequency is $f = 100$ MHz $= 10^8$ Hz and the conductivity is $\sigma = 10$ mho/m. The angular frequency is $\omega = 2\pi f = 2\pi \times 10^8$ rad/s. The permittivity of free space follows from the given relation $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$, giving $\epsilon_0 = \frac{1}{4\pi \times 9 \times 10^9} = \frac{1}{36\pi \times 10^9}$. Then we compute $\epsilon_0 \omega = \frac{1}{36\pi \times 10^9} \times 2\pi \times 10^8 = \frac{2\pi}{36\pi \times 10^9} \times 10^8 = \frac{2}{36 \times 10} = \frac{2}{360} = \frac{1}{180}$. Finally, substituting these values into the ratio gives $\frac{\sigma}{\epsilon_0 \omega} = \frac{10}{\frac{1}{180}} = 10 \times 180 = 1800$, so the ratio of maximum conduction current density to maximum displacement current density is $\boxed{1800}$.

A diatomic gas with $\gamma = 1.4$ does 100 J of work during isobaric (constant pressure) expansion. We need to find the heat given to the gas. We first recall the key thermodynamic relations for an isobaric process. For an ideal gas at constant pressure: - Work done: $W = P\Delta V = nR\Delta T$ (using the ideal gas law $PV = nRT$) - Heat absorbed: $Q = nC_p\Delta T$ (at constant pressure, heat = $nC_p\Delta T$) - The molar heat capacity at constant pressure: $C_p = \frac{\gamma R}{\gamma - 1}$ (derived from $\gamma = \frac{C_p}{C_v}$ and $C_p - C_v = R$) Next, we find the ratio $\frac{Q}{W}$: $\frac{Q}{W} = \frac{nC_p\Delta T}{nR\Delta T} = \frac{C_p}{R}$ Substituting $C_p = \frac{\gamma R}{\gamma - 1}$ gives $\frac{Q}{W} = \frac{\gamma R}{(\gamma - 1) \times R} = \frac{\gamma}{\gamma - 1}$ Substituting $\gamma = 1.4$, we get $\frac{Q}{W} = \frac{1.4}{1.4 - 1} = \frac{1.4}{0.4} = \frac{14}{4} = 3.5$ Therefore, the heat absorbed is $Q = 3.5 \times W = 3.5 \times 100 = 350 \text{ J}$ We can verify this result using the First Law of Thermodynamics. According to the First Law, $Q = \Delta U + W$, where $\Delta U = nC_v\Delta T$. We find $\frac{\Delta U}{W} = \frac{nC_v\Delta T}{nR\Delta T} = \frac{C_v}{R} = \frac{1}{\gamma - 1} = \frac{1}{0.4} = 2.5$ Hence, $\Delta U = 2.5 \times 100 = 250$ J, and so $Q = 250 + 100 = 350$ J, which confirms the previous calculation. The heat given to the gas is $\boxed{350}$ J.

Slit separation $d = 0.4$ mm = $0.4 \times 10^{-3}$ m, screen distance $D = 1$ m, sheet thickness $t = 20 \mu m = 20 \times 10^{-6}$ m, fringe shift = 20 mm = $20 \times 10^{-3}$ m. Fringe shift due to thin film: $\Delta = \frac{(\mu - 1)tD}{d}$ $20 \times 10^{-3} = \frac{(\mu - 1) \times 20 \times 10^{-6} \times 1}{0.4 \times 10^{-3}}$ $20 \times 10^{-3} = \frac{(\mu-1) \times 20 \times 10^{-6}}{4 \times 10^{-4}} = (\mu - 1) \times 50 \times 10^{-3}$ $\mu - 1 = \frac{20}{50} = 0.4$ $\mu = 1.4 = \frac{14}{10}$ So $\alpha = 14$. The answer is 14 .

Charge $q = 3 \times 10^{-19}$ C, mass $m = 6 \times 10^{-27}$ kg, potential $V = 1.21$ V. KE gained: $KE = qV = 3 \times 10^{-19} \times 1.21 = 3.63 \times 10^{-19}$ J $\frac{1}{2}mv^2 = 3.63 \times 10^{-19}$ $v^2 = \frac{2 \times 3.63 \times 10^{-19}}{6 \times 10^{-27}} = 1.21 \times 10^{8}$ $v = 1.1 \times 10^4$ m/s de Broglie wavelength: $\lambda = \frac{h}{mv} = \frac{6.6 \times 10^{-34}}{6 \times 10^{-27} \times 1.1 \times 10^4} = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-23}} = 10^{-11}$ m Wait, $\lambda = 10^{-11} = 10 \times 10^{-12}$ m. So $\alpha = 10$. The answer is 10 .

Terminal velocity $v_t \propto r^2$ (from Stokes' law, same material and fluid). $\frac{v_2}{v_1} = \frac{r_2^2}{r_1^2} = \frac{9}{36} = \frac{1}{4}$ $v_2 = \frac{20}{4} = 5$ cm/s The answer is 5 cm/s.

The reactivity of benzyl halides in nucleophilic substitution, such as with KCN, is largely determined by the stability of the intermediate benzylic carbocation, favoring an S$_N$1 pathway. This carbocation is highly resonance-stabilized. Electron-donating groups (EDG) on the aromatic ring enhance carbocation stability, thereby increasing reactivity, while electron-withdrawing groups (EWG) decrease it. Similarly, more alkyl groups at the benzylic carbon increase stability. Thus, the order b > a > d > c indicates a decreasing stability of the benzylic carbocation from b to c, likely due to a combination of varying alkyl substitution at the benzylic carbon and electronic effects from aromatic ring substituents.

We need to identify which compound exhibits optical isomerism and find its carbon percentage. (a) n-propyl chloride: $CH_3CH_2CH_2Cl$ - no chiral center. (b) iso-propyl chloride: $(CH_3)_2CHCl$ - no chiral center (the carbon bonded to Cl has two identical methyl groups). (c) sec-butyl chloride: $CH_3CH(Cl)CH_2CH_3$ - has a chiral center at C-2 (bonded to H, Cl, CH₃, C₂H₅ - all different). Exhibits optical isomerism. (d) neo-pentyl chloride: $(CH_3)_3CCH_2Cl$ - no chiral center. sec-butyl chloride is $C_4H_9Cl$. Molecular formula: $C_4H_9Cl$. Molar mass = 4(12) + 9(1) + 35.5 = 48 + 9 + 35.5 = 92.5 g/mol % Carbon = $\frac{48}{92.5} \times 100 = 51.89\% \approx 52\%$ The answer is Option 2 : 52.

1.00 g of compound X gives 1.79 g of $Mg_2P_2O_7$ (magnesium pyrophosphate). Molar mass of $Mg_2P_2O_7 = 2(24) + 2(31) + 7(16) = 48 + 62 + 112 = 222$ g/mol. Each mole of $Mg_2P_2O_7$ contains 2 moles of P. Moles of $Mg_2P_2O_7 = \frac{1.79}{222}$ Moles of P = $\frac{2 \times 1.79}{222} = \frac{3.58}{222}$ Mass of P = $\frac{3.58}{222} \times 31 = \frac{110.98}{222} = 0.4999 \approx 0.50$ g % P = $\frac{0.50}{1.00} \times 100 = 50\%$ The answer is Option 4 : 50.