JEE Main 2013 22nd April Morning Shift Question Paper with Solutions

Let's find the dimensions of angular momentum, latent heat, and capacitance step by step.

First, angular momentum is given by the formula: angular momentum = radius × linear momentum. Radius has the dimension of length, so its dimension is $L$. Linear momentum is mass × velocity. Mass has dimension $M$, and velocity is distance over time, so its dimension is $LT^{-1}$. Therefore, linear momentum has dimension $M \times LT^{-1} = MLT^{-1}$. Now, angular momentum = radius × linear momentum = $L \times MLT^{-1} = ML^2T^{-1}$.

Second, latent heat is the heat energy absorbed or released per unit mass during a phase change. Heat energy has the same dimension as work, which is force × distance. Force is mass × acceleration, so force has dimension $M \times LT^{-2} = MLT^{-2}$. Work = force × distance = $MLT^{-2} \times L = ML^2T^{-2}$. Latent heat = heat energy / mass = $ML^2T^{-2} / M = L^2T^{-2}$.

Third, capacitance is defined as charge divided by voltage. Charge has dimension $AT$ (since current I is charge per time, so charge = current × time). Voltage (potential difference) is work done per unit charge, so voltage = work / charge = $ML^2T^{-2} / AT = ML^2T^{-3}A^{-1}$. Therefore, capacitance = charge / voltage = $AT / (ML^2T^{-3}A^{-1}) = AT \times M^{-1}L^{-2}T^{3}A = M^{-1}L^{-2}T^{4}A^{2}$.

So, the dimensions are:

• Angular momentum: $ML^2T^{-1}$
• Latent heat: $L^2T^{-2}$
• Capacitance: $M^{-1}L^{-2}T^{4}A^{2}$

Now, comparing with the options:

• Option A: $ML^2T^1A^2$ (incorrect angular momentum exponent for T), $L^2T^{-2}$ (correct latent heat), $M^{-1}L^{-2}T^2$ (incorrect capacitance exponent for T)
• Option B: $ML^2T^{-2}$ (incorrect angular momentum exponent for T), $L^2T^2$ (incorrect latent heat exponent for T), $M^{-1}L^{-2}T^4A^2$ (correct capacitance)
• Option C: $ML^2T^{-1}$ (correct angular momentum), $L^2T^{-2}$ (correct latent heat), $ML^2TA^2$ (incorrect capacitance exponents and sign)
• Option D: $ML^2T^{-1}$ (correct angular momentum), $L^2T^{-2}$ (correct latent heat), $M^{-1}L^{-2}T^4A^2$ (correct capacitance)

Hence, the correct answer is Option D.

A ball is projected from the ground at an angle of 45°. The point of projection is 4 m from the foot of the wall, and the ball lands 6 m beyond the wall. The total horizontal distance covered (range) is the sum of these distances: 4 m + 6 m = 10 m.

For a projectile launched at an angle θ with initial velocity u, the range R is given by:

$R = \frac{u^2 \sin 2\theta}{g}$

Here, θ = 45°, so 2θ = 90° and sin 90° = 1. Thus:

$R = \frac{u^2}{g}$

Given R = 10 m:

$10 = \frac{u^2}{g}$

Solving for u²:

$u^2 = 10g \quad \text{(Equation 1)}$

The trajectory equation for a projectile is:

$y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$

With θ = 45°, tan 45° = 1 and cos 45° = $\frac{1}{\sqrt{2}}$, so cos² 45° = $\left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}$. Substituting these values:

$y = x \cdot 1 - \frac{g x^2}{2 u^2 \cdot \frac{1}{2}}$

Simplifying:

$y = x - \frac{g x^2}{u^2} \quad \text{(Equation 2)}$

The ball just clears the wall at x = 4 m. Substituting x = 4 into Equation 2:

$y = 4 - \frac{g (4)^2}{u^2}$ $y = 4 - \frac{16g}{u^2}$

From Equation 1, u² = 10g. Substituting this:

$y = 4 - \frac{16g}{10g}$

The g cancels out:

$y = 4 - \frac{16}{10}$

Simplifying the fraction:

$\frac{16}{10} = 1.6$

Thus:

$y = 4 - 1.6 = 2.4 \text{ m}$

Therefore, the height of the wall is 2.4 m.

Comparing with the options:

A. 4.4 m

B. 2.4 m

C. 3.6 m

D. 1.6 m

Hence, the correct answer is Option B.

Total Mass ($M_{total}$): $M_1 + M_2 + M_{rod} = 20\text{ kg} + 12\text{ kg} + 8\text{ kg} = 40\text{ kg}$.

Total Weight ($W$): $M_{total} \times g = 40\text{ kg} \times 10\text{ m/s}^2 = 400\text{ N}$.

$F_{net} = 480\text{ N} - 400\text{ N} = 80\text{ N}$

Acceleration ($a$): $a = \frac{F_{net}}{M_{total}} = \frac{80\text{ N}}{40\text{ kg}} = 2\text{ m/s}^2 \text{ (upwards)}$

To find the tension ($T$) at the midpoint of the rod, we isolate the part of the system below that point. This lower section includes block $M_2$ and the bottom half of the rod.

Mass of the lower section ($m'$): $m' = M_2 + \frac{M_{rod}}{2} = 12\text{ kg} + \frac{8\text{ kg}}{2} = 12 + 4 = 16\text{ kg}$
$T - m'g = m'a$

$T = m'(g + a)$

$T = 16 \times (10 + 2)$

$T = 16 \times 12 = 192\text{ N}$

A body starts from rest on an inclined plane with a slope of 45°. The coefficient of friction is given by $\mu = 0.3x$, where $x$ is the distance traveled down the plane. We need to find the distance $x$ at which the body achieves maximum speed, with $g = 10 \text{m/s}^2$.

First, consider the forces acting on the body. The gravitational force component pulling the body down the plane is $mg \sin 45^\circ$. Since $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$, this component is $mg \cdot \frac{1}{\sqrt{2}} = \frac{mg}{\sqrt{2}}$.

The frictional force opposes the motion and is given by $\mu \times \text{normal reaction}$. The normal reaction is $mg \cos 45^\circ = mg \cdot \frac{1}{\sqrt{2}} = \frac{mg}{\sqrt{2}}$. Substituting $\mu = 0.3x$, the frictional force is $0.3x \cdot \frac{mg}{\sqrt{2}}$.

The net force down the plane is the gravitational component minus the frictional force:

$F_{\text{net}} = \frac{mg}{\sqrt{2}} - 0.3x \cdot \frac{mg}{\sqrt{2}} = \frac{mg}{\sqrt{2}} (1 - 0.3x)$

By Newton's second law, $F_{\text{net}} = ma$, so:

$ma = \frac{mg}{\sqrt{2}} (1 - 0.3x)$

Dividing both sides by $m$:

$a = \frac{g}{\sqrt{2}} (1 - 0.3x)$

Substituting $g = 10 \text{m/s}^2$:

$a = \frac{10}{\sqrt{2}} (1 - 0.3x)$

Simplifying $\frac{10}{\sqrt{2}}$ by multiplying numerator and denominator by $\sqrt{2}$:

$\frac{10}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$

So:

$a = 5\sqrt{2} (1 - 0.3x) = 5\sqrt{2} - 1.5\sqrt{2} x$

Acceleration $a$ is the derivative of velocity with respect to time, but we need velocity as a function of distance $x$. Using the chain rule:

$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$

So:

$v \frac{dv}{dx} = 5\sqrt{2} - 1.5\sqrt{2} x$

Factor out $\sqrt{2}$:

$v \frac{dv}{dx} = \sqrt{2} (5 - 1.5x)$

To avoid decimals, note that $1.5 = \frac{3}{2}$, so:

$v \frac{dv}{dx} = \sqrt{2} \left(5 - \frac{3}{2}x\right)$

Separate variables:

$v dv = \sqrt{2} \left(5 - \frac{3}{2}x\right) dx$

Integrate both sides. The body starts from rest at $x = 0$, so initial conditions are $v = 0$ when $x = 0$:

$\int_0^v v dv = \sqrt{2} \int_0^x \left(5 - \frac{3}{2}x\right) dx$

Left side:

$\left[ \frac{v^2}{2} \right]_0^v = \frac{v^2}{2} - 0 = \frac{v^2}{2}$

Right side:

$\sqrt{2} \left[ 5x - \frac{3}{2} \cdot \frac{x^2}{2} \right]_0^x = \sqrt{2} \left[ 5x - \frac{3}{4} x^2 \right] - 0$

So:

$\frac{v^2}{2} = \sqrt{2} \left(5x - \frac{3}{4} x^2\right)$

Multiply both sides by 2:

$v^2 = 2\sqrt{2} \left(5x - \frac{3}{4} x^2\right) = 2\sqrt{2} \cdot 5x - 2\sqrt{2} \cdot \frac{3}{4} x^2 = 10\sqrt{2} x - \frac{6\sqrt{2}}{4} x^2 = 10\sqrt{2} x - \frac{3\sqrt{2}}{2} x^2$

Factor out $\sqrt{2}$:

$v^2 = \sqrt{2} \left(10x - \frac{3}{2} x^2\right)$

Since $v$ is non-negative, $v^2$ is maximum when $v$ is maximum. Let $s = v^2 = \sqrt{2} \left(10x - \frac{3}{2} x^2\right)$. To find the maximum, take the derivative of $s$ with respect to $x$ and set it to zero:

$\frac{ds}{dx} = \sqrt{2} \cdot \frac{d}{dx}\left(10x - \frac{3}{2} x^2\right) = \sqrt{2} \left(10 - \frac{3}{2} \cdot 2x\right) = \sqrt{2} (10 - 3x)$

Set $\frac{ds}{dx} = 0$:

$\sqrt{2} (10 - 3x) = 0 \implies 10 - 3x = 0 \implies 3x = 10 \implies x = \frac{10}{3} \approx 3.33 \text{m}$

To confirm it is a maximum, check the second derivative:

$\frac{d^2s}{dx^2} = \sqrt{2} \cdot (-3) = -3\sqrt{2} < 0$

Since the second derivative is negative, $s$ (and thus $v$) is maximum at $x = \frac{10}{3} \text{m}$.

Comparing with the options:

A. 9.8 m

B. 27 m

C. 12 m

D. 3.33 m

Hence, the correct answer is Option D.

Total energy at O is potential, and at A it is a mix of translational and rotational kinetic energy:

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{3}mR^2\right)\left(\frac{v}{R}\right)^2$

$mgh = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = \frac{5}{6}mv^2$

$v^2 = \frac{6gh}{5} = \frac{6 \times 10 \times 1.8}{5} = \frac{108}{5} = 21.6$

$v = \sqrt{21.6} \approx 4.65\text{ m/s}$

$R_{range} = \frac{v^2 \sin(2\theta)}{g}$

$AB = \frac{21.6 \times \sin(60^\circ)}{10}$

$AB = 2.16 \times 0.866 \approx 1.87\text{ m}$

Position 1: Solar Eclipse

The Moon is located between the Earth and the Sun. In this case, both the Sun and the Moon pull the Earth in the same direction (towards the Sun).

$a_1 = a_s + a_m$

Position 2: Other side (Full Moon)

The Moon is on the opposite side of the Earth, away from the Sun. The Sun pulls the Earth towards itself, while the Moon pulls the Earth in the opposite direction.

$a_2 = a_s - a_m$

The change in the value of acceleration ($\Delta a$) is $\Delta a = |a_1 - a_2|$

$\Delta a = (a_s + a_m) - (a_s - a_m)$

$\Delta a = 2 \cdot a_m$

The change is simply twice the acceleration produced by the Moon on the Earth.

The acceleration produced by the Moon on the Earth is given by Newton's Law of Universal Gravitation:

$a_m = \frac{G \cdot M_{moon}}{r^2}$

$a_m = \frac{6.6 \times 10^{-11} \cdot 7.36 \times 10^{22}}{(3.8 \times 10^8)^2}$
$a_m = \frac{48.576 \times 10^{11}}{14.44 \times 10^{16}}$
$a_m \approx 3.364 \times 10^{-5} \text{ m/s}^2$

$\Delta a = 2 \cdot a_m$

$\Delta a = 6.73 \times 10^{-5} \text{ m/s}^2$

A uniform wire is subjected to a longitudinal tensile stress, and we are given Young's modulus $Y = 2 \times 10^{11}$ N/m² and stress $\sigma = 5 \times 10^7$ N/m². The overall volume change is 0.02%, which means $\frac{\Delta V}{V} = \frac{0.02}{100} = 0.0002$. We need to find the fractional decrease in the radius, which is $\left| \frac{\Delta r}{r} \right|$.

When a wire is stretched, the volume change relates to Poisson's ratio ($\nu$) and the longitudinal strain. The formula for volume change is:

$\frac{\Delta V}{V} = (1 - 2\nu) \cdot \epsilon_{\text{long}}$

where $\epsilon_{\text{long}}$ is the longitudinal strain. Using Hooke's law, $\sigma = Y \cdot \epsilon_{\text{long}}$, so $\epsilon_{\text{long}} = \frac{\sigma}{Y}$.

First, calculate $\epsilon_{\text{long}}$:

$\epsilon_{\text{long}} = \frac{\sigma}{Y} = \frac{5 \times 10^7}{2 \times 10^{11}} = \frac{5}{2} \times 10^{7-11} = 2.5 \times 10^{-4}$

Now substitute into the volume change formula:

$\frac{\Delta V}{V} = (1 - 2\nu) \cdot \epsilon_{\text{long}}$

$0.0002 = (1 - 2\nu) \cdot (2.5 \times 10^{-4})$

Solve for $1 - 2\nu$:

$1 - 2\nu = \frac{0.0002}{2.5 \times 10^{-4}} = \frac{2 \times 10^{-4}}{2.5 \times 10^{-4}} = \frac{2}{2.5} = 0.8$

Now solve for $\nu$:

$1 - 2\nu = 0.8$

$-2\nu = 0.8 - 1 = -0.2$

$\nu = \frac{-0.2}{-2} = 0.1$

Poisson's ratio $\nu$ is 0.1. The fractional decrease in radius is given by the magnitude of the lateral strain. Poisson's ratio relates lateral strain ($\epsilon_{\text{lat}}$) to longitudinal strain:

$\nu = -\frac{\epsilon_{\text{lat}}}{\epsilon_{\text{long}}}$

So,

$\epsilon_{\text{lat}} = -\nu \cdot \epsilon_{\text{long}}$

The fractional change in radius is $\frac{\Delta r}{r} = \epsilon_{\text{lat}}$, so the fractional decrease is $\left| \frac{\Delta r}{r} \right| = \nu \cdot \epsilon_{\text{long}}$.

Substitute the values:

$\left| \frac{\Delta r}{r} \right| = \nu \cdot \epsilon_{\text{long}} = 0.1 \cdot (2.5 \times 10^{-4}) = 0.25 \times 10^{-4}$

This matches option C. Therefore, the fractional decrease in the radius is $0.25 \times 10^{-4}$.

Hence, the correct answer is Option C.

To solve this problem, we need to find the pressure difference between the upper and lower sides of the aeroplane wings. We are given that the density of air, ρ, is 1.2 kg m⁻³. The speed of air above the wing is 150 m s⁻¹, and below the wing is 100 m s⁻¹.

We use Bernoulli's theorem, which states that for an ideal fluid in steady flow, the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant along a streamline. Since the wings are horizontal, the height difference is negligible, so we can ignore the potential energy term. Therefore, Bernoulli's equation simplifies to:

$P + \frac{1}{2} \rho v^2 = \text{constant}$

Applying this equation to points above and below the wing:

$P_{\text{above}} + \frac{1}{2} \rho v_{\text{above}}^2 = P_{\text{below}} + \frac{1}{2} \rho v_{\text{below}}^2$

We rearrange this equation to find the pressure difference, $P_{\text{below}} - P_{\text{above}}$:

$P_{\text{below}} - P_{\text{above}} = \frac{1}{2} \rho v_{\text{above}}^2 - \frac{1}{2} \rho v_{\text{below}}^2$

Factor out $\frac{1}{2} \rho$:

$P_{\text{below}} - P_{\text{above}} = \frac{1}{2} \rho \left( v_{\text{above}}^2 - v_{\text{below}}^2 \right)$

Now substitute the given values: ρ = 1.2 kg m⁻³, $v_{\text{above}} = 150$ m s⁻¹, and $v_{\text{below}} = 100$ m s⁻¹.

First, calculate $v_{\text{above}}^2$:

$v_{\text{above}}^2 = (150)^2 = 22500$

Next, calculate $v_{\text{below}}^2$:

$v_{\text{below}}^2 = (100)^2 = 10000$

Now find the difference:

$v_{\text{above}}^2 - v_{\text{below}}^2 = 22500 - 10000 = 12500$

Substitute into the equation:

$P_{\text{below}} - P_{\text{above}} = \frac{1}{2} \times 1.2 \times 12500$

First, multiply 1.2 and 12500:

$1.2 \times 12500 = 1.2 \times (125 \times 100) = (1.2 \times 125) \times 100$

$1.2 \times 125 = 1.2 \times (100 + 25) = (1.2 \times 100) + (1.2 \times 25) = 120 + 30 = 150$

Then, $150 \times 100 = 15000$, so:

$1.2 \times 12500 = 15000$

Now multiply by $\frac{1}{2}$:

$\frac{1}{2} \times 15000 = 7500$

Therefore, the pressure difference is 7500 N m⁻².

Comparing with the options:

A. 60 Nm⁻²

B. 180 Nm⁻²

C. 7500 Nm⁻²

D. 12500 Nm⁻²

Hence, the correct answer is Option C.

To find the change in internal energy for 1 gram of water vaporizing at 373 K, we use the first law of thermodynamics. The first law states that the change in internal energy, denoted as $\Delta U$, is equal to the heat absorbed by the system, $Q$, minus the work done by the system, $W$. This is expressed as:

$\Delta U = Q - W$

In this process, water changes from liquid to vapor at constant temperature (373 K) and constant atmospheric pressure. The heat absorbed, $Q$, is the latent heat of vaporization. Given that the latent heat is 2256 J per gram and we have 1 gram of water, we have:

$Q = 2256 \text{J}$

Next, we calculate the work done by the system, $W$. When water vaporizes, it expands against the constant atmospheric pressure. The work done in an expansion at constant pressure is given by:

$W = P \Delta V$

where $P$ is the pressure and $\Delta V$ is the change in volume. The volume of 1 gram of water in the liquid phase is given as 1 cm³, and in the vapor phase, it is 1671 cm³. The change in volume is:

$\Delta V = V_{\text{vapour}} - V_{\text{liquid}} = 1671 \text{cm}^3 - 1 \text{cm}^3 = 1670 \text{cm}^3$

We need consistent SI units for the calculation. Convert volume from cm³ to m³. Since 1 m³ = 10^6 cm³, we have:

$\Delta V = 1670 \text{cm}^3 = 1670 \times 10^{-6} \text{m}^3 = 1.670 \times 10^{-3} \text{m}^3$

The pressure is atmospheric pressure. For simplicity in such problems, we often take 1 atmosphere as exactly $10^5$ Pa (Pascals). So,

$P = 10^5 \text{Pa}$

Now, calculate the work done:

$W = P \Delta V = (10^5 \text{Pa}) \times (1.670 \times 10^{-3} \text{m}^3)$

Multiply the numbers:

$10^5 \times 1.670 \times 10^{-3} = 1.670 \times 10^{5 - 3} = 1.670 \times 10^2 = 167 \text{J}$

So, the work done by the system is 167 J.

Now, substitute the values of $Q$ and $W$ into the first law equation to find the change in internal energy:

$\Delta U = Q - W = 2256 \text{J} - 167 \text{J} = 2089 \text{J}$

Hence, the change in internal energy for 1 gram of water vaporizing at 373 K is 2089 J.

Comparing with the options, 2089 J corresponds to option C.

Hence, the correct answer is Option C.

We are given an ideal gas initially at atmospheric pressure (1 atm) that undergoes adiabatic compression. After compression, the density becomes 32 times its initial value, and the final pressure is 128 atm. We need to find the value of $\gamma$, the adiabatic index (ratio of specific heats).

For an adiabatic process in an ideal gas, the relationship between pressure and volume is $P V^{\gamma} = \text{constant}$. Since density $\rho$ is mass per unit volume and mass is constant, volume $V$ is inversely proportional to density, i.e., $V \propto \frac{1}{\rho}$. Substituting this into the adiabatic relation gives:

$P \left( \frac{1}{\rho} \right)^{\gamma} = \text{constant}$

This simplifies to:

$P \rho^{-\gamma} = \text{constant}$

Therefore, for the initial state (1) and final state (2):

$P_1 \rho_1^{-\gamma} = P_2 \rho_2^{-\gamma}$

Rearranging this equation:

$\frac{P_1}{P_2} = \left( \frac{\rho_2}{\rho_1} \right)^{\gamma}$

Given that the final density $\rho_2$ is 32 times the initial density $\rho_1$, we have $\rho_2 = 32 \rho_1$, so:

$\frac{\rho_2}{\rho_1} = 32$

Substituting the known pressures: initial pressure $P_1 = 1$ atm and final pressure $P_2 = 128$ atm, we get:

$\frac{1}{128} = (32)^{\gamma}$

Note that $(32)^{\gamma}$ is equivalent to $32^{\gamma}$. We can rewrite the equation as:

$32^{\gamma} = 128$

Now, express both sides as powers of 2. Since $32 = 2^5$ and $128 = 2^7$, substitute:

$(2^5)^{\gamma} = 2^7$

Using the exponent rule $(a^m)^n = a^{m n}$, simplify:

$2^{5\gamma} = 2^7$

Since the bases are equal, set the exponents equal:

$5\gamma = 7$

Solving for $\gamma$:

$\gamma = \frac{7}{5} = 1.4$

Hence, the value of $\gamma$ is 1.4. Comparing with the options: A (1.5), B (1.4), C (1.3), D (1.6), the correct answer is Option B.

 Two Isobars: Processes BC and AD are horizontal lines.

One Isochore: Process AB is a vertical line.

One Isothermal: Process CD is a hyperbolic curve.

The mass $m$ tends to get detached from the pan when the normal force between them becomes zero. In simple harmonic motion (SHM), this occurs at the highest point of the oscillation where the downward acceleration of the pan exceeds the acceleration due to gravity $g$.

$a_{max} = \omega^2 A$

$a_{max} \geq g \implies \omega^2 A \geq g$

$\omega^2 = \frac{k}{m}$

$\omega^2 = \frac{500}{1.0} = 500 \text{ rad}^2/\text{s}^2$

$500 \times A \geq 10$

$A \geq \frac{10}{500} \text{ m}$

$A \geq 0.02 \text{ m} = 2.0 \text{ cm}$

The apparent frequency ($f$) heard by a stationary listener when a source moves is given by $f' = f \left( \frac{v}{v \mp v_s} \right)$

Case 1: A moves towards C

$f'_{A1} = 500 \left( \frac{340}{340 - 4} \right) = 500 \left( \frac{340}{336} \right) \approx \mathbf{506\text{ Hz}}$

Case 2: A moves away from C

$f'_{A2} = 500 \left( \frac{340}{340 + 4} \right) = 500 \left( \frac{340}{344} \right) \approx \mathbf{494\text{ Hz}}$

Beat frequency is the absolute difference between two frequencies: $f_{\text{beat}} = |f_A' - f_B|$

From Case 1 (6 beats): $|506 - f_B| = 6 \implies f_B = 506 + 6 = \mathbf{512\text{ Hz}} \text{ or } 506 - 6 = 500\text{ Hz}$

From Case 2 (18 beats): $|494 - f_B| = 18 \implies f_B = 494 + 18 = \mathbf{512\text{ Hz}} \text{ or } 494 - 18 = 476\text{ Hz}$

Comparing both scenarios, the only consistent value for the frequency of source B is $512\text{ Hz}$.

$V_{\text{center}} = V_q + V_{\text{induced}}$

For an uncharged spherical conductor, the net induced charge on the surface is zero. Due to the symmetry of the sphere, the potential at its center due to these induced surface charges is $V_{\text{induced}} = \int \frac{k \cdot dq_{\text{induced}}}{R} = \frac{k}{R} \int dq_{\text{induced}} = 0$

Thus, the potential at the center is solely due to the point charge:

$V_{\text{center}} = \frac{kq}{r} = \frac{9 \times 10^9 \times 10^{-6}}{0.04}$

$V_{\text{center}} = \frac{9 \times 10^3}{4 \times 10^{-2}} = 2.25 \times 10^5\text{ V}$

$\vec{E}_{\text{net}} = \vec{E}_q + \vec{E}_{\text{induced}} = 0$

$\vec{E}_{\text{induced}} = -\vec{E}_q$

$E_q = \frac{kq}{r^2} = \frac{9 \times 10^9 \times 10^{-6}}{(0.04)^2}$

$E_q = \frac{9 \times 10^3}{16 \times 10^{-4}} = 0.5625 \times 10^7 = 5.625 \times 10^6\text{ V/m}$

$E_{\text{induced}} = -5.625 \times 10^6\text{ V/m}$

Let distance between charges be

2r

So midpoint is at distance

r

from each charge.

Potential at midpoint due to one charge:

$V_1=\frac{kq}{r}$

From both charges,

$V=\frac{2kq}{r}$

Now find r.

For one charge in equilibrium:

Horizontal balance:

$T\sin\theta=F_e$

Vertical balance:

$T\cos\theta=mg$

Dividing,

$\tan\theta=\frac{F_e}{mg}$

Electrostatic repulsion between charges:

$F_e=\frac{kq^2}{(2r)^2}=\frac{kq^2}{4r^2}$

So

$mg\tan\theta=\frac{kq^2}{4r^2}$

Thus

$r^2=\frac{kq^2}{4mg\tan\theta}$

$r=\frac{q}{2}\sqrt{\frac{k}{mg\tan\theta}}$

Substitute into

$V=\frac{2kq}{r}$

$V=\frac{2kq}{\frac{q}{2}\sqrt{\frac{k}{mg\tan\theta}}}$

$=4k\sqrt{\frac{mg\tan\theta}{k}}$

$=4\sqrt{kmg\tan\theta}$

The relationship between the current through a capacitor and the rate of change of the potential difference across its plates is given by the formula:

$i = C \frac{dV}{dt}$

where:

• $i$ is the instantaneous current,
• $C$ is the capacitance,
• $\frac{dV}{dt}$ is the rate of change of potential difference with respect to time.

Given in the question:

• Instantaneous current, $i = 2$ A,
• Capacitance, $C = 1 \mu F = 1 \times 10^{-6}$ F (since 1 μF = 10^{-6} F).

Substitute these values into the formula:

$2 = (1 \times 10^{-6}) \times \frac{dV}{dt}$

To solve for $\frac{dV}{dt}$, rearrange the equation:

$\frac{dV}{dt} = \frac{2}{1 \times 10^{-6}}$

Dividing by $10^{-6}$ is equivalent to multiplying by $10^6$:

$\frac{dV}{dt} = 2 \times 10^6$

Therefore, the potential difference across the capacitor plates must be changed at the rate of $2 \times 10^6$ V/s.

Comparing with the options:

• A. $2 \times 10^4$ V/s
• B. $4 \times 10^6$ V/s
• C. $2 \times 10^6$ V/s
• D. $4 \times 10^4$ V/s

Hence, the correct answer is Option C.

We are given a DC source with emf $E_1 = 100$ V and internal resistance $r = 0.5 \, \Omega$, a storage battery with emf $E_2 = 90$ V, and an external resistance $R$. The goal is to find the value of $R$ such that no current passes through the battery. This means the battery $E_2$ is effectively inactive, and the voltage across its terminals must equal its emf, $90$ V.

In the circuit, the DC source $E_1$ with internal resistance $r$ is connected in series with the external resistance $R$. The battery $E_2$ is connected in parallel with $R$. Therefore, the voltage across $R$ must equal $E_2$ for no current to flow through the battery. So, we set the voltage across $R$ equal to $90$ V.

The total resistance in the series circuit is $r + R = 0.5 + R$ ohms. The current $I$ flowing through the circuit is given by Ohm's law:

$I = \frac{E_1}{r + R} = \frac{100}{0.5 + R}$

The voltage across $R$ is $V_R = I \times R$. Setting this equal to $E_2$:

$V_R = \frac{100}{0.5 + R} \times R = 90$

Now, solve for $R$:

$\frac{100R}{0.5 + R} = 90$

Multiply both sides by $0.5 + R$:

$100R = 90 \times (0.5 + R)$

Calculate $90 \times 0.5 = 45$:

$100R = 45 + 90R$

Subtract $90R$ from both sides:

$100R - 90R = 45$

$10R = 45$

Divide both sides by 10:

$R = \frac{45}{10} = 4.5 \, \Omega$

Thus, when $R = 4.5 \, \Omega$, no current flows through the battery. Comparing with the options:

A. $5.5 \, \Omega$

B. $3.5 \, \Omega$

C. $4.5 \, \Omega$

D. $2.5 \, \Omega$

Hence, the correct answer is Option C.

Initial state ($R_3 = 0$):

In this case, the galvanometer is in parallel with $R_2$. The total resistance of the circuit is $R_{total} = R_1 + \frac{G R_2}{G + R_2}$. Since $R_1 \gg R_2$, the current from the battery is $I \approx \frac{V}{R_1}$. 

The current through the galvanometer ($I_g$) corresponding to deflection $d$ is $I_g = I \left( \frac{R_2}{G + R_2} \right) = \frac{V}{R_1} \left( \frac{R_2}{G + R_2} \right) \quad \dots (1)$

Final state ($R_3 = 107\ \Omega$):

The new current through the galvanometer ($I_g'$) corresponding to half deflection $d/2$ is $I_g' = I \left( \frac{R_2}{G + R_3 + R_2} \right) = \frac{V}{R_1} \left( \frac{R_2}{G + 107 + R_2} \right) \quad \dots (2)$

Since $I_g' = \frac{1}{2} I_g$, $\frac{R_2}{G + 107 + R_2} = \frac{1}{2} \left( \frac{R_2}{G + R_2} \right)$

$2(G + R_2) = G + 107 + R_2$