30 Jan 2024 Shift 1

Coefficient of viscosity: From Newton's law of viscosity: $F=\eta A\frac{dv}{dx}$ $\eta=\frac{F}{A}\cdot\frac{dx}{dv}$ $[\eta]=\frac{MLT^{-2}}{L^2}\cdot\frac{L}{LT^{-1}}=[ML^{-1}T^{-1}]$ So, (A) matches with (III) Surface tension: $\text{Surface tension}=\frac{\text{Force}}{\text{length}}$ $\frac{MLT^{-2}}{L}=[MT^{-2}]$ So, (B) matches with (IV) Angular momentum: $L=r\times p=r\cdot mv$ $[L]=[L]\cdot\$[MLT^{-1}]\$=[ML^2T^{-1}]$ So, (C) matches with (II) Rotational kinetic energy: $E=\frac{1}{2}I\omega^2$ $[I]=ML^2,\quad[\omega]=T^{-1}$ $[E]=ML^2\cdot T^{-2}=[ML^2T^{-2}]$ So, (D) matches with (I) Final matching: (A)−(III), (B)−(IV), (C)−(II), (D)−(I)

A particle of mass $m$ is projected with velocity $u$ at angle $30°$ with the horizontal. We need the angular momentum about the point of projection at maximum height. Find velocity and position at maximum height: at maximum height, only the horizontal component of velocity remains: $v_x = u\cos 30° = \frac{u\sqrt{3}}{2}$ The maximum height is: $h = \frac{u^2 \sin^2 30°}{2g} = \frac{u^2 \cdot \frac{1}{4}}{2g} = \frac{u^2}{8g}$ Compute angular momentum: the angular momentum formula about the point of projection is $L = m \cdot v_x \cdot h$ (since the velocity is purely horizontal and the perpendicular distance from the projection point to the velocity line is $h$). $L = m \cdot \frac{u\sqrt{3}}{2} \cdot \frac{u^2}{8g} = \frac{\sqrt{3}}{16} \cdot \frac{\mu^3}{g}$ The correct answer is Option (1): $\frac{\sqrt{3}}{16} \frac{\mu^3}{g}$.

You have: a 2 kg block on a 30° incline connected by a string over a pulley then a movable pulley holding a 4 kg mass Key idea: movable pulley changes both force and motion relation Step 1: understand tension Let tension in string = T For the 4 kg block: It is supported by two segments of the string So upward force = 2T Equation: $2T−4g=4a₄.$ Step 2: relation of acceleration Very important part If the 2 kg block moves up the incline by x the movable pulley (and 4 kg mass) moves down by x/2 So: $a₂=2a₄$ Now lets write equation for the 2 Kg block $m_2g\sin\ \theta\ -T=m_2a_2$ $2g\times \ \frac{1}{2}-T=2a_2$ $g-T=4a_4$ $2T−4g=4a₄.$ now on multiply and adding both so that T gets cancelled $-2g=12a_4$ $\ a_4=-\frac{g}{6}\left(negative\ as\ we\ considered\ opposite\ direction\right)$ so now we know $a₂=2a₄$ so $a₂=2\times \ \frac{g}{6}$ $a₂=\ \frac{g}{3}$

As the track is frictionless, the total mechanical energy of the particle is conserved. We apply the principle of conservation of energy between points A and B. The total energy at A must equal the total energy at B.

The energy conservation equation is:
$\frac{1}{2}mv_A^2 + mgh_A = \frac{1}{2}mv_B^2 + mgh_B$

The particle is gently pushed from rest at A, so its initial speed $v_A = 0$. From the figure, the height at A is $h_A = 1 \text{ m}$ and the height at B is $h_B = 0.5 \text{ m}$.

Substituting these values into the equation:
$0 + mg(1) = \frac{1}{2}mv_B^2 + mg(0.5)$

Simplifying by dividing by $m$ and rearranging for $v_B^2$:
$g - 0.5g = \frac{1}{2}v_B^2 \implies 0.5g = \frac{1}{2}v_B^2 \implies v_B^2 = g$

Using the given value $g = 10 \text{ m s}^{-2}$:
$v_B = \sqrt{10} \text{ m s}^{-1}$

A spherical body of mass $m = 100 \text{ g} = 0.1 \text{ kg}$ is dropped from height $10 \text{ m}$ and rebounds to height $5 \text{ m}$. Find the velocity just before hitting the ground: using $v = \sqrt{2gh}$: $v_1 = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14 \text{ m/s}$ (downward) Find the velocity just after rebounding: $v_2 = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} = 7\sqrt{2} \text{ m/s}$ (upward) Calculate the impulse: impulse equals the change \in momentum. Taking upward as positive: $J = m(v_2 - (-v_1)) = m(v_2 + v_1)$ $J = 0.1 \times (7\sqrt{2} + 14) = 0.1 \times (9.899 + 14) = 0.1 \times 23.899$ $J \approx 2.39 \text{ kg m s}^{-1}$ The correct answer is Option (4): $2.39 \text{ kg m s}^{-1}$.

Given: Gravitational potential $V = -5.12 \times 10^7 \text{ J kg}^{-1}$, acceleration due to gravity $g = 6.4 \text{ m s}^{-2}$, and mean radius of Earth $R = 6400 \text{ km}$. Key formulas: At distance $r$ from the centre of Earth: $V = -\frac{GM}{r}$ and $g = \frac{GM}{r^2}$ Find $r$ from the ratio $V/g$: $\frac{|V|}{g} = \frac{GM/r}{GM/r^2} = r$ $r = \frac{5.12 \times 10^7}{6.4} = 8 \times 10^6 \text{ m} = 8000 \text{ km}$ Find the height above the surface: $h = r - R = 8000 - 6400 = 1600 \text{ km}$ The correct answer is Option (1): $1600 \text{ km}$.

Young's modulus is a material property that depends only on the nature of the material, not on the dimensions of the wire. By definition, Young's modulus is: $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$ While stress and strain depend on the dimensions (length $L$ and cross-sectional area $A$), the ratio $Y$ is an intrinsic property of the material itself. Therefore, if the length is doubled and the cross-sectional area is halved, the Young's modulus remains unchanged at $Y$. Option Y

The r.m.s. velocity of a gas molecule is given by: $v_{rms} = \sqrt{\frac{3RT}{M}}$ where $R$ is the gas constant, $T$ is the absolute temperature, and $M$ is the molar mass. We need the r.m.s. velocity of hydrogen ($M_{H_2} = 2$ g/mol) to equal that of oxygen ($M_{O_2} = 32$ g/mol) at $47°C = 320$ K. Setting the two velocities equal: $\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$ $\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$ $T_{H_2} = T_{O_2} \times \frac{M_{H_2}}{M_{O_2}} = 320 \times \frac{2}{32}$ $T_{H_2} = 320 \times \frac{1}{16} = 20 \text{ K}$ The correct answer is $20 \text{ K}$.

In the graph, the lines have positive slope in the (log⁡V,log⁡P) plane, so they represent $\log P=n\log V$ or $P=V^n$ which can be written as $PV^{-n}=\text{constant}$ This is a polytropic process with index m=−n For a polytropic process, molar heat capacity is $C=C_V+\frac{R}{1-m}$ Substituting m=−n, $C=C_V+\frac{R}{1+n}$ For process A, slope is given by $\tan⁡\theta =\gamma$ so $n=\gamma$ Hence $C_A=C_V+\frac{R}{1+\gamma}$ For process BBB, angle is $45^{\circ}$, so n=1 and $C_B=C_V+\frac{R}{2}$ Also, $C_P=C_V+R$ Now comparing: $C_P>C_B$ because $R>\frac{R}{2}$ Also since \gamma >1, $\frac{R}{2}>\frac{R}{1+\gamma}$ so $C_B>C_A$ and clearly $C_A>C_V$ Therefore, $CP>CB>CA>CV$

The electrostatic potential due to an electric dipole at a point at distance $r$ from the center of the dipole is given by: $V = \frac{1}{4\pi\epsilon_0} \cdot \frac{p\cos\theta}{r^2}$ where $p$ is the dipole moment and $\theta$ is the angle between the dipole axis and the line joining the center of the dipole to the point. From this expression, we can see that: $V \propto \frac{1}{r^2}$ Note: This is different from a point charge, where $V \propto \frac{1}{r}$, and from the electric field of a dipole, where $E \propto \frac{1}{r^3}$. The correct answer is $\frac{1}{r^2}$.

Since all resistors are connected in series, $R_{total} = 3300\Omega + (7 \times 100\Omega)$ $R_{total} = 3300\Omega + 700\Omega = 4000\Omega$ The measurement for $V_0$ begins after the $3.3\text{ k}\Omega$ and the first two $100\Omega$ resistors. The measurement includes the remaining five $100\Omega$ resistors. $R_{out} = 5 \times 100\Omega = 500\Omega$ $V_0 = V_{\in} \cdot \left( \frac{R_{out}}{R_{total}} \right)$ $V_0 = 4\text{ V} \cdot \left( \frac{500\Omega}{4000\Omega} \right)$ $V_0 = 0.5\text{ V}$

The resistance of a conductor varies with temperature according to: $R = R_0(1 + \alpha \Delta T)$ where $R_0$ is the resistance at the reference temperature, $\alpha$ is the temperature coefficient of resistance, and $\Delta T$ is the change \in temperature. Given: $R_0 = 60 \text{ } \Omega$ at $27°C$, supply voltage $= 220 \text{ V}$, current $= 2.75 \text{ A}$, $\alpha = 2 \times 10^{-4} \text{ °C}^{-1}$. First, find the resistance at the operating temperature: $R = \frac{V}{I} = \frac{220}{2.75} = 80 \text{ } \Omega$ Now apply the temperature-resistance relation: $80 = 60(1 + 2 \times 10^{-4} \times \Delta T)$ $\frac{80}{60} = 1 + 2 \times 10^{-4} \times \Delta T$ $\frac{4}{3} = 1 + 2 \times 10^{-4} \times \Delta T$ $\frac{1}{3} = 2 \times 10^{-4} \times \Delta T$ $\Delta T = \frac{1}{3 \times 2 \times 10^{-4}} = \frac{10^4}{6} \approx 1667°C$ The final temperature is: $T = 27 + 1667 = 1694°C$ The correct answer is $1694°C$.

The magnetic field at the center of a single circular loop of radius $a$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2a}$. Its direction is perpendicular to the plane of the loop, determined by the right-hand rule.

The two loops, A and B, are in perpendicular planes. Therefore, the magnetic fields they produce at the common center, $\vec{B}_A$ and $\vec{B}_B$, are also perpendicular to each other.

The magnitude of the magnetic field from each loop is the same:
$B_A = B_B = \frac{\mu_0 I}{2a}$
The net magnetic field, $\vec{B}_{net}$, is the vector \sum of the individual fields. Since $\vec{B}_A$ and $\vec{B}_B$ are perpendicular, we find the magnitude using the Pythagorean theorem:
$B_{net} = \sqrt{B_A^2 + B_B^2} = \sqrt{\left(\frac{\mu_0 I}{2a}\right)^2 + \left(\frac{\mu_0 I}{2a}\right)^2}$
$B_{net} = \sqrt{2 \left(\frac{\mu_0 I}{2a}\right)^2} = \frac{\sqrt{2} \mu_0 I}{2a} = \frac{\mu_0 I}{\sqrt{2}a}$

For a series LR circuit, the power factor is given by: $\cos\phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + \omega^2 L^2}}$ Case 1: $\omega_1 = 1000 \text{ rad/s}$, power factor $= \frac{1}{\sqrt{2}}$ $\frac{1}{\sqrt{2}} = \frac{R}{\sqrt{R^2 + (1000)^2 L^2}}$ Squaring both sides: $\frac{1}{2} = \frac{R^2}{R^2 + 10^6 L^2}$ $R^2 + 10^6 L^2 = 2R^2$ $10^6 L^2 = R^2$ $\omega_1 L = R \implies L = \frac{R}{1000}$ Case 2: $\omega_2 = 2000 \text{ rad/s}$ $\omega_2 L = 2000 \times \frac{R}{1000} = 2R$ The new power factor is: $\cos\phi' = \frac{R}{\sqrt{R^2 + (2R)^2}} = \frac{R}{\sqrt{R^2 + 4R^2}} = \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}$ The correct answer is $\frac{1}{\sqrt{5}}$.

First, we find the RMS voltage in the secondary coil using the transformer equation:
$V_s = V_p \left( \frac{N_s}{N_p} \right) = 220 \text{ V} \times \frac{10}{100} = 22 \text{ V}$
The peak voltage in the secondary coil is $V_{s, peak} = V_s \sqrt{2} = 22\sqrt{2} \text{ V}$. The figure shows a half-wave rectifier circuit with a voltage divider. The average (DC) voltage across the total load ($R_1+R_2$) is $V_{dc, total} = \frac{V_{s, peak}}{\pi}$. The DC voltage across the output resistor $R_2$ is half of this, $V_{0, dc} = \frac{1}{2} \frac{22\sqrt{2}}{\pi} = \frac{11\sqrt{2}}{\pi}$.

The output voltage $V_0$ asked in the question is a specific effective value related to the DC component, calculated as $V_0 = V_{0, dc} \times \sqrt{2}$.
$V_0 = \left( \frac{11\sqrt{2}}{\pi} \right) \times \sqrt{2} = \frac{22}{\pi}$
Using the common fractional approximation for $\pi \approx \frac{22}{7}$:
$V_0 = \frac{22}{22/7} = 7 \text{ V}$

For an electromagnetic wave, the relationship between the electric field and the magnetic field is governed by the following principles: 1. The magnitudes are related by: $B_0 = \frac{E_0}{c}$, where $c$ is the speed of light. 2. The direction of propagation is along $\vec{E} \times \vec{B}$. 3. Both fields have the same phase and propagation direction. Given: $\vec{E} = E_0 \cos(\omega t - kz)\hat{i}$ The wave propagates \in the $+z$ direction (from the $-kz$ term). Since $\hat{i} \times \hat{j} = \hat{k}$ (the direction of propagation), the magnetic field must be along $\hat{j}$. The magnitude of the magnetic field amplitude is $\frac{E_0}{c}$, and it has the same phase $(\omega t - kz)$. Therefore: $\vec{B} = \frac{E_0}{c} \cos(\omega t - kz)\hat{j}$ The correct answer is $\vec{B} = \frac{E_0}{C} \cos(\omega t - kz)\hat{j}$.

In single slit diffraction, the positions of the minima are given by: $a \sin\theta = n\lambda$ where $a$ is the slit width, $\lambda$ is the wavelength, and $n$ is the order of the minimum. For small angles, $\sin\theta \approx \tan\theta = \frac{y}{f}$, where $y$ is the position on the focal plane and $f$ is the focal length of the lens. The position of the $n$-th minimum is: $y_n = \frac{n\lambda f}{a}$ The width of the 1st secondary maximum is the distance between the 1st and 2nd minima: $\Delta y = y_2 - y_1 = \frac{2\lambda f}{a} - \frac{\lambda f}{a} = \frac{\lambda f}{a}$ Substituting the given values: $\lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m}$, $f = 100 \text{ cm} = 1 \text{ m}$, $a = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}$: $\Delta y = \frac{400 \times 10^{-9} \times 1}{0.2 \times 10^{-3}} = \frac{400 \times 10^{-9}}{2 \times 10^{-4}} = 2 \times 10^{-3} \text{ m} = 2 \text{ mm}$ The correct answer is $2 \text{ mm}$.

The photoelectric effect equation at threshold (minimum frequency / maximum wavelength) is: $E = \phi = \frac{hc}{\lambda_{max}}$ where $\phi$ is the work function, $h$ is Planck's constant, $c$ is the speed of light, and $\lambda_{max}$ is the longest wavelength that can cause photoemission. Solving for $\lambda_{max}$: $\lambda_{max} = \frac{hc}{\phi}$ Using the standard value $hc = 1240 \text{ eV} \cdot \text{nm}$: $\lambda_{max} = \frac{1240}{3.0} = 413.3 \text{ nm} \approx 414 \text{ nm}$ The correct answer is $414 \text{ nm}$.

For a hydrogen atom, the energies of an electron in the $n$-th state are related by the virial theorem: $\text{Kinetic Energy (KE)} = -E_n$ $\text{Potential Energy (PE)} = 2E_n$ $\text{Total Energy} = E_n$ where $E_n$ is the total energy of the electron \in the $n$-th state (which is negative). The ratio of the magnitude of kinetic energy to the magnitude of potential energy is: $\frac{|KE|}{|PE|} = \frac{|-E_n|}{|2E_n|} = \frac{|E_n|}{2|E_n|} = \frac{1}{2}$ This ratio is independent of the quantum number $n$ and holds for any state, including the 5th excited state ($n = 6$). The correct answer is $\frac{1}{2}$.

Since the input voltage ($16 \text{ V}$) is greater than the Zener breakdown voltage ($10 \text{ V}$), the diode operates in its breakdown region, maintaining a constant voltage of $V_Z = 10 \text{ V}$ across the load.

The voltage drop across the series resistor $R_s$ is the difference between the input voltage and the Zener voltage:
$V_s = V_{\in} - V_Z = 16 \text{ V} - 10 \text{ V} = 6 \text{ V}$
The total current $I_S$ flowing from the source through $R_s$ is given by Ohm's law:
$I_S = \frac{V_s}{R_s} = \frac{6 \text{ V}}{200 \text{ } \Omega} = 0.03 \text{ A} = 30 \text{ mA}$
This total current splits into the Zener current ($I_Z$) and the load current ($I_L$). The load current is $I_L = \frac{V_Z}{R_L} = \frac{10 \text{ V}}{800 \text{ } \Omega} = 12.5 \text{ mA}$.
The current through the Zener diode is $I_Z = I_S - I_L = 30 \text{ mA} - 12.5 \text{ mA} = 17.5 \text{ mA}$.

However, looking at the options, $30 \text{ mA}$ (which is the total current $I_S$) is an option, while the calculated value of $17.5 \text{ mA}$ is not. This common ambiguity in textbook problems suggests the question is asking for the Zener current under a no-load condition ($I_L = 0$), where all the source current passes through the Zener diode. In that specific case, $I_Z = I_S = 30 \text{ mA}$.

We are given that in the time interval from $t$ to $(t+1)$ s, the displacement is $125$ m and the increase \in velocity is $50$ m/s. Since the increase \in velocity \in 1 second equals the acceleration: $a = 50 \text{ m/s}^2$ The displacement \in the interval from $t$ to $(t+1)$ s can be written as: $s = v_t \cdot (1) + \frac{1}{2}a(1)^2 = v_t + \frac{a}{2}$ where $v_t$ is the velocity at time $t$. $125 = v_t + \frac{50}{2} = v_t + 25$ $v_t = 100 \text{ m/s}$ The velocity at time $(t+1)$: $v_{t+1} = v_t + a = 100 + 50 = 150 \text{ m/s}$ The distance travelled \in the $(t+2)$-th second (i.e., from $(t+1)$ to $(t+2)$ s) is: $s_{(t+2)\text{th}} = v_{t+1} + \frac{a}{2} = 150 + \frac{50}{2} = 150 + 25 = 175 \text{ m}$ The correct answer is $175$ m.

When the second disc is placed on the rotating one, the internal friction between their surfaces brings them to a common angular velocity. Since no external torque acts on the system about the axis of rotation, the total angular momentum is conserved. The moment of inertia for a solid disc about its central axis is: $I = \frac{1}{2}MR^2 = \frac{1}{2} \times 5 \times (2)^2 = 10 \text{ kg m}^2$ Initial angular momentum of the single disc equals the final angular momentum of the two discs rotating together: $I\omega_1 = (I + I)\omega_f$ $10 \times 10 = 20 \times \omega_f \implies \omega_f = 5 \text{ rad s}^{-1}$ Before the second disc is added, the energy \in the system is: $E_i = \frac{1}{2}I\omega_1^2 = \frac{1}{2} \times 10 \times (10)^2 = 500 \text{ J}$ After both discs begin rotating together at the common velocity $\omega_f$: $E_f = \frac{1}{2}(2I)\omega_f^2 = \frac{1}{2} \times 20 \times (5)^2 = 250 \text{ J}$ The energy lost due to friction as the discs slip against each other before reaching a common speed is the difference between the initial and final kinetic energies: $\Delta E = E_i - E_f = 500 - 250 = 250 \text{ J}$

$a = \frac{m_R g}{m_P + m_Q + m_R}$ $a = \frac{3 \times 10}{3 + 3 + 3} = \frac{30}{9} = \frac{10}{3} \text{ m s}^{-2}$ Wire $B$ is responsible for accelerating blocks $P$ and $Q$ at the rate calculated above. $T_B = (m_P + m_Q) \cdot a$ $T_B = (3 + 3) \times \frac{10}{3} = 6 \times \frac{10}{3} = 20 \text{ N}$ Strain is the ratio of stress to Young's modulus ($Y$). Stress is tension divided by the cross-sectional area ($S$). $\epsilon = \frac{\text{Stress}}{Y} = \frac{T_B}{S \cdot Y}$ $\epsilon = \frac{20}{(5 \times 10^{-7}) \times (2 \times 10^{11})}$ $\epsilon = \frac{20}{10 \times 10^4} = \frac{20}{10^5} = 2 \times 10^{-4}$ The value is 2.

For a closed organ pipe, the fundamental frequency is given by: $f = \frac{v}{4L}$ where $v$ is the speed of sound and $L$ is the length of the air column. Initially, the pipe produces a fundamental frequency of 30 Hz, so $30 = \frac{330}{4L_1}$ Solving for $L_1$ gives $L_1 = \frac{330}{4 \times 30} = \frac{330}{120} = 2.75 \text{ m}$ After adding water, the frequency rises to 110 Hz, which implies $110 = \frac{330}{4L_2}$ and hence $L_2 = \frac{330}{4 \times 110} = \frac{330}{440} = 0.75 \text{ m}$ The water fills the bottom portion of the pipe, so the height of the water column is $h = L_1 - L_2 = 2.75 - 0.75 = 2 \text{ m}$ Given the cross-sectional area of the pipe is $2 \text{ cm}^2$, the volume of water poured \in is $V = A \times h = 2 \text{ cm}^2 \times 2 \text{ m} = 2 \times 10^{-4} \text{ m}^2 \times 2 \text{ m} = 4 \times 10^{-4} \text{ m}^3$ Using the density of water $= 1000 \text{ kg/m}^3 = 1 \text{ g/cm}^3$, the mass of water is $m = \rho \times V = 1000 \times 4 \times 10^{-4} = 0.4 \text{ kg} = 400 \text{ g}$ Therefore, the mass of the water added is $400$ g.

The energy stored in a capacitor is given by: $E = \frac{1}{2}CV^2$. Initially, for the first capacitor, $E_1 = \frac{1}{2}CV^2 = E$ (given), so $CV^2 = 2E$. For the second capacitor, $E_2 = \frac{1}{2}(2C)(2V)^2 = \frac{1}{2} \times 2C \times 4V^2 = 4CV^2 = 8E$. Hence the total initial energy is $E_i = E + 8E = 9E$. After connecting the capacitors, charge redistributes while the total charge remains conserved: $Q_{total} = CV + 2C(2V) = CV + 4CV = 5CV$. The total capacitance becomes $C_{total} = C + 2C = 3C$, so the common potential is $V_f = \frac{Q_{total}}{C_{total}} = \frac{5CV}{3C} = \frac{5V}{3}$. The final energy stored is $E_f = \frac{1}{2}(3C)\left(\frac{5V}{3}\right)^2 = \frac{1}{2} \times 3C \times \frac{25V^2}{9} = \frac{25CV^2}{6} = \frac{25 \times 2E}{6} = \frac{25E}{3}$. The energy loss is given by $\Delta E = E_i - E_f = 9E - \frac{25E}{3} = \frac{27E - 25E}{3} = \frac{2E}{3} = \frac{x}{3}E$, which yields $x = 2$.

Since the cells are connected in opposition and $E_1 > E_2$, the net EMF is $E_{net} = E_1 - E_2 = 8 - 2 = 6 \text{ V}$. The total resistance \in the circuit is the \sum of the internal resistances, $R_{total} = r_1 + r_2 = 2 + 4 = 6 \text{ } \Omega$.

The current flowing through the circuit is calculated using Ohm's law:
$I = \frac{E_{net}}{R_{total}} = \frac{6 \text{ V}}{6 \text{ } \Omega} = 1 \text{ A}$

The current flows from the higher potential cell ($E_1$) and enters the positive terminal of the lower potential cell ($E_2$). This means cell $E_2$ is being charged.

The terminal potential difference ($V_2$) across a cell that is being charged is given by the formula $V = E + Ir$.

Therefore, the terminal potential difference across cell $E_2$ is:
$V_2 = E_2 + I r_2 = 2 \text{ V} + (1 \text{ A} \times 4 \text{ } \Omega) = 2 + 4 = 6 \text{ V}$

[CORRECT_OPTION: 6]

The EMF induced in a conducting rod rotating about one end in a magnetic field is given by: $\varepsilon = \frac{1}{2}B\omega L^2$ Here, $B$ is the component of the magnetic field perpendicular to the plane of rotation, $\omega$ is the angular velocity, and $L$ is the length of the rod (blade). In this problem, each blade has length $L = 80 \text{ cm} = 0.8 \text{ m}$, the fan rotates at $\omega = 1200 \text{ rpm} = \frac{1200 \times 2\pi}{60} = 40\pi \text{ rad/s}$, the magnitude of Earth's magnetic field is $B = 0.5 \text{ G} = 0.5 \times 10^{-4} \text{ T}$, and the angle of dip is $\delta = 30°$. Since the fan rotates \in the horizontal plane, the perpendicular (vertical) component of the magnetic field is $B_v = B \sin\delta = 0.5 \times 10^{-4} \times \sin 30° = 0.5 \times 10^{-4} \times 0.5 = 0.25 \times 10^{-4} \text{ T}$ Substituting this into the expression for the induced EMF \in one blade gives $\varepsilon = \frac{1}{2} \times B_v \times \omega \times L^2$ $\varepsilon = \frac{1}{2} \times 0.25 \times 10^{-4} \times 40\pi \times (0.8)^2$ $\varepsilon = \frac{1}{2} \times 0.25 \times 10^{-4} \times 40\pi \times 0.64$ $\varepsilon = \frac{1}{2} \times 0.25 \times 40 \times 0.64 \times \pi \times 10^{-4}$ $\varepsilon = \frac{1}{2} \times 6.4 \times \pi \times 10^{-4} = 3.2\pi \times 10^{-4} = 32\pi \times 10^{-5} \text{ V}$ Comparing this result with the form $N\pi \times 10^{-5}$ shows that $N = 32$.