WBJEE 2016 Physics Question Paper with Answer and Solution

(A) Let the mass $m_{1}$ be placed at the origin $(0, 0)$ and the mass $m_{2}$ be placed at a distance $R$ along the x-axis at $(R, 0)$.
The formula for the x-coordinate of the centre of mass is given by:
$X_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$
Substituting the values $x_{1} = 0$ and $x_{2} = R$:
$X_{cm} = \frac{m_{1} \times 0 + m_{2} \times R}{m_{1} + m_{2}}$
$X_{cm} = \frac{m_{2} R}{m_{1} + m_{2}}$
Thus,the distance of the centre of mass from $m_{1}$ is $\frac{m_{2} R}{m_{1} + m_{2}}$.

(C) The root mean square (rms) velocity of a gas is given by the formula: $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$,where $R$ is the gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
Since $R$ and $T$ are constant for both gases,$v_{\text{rms}} \propto \frac{1}{\sqrt{M}}$.
For hydrogen gas $(H_2)$,$M_{H_2} = 2 \text{ g/mol}$. Given $v_{\text{rms}, H_2} = c$.
For oxygen gas $(O_2)$,$M_{O_2} = 32 \text{ g/mol}$.
Taking the ratio: $\frac{v_{\text{rms}, H_2}}{v_{\text{rms}, O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.
Therefore,$v_{\text{rms}, O_2} = \frac{v_{\text{rms}, H_2}}{4} = \frac{c}{4}$.

(A) Given: Mass $m = 1 \ kg$,acceleration $a = 4.9 \ ms^{-2}$,acceleration due to gravity $g \approx 9.8 \ ms^{-2}$.
$(i)$ When the body is lifted up with acceleration $a$,the equation of motion is:
$T_1 - mg = ma$
$T_1 = m(g + a) = 1 \times (9.8 + 4.9) = 14.7 \ N$
(ii) When the body is lowered with acceleration $a$,the equation of motion is:
$mg - T_2 = ma$
$T_2 = m(g - a) = 1 \times (9.8 - 4.9) = 4.9 \ N$
The ratio of tension in the first and second case is:
$\frac{T_1}{T_2} = \frac{14.7}{4.9} = \frac{3}{1} = 3:1$

$(A)$ The work done in stretching a spring by a distance $x$ is given by $W = \frac{1}{2} K x^2$, where $K$ is the spring constant.
For the first stretch, $x_1 = 1 \text{ mm}$, the work done is $W_1 = \frac{1}{2} K (1)^2 = 10 \text{ J}$.
To stretch it further by $1 \text{ mm}$, the total extension becomes $x_2 = 1 \text{ mm} + 1 \text{ mm} = 2 \text{ mm}$.
The total work done for this extension is $W_2 = \frac{1}{2} K (x_2)^2 = \frac{1}{2} K (2)^2 = 4 \times (\frac{1}{2} K (1)^2) = 4 \times W_1$.
Substituting $W_1 = 10 \text{ J}$, we get $W_2 = 4 \times 10 \text{ J} = 40 \text{ J}$.
The additional work required to stretch it further is $W_{\text{extra}} = W_2 - W_1 = 40 \text{ J} - 10 \text{ J} = 30 \text{ J}$.

(A) Let the radius of the small drop be $r = 1 \ mm = 1 \times 10^{-3} \ m$.
Let the radius of the large drop be $R$.
The volume of the large drop is equal to the sum of the volumes of $1000$ small drops:
$\frac{4}{3} \pi R^3 = 1000 \times \frac{4}{3} \pi r^3$
$R^3 = 1000 r^3 \implies R = 10r = 10 \times 10^{-3} \ m = 10^{-2} \ m$.
The initial surface area of $1000$ drops is $A_i = 1000 \times 4 \pi r^2$.
The final surface area of the single drop is $A_f = 4 \pi R^2 = 4 \pi (10r)^2 = 400 \pi r^2$.
The change in surface area is $\Delta A = A_f - A_i = 400 \pi r^2 - 1000 \times 4 \pi r^2 = -3600 \pi r^2$.
The energy loss is $\Delta E = S \times |\Delta A| = S \times 3600 \pi r^2$.
Substituting the values: $\Delta E = 0.072 \times 3600 \times \pi \times (10^{-3})^2$.
$\Delta E = 0.072 \times 3600 \times 3.14159 \times 10^{-6} \approx 8.143 \times 10^{-4} \ J$.
Rounding to the given option,the energy loss is $8.146 \times 10^{-4} \ J$.

(D) Let the mass of the drop be $m$.
The weight of the drop is $W = mg$,which acts in the downward direction.
The force due to surface tension acting on the drop at the exit of the tap is $F_s = \sigma \cdot 2\pi R$,where $R$ is the radius of the tap exit. This force acts in the upward direction.
The drop of water will detach when the weight of the drop exceeds the upward force due to surface tension:
$mg > \sigma \cdot 2\pi R$
Since the mass $m$ of the spherical drop is given by $m = V \cdot \rho = \frac{4}{3} \pi r^3 \rho$,where $r$ is the radius of the drop,we substitute this into the inequality:
$\frac{4}{3} \pi r^3 \rho g > \sigma \cdot 2\pi R$
Solving for $r$:
$r^3 > \frac{2\pi R \sigma \cdot 3}{4\pi \rho g}$
$r^3 > \frac{3 R \sigma}{2 \rho g}$
$r > \left( \frac{3 R \sigma}{2 \rho g} \right)^{1/3}$
Comparing this result with the given options,none of the options $A$,$B$,or $C$ match the derived expression. Therefore,the correct option is $D$.

(D) The gas bubble is rising through the liquid of density $\rho = 1.75 \ g/cm^3$ with a constant terminal velocity $v_T = 0.35 \ cm/s$.
Since the density of the gas is negligible,the forces acting on the bubble are the buoyancy force $F_b$ (upward) and the viscous drag force $F_V$ (downward).
At terminal velocity,the net force is zero,so $F_V = F_b$.
Using Stokes' Law,the viscous force is $F_V = 6 \pi \eta r v_T$,where $\eta$ is the coefficient of viscosity and $r$ is the radius of the bubble.
The buoyancy force is $F_b = V \rho g = \frac{4}{3} \pi r^3 \rho g$.
Equating the two: $6 \pi \eta r v_T = \frac{4}{3} \pi r^3 \rho g$.
Solving for $\eta$: $\eta = \frac{2}{9} \frac{r^2 \rho g}{v_T}$.
Given: $r = 1 \ cm = 10^{-2} \ m$,$\rho = 1.75 \ g/cm^3 = 1750 \ kg/m^3$,$v_T = 0.35 \ cm/s = 3.5 \times 10^{-3} \ m/s$,$g = 9.8 \ m/s^2$.
Using $CGS$ units for convenience: $r = 1 \ cm$,$\rho = 1.75 \ g/cm^3$,$v_T = 0.35 \ cm/s$,$g = 980 \ cm/s^2$.
$\eta = \frac{2}{9} \times \frac{(1)^2 \times 1.75 \times 980}{0.35} = \frac{2}{9} \times \frac{1715}{0.35} = \frac{2}{9} \times 4900 \approx 1088.8 \ \text{poise} \approx 1089 \ \text{poise}$.

(C) In Simple Harmonic Motion $(SHM)$,the acceleration $a$ of a vibrating particle is given by the relation $a = \omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement from the mean position.
Given: Acceleration $a = 16 \ cm/s^2$ and displacement $x = 4 \ cm$.
Substituting these values into the formula:
$16 = \omega^2 \times 4$
$\omega^2 = \frac{16}{4} = 4$
$\omega = 2 \ rad/s$
The time period $T$ is related to angular frequency by the formula $T = \frac{2\pi}{\omega}$.
$T = \frac{2\pi}{2} = \pi \ s$
Using $\pi \approx 3.142$,we get $T = 3.142 \ s$.

(D) Let $A$ be the surface area of the pond and $x$ be the thickness of the ice layer at any time $t$. The temperature difference across the ice layer is $\Delta T = 0^{\circ} C - (-20^{\circ} C) = 20^{\circ} C$.
The rate of heat flow through the ice layer is given by conduction: $\frac{dQ}{dt} = \frac{kA \Delta T}{x} = \frac{kA(20)}{x}$.
As the ice layer thickens by $dx$ in time $dt$,the heat released is $dQ = L dm = L(\rho A dx)$.
Equating the two expressions for $\frac{dQ}{dt}$:
$\rho A L \frac{dx}{dt} = \frac{20 k A}{x}$
$\int_{0}^{Z} x dx = \int_{0}^{t} \frac{20 k}{\rho L} dt$
$\frac{Z^2}{2} = \frac{20 k}{\rho L} t$
$Z^2 = \frac{40 k}{\rho L} t$.

(D) According to the Stefan-Boltzmann law,the total energy radiated per unit time (power) by a black body is proportional to the fourth power of its absolute temperature $(T)$:
$P \propto T^4$
Let the initial temperature of the Sun be $T_1 = T$ and the initial power received be $P_1$.
When the temperature is doubled,the new temperature is $T_2 = 2T$.
The new power received $P_2$ is given by:
$P_2 \propto (T_2)^4$
$P_2 \propto (2T)^4$
$P_2 \propto 16T^4$
Therefore,the ratio of the new power to the initial power is:
$\frac{P_2}{P_1} = \frac{16T^4}{T^4} = 16$
Thus,the rate of energy received on the Earth will increase by a factor of $16$.

(D) The energy density $u$ of blackbody radiation at temperature $T$ is given by $u = aT^4$,where $a$ is the radiation constant.
The total energy $E$ in a container of volume $V$ is $E = uV = aVT^4$.
The heat required to increase the temperature from $T_1$ to $T_2$ is $\Delta E = aV(T_2^4 - T_1^4)$.
Since the volume $V$ of the container is not provided in the problem statement,it is impossible to calculate the exact numerical value of the heat required.
Therefore,the information provided is insufficient.

(B) The mass of the hydrogen gas is given as $m = 4 \ g$.
The molar mass of hydrogen gas $(H_2)$ is $M = 2 \ g/mol$.
The number of moles $n$ is calculated as $n = \frac{m}{M} = \frac{4 \ g}{2 \ g/mol} = 2 \ mol$.
The ideal gas equation is given by $p V = n R T$.
Substituting the value of $n = 2$,we get $p V = 2 R T$.

(D) The tube is closed at one end,so it acts as a closed organ pipe. The lowest frequency (fundamental frequency) produced in the tube corresponds to the condition where the length of the tube $l = \frac{\lambda}{4}$.
Given: $l = 1.0 \text{ m}$,$f = 84 \text{ Hz}$.
Therefore,the wavelength $\lambda = 4l = 4 \times 1.0 = 4 \text{ m}$.
The velocity of sound in air is given by $v = f \lambda = 84 \times 4 = 336 \text{ m/s}$.
The velocity of sound is also given by the formula $v = \sqrt{\frac{\gamma P}{\rho}}$,where $P$ is the atmospheric pressure and $\rho$ is the density of air.
Squaring both sides,we get $v^{2} = \frac{\gamma P}{\rho}$.
Rearranging for $\gamma$,we get $\gamma = \frac{v^{2} \rho}{P}$.
Substituting the values: $\gamma = \frac{(336)^{2} \times 1.2}{1.0 \times 10^{5}} = \frac{112896 \times 1.2}{100000} = \frac{135475.2}{100000} \approx 1.355$.
Rounding to the nearest option,we get $\gamma = 1.4$.

(B) The velocity of sound in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$,where $\gamma$ is the adiabatic index,$R$ is the universal gas constant,$T$ is the absolute temperature in Kelvin,and $M$ is the molar mass of the gas.
Note that the velocity of sound is independent of pressure for an ideal gas.
Given:
$T_1 = 273 + 20 = 293 \text{ K}$
$T_2 = 273 + 40 = 313 \text{ K}$
$v_1 = 344.2 \text{ m/s}$
Using the ratio:
$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$
$v_2 = v_1 \sqrt{\frac{313}{293}}$
$v_2 = 344.2 \times \sqrt{1.06826}$
$v_2 \approx 344.2 \times 1.03356$
$v_2 \approx 355.75 \text{ m/s} \approx 356 \text{ m/s}$.

(B) According to Bohr's quantization condition,the angular momentum of an electron in an orbit is given by $L = mvr = \frac{nh}{2\pi}$.
From the de-Broglie hypothesis,the wavelength of an electron is $\lambda = \frac{h}{mv}$.
Substituting $mv = \frac{h}{\lambda}$ into the quantization condition,we get $\frac{h}{\lambda} r = \frac{nh}{2\pi}$.
This simplifies to $2\pi r = n\lambda$,where $2\pi r$ is the circumference of the $n^{th}$ orbit.
For the second Bohr orbit,$n = 2$.
Therefore,the circumference of the second orbit is $2\pi r = 2\lambda$.
This implies that the number of de-Broglie wavelengths contained in the second Bohr orbit is $2$.

(D) The Rydberg formula is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the Balmer series,$n_1 = 2$. The first line corresponds to $n_2 = 3$,and the second line corresponds to $n_2 = 4$.
$\frac{1}{\lambda_B} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{4} - \frac{1}{16} \right] = R \left( \frac{3}{16} \right)$.
Given $\lambda_B = 600 \ nm$,so $\frac{1}{600} = R \left( \frac{3}{16} \right) \implies R = \frac{16}{1800} \ nm^{-1}$.
For the Lyman series,$n_1 = 1$. The third line corresponds to $n_2 = 4$ (since $n_2 = 2, 3, 4, \dots$).
$\frac{1}{\lambda_L} = R \left[ \frac{1}{1^2} - \frac{1}{4^2} \right] = R \left[ 1 - \frac{1}{16} \right] = R \left( \frac{15}{16} \right)$.
Substituting the value of $R$:
$\frac{1}{\lambda_L} = \left( \frac{16}{1800} \right) \left( \frac{15}{16} \right) = \frac{15}{1800} = \frac{1}{120}$.
Therefore,$\lambda_L = 120 \ nm$.
Since $120 \ nm$ is not among the options,the correct choice is $D$.

(B) The circuit consists of a bridge-like structure. Let us simplify it.
$1$. The two $4 \mu F$ capacitors in the top branch (between $A$ and $B$) are in series. Their equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$,so $C_1 = 2 \mu F$.
$2$. The two $4 \mu F$ capacitors in the bottom branches (connected to $C$) are also in series with each other. Their equivalent capacitance $C_2$ is $\frac{1}{C_2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$,so $C_2 = 2 \mu F$.
$3$. This $C_2$ is in parallel with the middle $4 \mu F$ capacitor. Thus,the equivalent capacitance of this middle section is $C_3 = C_2 + 4 \mu F = 2 \mu F + 4 \mu F = 6 \mu F$.
$4$. Finally,$C_1$ and $C_3$ are in parallel between points $A$ and $B$. Therefore,the total equivalent capacitance $C_{eq} = C_1 + C_3 = 2 \mu F + 6 \mu F = 8 \mu F$.

(B) Let the initial resistances be $R_1 = 400 \Omega$ and $R_2 = 400 \Omega$. The total resistance is $R_{eq} = R_1 + R_2 = 800 \Omega$. The current in the circuit is $I = V / R_{eq} = 8 / 800 = 0.01 A$. The potential difference across the second resistor is $V_2 = I R_2 = 0.01 \times 400 = 4 V$.
When $R_1$ increases by $0.5 \%$,the new resistance $R_1' = 400 + (0.5 / 100) \times 400 = 400 + 2 = 402 \Omega$.
To keep the potential difference $V_2$ across $R_2$ constant at $4 V$,the current $I'$ must remain $0.01 A$. Since $I' = V / (R_1' + R_2')$,we have $0.01 = 8 / (402 + R_2')$.
Solving for $R_2'$,we get $402 + R_2' = 8 / 0.01 = 800$,which gives $R_2' = 800 - 402 = 398 \Omega$.
The change in $R_2$ is $\Delta R_2 = R_2' - R_2 = 398 - 400 = -2 \Omega$.
Wait,let's re-evaluate: If $R_1$ increases,the total resistance increases,causing the current to decrease. To keep $V_2$ constant,$R_2$ must be adjusted. If $V_2 = I R_2 = (V / (R_1 + R_2)) R_2$,for $V_2$ to be constant,the ratio $R_2 / (R_1 + R_2)$ must be constant. This implies $R_2 / R_1$ must remain constant. Since $R_1$ increased by $0.5 \%$,$R_2$ must also increase by $0.5 \%$.
Increase in $R_2 = 0.5 \% \text{ of } 400 = 2 \Omega$.

(C) Let the resistance of the link $AB$ be $R_{AB} = 1 \Omega$. Let $R_{eq}$ be the effective resistance of the remaining part of the cube when the link $AB$ is removed.
When the link $AB$ is present,it is in parallel with the rest of the cube's network. The equivalent resistance is given by:
$\frac{1}{R_{total}} = \frac{1}{R_{AB}} + \frac{1}{R_{eq}}$
Given $R_{total} = \frac{7}{12} \Omega$ and $R_{AB} = 1 \Omega$:
$\frac{12}{7} = \frac{1}{1} + \frac{1}{R_{eq}}$
$\frac{1}{R_{eq}} = \frac{12}{7} - 1 = \frac{5}{7}$
Therefore,$R_{eq} = \frac{7}{5} \Omega$.

(C) The time constant $\tau$ of an $RC$ circuit is given by $\tau = R \cdot C$.
Given:
$C = 1 \mu F = 1 \times 10^{-6} F$
$R = 1 \text{ M}\Omega = 1 \times 10^{6} \Omega$
Therefore, $\tau = (1 \times 10^{6} \Omega) \times (1 \times 10^{-6} F) = 1 \text{ s}$.
The voltage across a charging capacitor at time $t$ is given by $V(t) = V_0(1 - e^{-t/\tau})$.
Substituting $t = 1 \text{ s}$ and $\tau = 1 \text{ s}$:
$V(1) = 10(1 - e^{-1/1}) = 10(1 - e^{-1})$.
Since $e^{-1} \approx 0.37$, we have:
$V(1) \approx 10(1 - 0.37) = 10(0.63) = 6.3 V$.
Thus, the voltage across the capacitor after $1 \text{ s}$ is approximately $6.3 V$.

(A) The current is given by $I = I_{0} e^{-\lambda t}$.
We know that current $I$ is the rate of flow of charge,so $I = \frac{dQ}{dt}$.
Therefore,$dQ = I dt = I_{0} e^{-\lambda t} dt$.
To find the total charge $Q$ over the entire pulse period,we integrate from $t = 0$ to $t = \infty$:
$Q = \int_{0}^{\infty} I_{0} e^{-\lambda t} dt$
$Q = I_{0} \left[ \frac{e^{-\lambda t}}{-\lambda} \right]_{0}^{\infty}$
$Q = \frac{I_{0}}{-\lambda} [e^{-\infty} - e^{0}]$
Since $e^{-\infty} = 0$ and $e^{0} = 1$,we get:
$Q = \frac{I_{0}}{-\lambda} [0 - 1] = \frac{I_{0}}{\lambda}$.

(B) The resistance of a wire is given by $R = \frac{\rho l}{A}$,where $\rho$ is resistivity,$l$ is length,and $A$ is the cross-sectional area.
When two wires are connected in series,the total resistance $R_{eq}$ is the sum of individual resistances:
$R_{eq} = R_{1} + R_{2}$
Since the wires have the same radius,their cross-sectional areas are equal $(A_{1} = A_{2} = A)$.
$R_{eq} = \frac{\rho_{1} l_{1}}{A} + \frac{\rho_{2} l_{2}}{A} = \frac{\rho_{1} l_{1} + \rho_{2} l_{2}}{A}$
For the equivalent wire,the total length is $L = l_{1} + l_{2}$ and the area is $A$. The equivalent resistivity $\rho_{eq}$ is defined as:
$R_{eq} = \frac{\rho_{eq} (l_{1} + l_{2})}{A}$
Equating the two expressions for $R_{eq}$:
$\frac{\rho_{eq} (l_{1} + l_{2})}{A} = \frac{\rho_{1} l_{1} + \rho_{2} l_{2}}{A}$
$\rho_{eq} = \frac{\rho_{1} l_{1} + \rho_{2} l_{2}}{l_{1} + l_{2}}$

(C) The de-Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula:
$\lambda = \frac{h}{\sqrt{2meV}}$
Squaring both sides, we get:
$\lambda^2 = \frac{h^2}{2meV}$
Rearranging for $V$:
$V = \frac{h^2}{2me\lambda^2}$
Using the standard approximation formula for an electron:
$V \approx \frac{150}{\lambda^2} \text{ V}$, where $\lambda$ is in $\text{Å}$.
Given $\lambda = 1 \text{ Å}$:
$V = \frac{150}{(1)^2} = 150 \text{ V}$.
Thus, the required potential difference is $150 \text{ V}$.

(D) The work function of Cesium is $\phi = 2.27 eV$.
The wavelength of the incident light is $\lambda = 600 nm = 600 \times 10^{-9} m$.
The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Substituting the values ($h = 6.63 \times 10^{-34} Js$,$c = 3 \times 10^8 m/s$):
$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}} J = 3.315 \times 10^{-19} J$.
Converting this energy into electron-volts $(eV)$:
$E = \frac{3.315 \times 10^{-19}}{1.6 \times 10^{-19}} eV \approx 2.07 eV$.
Since the energy of the incident photon $(2.07 eV)$ is less than the work function of Cesium $(2.27 eV)$,the photoelectric effect will not occur.
Therefore,no electrons are emitted,and no cut-off voltage is required. Thus,the correct answer is $D$.

(A) The intensity $I$ of light at a distance $r$ from a point source is given by $I = \frac{P}{4\pi r^2}$,so $I \propto \frac{1}{r^2}$.
When the distance is changed from $d$ to $\frac{d}{2}$,the new intensity $I'$ becomes $I' = \frac{1}{(d/2)^2} = 4 \times \frac{1}{d^2} = 4I$.
Since the number of photoelectrons emitted per second is directly proportional to the intensity of incident light,the emission rate becomes $4$ times the original value.
According to Einstein's photoelectric equation,$KE_{max} = h\nu - \phi$,where $\nu$ is the frequency of incident light and $\phi$ is the work function.
Since the frequency $\nu$ remains unchanged,the maximum kinetic energy $(KE_{max})$ and the stopping potential $(V_s)$ remain unchanged,as $eV_s = KE_{max}$.

(C) Given, $L_{1} = 6 mH = 6 \times 10^{-3} H$ and $L_{2} = 8 mH = 8 \times 10^{-3} H$.
For two coils connected in series with the highest coefficient of coupling $(k = 1)$, the equivalent self-inductance $L$ is given by the formula:
$L = L_{1} + L_{2} + 2M$
where $M$ is the mutual inductance, $M = k \sqrt{L_{1} L_{2}}$.
For the highest coupling, $k = 1$, so $M = \sqrt{L_{1} L_{2}}$.
Substituting the values:
$L = L_{1} + L_{2} + 2 \sqrt{L_{1} L_{2}} \text{ (in } mH)$
$L = 6 + 8 + 2 \sqrt{6 \times 8}$
$L = 14 + 2 \sqrt{48}$
$L = 14 + 2 \times 6.928$
$L = 14 + 13.856 = 27.856 mH$
Rounding to the nearest integer, we get $L \approx 28 mH$.

(B) Inside a hollow charged metal sphere,the electric field intensity is always zero because there is no charge enclosed within the Gaussian surface inside the sphere.
$\therefore E = 0$.
However,the electric potential inside the sphere is constant and equal to the potential at its surface.
The potential on the surface of the hollow sphere is given by:
$V = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} = \frac{Q}{4 \pi \varepsilon_{0} R}$.
Thus,inside the sphere,the potential is $\frac{Q}{4 \pi \varepsilon_{0} R}$ and the electric field intensity is zero.

(B) An equipotential surface is a surface where the electric potential is the same at every point.
By definition,the work done in moving a charge $q$ along an equipotential surface is zero,because $W = q \Delta V$ and $\Delta V = 0$.
Since the work done is $W = \int \vec{F} \cdot d\vec{l} = q \int \vec{E} \cdot d\vec{l} = 0$,the electric field $\vec{E}$ must be perpendicular to the displacement $d\vec{l}$ along the surface.
Therefore,the electric lines of force are always perpendicular to the equipotential surface,meaning the angle between them is $90^{\circ}$.

(A) When a current-carrying rectangular coil is placed in a non-uniform magnetic field,the magnetic field strength varies at different points along the coil.
Since the magnetic field is non-uniform,the magnetic forces acting on different segments of the coil do not cancel each other out,resulting in a non-zero net force.
Additionally,because the forces are distributed such that they do not act through a single point or are not balanced,a non-zero net torque is also produced.
Therefore,both the net force and the net torque acting on the coil are non-zero.
Given the standard nature of such multiple-choice questions where one answer is expected,both $(a)$ and $(d)$ are physically correct statements.

(D) The total Lorentz force on the electron is given by $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.
Given: $\vec{E} = 3\hat{i} + 6\hat{j} + 2\hat{k} \text{ V m}^{-1}$,$\vec{B} = 2\hat{i} + 3\hat{j} \text{ T}$,$\vec{v} = 2\hat{i} + 3\hat{j} \text{ m s}^{-1}$,and $q = -1.6 \times 10^{-19} \text{ C}$.
First,calculate the magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$.
Since $\vec{v} = 2\hat{i} + 3\hat{j}$ and $\vec{B} = 2\hat{i} + 3\hat{j}$,the vectors are parallel $(\vec{v} \parallel \vec{B})$.
Therefore,$\vec{v} \times \vec{B} = 0$,which implies $\vec{F}_m = 0$.
Now,calculate the electric force $\vec{F}_e = q\vec{E}$.
$\vec{F}_e = (-1.6 \times 10^{-19}) (3\hat{i} + 6\hat{j} + 2\hat{k}) = -4.8 \times 10^{-19}\hat{i} - 9.6 \times 10^{-19}\hat{j} - 3.2 \times 10^{-19}\hat{k} \text{ N}$.
The magnitude of the force is $|\vec{F}| = |\vec{F}_e| = 1.6 \times 10^{-19} \times \sqrt{3^2 + 6^2 + 2^2} = 1.6 \times 10^{-19} \times \sqrt{9 + 36 + 4} = 1.6 \times 10^{-19} \times \sqrt{49} = 1.6 \times 10^{-19} \times 7 = 1.12 \times 10^{-18} \text{ N}$.
Since $1.12 \times 10^{-18} \text{ N}$ is not among the options,the correct choice is $D$.

(C) The necessary centripetal force for circular motion is provided by the electrostatic Coulomb force between the two charges.
Equating the centripetal force to the Coulomb force:
$\frac{m_{1} v^{2}}{r} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{|q_{1} q_{2}|}{r^{2}}$
Solving for velocity $v$:
$v^{2} = \frac{|q_{1} q_{2}|}{4 \pi \varepsilon_{0} m_{1} r}$
$v = \sqrt{\frac{|q_{1} q_{2}|}{4 \pi \varepsilon_{0} m_{1} r}}$
Now,the time period $T$ is given by $T = \frac{2 \pi r}{v}$.
Substituting the expression for $v$:
$T = 2 \pi r \sqrt{\frac{4 \pi \varepsilon_{0} m_{1} r}{|q_{1} q_{2}|}}$
$T = \sqrt{4 \pi^{2} r^{2} \cdot \frac{4 \pi \varepsilon_{0} m_{1} r}{|q_{1} q_{2}|}}$
$T = \sqrt{\frac{16 \pi^{3} \varepsilon_{0} m_{1} r^{3}}{|q_{1} q_{2}|}}$
Comparing this with the given options,option $C$ is the correct expression for the time period $T$ (assuming $q_{1}$ and $q_{2}$ have opposite signs for attraction).

(D) Given that the angle of incidence $i = 60^{\circ}$.
According to Brewster's law,when the reflected and refracted rays are perpendicular to each other,the angle of incidence is the polarizing angle $(i_p)$.
From Snell's law,$\mu = \frac{\sin i}{\sin r}$.
Since the reflected and refracted rays are perpendicular,the angle between them is $90^{\circ}$.
From the geometry of the setup,$i + 90^{\circ} + r = 180^{\circ}$,which gives $r = 90^{\circ} - i$.
Substituting this into Snell's law:
$\mu = \frac{\sin i}{\sin(90^{\circ} - i)} = \frac{\sin i}{\cos i} = \tan i$.
Given $i = 60^{\circ}$,we have $\mu = \tan 60^{\circ} = \sqrt{3}$.

(C) The thickness of the glass plate is given as $t$.
The refractive index of the glass is $\mu$.
The velocity of light in vacuum is $c$.
The velocity of light in a medium $(v)$ is related to the refractive index $(\mu)$ and the velocity of light in vacuum $(c)$ by the formula: $\mu = \frac{c}{v}$.
Therefore,the speed of light in the glass plate is $v = \frac{c}{\mu}$.
The time $(T)$ taken to travel a distance $(t)$ at a constant speed $(v)$ is given by $T = \frac{\text{distance}}{\text{speed}} = \frac{t}{v}$.
Substituting the value of $v$: $T = \frac{t}{c / \mu} = \frac{\mu t}{c}$.
Thus,the time taken is $\frac{\mu t}{c}$.

(D) The Zener diode is connected in series with a resistor $R = 100 \ \Omega$ and a battery of emf $V = 10 \ V$.
When the Zener diode is in the breakdown region,it maintains a constant voltage across it equal to its breakdown voltage,$V_Z = 5.6 \ V$.
The potential difference across the resistor $R$ is given by $\Delta V = V - V_Z = 10 \ V - 5.6 \ V = 4.4 \ V$.
Using Ohm's law,the current $I$ flowing through the circuit (and thus through the Zener diode) is:
$I = \frac{\Delta V}{R} = \frac{4.4 \ V}{100 \ \Omega} = 0.044 \ A$.
Converting this to milliamperes:
$I = 0.044 \times 1000 \ mA = 44 \ mA$.

(C) Given: Current gain $\beta = 45$,Collector resistance $R_C = 1 \ k\Omega = 1000 \ \Omega$,Potential drop across collector resistance $V_C = 5 \ V$.
We know that the collector current $I_C$ is given by $I_C = \frac{V_C}{R_C}$.
Substituting the values: $I_C = \frac{5 \ V}{1000 \ \Omega} = 5 \times 10^{-3} \ A = 5 \ mA$.
The relationship between collector current $I_C$ and base current $I_B$ is given by $\beta = \frac{I_C}{I_B}$.
Therefore,$I_B = \frac{I_C}{\beta}$.
Substituting the values: $I_B = \frac{5 \times 10^{-3} \ A}{45} = \frac{1}{9} \times 10^{-3} \ A \approx 0.111 \times 10^{-3} \ A$.
Converting to microamperes: $I_B \approx 111 \ \mu A$.

(B) In Fraunhofer diffraction,the source of light and the screen are effectively at an infinite distance from the diffracting aperture or obstacle.
This ensures that the wavefronts incident on the aperture are plane wavefronts,and the diffracted rays reaching the screen are parallel to each other.
Therefore,both the source and the screen must be at infinity relative to the diffracting element.