WBJEE 2015 Physics Question Paper with Answer and Solution

(B) Let the positions of the particles be represented by vectors $\vec{r}_i$ such that $|\vec{r}_i| = R$ for all $i = 1, 2, ..., n$.
The position of the center of mass $\vec{R}_{cm}$ is given by $\vec{R}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$.
Using the triangle inequality for vectors,the magnitude of the center of mass is $|\vec{R}_{cm}| = \frac{|\sum m_i \vec{r}_i|}{\sum m_i} \le \frac{\sum m_i |\vec{r}_i|}{\sum m_i}$.
Since $|\vec{r}_i| = R$ for all particles,we have $|\vec{R}_{cm}| \le \frac{\sum m_i R}{\sum m_i} = R$.
Thus,the distance of the center of mass from the origin is always less than or equal to $R$.

(A) For a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $a$ (orbital radius),the energies are given by:
Kinetic energy,$K = \frac{GMm}{2a}$
Potential energy,$V = -\frac{GMm}{a}$
Total energy,$E = K + V = \frac{GMm}{2a} - \frac{GMm}{a} = -\frac{GMm}{2a}$
Comparing these expressions:
$V = -\frac{GMm}{a} = -2 \left( \frac{GMm}{2a} \right) = -2K$
Therefore,$K = -V / 2$.

(C) The gravitational force acting between the two masses $m_{1}$ and $m_{2}$ separated by a distance $r$ is given by Newton's law of gravitation:
$F = \frac{G m_{1} m_{2}}{r^{2}}$
According to Newton's second law of motion,the force on mass $m_{1}$ is $F = m_{1} a_{1}$,where $a_{1}$ is the acceleration of mass $m_{1}$.
Equating the two expressions:
$m_{1} a_{1} = \frac{G m_{1} m_{2}}{r^{2}}$
$a_{1} = \frac{G m_{2}}{r^{2}}$
Since $G$ and $r$ are constant for the system at any given instant,we find that $a_{1} \propto m_{2}$.
Similarly,for mass $m_{2}$,$a_{2} = \frac{G m_{1}}{r^{2}}$,which implies $a_{2} \propto m_{1}$.

(C) The root mean square (rms) speed of a gas is given by the formula $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass.
From this,we can see that $v_{\text{rms}} \propto \sqrt{\frac{T}{M}}$.
Let the initial state be $T_1 = T$ and $M_1 = M$ (for $O_2$ molecules). Then $v_1 = v \propto \sqrt{\frac{T}{M}}$.
In the final state,the temperature is doubled,so $T_2 = 2T$. The oxygen molecules dissociate into atoms,so the molar mass becomes half,$M_2 = M/2$.
The new rms speed $v_2$ is proportional to $\sqrt{\frac{T_2}{M_2}} = \sqrt{\frac{2T}{M/2}} = \sqrt{\frac{4T}{M}} = 2 \sqrt{\frac{T}{M}}$.
Comparing the two,$v_2 = 2 \times v_1 = 2v$.

(B) Let the weight of block $A$ be $W^{\prime}$.
For the system to be in equilibrium,the tension $T_1$ in the horizontal cord connected to block $B$ must be equal to the limiting friction force,so $T_1 = \mu W$.
Now,consider the equilibrium of the knot. Let $T_2$ be the tension in the cord inclined at an angle $\theta$ to the horizontal,and $T_3$ be the tension in the vertical cord connected to block $A$. Thus,$T_3 = W^{\prime}$.
Resolving the forces at the knot:
Horizontal component: $T_2 \cos \theta = T_1 = \mu W$
Vertical component: $T_2 \sin \theta = T_3 = W^{\prime}$
Dividing the two equations:
$\frac{T_2 \sin \theta}{T_2 \cos \theta} = \frac{W^{\prime}}{\mu W}$
$\tan \theta = \frac{W^{\prime}}{\mu W}$
$W^{\prime} = \mu W \tan \theta$

(B) The density of the metal is $\rho$. The density of water is $\rho_w = 1 \text{ g/cm}^3$ (or $1000 \text{ kg/m}^3$).
For the hollow sphere to float,its weight must be less than or equal to the weight of the water displaced by the total volume of the sphere.
The volume of the metal shell is $V_m = 4 \pi R^2 t$ (since $t \ll R$).
The mass of the sphere is $m_s = V_m \times \rho = 4 \pi R^2 t \rho$.
The volume of water displaced by the sphere is $V_w = \frac{4}{3} \pi R^3$.
The mass of the displaced water is $m_w = V_w \times \rho_w = \frac{4}{3} \pi R^3 \times 1$ (assuming $\rho$ is relative to water density).
For floatation,$m_s \leq m_w$:
$4 \pi R^2 t \rho \leq \frac{4}{3} \pi R^3$
Dividing both sides by $4 \pi R^2$:
$t \rho \leq \frac{R}{3}$
$t \leq \frac{R}{3 \rho}$

(C) Let the total height of the water level from the plane $PQ$ be $H$. The cylinder has height $h$ and is placed on a block of height $h/2$. Thus, $H = h + h/2 = 3h/2$.
Let $y$ be the height of a hole from the base of the cylinder. The depth of this hole from the free water surface is $d = H - y = 3h/2 - y$.
The horizontal range $R$ of the water jet is given by $R = 2\sqrt{d \cdot y_{ground}}$, where $y_{ground}$ is the height of the hole from the plane $PQ$. Here, $y_{ground} = y + h/2$.
So, $R = 2\sqrt{(3h/2 - y)(y + h/2)}$.
To maximize $R$, we maximize the product $(3h/2 - y)(y + h/2)$. Let $f(y) = (3h/2 - y)(y + h/2)$.
Taking the derivative with respect to $y$ and setting it to zero: $f'(y) = -(y + h/2) + (3h/2 - y) = 0$.
$2y = h$, which gives $y = h/2$.
Comparing the given hole heights: Hole $1$ is at $y=0$, Hole $2$ is at $y=h/4$, Hole $3$ is at $y=h/2$, and Hole $4$ is at $y=3h/4$.
Thus, Hole $3$ is at the height $y=h/2$ which provides the maximum range.

(D) The height of the liquid column in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$.
In a freely falling platform,the effective acceleration due to gravity $g_{eff}$ becomes $0$ because the system is in a state of weightlessness.
As $g_{eff} \to 0$,the height $h$ tends to infinity $(h \propto \frac{1}{g_{eff}})$.
However,the liquid cannot rise beyond the physical length of the capillary tube.
Therefore,the water will rise to fill the entire length of the capillary tube,which is $20 cm$.

(C) Let the unstretched length of the metal wire be $L$ and its cross-sectional area be $A$.
According to Hooke's Law,the Young's modulus $Y$ is given by $Y = \frac{T/A}{\Delta L/L}$,where $\Delta L$ is the change in length.
For tension $T_1$,the length is $L_1$,so the extension is $\Delta L_1 = L_1 - L$. Thus,$Y = \frac{T_1 L}{A(L_1 - L)}$.
For tension $T_2$,the length is $L_2$,so the extension is $\Delta L_2 = L_2 - L$. Thus,$Y = \frac{T_2 L}{A(L_2 - L)}$.
Equating the two expressions for $Y$:
$\frac{T_1 L}{A(L_1 - L)} = \frac{T_2 L}{A(L_2 - L)}$
$\frac{T_1}{L_1 - L} = \frac{T_2}{L_2 - L}$
$T_1(L_2 - L) = T_2(L_1 - L)$
$T_1 L_2 - T_1 L = T_2 L_1 - T_2 L$
$T_2 L - T_1 L = T_2 L_1 - T_1 L_2$
$L(T_2 - T_1) = T_2 L_1 - T_1 L_2$
$L = \frac{T_2 L_1 - T_1 L_2}{T_2 - T_1}$

(B) The velocity of particle $A$ is given by $\vec{v}_A = 10 \hat{i} \ m/s$.
The velocity of particle $B$ can be resolved into components. Its magnitude is $20 \ m/s$ at an angle of $60^{\circ}$ with the $X$-axis:
$\vec{v}_B = (20 \cos 60^{\circ}) \hat{i} + (20 \sin 60^{\circ}) \hat{j}$
$\vec{v}_B = (20 \times 0.5) \hat{i} + (20 \times \frac{\sqrt{3}}{2}) \hat{j} = 10 \hat{i} + 10 \sqrt{3} \hat{j} \ m/s$.
The relative velocity of $B$ with respect to $A$ is $\vec{v}_{BA} = \vec{v}_B - \vec{v}_A$.
$\vec{v}_{BA} = (10 \hat{i} + 10 \sqrt{3} \hat{j}) - (10 \hat{i}) = 10 \sqrt{3} \hat{j} \ m/s$.
This result indicates a velocity of magnitude $10 \sqrt{3} \ m/s$ directed along the positive $Y$-axis,which is perpendicular to the $X$-axis.

(B) For a bob of mass $m$ moving in a vertical circular path of radius $L$ with speed $v$,the forces acting on the bob along the radial direction are the tension $T$ (towards the center) and the component of weight $mg \cos \theta$ (away from the center).
The net centripetal force required for circular motion is provided by the resultant of these radial forces:
$T - mg \cos \theta = \frac{mv^2}{L}$
Rearranging the equation to solve for the tension $T$:
$T = \frac{mv^2}{L} + mg \cos \theta$

(A) The angular momentum $L$ of a particle about a point $O$ is given by $L = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$. The magnitude is $L = mvr \sin \theta$,where $r \sin \theta$ is the perpendicular distance from $O$ to the line of motion.
For particle $A$: $m_A = 6.5 \ kg$,$v_A = 2.2 \ m/s$,and the perpendicular distance $r_A = 1.5 \ m$. The motion is clockwise,so $L_A = -m_A v_A r_A = -(6.5 \times 2.2 \times 1.5) = -21.45 \ kg \ m^2/s$.
For particle $B$: $m_B = 3.1 \ kg$,$v_B = 3.6 \ m/s$,and the perpendicular distance $r_B = 2.8 \ m$. The motion is counter-clockwise,so $L_B = +m_B v_B r_B = +(3.1 \times 3.6 \times 2.8) = +31.248 \ kg \ m^2/s$.
The total angular momentum is $L = L_A + L_B = -21.45 + 31.248 = 9.798 \ kg \ m^2/s \approx 9.8 \ kg \ m^2/s$.

(A, C, D) For a circular disc rolling without slipping on a horizontal floor, the velocity of any point on the rim is the vector sum of the translational velocity of the centre $(v)$ and the tangential velocity due to rotation $(\omega R)$.
Since the disc rolls without slipping, $v = \omega R$.
$1$. At the point of contact $(P)$ with the floor, the velocity is $v_P = v - \omega R = v - v = 0$.
$2$. At the top point $(S)$, the velocity is $v_S = v + \omega R = v + v = 2v$.
$3$. At any other point on the rim, the magnitude of the velocity lies between $0$ and $2v$. Specifically, at the level of the centre, the velocity is $\sqrt{v^2 + v^2} = \sqrt{2}v$.
Thus, the possible values for the velocity of a point on the rim include $0$, $v$, $\sqrt{2}v$, and $2v$. Among the given options, $v$, $2v$, and $0$ are valid.

(B) Let the final equilibrium temperature be $\theta$. According to the principle of calorimetry,the heat lost by the hotter bodies must equal the heat gained by the colder bodies.
Heat lost by the body at $60^{\circ} C$ (mass $3m$) = $3m \cdot s \cdot (60 - \theta)$
Heat gained by the body at $50^{\circ} C$ (mass $m$) = $m \cdot s \cdot (\theta - 50)$
Heat gained by the body at $40^{\circ} C$ (mass $m$) = $m \cdot s \cdot (\theta - 40)$
Equating heat lost and heat gained:
$3ms(60 - \theta) = ms(\theta - 50) + ms(\theta - 40)$
Dividing by $ms$:
$3(60 - \theta) = (\theta - 50) + (\theta - 40)$
$180 - 3\theta = 2\theta - 90$
$5\theta = 270$
$\theta = 54^{\circ} C$

(B) According to the Stefan-Boltzmann Law,the rate of heat energy radiated by a body at absolute temperature $T$ is given by $P = \sigma A e T^{4}$.
When a body at absolute temperature $T_{1}$ is placed in an enclosure at absolute temperature $T_{2}$,the net rate of heat exchange is given by $P_{net} = \sigma A e (T_{2}^{4} - T_{1}^{4})$.
Here,the absolute temperatures are $T_{1} = (t_{1} + 273) \ K$ and $T_{2} = (t_{2} + 273) \ K$.
Since the body is absorbing heat from the chamber,the rate of heat absorption is proportional to the difference of the fourth powers of their absolute temperatures.
Therefore,the rate of heat absorbed is proportional to $(t_{2} + 273)^{4} - (t_{1} + 273)^{4}$.

(A) Given the equation of state: $PV = AT - BT^2$.
Since the pressure $P$ is constant,we differentiate the equation with respect to $T$:
$P dV = A dT - B(2T) dT$.
Work done $W = \int P dV$.
Substituting $P dV$ into the integral:
$W = \int_{T_1}^{T_2} (A - 2BT) dT$.
$W = A \int_{T_1}^{T_2} dT - 2B \int_{T_1}^{T_2} T dT$.
$W = A(T_2 - T_1) - 2B \left[ \frac{T^2}{2} \right]_{T_1}^{T_2}$.
$W = A(T_2 - T_1) - B(T_2^2 - T_1^2)$.

(A) The apparent frequency $f'$ heard by an observer is given by the Doppler effect formula:
$f' = f_0 \left( \frac{v + v_o}{v - v_s} \right)$
Where:
$f_0 = 1000 \ Hz$ (source frequency)
$v = 333 \ m/s$ (velocity of sound)
$v_o = 33 \ m/s$ (velocity of observer,positive as it approaches the source)
$v_s = 33 \ m/s$ (velocity of source,positive as it moves towards the observer)
Substituting the values:
$f' = 1000 \left( \frac{333 + 33}{333 - 33} \right)$
$f' = 1000 \left( \frac{366}{300} \right)$
$f' = 1000 \times 1.22 = 1220 \ Hz$

(C) Let the length of the closed organ pipe be $l$. The fundamental frequency of an open organ pipe of length $L = 2l$ is given by $f_{open} = \frac{v}{2L} = \frac{v}{2(2l)} = \frac{v}{4l} = 100 \ Hz$.
The frequencies of a closed organ pipe of length $l$ are given by $f_n = (2n - 1) \frac{v}{4l}$,where $n = 1, 2, 3, ...$.
The first harmonic (fundamental) is $f_1 = \frac{v}{4l} = 100 \ Hz$.
The third harmonic corresponds to $n = 2$ (since the harmonics of a closed pipe are odd multiples of the fundamental: $f_1, 3f_1, 5f_1, ...$).
Therefore,the frequency of the third harmonic is $f_3 = 3 \times f_1 = 3 \times 100 \ Hz = 300 \ Hz$.

(D) When the switch is in position $a$,the capacitor is charged to a potential $E$. The charge on the capacitor is $q = CE$.
The energy stored in the capacitor is $U = \frac{q^2}{2C} = \frac{(CE)^2}{2C} = \frac{1}{2} CE^2$.
When the switch is thrown to position $b$,the circuit becomes an $LC$ oscillator. The total energy is conserved,oscillating between the electric field of the capacitor and the magnetic field of the inductor.
Let $I_0$ be the amplitude of the oscillating current. The maximum magnetic energy in the inductor is $\frac{1}{2} L I_0^2$.
By the law of conservation of energy,the maximum electrical energy equals the maximum magnetic energy:
$\frac{1}{2} CE^2 = \frac{1}{2} L I_0^2$
$CE^2 = L I_0^2$
$I_0^2 = \frac{C}{L} E^2$
$I_0 = E \sqrt{\frac{C}{L}}$

(C) The capacitors are connected in series. The equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{5 \mu F} + \frac{1}{10 \mu F} = \frac{2+1}{10 \mu F} = \frac{3}{10 \mu F}$
$C_{eq} = \frac{10}{3} \mu F = \frac{10}{3} \times 10^{-6} \ F$
The total energy $U$ stored in the series combination is given by:
$U = \frac{1}{2} C_{eq} V^2$
$U = \frac{1}{2} \times (\frac{10}{3} \times 10^{-6} \ F) \times (300 \ V)^2$
$U = \frac{1}{2} \times \frac{10}{3} \times 10^{-6} \times 90000$
$U = \frac{1}{2} \times \frac{10}{3} \times 10^{-6} \times 9 \times 10^4$
$U = \frac{1}{2} \times 30 \times 10^{-2} = 15 \times 10^{-2} \ J = 0.15 \ J$

(C) The current $i$ flowing through the circuit is given by the net emf divided by the total resistance of the circuit.
Since the cells are connected in opposition,the net emf is $\varepsilon_{net} = 2 \text{ V} - 1.5 \text{ V} = 0.5 \text{ V}$.
The total resistance of the circuit is $R_{total} = 5 \text{ } \Omega + 5 \text{ } \Omega + 10 \text{ } \Omega = 20 \text{ } \Omega$.
Therefore,the current $i = \frac{0.5 \text{ V}}{20 \text{ } \Omega} = 0.025 \text{ A}$.
For cell $A$,which is discharging,the terminal potential difference is $V_{A} = \varepsilon_{A} - i r_{A} = 2 \text{ V} - (0.025 \text{ A} \times 5 \text{ } \Omega) = 2 - 0.125 = 1.875 \text{ V}$.
For cell $B$,which is charging,the terminal potential difference is $V_{B} = \varepsilon_{B} + i r_{B} = 1.5 \text{ V} + (0.025 \text{ A} \times 5 \text{ } \Omega) = 1.5 + 0.125 = 1.625 \text{ V}$.

(B) Let the currents in the left and right branches be $I_1$ and $I_2$ respectively. The middle branch carries current $I = I_1 + I_2$.
Applying Kirchhoff's Voltage Law $(KVL)$ to the left loop:
$2 - 2I_1 - 2I = 0 \implies 2 - 2I_1 - 2(I_1 + I_2) = 0 \implies 2 - 4I_1 - 2I_2 = 0 \implies 2I_1 + I_2 = 1$ ---$(i)$
Applying $KVL$ to the right loop:
$-2 - 2I_2 - 2I = 0 \implies -2 - 2I_2 - 2(I_1 + I_2) = 0 \implies -2 - 2I_1 - 4I_2 = 0 \implies I_1 + 2I_2 = -1$ ---(ii)
Solving equations $(i)$ and (ii):
From $(i)$,$I_2 = 1 - 2I_1$. Substituting into (ii):
$I_1 + 2(1 - 2I_1) = -1$
$I_1 + 2 - 4I_1 = -1$
$-3I_1 = -3 \implies I_1 = 1 A$
Then $I_2 = 1 - 2(1) = -1 A$.
The current $I$ in the middle branch is $I = I_1 + I_2 = 1 + (-1) = 0 A$.

(C) The resistance of a wire is given by $R = \rho \frac{l}{A}$.
Since the volume $V = A \cdot l$ remains constant,we can write $A = \frac{V}{l}$.
Substituting this into the resistance formula,we get $R = \rho \frac{l}{V/l} = \rho \frac{l^2}{V}$.
Since the resistivity $\rho$ and volume $V$ remain constant,we have $R \propto l^2$.
Given that the length is doubled,$l_2 = 2l_1$.
Therefore,$\frac{R_2}{R_1} = \left( \frac{l_2}{l_1} \right)^2 = \left( \frac{2l_1}{l_1} \right)^2 = 2^2 = 4$.
Thus,$R_2 : R_1 = 4 : 1$.

(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.
Case $A$: For free fall,$v = \sqrt{2gh}$. Since $v$ is independent of mass,$\lambda = \frac{h}{m\sqrt{2gh}} \propto \frac{1}{m}$. Thus,the heavier particle has a smaller wavelength.
Case $B$: With same kinetic energy $K$,$\lambda = \frac{h}{\sqrt{2mK}}$. Since $\lambda \propto \frac{1}{\sqrt{m}}$,the heavier particle has a smaller wavelength.
Case $C$: With same momentum $p$,$\lambda = \frac{h}{p}$. Since $p$ is the same,$\lambda$ is the same for both.
Case $D$: With same speed $v$,$\lambda = \frac{h}{mv}$. Since $\lambda \propto \frac{1}{m}$,the heavier particle has a smaller wavelength.
Note: Options $A$,$B$,and $D$ all result in the heavier particle having a smaller de-Broglie wavelength. However,in standard physics problems of this type,the most direct relationship is often associated with constant speed or constant kinetic energy. Given the options,$D$ is the most fundamental case where $\lambda \propto 1/m$ is directly observed.

(D) The energy of the photon is given by $E = \frac{hc}{\lambda}$.
Substituting the given values:
$E = \frac{4 \times 10^{-15} \ eV \cdot s \times 3 \times 10^{8} \ m/s}{300 \times 10^{-9} \ m} = \frac{12 \times 10^{-7}}{300 \times 10^{-9}} \ eV = \frac{1200}{300} \ eV = 4 \ eV$.
The energy required to excite a hydrogen atom from the ground state $(n=1)$ to the first excited state $(n=2)$ is $\Delta E = 13.6 \ eV \times (1 - \frac{1}{4}) = 13.6 \times 0.75 = 10.2 \ eV$.
Since the energy of the photon $(4 \ eV)$ is less than the energy required for the first excitation $(10.2 \ eV)$ and also less than the ionization energy $(13.6 \ eV)$,the electron cannot absorb this energy to transition to a higher state.
Therefore,the electron will remain in the ground state.

(B) The photoelectric effect occurs when the energy of incident photons $E = \frac{hc}{\lambda}$ is greater than or equal to the work function $\phi$ of the metal.
Given the range of work function $2 eV \leq \phi \leq 5 eV$,the threshold wavelength $\lambda$ must satisfy $\lambda \leq \frac{hc}{\phi}$.
For $\phi = 2 eV$,$\lambda_{\max} = \frac{4 \times 10^{-15} eVs \times 3 \times 10^{8} m/s}{2 eV} = 6 \times 10^{-7} m = 600 nm$.
For $\phi = 5 eV$,$\lambda_{\min} = \frac{4 \times 10^{-15} eVs \times 3 \times 10^{8} m/s}{5 eV} = 2.4 \times 10^{-7} m = 240 nm$.
Thus,the range of wavelengths that can cause the photoelectric effect is $240 nm \leq \lambda \leq 600 nm$.
Comparing the given options,$650 nm$ is greater than $600 nm$,so it cannot cause the photoelectric effect.

(B, D) According to Faraday's law of electromagnetic induction, an emf is induced in a loop whenever there is a change in the magnetic flux $(\Phi_B = \vec{B} \cdot \vec{A} = BA \cos \theta)$ linked with it.
$1$. If the loop is translated parallel to itself in a uniform magnetic field, the flux remains constant, so no emf is induced.
$2$. If the loop is rotated about one of its diameters, the angle $\theta$ between the area vector and the magnetic field changes, causing a change in flux, thus inducing an emf.
$3$. If the loop is rotated about its own axis (which is parallel to the field), the angle $\theta$ remains constant, so no emf is induced.
$4$. If the loop is deformed, the area $A$ changes, which also changes the magnetic flux, thus inducing an emf.
Therefore, options $B$ and $D$ both lead to an induced emf.

(A) The electric field $E$ due to a charged infinite conducting sheet is given by $E = \frac{\sigma}{\varepsilon_{0}}$.
However,for a non-conducting sheet,$E = \frac{\sigma}{2\varepsilon_{0}}$. Given the context of the options and the provided solution image,the force acting on the ball is $F = qE = \frac{q\sigma}{2\varepsilon_{0}}$.
Considering the forces acting on the ball $B$: the electric force $F$ acts horizontally,and the gravitational force $mg$ acts vertically downwards.
In equilibrium,the tension in the string balances these forces.
Therefore,$\tan \theta = \frac{F}{mg} = \frac{q\sigma / 2\varepsilon_{0}}{mg} = \frac{q\sigma}{2\varepsilon_{0}mg}$.

(B, C) In a uniform electric field $\vec{E}$,the force on charge $+q$ is $\vec{F}_+ = q\vec{E}$ and the force on charge $-q$ is $\vec{F}_- = -q\vec{E}$.
The net force on the dipole is $\vec{F}_{net} = \vec{F}_+ + \vec{F}_- = q\vec{E} - q\vec{E} = 0$. Thus,the net force on the dipole always vanishes.
The torque on the dipole is given by $\vec{\tau} = \vec{p} \times \vec{E}$. Since the torque is defined as the cross product of the dipole moment and the electric field,it is a physical quantity and is independent of the choice of the coordinate system.
Therefore,statements $(b)$ and $(c)$ are correct.

(D) According to Gauss's Law,the total electric flux through a closed surface is $\phi_{total} = q / \varepsilon_{0}$.
When a charge $q$ is placed at one corner of a cube,the cube can be considered as part of a larger Gaussian surface consisting of $8$ identical cubes to enclose the charge completely at the center.
Thus,the total flux through the entire cube is $\phi_{cube} = q / (8 \varepsilon_{0})$.
The charge $q$ lies on three faces of the cube. The electric field lines are parallel to these three faces,so the electric flux through these three faces is zero.
The remaining flux $\phi_{cube}$ is distributed equally among the other three faces of the cube.
Therefore,the flux through any one of the other three faces is $\phi = \frac{1}{3} \times \phi_{cube} = \frac{1}{3} \times \frac{q}{8 \varepsilon_{0}} = \frac{q}{24 \varepsilon_{0}}$.

(B) Given:
Length of the conductor,$l = 0.1 \ m$
Magnetic field,$B = 0.1 \ T$
Velocity of the conductor,$v = 15 \ m/s$
The angle between the velocity vector $v$ and the magnetic field $B$ is $\theta = 90^{\circ}$.
When a conductor moves perpendicular to a uniform magnetic field,the motional emf induced is given by the formula:
$\varepsilon = B \cdot l \cdot v \cdot \sin(\theta)$
Substituting the given values:
$\varepsilon = 0.1 \times 0.1 \times 15 \times \sin(90^{\circ})$
Since $\sin(90^{\circ}) = 1$:
$\varepsilon = 0.01 \times 15 = 0.15 \ V$
Therefore,the induced emf is $0.15 \ V$.

(B) The radius $R$ of the circular path followed by a charged particle in a magnetic field is given by $R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$,where $K$ is the kinetic energy.
Since the particle is accelerated through a potential difference $V$,its kinetic energy is $K = qV$.
Substituting this into the radius formula,we get $R = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.
Given that $q$,$V$,and $B$ are the same for both particles,we have $R \propto \sqrt{m}$.
Therefore,$\frac{R_{1}}{R_{2}} = \sqrt{\frac{m_{A}}{m_{B}}}$.
Squaring both sides,we get $\frac{m_{A}}{m_{B}} = \left(\frac{R_{1}}{R_{2}}\right)^{2}$.

(B) For the first lens:
Using the lens formula $\frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1}$,where $f_1 = 2 \ m$ and $u_1 = -4 \ m$:
$\frac{1}{2} = \frac{1}{v_1} - \frac{1}{-4} \Rightarrow \frac{1}{v_1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \Rightarrow v_1 = 4 \ m$.
This image is formed $4 \ m$ to the right of the first lens.
For the second lens:
The distance between the lenses is $3 \ m$. The image formed by the first lens acts as an object for the second lens.
Since the image is $4 \ m$ to the right of the first lens and the second lens is $3 \ m$ to the right of the first,the image is $1 \ m$ to the right of the second lens.
Thus,the object distance for the second lens is $u_2 = +1 \ m$ (virtual object).
Using the lens formula $\frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2}$,where $f_2 = 1 \ m$ and $u_2 = +1 \ m$:
$\frac{1}{1} = \frac{1}{v_2} - \frac{1}{1} \Rightarrow \frac{1}{v_2} = 1 + 1 = 2 \Rightarrow v_2 = 0.5 \ m$.
The final image is $0.5 \ m$ to the right of the second lens.
The total distance from the source is $4 \ m$ (to first lens) $+ 3 \ m$ (between lenses) $+ 0.5 \ m$ (from second lens) $= 7.5 \ m$.

(B) According to the law of reflection,the angle of incidence $i$ is equal to the angle of reflection $\theta$. So,$i = \theta$.
The sum of the angles on a straight line is $180^{\circ}$. Therefore,the angle of reflection $\theta$,the angle between the reflected and refracted rays $(90^{\circ})$,and the angle of refraction $r$ must satisfy:
$\theta + 90^{\circ} + r = 180^{\circ}$
Substituting $\theta = i$:
$i + 90^{\circ} + r = 180^{\circ}$
$r = 90^{\circ} - i$
According to Snell's law,$\mu = \frac{\sin i}{\sin r}$.
Substituting $r = 90^{\circ} - i$:
$\mu = \frac{\sin i}{\sin(90^{\circ} - i)}$
Since $\sin(90^{\circ} - i) = \cos i$,we get:
$\mu = \frac{\sin i}{\cos i} = \tan i$
Thus,$\tan i = \mu$.

(C) When light travels from one medium to another,its speed and wavelength change because the optical density of the medium changes. However,the frequency of light is determined by the source of the light and remains constant during refraction. Therefore,the frequency does not change.

(D) In the given circuit,the top diode is forward-biased,while the middle diode is reverse-biased.
The reverse-biased diode acts as an open circuit (infinite resistance),so no current flows through the middle branch.
The circuit simplifies to a series combination of the $10 \text{ V}$ battery,the $150 \Omega$ resistor,the top diode (with $50 \Omega$ forward resistance),and the $50 \Omega$ resistor in that branch.
The total resistance of the circuit is $R_{total} = R_{diode} + R_{top} + R_{series} = 50 \Omega + 50 \Omega + 150 \Omega = 250 \Omega$.
The current $I$ flowing through the $150 \Omega$ resistor is given by Ohm's law: $I = \frac{V}{R_{total}} = \frac{10 \text{ V}}{250 \Omega} = 0.04 \text{ A}$.

(C) The given digital circuit consists of a $NAND$ gate and a $NOT$ gate,whose outputs are fed into an $OR$ gate.
$1$. The output of the $NAND$ gate with inputs $A$ and $B$ is $\overline{AB}$.
$2$. The output of the $NOT$ gate with input $C$ is $\bar{C}$.
$3$. These two outputs are fed into an $OR$ gate,so the final output $Y$ is $Y = \overline{AB} + \bar{C}$.
$4$. According to De Morgan's theorem,$\overline{AB} = \bar{A} + \bar{B}$.
$5$. Substituting this into the expression for $Y$,we get $Y = \bar{A} + \bar{B} + \bar{C}$.

(C) In Fraunhofer diffraction,the source of light and the screen are effectively at an infinite distance from the aperture (slit).
This implies that the wavefronts incident on the slit are plane wavefronts,which means the incident beam must be parallel.
This condition is practically achieved by placing the light source at the focus of a converging lens,or by placing the source at an infinite distance from the slit.
Therefore,the correct condition is that the source is at infinity and the incident beam is parallel.

(A) When a thin sheet of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits, the path difference introduced is $\Delta x = (\mu - 1)t$.
Given that the central point on the screen is now occupied by the $7^{th}$ bright fringe, the shift in the fringe pattern is equal to $7$ fringe widths.
The condition for the shift is $(\mu - 1)t = n\lambda$, where $n = 7$ and $\lambda = 600 \ nm = 0.6 \ \mu m$.
Substituting the values: $(1.6 - 1)t = 7 \times 0.6 \ \mu m$.
$0.6 \times t = 4.2 \ \mu m$.
$t = \frac{4.2}{0.6} \ \mu m = 7 \ \mu m$.