JEE Main 2014 9th April Morning Shift Question Paper with Solutions

The formula for the time period $T$ of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$. Squaring both sides gives $T^2 = 4\pi^2 \frac{L}{g}$, and rearranging for $g$ yields $g = 4\pi^2 \frac{L}{T^2}$.

To find the error in $g$, we use the relative error formula. The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.

First, determine the error in length $L$. The length $L = 20.0 \text{cm}$ is measured with a meter scale of least count 1 mm = 0.1 cm. The absolute error $\Delta L$ is the least count, so $\Delta L = 0.1 \text{cm}$. The relative error in $L$ is $\frac{\Delta L}{L} = \frac{0.1}{20.0} = 0.005$.

Next, determine the error in the time period $T$. The time for 100 oscillations is $t = 90.0 \text{seconds}$, measured with a watch of least count 1 second. The absolute error in the time for 100 oscillations $\Delta t$ is the least count, so $\Delta t = 1 \text{second}$. The time period $T$ is the time for one oscillation, so $T = \frac{t}{100} = \frac{90.0}{100} = 0.9 \text{seconds}$. The absolute error in $T$ is $\Delta T = \frac{\Delta t}{100} = \frac{1}{100} = 0.01 \text{seconds}$. The relative error in $T$ is $\frac{\Delta T}{T} = \frac{0.01}{0.9} = \frac{1}{90} \approx 0.011111$.

Now substitute into the relative error formula for $g$:

$\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} = 0.005 + 2 \times \frac{1}{90}$

Calculate the terms:

$2 \times \frac{1}{90} = \frac{2}{90} = \frac{1}{45} \approx 0.022222$

$0.005 + 0.022222 = 0.027222$

To express as a percentage, multiply by 100:

$0.027222 \times 100 = 2.7222\%$

Rounding to one decimal place gives 2.7%.

Hence, the error in the determination of $g$ is 2.7%, which corresponds to Option D.

So, the answer is Option D.

The position vector at time $t = 2$ seconds is given as $\vec{r} = 40\hat{i} + 50\hat{j}$ meters. This means the horizontal displacement $x = 40$ m and the vertical displacement $y = 50$ m.

For a projectile launched from the origin with initial velocity $u$ at an angle $\theta$ to the horizontal, the equations of motion are:

Horizontal motion (no acceleration): $x = u_x t$, where $u_x = u \cos \theta$ is the horizontal component of initial velocity.

Vertical motion (constant acceleration due to gravity $g = 10$ m/s² downward): $y = u_y t - \frac{1}{2} g t^2$, where $u_y = u \sin \theta$ is the vertical component of initial velocity.

At $t = 2$ s, $x = 40$ m. Substituting into the horizontal equation:

$40 = u_x \cdot 2$

Solving for $u_x$:

$u_x = \frac{40}{2} = 20 \text{ m/s}$

At $t = 2$ s, $y = 50$ m. Substituting into the vertical equation:

$50 = u_y \cdot 2 - \frac{1}{2} \cdot 10 \cdot (2)^2$

First, compute the term $\frac{1}{2} \cdot 10 \cdot 4$:

$\frac{1}{2} \cdot 10 = 5, \quad 5 \cdot 4 = 20$

So the equation becomes:

$50 = 2u_y - 20$

Add 20 to both sides:

$50 + 20 = 2u_y$

$70 = 2u_y$

Solving for $u_y$:

$u_y = \frac{70}{2} = 35 \text{ m/s}$

The angle $\theta$ is related to the components by:

$\tan \theta = \frac{u_y}{u_x}$

Substituting the values:

$\tan \theta = \frac{35}{20} = \frac{7}{4}$

Therefore, $\theta = \tan^{-1} \left( \frac{7}{4} \right)$.

Comparing with the options:

A. $\tan^{-1} \frac{3}{2}$

B. $\tan^{-1} \frac{2}{3}$

C. $\tan^{-1} \frac{7}{4}$

D. $\tan^{-1} \frac{4}{5}$

Option C matches our result.

Hence, the correct answer is Option C.

To solve this problem, we need to find the minimum force required to stop the flow of water using your palm. The water flows at a speed of 1.5 m/s through a tube with a cross-sectional area of $10^{-2}$ m², and it stops immediately upon hitting the palm. The density of water is $10^3$ kg/m³.

The force exerted by the palm must counteract the momentum of the incoming water per unit time. Force is defined as the rate of change of momentum. When the water hits the palm and stops, its velocity changes from 1.5 m/s to 0 m/s. The change in velocity ($\Delta v$) is therefore $1.5 - 0 = 1.5$ m/s.

First, we calculate the volume flow rate, which is the volume of water passing through the tube per second. This is given by the product of the cross-sectional area ($A$) and the flow velocity ($v$):

Volume flow rate = $A \times v = 10^{-2} \, \text{m}^2 \times 1.5 \, \text{m/s} = 0.015 \, \text{m}^3/\text{s}$.

Next, we find the mass flow rate, which is the mass of water flowing per second. This is obtained by multiplying the volume flow rate by the density of water ($\rho$):

Mass flow rate = $\rho \times \text{volume flow rate} = 10^3 \, \text{kg/m}^3 \times 0.015 \, \text{m}^3/\text{s} = 15 \, \text{kg/s}$.

The rate of change of momentum is equal to the mass flow rate multiplied by the change in velocity ($\Delta v$), since the water comes to a complete stop:

Rate of change of momentum = mass flow rate $\times \Delta v = 15 \, \text{kg/s} \times 1.5 \, \text{m/s} = 22.5 \, \text{kg} \cdot \text{m/s}^2$.

Since $1 \, \text{N} = 1 \, \text{kg} \cdot \text{m/s}^2$, the force required is 22.5 N. This is the minimum force needed to stop the water flow, assuming an immediate stop with no other losses.

Comparing with the options: A. 33.7 N, B. 45 N, C. 15 N, D. 22.5 N. Hence, the correct answer is Option D.

$f = f_k = \mu_k R = \mu_k m_1\text{ }g$

$12 = f_k = \mu_k \times 4\text{ }g$

$\therefore \mu_k = \frac{12}{4\text{ }g} = \frac{3}{g}$

As block $B$ is on smooth surface, therefore to move $A$ and $B$ together, maximum force $F$ required to be applied on $B =$ frictional force applied on A by B

$F = \frac{3}{g}(4 + 5)\text{ }g = 27\text{ }N$

The linear acceleration $a$ of a rigid body rolling down an inclined plane of angle $\theta$ without slipping is given by:

$a = \frac{g \sin \theta}{1 + \frac{I}{M R^2}}$

Assuming a standard solid cylinder, its moment of inertia is $I = \frac{1}{2} M_C R^2$,

$a_C = \frac{g \sin \theta_C}{1 + \frac{1}{2}} = \frac{g \sin \theta_C}{\frac{3}{2}} = \frac{2}{3} g \sin \theta_C$

Assuming a standard solid sphere, its moment of inertia is $I = \frac{2}{5} M_S R^2$,

$a_S = \frac{g \sin \theta_S}{1 + \frac{2}{5}} = \frac{g \sin \theta_S}{\frac{7}{5}} = \frac{5}{7} g \sin \theta_S$

$\frac{2}{3} g \sin \theta_C = \frac{5}{7} g \sin \theta_S$

$\frac{\sin \theta_C}{\sin \theta_S} = \frac{\frac{5}{7}}{\frac{2}{3}} = \frac{5}{7} \times \frac{3}{2} = \frac{15}{14}$

To estimate the time taken for Mangalyan to travel from Earth to Mars using a transfer orbit around the sun, we apply Kepler's laws. The transfer orbit is elliptical, with the sun at one focus. Mangalyan leaves Earth at perihelion (closest point to the sun) and meets Mars at aphelion (farthest point). The semi-major axis of Earth's orbit is given as $a_e = 1.5 \times 10^{11}$ m, and for Mars, $a_m = 2.28 \times 10^{11}$ m.

For the transfer orbit, the perihelion distance equals Earth's orbital radius ($a_e$), and the aphelion distance equals Mars' orbital radius ($a_m$). The semi-major axis of the transfer orbit ($a_t$) is the average of these distances:

$a_t = \frac{a_e + a_m}{2} = \frac{1.5 \times 10^{11} + 2.28 \times 10^{11}}{2} = \frac{3.78 \times 10^{11}}{2} = 1.89 \times 10^{11} \text{ m}$

Kepler's third law states that the square of the orbital period ($T$) is proportional to the cube of the semi-major axis ($a$): $T^2 \propto a^3$. For Earth, the orbital period $T_e$ is 1 year (365 days), and the semi-major axis is $a_e$. For the transfer orbit, the period $T_t$ relates to $a_t$ by the same proportionality constant. Therefore, we can write:

$\frac{T_t^2}{T_e^2} = \frac{a_t^3}{a_e^3}$

Solving for $T_t$:

$T_t = T_e \left( \frac{a_t}{a_e} \right)^{3/2}$

The travel time from Earth to Mars is half the period of the transfer orbit, as Mangalyan travels from perihelion to aphelion, covering half the ellipse. Thus, the travel time $\tau$ is:

$\tau = \frac{T_t}{2} = \frac{T_e}{2} \left( \frac{a_t}{a_e} \right)^{3/2}$

Substitute the values $T_e = 365$ days, $a_t = 1.89 \times 10^{11}$ m, and $a_e = 1.5 \times 10^{11}$ m:

$\frac{a_t}{a_e} = \frac{1.89 \times 10^{11}}{1.5 \times 10^{11}} = \frac{1.89}{1.5} = 1.26$

Now compute $\left( \frac{a_t}{a_e} \right)^{3/2} = (1.26)^{3/2}$:

First, find $(1.26)^{3}$:

$1.26 \times 1.26 = 1.5876$

$1.5876 \times 1.26 = 1.5876 \times 1.2 + 1.5876 \times 0.06 = 1.90512 + 0.095256 = 2.000376$

Now, $(1.26)^{3/2} = \sqrt{(1.26)^3} = \sqrt{2.000376}$. Using $\sqrt{2} \approx 1.414213562$ and approximating $\sqrt{2.000376} \approx \sqrt{2} \times \sqrt{1.000188} \approx 1.414213562 \times (1 + \frac{0.000188}{2}) = 1.414213562 \times 1.000094 \approx 1.414213562 + 1.414213562 \times 0.000094$:

$1.414213562 \times 0.000094 = 0.000132936074828$

$\sqrt{2.000376} \approx 1.414213562 + 0.000132936074828 = 1.414346498074828 \approx 1.4143465$

Now substitute into the formula for $\tau$:

$\tau = \frac{365}{2} \times 1.4143465 = 182.5 \times 1.4143465$

Compute step by step:

$182.5 \times 1.4 = 255.5$

$182.5 \times 0.014 = 2.555$

$182.5 \times 0.0003465 \approx 182.5 \times 0.0003 = 0.05475 \quad \text{and} \quad 182.5 \times 0.0000465 \approx 0.00848625$

$0.0003465 \approx 0.0003 + 0.0000465 \quad \text{so} \quad 0.05475 + 0.00848625 = 0.06323625$

Sum the parts:

$255.5 + 2.555 = 258.055$

$258.055 + 0.06323625 = 258.11823625 \approx 258.12 \text{ days}$

The calculated travel time is approximately 258.12 days. Comparing with the options:

A. 220 days

B. 500 days

C. 260 days

D. 320 days

The value 258.12 days is closest to 260 days.

Hence, the correct answer is Option C.

Since it takes significantly less effort to twist or slide a metal block than it does to stretch it or compress its entire volume, the energy required scales accordingly. Therefore, the resistance to changing shape is the lowest, while the resistance to changing volume is the highest, giving the order: Shear moduli < Young's moduli < bulk moduli

The height of the water column in a capillary tube is given by the formula:

where $S$ is the surface tension, $\theta$ is the contact angle, $\rho$ is the density of water, $g$ is the acceleration due to gravity, and $r$ is the radius of the capillary tube.

Since the capillary tube and the liquid (water) are the same in both scenarios, $S$, $\theta$, $\rho$, and $r$ are constant. Therefore, the height $h$ is inversely proportional to $g$:

At the Earth's surface, the height is $x$, so:

In the mine at depth $d$, the height is $y$, so:

The ratio $\frac{x}{y}$ is:

Now, we need to find how $g$ changes with depth. The acceleration due to gravity at a depth $d$ below the Earth's surface is given by:

where $R$ is the radius of the Earth. This is because, at depth $d$, only the mass of the Earth within the sphere of radius $R - d$ contributes to gravity, and under the assumption of uniform density, $g$ decreases linearly with depth.

Therefore, substituting this into the ratio:

So, the ratio $\frac{x}{y}$ is:

Comparing with the options:

A. $\left(\frac{R-d}{R+d}\right)$

B. $\left(\frac{R+d}{R-d}\right)$

C. $\left(1 - \frac{2d}{R}\right)$

D. $\left(1 - \frac{d}{R}\right)$

Option D matches the derived expression.

Hence, the correct answer is Option D.

We are given that the volume of water is 2 liters. Since the density of water is 1 kg/L, the mass of water is 2 kg.

The heating coil provides energy at a rate of 1 kW. Since 1 kW equals 1000 W and 1 W is 1 J/s, the coil supplies 1000 J/s.

The container loses energy at a rate of 160 J/s. Therefore, the net energy absorbed by the water per second is the energy supplied by the coil minus the energy lost by the container:

$\text{Net energy per second} = 1000 \text{J/s} - 160 \text{J/s} = 840 \text{J/s}$

The temperature needs to rise from 27°C to 77°C. The change in temperature is:

$\Delta T = 77^\circ \text{C} - 27^\circ \text{C} = 50^\circ \text{C}$

The specific heat of water is given as 4.2 kJ/kg·°C. Converting to joules: 4.2 kJ/kg·°C = 4.2 × 1000 = 4200 J/kg·°C.

The total heat energy required to raise the temperature of the water is given by:

$Q = m \cdot c \cdot \Delta T$

Substituting the values:

$Q = 2 \text{kg} \times 4200 \text{J/kg·°C} \times 50^\circ \text{C}$

First, multiply the mass and specific heat:

$2 \times 4200 = 8400$

Then multiply by the temperature change:

$8400 \times 50 = 420000 \text{J}$

So, the total heat required is 420,000 J.

The net energy supplied per second is 840 J/s. The time $t$ in seconds required to supply 420,000 J is:

$t = \frac{\text{Total heat}}{\text{Net energy per second}} = \frac{420000}{840}$

Simplifying the fraction:

$\frac{420000}{840} = \frac{420000 \div 120}{840 \div 120} = \frac{3500}{7} = 500 \text{seconds}$

Alternatively, note that 840 × 500 = 420,000, so $t = 500$ seconds.

Converting 500 seconds to minutes and seconds: since 1 minute = 60 seconds,

$\text{Minutes} = \left\lfloor \frac{500}{60} \right\rfloor = \left\lfloor 8.333\ldots \right\rfloor = 8 \text{minutes}$

The remaining seconds are:

$500 - (8 \times 60) = 500 - 480 = 20 \text{seconds}$

Thus, the time is 8 minutes and 20 seconds.

Comparing with the options, Option A is 8 min 20 s.

Hence, the correct answer is Option A.

$PV = nRT + \alpha V$

$PV = RT + \alpha V$ ($n = 1$)

$PV - \alpha V = RT$

$V(P - \alpha) = RT \implies V = \frac{RT}{P - \alpha}$

$V_{\text{initial}} = \frac{RT_0}{P_0 - \alpha}$

$V_{\text{final}} = \frac{R(2T_0)}{P_0 - \alpha} = \frac{2RT_0}{P_0 - \alpha}$

$W = P_0 \cdot \Delta V = P_0 (V_{\text{final}} - V_{\text{initial}})$

$W = P_0 \left( \frac{2RT_0}{P_0 - \alpha} - \frac{RT_0}{P_0 - \alpha} \right)$

$W = P_0 \left( \frac{RT_0}{P_0 - \alpha} \right)$

$W = \frac{P_0 T_0 R}{P_0 - \alpha}$

To find the mean distance between molecules in the evacuated vessel, we start by using the ideal gas law and the given parameters. The pressure is $4.0 \times 10^{-15}$ atm, but we need it in Pascals since other units are in SI. Given that 1 atm = $10^5$ Pa, we convert:

Pressure $P = 4.0 \times 10^{-15} \times 10^5 = 4.0 \times 10^{-10}$ Pa.

The temperature $T = 300$ K, gas constant $R = 8.3$ J K$^{-1}$ mol$^{-1}$, and Avogadro's number $N_A = 6 \times 10^{23}$ mol$^{-1}$. The mean distance between molecules is related to the number density $n_v$, which is the number of molecules per cubic meter. From the ideal gas law:

$PV = nRT$

where $n$ is the number of moles. Since the number of molecules $N = n \times N_A$, we substitute:

$PV = \left(\frac{N}{N_A}\right) RT$

Rearranging for pressure:

$P = \frac{N}{V} \times \frac{R}{N_A} \times T$

Here, $\frac{N}{V} = n_v$ is the number density, and $\frac{R}{N_A} = k$ is Boltzmann's constant. So:

$P = n_v k T$

Solving for $n_v$:

$n_v = \frac{P}{k T}$

But $k = \frac{R}{N_A}$, so:

$n_v = \frac{P N_A}{R T}$

Now plug in the values:

$n_v = \frac{(4.0 \times 10^{-10}) \times (6 \times 10^{23})}{8.3 \times 300}$

First, calculate the numerator:

$4.0 \times 10^{-10} \times 6 \times 10^{23} = 24 \times 10^{13} = 2.4 \times 10^{14}$

Next, the denominator:

$8.3 \times 300 = 2490$

So:

$n_v = \frac{2.4 \times 10^{14}}{2490}$

Dividing:

$2490 = 2.49 \times 10^3$

$n_v = \frac{2.4 \times 10^{14}}{2.49 \times 10^3} = \frac{2.4}{2.49} \times 10^{11} \approx 0.964 \times 10^{11} = 9.64 \times 10^{10}$ molecules per cubic meter.

The volume per molecule $V_m$ is the reciprocal of $n_v$:

$V_m = \frac{1}{n_v} = \frac{1}{9.64 \times 10^{10}} \approx 1.037 \times 10^{-11}$ cubic meters.

The mean distance $d$ between molecules is approximately the cube root of the volume per molecule, since each molecule occupies a cubic space:

$d = \left(V_m\right)^{1/3} = \left(1.037 \times 10^{-11}\right)^{1/3}$

We can separate the terms:

$d = \left(1.037\right)^{1/3} \times \left(10^{-11}\right)^{1/3}$

First, $\left(10^{-11}\right)^{1/3} = 10^{-11/3} = 10^{-3.6667}$. Since $10^{-3} = 0.001$ and $10^{-0.6667} \approx 0.2154$ (because $10^{2/3} \approx 4.6416$, so $10^{-2/3} \approx 1/4.6416 \approx 0.2154$), we get:

$10^{-3.6667} \approx 0.001 \times 0.2154 = 2.154 \times 10^{-4}$

Next, $\left(1.037\right)^{1/3}$. Using approximation $(1 + x)^n \approx 1 + nx$ for small $x$, where $x = 0.037$ and $n = 1/3$:

$\left(1.037\right)^{1/3} \approx 1 + \frac{0.037}{3} = 1 + 0.012333 \approx 1.012333$

But a more accurate calculation shows that $1.012^3 = 1.03643488$, which is slightly less than 1.037. The difference is $1.037 - 1.03643488 = 0.00056512$. The derivative of $x^3$ is $3x^2$, so at $x = 1.012$, $3 \times (1.012)^2 \approx 3 \times 1.024144 = 3.072432$. The increment is:

$\Delta x \approx \frac{0.00056512}{3.072432} \approx 0.000184$

Thus:

$\left(1.037\right)^{1/3} \approx 1.012 + 0.000184 = 1.012184$

Now multiply:

$d \approx 1.012184 \times 2.154 \times 10^{-4} \approx 2.1807 \times 10^{-4}$ meters.

Converting to millimeters (since 1 meter = 1000 mm):

$2.1807 \times 10^{-4} \text{ m} = 0.21807 \text{ mm}$

This value is approximately 0.22 mm, which is of the same order as 0.2 mm (both are $10^{-4}$ meters). The options are in different units, so comparing:

• A. 0.2 μm = $0.2 \times 10^{-6}$ m = $2 \times 10^{-7}$ m
• B. 0.2 mm = $0.2 \times 10^{-3}$ m = $2 \times 10^{-4}$ m
• C. 0.2 cm = $0.2 \times 10^{-2}$ m = $2 \times 10^{-3}$ m
• D. 0.2 nm = $0.2 \times 10^{-9}$ m = $2 \times 10^{-10}$ m

Our calculated distance $2.18 \times 10^{-4}$ m is closest to $2 \times 10^{-4}$ m, which is 0.2 mm.

Hence, the correct answer is Option B.

The force exerted by the system on the floor is the sum of the weight of the $4\text{ kg}$ mass and the downward force exerted by the spring.

The spring constant $k$ can be determined from the angular frequency $\omega$ and the oscillating mass $m_1 = 1\text{ kg}$:

$k = m_1 \omega^2 = 1 \times (25)^2 = 625\text{ N/m}$

$k x_0 = m_1 g \implies x_0 = \frac{1 \times 10}{625}\text{ m}$

$F_{s,max} = k(x_0 + A) = k x_0 + k A$

$F_{s,max} = m_1 g + k A = 10 + (625 \times 0.016) = 10 + 10 = 20\text{ N}$

$F_{floor} = m_2 g + F_{s,max}$

$F_{floor} = (4 \times 10) + 20 = 40 + 20 = 60\text{ N}$

For a damped pendulum oscillating in a medium, the amplitude decays exponentially as $A = A_0 e^{-bt/2m}$, where $b$ is the damping coefficient proportional to the viscosity $\eta$ of the medium (by Stokes law, $b \propto \eta$).

In air, the amplitude decreases from 10 cm to 8 cm in 40 seconds. So $8 = 10\, e^{-b_{\text{air}} \cdot 40 / 2m}$, which gives $e^{-20\, b_{\text{air}}/m} = 4/5$. Taking the natural logarithm, $\frac{20\, b_{\text{air}}}{m} = \ln\!\left(\frac{5}{4}\right) = \ln 5 - 2\ln 2 = 1.601 - 1.386 = 0.215$.

In carbon dioxide, we need the time $t$ for the amplitude to reduce from 10 cm to 5 cm. So $5 = 10\, e^{-b_{\text{CO}_2}\, t / 2m}$, which gives $\frac{b_{\text{CO}_2}\, t}{2m} = \ln 2 = 0.693$.

Since $\eta_{\text{air}} / \eta_{\text{CO}_2} = 1.3$, we have $b_{\text{CO}_2} = b_{\text{air}} / 1.3$. Substituting, $t = \frac{2m \times 0.693}{b_{\text{CO}_2}} = \frac{2m \times 0.693 \times 1.3}{b_{\text{air}}}$. From the air data, $b_{\text{air}}/m = 0.215/20 = 0.01075$, so $m/b_{\text{air}} = 1/0.01075$.

Therefore, $t = \frac{2 \times 0.693 \times 1.3}{0.01075} = \frac{1.8018}{0.01075} \approx 161 \text{ s}$. The closest answer among the given options is $161$ s.

$x = a_1 \cos \omega t \implies \cos \omega t = \frac{x}{a_1}$

$y = a_2 \cos 2\omega t$

$y = a_2 (2 \cos^2 \omega t - 1)$

$y = a_2 \left[ 2 \left( \frac{x}{a_1} \right)^2 - 1 \right]$

$y = \frac{2a_2}{a_1^2} x^2 - a_2$

This equation is of the form $y = kx^2 - c$, which represents a parabola symmetric about the $y$-axis with its vertex at $(0, -a_2)$.

The curve opens upwards because the coefficient of $x^2$ is positive. At $x = \pm a_1$, $y = a_2 [2(1) - 1] = a_2$. At $x = 0$, $y = -a_2$.

This matches the plot shown in option D.

From the wave equation $y = \frac{10}{\pi} \sin \left( \frac{2\pi}{T}t - \frac{2\pi}{\lambda}x \right)$, amplitude $A = \frac{10}{\pi}$ and the wave velocity $v_w = \frac{\omega}{k} = \frac{\lambda}{T}$.

The maximum particle velocity is given by $v_{p(max)} = A\omega = \left( \frac{10}{\pi} \right) \left( \frac{2\pi}{T} \right) = \frac{20}{T}$.
Substituting these into the given condition $v_w = 2 v_{p(max)}$ yields $\frac{\lambda}{T} = 2 \times \frac{20}{T}$, which simplifies to $\lambda = 40$ cm.

The electric field is directed inward toward the center of the Earth, meaning it points in the direction opposite to the outward normal of the Earth's surface. By Gauss's law, the total charge enclosed is $Q = \epsilon_0 \oint \vec{E} \cdot d\vec{A}$. Since $\vec{E}$ is directed inward and $d\vec{A}$ points outward, $\vec{E} \cdot d\vec{A} = -E\, dA$, giving $Q = -\epsilon_0 E \times 4\pi R_E^2$.

Substituting the given values: $Q = -8.85 \times 10^{-12} \times 150 \times 4\pi \times (6.37 \times 10^6)^2$. Computing $R_E^2 = (6.37)^2 \times 10^{12} = 40.58 \times 10^{12}$ m$^2$, and $4\pi R_E^2 = 4\pi \times 40.58 \times 10^{12} = 5.099 \times 10^{14}$ m$^2$.

So $Q = -8.85 \times 10^{-12} \times 150 \times 5.099 \times 10^{14} = -8.85 \times 150 \times 5.099 \times 10^{2} = -8.85 \times 76485 \approx -6.77 \times 10^5$ C $= -677$ kC, which is approximately $-680$ kC.

Each capacitor has capacitance $C = 3\,\mu\text{F}$.
To obtain an equivalent capacitance we must connect all three capacitors, using any series-parallel arrangement. With three identical capacitors there are only four essentially different ways to combine them.

Case 1:

All three in series.
$C_{\text{eq}} = \frac{C}{3} = \frac{3}{3} = 1\,\mu\text{F}$

Case 2:

All three in parallel.
$C_{\text{eq}} = 3C = 3 \times 3 = 9\,\mu\text{F}$

Case 3:

Two in series, that pair in parallel with the third.
Series pair: $C_s = \frac{C}{2} = \frac{3}{2} = 1.5\,\mu\text{F}$
Putting this in parallel with the remaining $3\,\mu\text{F}$ capacitor:
$C_{\text{eq}} = C_s + C = 1.5 + 3 = 4.5\,\mu\text{F}$

Case 4:

Two in parallel, that pair in series with the third.
Parallel pair: $C_p = 2C = 2 \times 3 = 6\,\mu\text{F}$
Putting this in series with the remaining $3\,\mu\text{F}$ capacitor:
$\frac{1}{C_{\text{eq}}} = \frac{1}{C_p} + \frac{1}{C} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6}$
so $C_{\text{eq}} = \frac{6}{3} = 2\,\mu\text{F}$

Thus the only equivalent capacitances obtainable with all three $3\,\mu\text{F}$ capacitors are

$1\,\mu\text{F},\;2\,\mu\text{F},\;4.5\,\mu\text{F},\;9\,\mu\text{F}$

From the given options the value that cannot be produced is
Option B: $6\,\mu\text{F}$.

The circuit consists of a DC main supply of 220 V connected across a storage battery of 200 V through a 1 Ω resistance. The battery terminals are also connected to an external resistance $R$. We need to find the minimum value of $R$ such that current flows through the battery to charge it, meaning current enters the positive terminal of the battery.

First, we assume the battery has no internal resistance, as it is not specified in the problem. Therefore, the voltage across the battery terminals is always equal to its EMF, which is 200 V. Let the common negative terminal be point C, and the positive terminal of the supply be point A. The supply is connected to a 1 Ω resistor, and the other end of this resistor is connected to the positive terminal of the battery (point B) and one end of $R$. The negative terminal of the battery and the other end of $R$ are connected to point C.

Since the voltage at point B with respect to point C is the same as the battery voltage, $V_{BC} = 200$ V. The supply voltage is 220 V between A and C, so $V_{AC} = 220$ V. The voltage drop across the 1 Ω resistor (between A and B) is $V_{AB} = V_{AC} - V_{BC} = 220 - 200 = 20$ V.

The current through the 1 Ω resistor, flowing from A to B, is given by Ohm's law: $I = \frac{V_{AB}}{1 \Omega} = \frac{20}{1} = 20$ A.

At junction B, this current $I$ splits into two parts: $I_b$ through the battery and $I_R$ through the external resistance $R$. Applying Kirchhoff's current law: $I = I_b + I_R$, so $20 = I_b + I_R$.

The voltage across $R$ is the same as $V_{BC} = 200$ V, so by Ohm's law, $I_R = \frac{V_{BC}}{R} = \frac{200}{R}$. Substituting this in: $20 = I_b + \frac{200}{R}$.

For the battery to charge, current must enter its positive terminal, so $I_b > 0$. Therefore, $20 - \frac{200}{R} > 0$. Solving this inequality:

$20 > \frac{200}{R}$

Since $R$ is positive, multiply both sides by $R$: $20R > 200$.

Divide both sides by 20: $R > 10 \Omega$.

Thus, $R$ must be greater than 10 Ω for charging to occur. The minimum value of $R$ in the given options that satisfies this condition is 11 Ω.

Hence, the correct answer is Option C.

The magnetic field $B$ produced by the first dipole at the center of the second dipole along its axis is $B = \frac{\mu_0}{4\pi} \frac{2M_1}{x^3}$

$U = -\vec{M_2} \cdot \vec{B} = -M_2 B \cos(0^\circ) = -\frac{\mu_0}{4\pi} \frac{2M_1 M_2}{x^3}$

$F = -\frac{dU}{dx} = -\frac{d}{dx} \left( -\frac{\mu_0}{4\pi} \frac{2M_1 M_2}{x^3} \right)$

$F = \frac{\mu_0}{4\pi} (2M_1 M_2) \frac{d}{dx} (x^{-3})$

$F = \frac{\mu_0}{4\pi} (2M_1 M_2) (-3x^{-4})$

$|F| \propto x^{-4}$

$n = 4$