JEE Main 2014 11th April Morning Shift Question Paper with Solutions

The characteristic impedance $Z$ of a medium is defined by the ratio of permeability $\mu$ to permittivity $\varepsilon$:

$Z = \sqrt{\frac{\mu}{\varepsilon}}$

$[Z] = [R]$

$\left[ \frac{\mu}{\varepsilon} \right] = [R^2]$

The bullet is fired with an initial speed of 630 m/s towards a target 700 m away at the same level. Since the rifle is aimed directly at the center of the target, the initial velocity is horizontal. This means the initial vertical velocity is 0 m/s, and the initial horizontal velocity is 630 m/s.

To find how far above or below the center the bullet hits, we need to determine the vertical displacement when the bullet has traveled 700 m horizontally. The horizontal motion has no acceleration, so the time taken to cover 700 m is given by the horizontal distance divided by the horizontal velocity:

$t = \frac{\text{horizontal distance}}{\text{horizontal velocity}} = \frac{700}{630}$

Simplifying the fraction:

$\frac{700}{630} = \frac{700 \div 70}{630 \div 70} = \frac{10}{9} \text{ seconds}$

In the vertical direction, the initial velocity is 0 m/s, and acceleration due to gravity is $g = 9.8 \text{m/s}^2$ downward. The vertical displacement $h$ is given by the equation of motion:

$h = u_y t + \frac{1}{2} a t^2$

Since $u_y = 0$ and $a = g$ (taking downward as positive for displacement magnitude), this becomes:

$h = \frac{1}{2} g t^2$

Substituting $g = 9.8$ and $t = \frac{10}{9}$:

$h = \frac{1}{2} \times 9.8 \times \left( \frac{10}{9} \right)^2$

First, compute $\left( \frac{10}{9} \right)^2$:

$\left( \frac{10}{9} \right)^2 = \frac{100}{81}$

Now, multiply:

$h = \frac{1}{2} \times 9.8 \times \frac{100}{81} = \frac{1}{2} \times \frac{9.8 \times 100}{81}$

Calculate $9.8 \times 100 = 980$:

$h = \frac{1}{2} \times \frac{980}{81} = \frac{980}{2 \times 81} = \frac{980}{162}$

Simplify by dividing numerator and denominator by 2:

$\frac{980 \div 2}{162 \div 2} = \frac{490}{81}$

Now, compute the numerical value:

$\frac{490}{81} \approx 6.04938 \text{ m}$

The negative sign indicates the displacement is downward, meaning the bullet hits below the center. The magnitude of the vertical displacement is approximately 6.05 m. Comparing with the options:

A. 1.0 m
B. 4.2 m
C. 6.1 m
D. 9.8 m

The value 6.05 m is closest to 6.1 m.

Hence, the correct answer is Option C.

We are given the mass of the body, $m = 5$ kg, the initial velocity at $t = 0$ s as $\vec{u} = (6\hat{i} - 2\hat{j})$ m/s, and the final velocity at $t = 10$ s as $\vec{v} = +6\hat{j}$ m/s. The force $\vec{F} = F_x\hat{i} + F_y\hat{j}$ is constant. We need to find $\vec{F}$.

According to Newton's second law, the force is given by $\vec{F} = m\vec{a}$, where $\vec{a}$ is the acceleration. Acceleration is the rate of change of velocity, so $\vec{a} = \frac{\Delta \vec{v}}{\Delta t}$.

First, calculate the change in velocity $\Delta \vec{v}$:

$\Delta \vec{v} = \vec{v} - \vec{u} = (6\hat{j}) - (6\hat{i} - 2\hat{j})$

Distribute the negative sign:

$\Delta \vec{v} = 6\hat{j} - 6\hat{i} + 2\hat{j}$

Combine like terms:

$\Delta \vec{v} = -6\hat{i} + (6 + 2)\hat{j} = -6\hat{i} + 8\hat{j} \text{ m/s}$

The time interval $\Delta t = t_{\text{final}} - t_{\text{initial}} = 10 - 0 = 10$ s.

Now, find the acceleration $\vec{a}$:

$\vec{a} = \frac{\Delta \vec{v}}{\Delta t} = \frac{-6\hat{i} + 8\hat{j}}{10}$

Separate the components:

$\vec{a} = \left( \frac{-6}{10} \right) \hat{i} + \left( \frac{8}{10} \right) \hat{j}$

Simplify the fractions:

$\vec{a} = -\frac{6}{10}\hat{i} + \frac{8}{10}\hat{j} = -\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j} \text{ m/s}^2$

Now, use Newton's second law to find the force:

$\vec{F} = m \vec{a} = 5 \times \left( -\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j} \right)$

Distribute the mass to each component:

$\vec{F} = 5 \times \left( -\frac{3}{5} \right) \hat{i} + 5 \times \left( \frac{4}{5} \right) \hat{j}$

Simplify each term:

$\vec{F} = \left( 5 \times -\frac{3}{5} \right) \hat{i} + \left( 5 \times \frac{4}{5} \right) \hat{j}$

$\vec{F} = \left( -3 \right) \hat{i} + \left( 4 \right) \hat{j}$

So, $\vec{F} = (-3\hat{i} + 4\hat{j})$ N.

Comparing with the options:

• A. $(-3\hat{i} + 4\hat{j})$ N
• B. $\left( -\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j} \right)$ N
• C. $(3\hat{i} - 4\hat{j})$ N
• D. $\left( \frac{3}{5}\hat{i} - \frac{4}{5}\hat{j} \right)$ N

Our result matches option A.

Hence, the correct answer is Option A.

$E_A = mgh + \frac{1}{2}mv_o^2$

$E_C = 0$ (ball comes to stop)

$W_{friction} = -f \cdot L = -\mu mgL$

$mgh + \frac{1}{2}mv_o^2 = \mu mgL$ (The initial energy is dissipated by the work done by friction)

$L = \frac{mgh}{\mu mg} + \frac{\frac{1}{2}mv_o^2}{\mu mg}$

$L = \frac{h}{\mu} + \frac{v_o^2}{2\mu g}$

$\lambda(x) = \lambda_0 + kx$

$M = \int_0^L \lambda(x) dx = \int_0^L (\lambda_0 + kx) dx = \left[ \lambda_0 x + \frac{kx^2}{2} \right]_0^L$

$M = \lambda_0 L + \frac{kL^2}{2} \implies kL^2 = 2(M - \lambda_0 L) \quad \dots (1)$

$x_{cm} = \frac{1}{M} \int_0^L x \lambda(x) dx$

$x_{cm} = \frac{1}{M} \int_0^L (x\lambda_0 + kx^2) dx = \frac{1}{M} \left[ \frac{\lambda_0 x^2}{2} + \frac{kx^3}{3} \right]_0^L$

$x_{cm} = \frac{1}{M} \left[ \frac{\lambda_0 L^2}{2} + \frac{k L^3}{3} \right]$

$x_{cm} = \frac{1}{M} \left[ \frac{\lambda_0 L^2}{2} + \frac{L}{3} (kL^2) \right]$

$x_{cm} = \frac{1}{M} \left[ \frac{\lambda_0 L^2}{2} + \frac{L}{3} (2M - 2\lambda_0 L) \right]$

$x_{cm} = \frac{1}{M} \left[ \frac{\lambda_0 L^2}{2} + \frac{2ML}{3} - \frac{2\lambda_0 L^2}{3} \right]$

$x_{cm} = \frac{2L}{3} + \frac{\lambda_0 L^2}{M} \left( \frac{1}{2} - \frac{2}{3} \right)$

$x_{cm} = \frac{2L}{3} - \frac{\lambda_0 L^2}{6M}$

The mass of the removed sphere ($m$) is $m = M \times \left( \frac{R/2}{R} \right)^3 = \frac{M}{8}$

Center of the Original Sphere ($O$): $(0,0)$

Center of the Cavity ($C$): Since the cavity is carved between the center and periphery, its center is at $(R/2, 0)$.

Center of the External removed sphere ($S$): The distance between the center of the original sphere and the removed sphere is $3R$. Its position is $(3R, 0)$.

Distance from $O$ to $S$ ($r_1$) = $3R$. Distance from $C$ to $S$ ($r_2$) = $3R - R/2 = 2.5R = \frac{5R}{2}$

$F_{net} = F_{original} - F_{cavity}$

$F_{net} = G \frac{M \cdot m}{r_1^2} - G \frac{m \cdot m}{r_2^2}$

$F_{net} = G \frac{M(M/8)}{(3R)^2} - G \frac{(M/8)(M/8)}{(5R/2)^2}$

$F_{net} = \frac{GM^2}{8 \cdot 9R^2} - \frac{GM^2}{64 \cdot \frac{25R^2}{4}}$

$F_{net} = \frac{GM^2}{72R^2} - \frac{GM^2}{16 \cdot 25R^2} = \frac{GM^2}{72R^2} - \frac{GM^2}{400R^2}$

$F_{net} = \frac{50GM^2}{3600R^2} - \frac{9GM^2}{3600R^2} = \frac{41GM^2}{3600R^2}$

The bulk modulus $K$ is defined as the ratio of the applied pressure change $\Delta P$ to the fractional decrease in volume $-\frac{\Delta V}{V}$. The formula is:

$K = - \frac{\Delta P}{\frac{\Delta V}{V}}$

Rearranging for the fractional compression $\frac{\Delta V}{V}$, we get:

$\frac{\Delta V}{V} = - \frac{\Delta P}{K}$

Since the fractional compression is typically considered as a magnitude for comparison, we take the absolute value:

$\left| \frac{\Delta V}{V} \right| = \frac{\Delta P}{K}$

For a given pressure change $\Delta P$ (same for all liquids), the fractional compression $\left| \frac{\Delta V}{V} \right|$ is inversely proportional to the bulk modulus $K$. Therefore, a smaller bulk modulus results in a larger fractional compression.

Given the bulk moduli in units of $10^9$ Nm$^{-2}$:

• Ethanol: $K_e = 0.9$
• Mercury: $K_m = 25$
• Water: $K_w = 2.2$

Comparing the values:

$K_e = 0.9, \quad K_w = 2.2, \quad K_m = 25$

So, $K_e < K_w < K_m$.

Since fractional compression is inversely proportional to $K$:

$\left| \frac{\Delta V}{V} \right| \propto \frac{1}{K}$

The order of fractional compression from largest to smallest is:

Ethanol (smallest $K$) > Water (next smallest $K$) > Mercury (largest $K$)

Thus, $\frac{\Delta V}{V}$ is largest for ethanol, followed by water, and smallest for mercury.

Now, comparing with the options:

A. Ethanol > Water > Mercury

B. Water > Ethanol > Mercury

C. Mercury > Ethanol > Water

D. Ethanol > Mercury > Water

Option A matches our result.

Hence, the correct answer is Option A.

The average mass of each rain drop is $m = 3.0 \times 10^{-5}$ kg, and the average terminal velocity is $v = 9$ m/s. The energy transferred by each rain drop when it hits the surface is due to its kinetic energy. The kinetic energy (KE) of one rain drop is given by:

$KE = \frac{1}{2} m v^2$

Substituting the values:

$KE = \frac{1}{2} \times (3.0 \times 10^{-5}) \times (9)^2$

First, calculate $9^2$:

$9^2 = 81$

Then multiply by the mass:

$m \times 81 = 3.0 \times 10^{-5} \times 81 = 243 \times 10^{-5} = 2.43 \times 10^{-3}$

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

$KE = \frac{1}{2} \times 2.43 \times 10^{-3} = 1.215 \times 10^{-3} \text{ J}$

So, each rain drop transfers $1.215 \times 10^{-3}$ J of energy.

The place receives 100 cm of rain in a year. Convert this to meters:

$100 \text{ cm} = 1 \text{ m}$

This is the height of the water column over an area of 1 square meter. The volume of rain falling on 1 m² area is:

$\text{Volume} = \text{area} \times \text{height} = 1 \times 1 = 1 \text{ m}^3$

The density of water is 1000 kg/m³, so the total mass of rain falling on 1 m² area is:

$\text{Mass} = \text{volume} \times \text{density} = 1 \times 1000 = 1000 \text{ kg}$

The mass of one rain drop is $m = 3.0 \times 10^{-5}$ kg. The number of rain drops falling on 1 m² area is:

$\text{Number of drops} = \frac{\text{Total mass}}{\text{Mass per drop}} = \frac{1000}{3.0 \times 10^{-5}}$

Calculate:

$\frac{1000}{3.0 \times 10^{-5}} = \frac{1000}{0.00003} = \frac{1000 \times 10^5}{3} = \frac{10^8}{3} = 3.333 \times 10^7$

The total energy transferred to 1 m² area is the number of drops multiplied by the energy per drop:

$E_{\text{total}} = \text{Number of drops} \times KE = \left( \frac{1000}{3.0 \times 10^{-5}} \right) \times (1.215 \times 10^{-3})$

Alternatively, we can use the formula for total kinetic energy since all drops have the same terminal velocity:

$E_{\text{total}} = \frac{1}{2} \times \text{Total mass} \times v^2 = \frac{1}{2} \times M \times v^2$

Substituting $M = 1000$ kg and $v = 9$ m/s:

$E_{\text{total}} = \frac{1}{2} \times 1000 \times (9)^2 = \frac{1}{2} \times 1000 \times 81$

First, $\frac{1}{2} \times 1000 = 500$:

$500 \times 81 = 40500 \text{ J}$

Which is $4.05 \times 10^4$ J.

Hence, the energy transferred by rain to each square metre of the surface is $4.05 \times 10^4$ J.

Comparing with the options:

A. $3.5 \times 10^5$ J

B. $4.05 \times 10^4$ J

C. $3.0 \times 10^5$ J

D. $9.0 \times 10^4$ J

So, the answer is Option B.

Pressure from Kerosene ($P_k$): $\rho_k g h_k = (0.8 \times 10^3) \times 10 \times 2 = 16,000\text{ Pa}$

Pressure from Water ($P_w$): $\rho_w g h_w = 10^3 \times 10 \times 3 = 30,000\text{ Pa}$

Total Gauge Pressure: $P_{total} = 16,000 + 30,000 = 46,000\text{ Pa}$

$P_{total} = \frac{1}{2} \rho_w v^2$

$46,000 = \frac{1}{2} \times 1000 \times v^2$

$v = \sqrt{92} \approx 9.59\text{ m/s}$

We are given:

• Radius of air bubble, $r = 0.1 \, \text{cm}$. Convert to meters: $0.1 \, \text{cm} = 0.1 \times 10^{-2} \, \text{m} = 0.001 \, \text{m}$.
• Surface tension, $S = 0.06 \, \text{N/m}$.
• Density of liquid, $\rho = 10^3 \, \text{kg/m}^3 = 1000 \, \text{kg/m}^3$.
• Excess pressure inside bubble over atmospheric pressure, $\Delta P = 1100 \, \text{N/m}^2$.
• Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$.

We need to find the depth $h$ (in meters) of the bubble below the surface of the liquid.

For an air bubble inside a liquid, the excess pressure inside the bubble over the atmospheric pressure is the sum of two components:

1. The pressure due to the depth of the bubble: $\rho g h$.
2. The pressure due to surface tension: $\frac{2S}{r}$.

Therefore, the equation for excess pressure is:

$\Delta P = \rho g h + \frac{2S}{r}$

Substitute the given values:

$1100 = (1000) \times (9.8) \times h + \frac{2 \times 0.06}{0.001}$

First, calculate the surface tension term:

$\frac{2 \times 0.06}{0.001} = \frac{0.12}{0.001} = 120 \, \text{N/m}^2$

So the equation becomes:

$1100 = 9800h + 120$

Now, solve for $h$. Subtract 120 from both sides:

$1100 - 120 = 9800h$

$980 = 9800h$

Divide both sides by 9800:

$h = \frac{980}{9800}$

Simplify the fraction:

$h = \frac{98}{980} = \frac{98 \div 98}{980 \div 98} = \frac{1}{10} = 0.1 \, \text{m}$

Therefore, the depth of the bubble is 0.1 m.

Comparing with the options:

• A. 0.1 m
• B. 0.15 m
• C. 0.20 m
• D. 0.25 m

Hence, the correct answer is Option A.

Newton's law of cooling states that the rate of cooling of a body is proportional to the difference between its temperature and the ambient temperature. The ambient temperature is given as 30°C. The differential equation is:

$\frac{dT}{dt} = -k(T - T_s)$

where $T$ is the temperature of the body at time $t$, $T_s = 30^\circ \text{C}$ is the ambient temperature, and $k$ is a positive constant. Separating variables and integrating:

$\int \frac{dT}{T - T_s} = -k \int dt$

$\ln|T - T_s| = -kt + C$

where $C$ is the integration constant. Since the body's temperature is above the ambient temperature, we can write:

$T - T_s = e^{-kt + C} = e^C e^{-kt}$

Let $A = e^C$, so:

$T - T_s = A e^{-kt}$

$T = T_s + A e^{-kt} = 30 + A e^{-kt}$

At time $t = 0$, the temperature is 80°C:

$80 = 30 + A e^{0} \Rightarrow 80 = 30 + A \Rightarrow A = 50$

Thus, the temperature equation is:

$T = 30 + 50 e^{-kt}$

It takes 5 minutes for the temperature to drop from 80°C to 40°C. At $t = 5$ minutes, $T = 40^\circ \text{C}$:

$40 = 30 + 50 e^{-k \cdot 5}$

$10 = 50 e^{-5k}$

$\frac{10}{50} = e^{-5k}$

$0.2 = e^{-5k}$

Taking natural logarithm on both sides:

$\ln(0.2) = \ln(e^{-5k})$

$\ln(0.2) = -5k$

Since $\ln(0.2) = \ln\left(\frac{1}{5}\right) = -\ln(5)$ and given $\ln(5) = 1.609$:

$-1.609 = -5k$

$k = \frac{1.609}{5} = 0.3218$

Now, we need to find the time taken to cool from 62°C to 32°C. Let $t_1$ be the time when the temperature is 62°C and $t_2$ be the time when it is 32°C. The time required is $t_2 - t_1$. Using the temperature equation:

At $T = 62^\circ \text{C}$:

$62 = 30 + 50 e^{-k t_1}$

$32 = 50 e^{-k t_1}$

$e^{-k t_1} = \frac{32}{50} = 0.64$

At $T = 32^\circ \text{C}$:

$32 = 30 + 50 e^{-k t_2}$

$2 = 50 e^{-k t_2}$

$e^{-k t_2} = \frac{2}{50} = 0.04$

Dividing the two expressions:

$\frac{e^{-k t_1}}{e^{-k t_2}} = \frac{0.64}{0.04} = 16$

$\frac{e^{-k t_1}}{e^{-k t_2}} = e^{-k(t_1 - t_2)} = e^{k(t_2 - t_1)}$

So:

$e^{k(t_2 - t_1)} = 16$

Taking natural logarithm:

$k(t_2 - t_1) = \ln(16)$

$\ln(16) = \ln(2^4) = 4 \ln 2 = 4 \times 0.693 = 2.772$

Substituting $k = \frac{\ln(5)}{5} = \frac{1.609}{5}$:

$t_2 - t_1 = \frac{\ln(16)}{k} = \frac{2.772}{0.3218} \approx 8.614$

Alternatively, using the exact expression:

$t_2 - t_1 = \frac{\ln(16)}{k} = \frac{4 \ln 2}{\frac{\ln 5}{5}} = \frac{4 \ln 2 \times 5}{\ln 5} = \frac{20 \ln 2}{\ln 5}$

Substituting the given values:

$t_2 - t_1 = \frac{20 \times 0.693}{1.609} = \frac{13.86}{1.609} \approx 8.614$

Rounding to one decimal place, the time is 8.6 minutes.

Hence, the correct answer is Option B.

In this problem, we are dealing with an adiabatic compression of a diatomic ideal gas. The work done on the gas is 830 J, the number of moles is 2, and the volume is reduced by 50%. The gas constant R is given as 8.3 J K⁻¹ mol⁻¹. We need to find the change in temperature.

First, recall that an adiabatic process means there is no heat exchange, so Q = 0. According to the first law of thermodynamics, the change in internal energy, ΔU, is related to the heat added to the system and the work done by the system by the equation:

$\Delta U = Q - W$

Since Q = 0, this simplifies to:

$\Delta U = -W$

Now, the work done on the gas is given as 830 J. This means the work done by the system is -830 J (because work done on the system is positive, so work done by the system is negative). Substituting W = -830 J:

$\Delta U = -(-830) = 830 \text{ J}$

So, the change in internal energy is 830 J.

For an ideal gas, the internal energy depends only on temperature. For a diatomic gas, the molar specific heat at constant volume, C_V, is (5/2)R, because diatomic gases have 5 degrees of freedom (3 translational and 2 rotational). The change in internal energy for n moles is given by:

$\Delta U = n C_V \Delta T = n \left( \frac{5}{2} R \right) \Delta T$

We have n = 2 moles, R = 8.3 J K⁻¹ mol⁻¹, and ΔU = 830 J. Substituting these values:

$830 = 2 \times \left( \frac{5}{2} \times 8.3 \right) \times \Delta T$

Simplify the expression inside the parentheses:

$830 = 2 \times \frac{5}{2} \times 8.3 \times \Delta T$

The factor of 2 and the denominator 2 cancel each other:

$830 = 5 \times 8.3 \times \Delta T$

Now, compute 5 times 8.3:

$5 \times 8.3 = 41.5$

So the equation becomes:

$830 = 41.5 \times \Delta T$

Solving for ΔT:

$\Delta T = \frac{830}{41.5}$

Performing the division:

$\Delta T = 20 \text{ K}$

Therefore, the change in temperature is 20 K. Note that the volume reduction by 50% was not directly used in this calculation because the change in internal energy depends only on the change in temperature for an ideal gas, and we were able to find ΔU from the work done.

Hence, the correct answer is Option C.

$V_1 = 2.4 \times 10^{-3} \text{ m}^3,$

$P_1 = P_0 = 10^5 \text{ }\frac{\text{N}}{\text{m}^2} \quad \text{and}$ $T_1 = 300\text{ K} \qquad \text{(given)}$

If area of cross-section of piston is $A$ and it moves through distance $x$ then increment in volume of the gas $= Ax$ and if force constant of a spring is $k$ then force $F = kx$ and pressure $= F/A = kx/A$

$V_2 = V_1 + Ax = 2.4 \times 10^{-3} + 8 \times 10^{-3} \times 0.1 = 3.2 \times 10^{-3}$

$\text{and } P_2 = P_0 + \frac{kx}{A} = 10^5 + \frac{8000 \times 0.1}{8 \times 10^{-3}} = 2 \times 10^5$

$\text{From ideal gas equation } \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

$\implies \frac{10^5 \times 2.4 \times 10^{-3}}{300} = \frac{2 \times 10^5 \times 3.2 \times 10^{-3}}{T_2}$

$\implies T_2 = 800\text{ K}$

The angular frequency of the undamped oscillator is $\omega_0 = \sqrt{\frac{k}{m}}$.

$\omega = \sqrt{\frac{k}{m} \left( 1 - \frac{r^2}{4mk} \right)} = \omega_0 \left( 1 - \frac{r^2}{4mk} \right)^{1/2}$

$T = \frac{2\pi}{\omega_0 \left( 1 - \frac{r^2}{4mk} \right)^{1/2}} = T_0 \left( 1 - \frac{r^2}{4mk} \right)^{-1/2}$, where $T_0$ is the time period of the undamped oscillator.

Given that the ratio $\frac{r^2}{mk} = 8\% = 0.08$, the term $\frac{r^2}{4mk} = \frac{0.08}{4} = 0.02$. Since $0.02 \ll 1$, we can use the binomial approximation $(1 - x)^n \approx 1 + nx$: $T \approx T_0 \left[ 1 + \left( -\frac{1}{2} \right) \left( -\frac{r^2}{4mk} \right) \right]$

$T \approx T_0 \left( 1 + \frac{r^2}{8mk} \right)$

$\frac{\Delta T}{T_0} = \frac{T - T_0}{T_0} = \frac{r^2}{8mk}$

$\text{Percentage change} = \left( \frac{r^2}{8mk} \right) \times 100 = \frac{1}{8} \left( \frac{r^2}{mk} \times 100 \right)$

$\frac{r^2}{mk} \times 100 = 8\%$

$\text{Percentage change} = \frac{1}{8} \times 8\% = 1\%$

Both factories emit sound at frequency $f_0 = 800$ Hz. The man walks from one factory toward the other at speed $v_o = 2$ m/s, while the velocity of sound is $v = 320$ m/s. Since both sources are stationary and the observer moves, the Doppler-shifted frequencies are as follows.

For the factory the man approaches, the observed frequency is $f_1 = f_0\left(\dfrac{v + v_o}{v}\right) = 800 \times \dfrac{322}{320}$.

For the factory the man moves away from, the observed frequency is $f_2 = f_0\left(\dfrac{v - v_o}{v}\right) = 800 \times \dfrac{318}{320}$.

The beat frequency is $f_1 - f_2 = 800 \times \dfrac{322 - 318}{320} = 800 \times \dfrac{4}{320} = 800 \times \dfrac{1}{80} = 10$ beats per second.

The cone has base radius $R$ and height $h$, and sits in a uniform electric field $\vec{E}$ directed parallel to the base. Since there is no enclosed charge, Gauss's law gives zero net flux through the entire closed surface of the cone (curved surface plus the circular base). Therefore the flux entering one side of the cone equals the flux leaving the other side.

To find the entering flux, consider the projection of the cone onto a plane perpendicular to $\vec{E}$. The cone's silhouette when viewed along the direction of $\vec{E}$ is a triangle with base equal to the diameter $2R$ and height equal to $h$. The area of this triangle is $\dfrac{1}{2} \times 2R \times h = Rh$.

Since the field is uniform, the flux through any closed surface equals the flux through the projected cross-sectional area perpendicular to the field. The entering flux (which equals the leaving flux) is therefore $\Phi = E \times Rh = EhR$.

Capacitance with dielectric ($C_1$): $C_1 = \frac{K\epsilon_0(xw)}{d}$

Capacitance with air ($C_2$): $C_2 = \frac{\epsilon_0(l-x)w}{d}$

Total Capacitance ($C$): $C = C_1 + C_2 = \frac{\epsilon_0 w}{d} [Kx + (l-x)] = \frac{\epsilon_0 w}{d} [l + x(K-1)]$

$U = \frac{Q^2}{2C} = \frac{Q^2 d}{2 \epsilon_0 w [l + x(K-1)]}$

$F = -\frac{\partial U}{\partial x}$

$F = -\frac{Q^2 d}{2 \epsilon_0 w} \cdot \frac{d}{dx} [l + x(K-1)]^{-1}$

$F = -\frac{Q^2 d}{2 \epsilon_0 w} \cdot (-1)[l + x(K-1)]^{-2} \cdot (K-1)$

$F = \frac{Q^2 d (K-1)}{2 \epsilon_0 w [l + x(K-1)]^2}$

When the slab is near the edge ($x \approx 0$): $F = \frac{Q^2 d}{2 \epsilon_0 w l^2} (K-1)$

$I_{20\Omega} = \frac{50\text{ V}}{20\ \Omega} = 2.5\text{ A (downward)}$

$I_{10\Omega} = \frac{30\text{ V}}{10\ \Omega} = 3\text{ A (downward)}$

$I_{horizontal} = \frac{50 - 30}{5 + 5} = \frac{20}{10} = 2\text{ A (left to right)}$

$I_{50V} = I_{20\Omega} + I_{horizontal} = 2.5\text{ A} + 2\text{ A} = 4.5\text{ A}$

$2\text{ A} + I_{30V} = 3\text{ A} \implies I_{30V} = 1\text{ A}$

In Circuit (d), the high resistance $R$ is placed in series with the main circuit branch. The low resistance shunt $S$ (along with key $K_2$) is connected in parallel directly across the galvanometer $G$. This is the correct configuration.

$I_g = \frac{V}{R + G}$