KCET 2013 Physics Question Paper with Answer and Solution

(C) In central force motion,the angular momentum of the body remains conserved.
Angular momentum is given by $L = \vec{r} \times \vec{p} = mvr \sin \theta$.
At points $A$ and $B$,the velocity vector is perpendicular to the position vector,so $\theta = 90^\circ$ and $\sin 90^\circ = 1$.
Thus,$L = mvr$.
Let $v_A$ and $v_B$ be the speeds of the Earth at $A$ and $B$,respectively,and $r_A = OA$ and $r_B = OB$ be the distances from the Sun.
By conservation of angular momentum:
$m v_A r_A = m v_B r_B$
$\Rightarrow \frac{v_A}{v_B} = \frac{r_B}{r_A} = \frac{OB}{OA}$
Given that $\frac{OB}{OA} = R$,we have $\frac{v_A}{v_B} = R$.
Therefore,the ratio of Earth's velocities at $A$ and $B$ is $R$.

(B) According to the law of floatation,for a body floating in equilibrium,the weight of the body is equal to the buoyant force (weight of the displaced liquid).
Let $V$ be the total volume of each solid,$\rho_P$ and $\rho_Q$ be their densities,and $\rho_w$ be the density of water.
For solid $P$: $V \rho_P g = (V/2) \rho_w g \Rightarrow \rho_P = \frac{1}{2} \rho_w$.
For solid $Q$: $V \rho_Q g = (2V/3) \rho_w g \Rightarrow \rho_Q = \frac{2}{3} \rho_w$.
The ratio of densities is $\frac{\rho_P}{\rho_Q} = \frac{\frac{1}{2} \rho_w}{\frac{2}{3} \rho_w} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$.

(D) When the lift is moving upwards with an acceleration $a$, the effective acceleration of the ball relative to the lift is $a_{eff} = g + a$.
Given: $g = 10 \,m/s^2$, $a = 5 \,m/s^2$, $s = 1.25 \,m$, and initial velocity $u = 0$.
Therefore, $a_{eff} = 10 + 5 = 15 \,m/s^2$.
Using the equation of motion $s = ut + \frac{1}{2} a_{eff} t^2$:
$1.25 = 0 \times t + \frac{1}{2} \times 15 \times t^2$
$1.25 = 7.5 \times t^2$
$t^2 = \frac{1.25}{7.5} = \frac{1}{6} \approx 0.166 \,s^2$
$t = \sqrt{0.166} \approx 0.4 \,s$.

(A) In projectile motion,the horizontal component of velocity remains constant throughout the flight because there is no acceleration in the horizontal direction.
Initial horizontal component of velocity: $u_x = u \cos 60^{\circ} = 10 \times \frac{1}{2} = 5 \ m/s$.
Let the speed at the later instant be $v$. The horizontal component of velocity at this instant is $v_x = v \cos 30^{\circ}$.
Since $u_x = v_x$,we have:
$5 = v \cos 30^{\circ}$
$5 = v \times \frac{\sqrt{3}}{2}$
$v = \frac{10}{\sqrt{3}} \ m/s$.

(B) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$.
Given $l = 1 \,m$ and taking $g = 10 \,m/s^2$ (or $\pi^2 \approx 10$),we have $T = 2 \pi \sqrt{\frac{1}{10}} \approx 2 \,s$.
For a simple pendulum,the time taken to travel from the extreme position to the mean position $(B)$ is $\frac{T}{4}$.
Since both spheres are released from their respective extreme positions,they will both reach the mean position $B$ at time $t = \frac{T}{4} = \frac{2}{4} = 0.5 \,s$.
Therefore,the two spheres will collide at the mean position $B$ after $0.5 \,s$.

(D) The system consists of two parallel paths for heat flow: the inner solid cylinder and the outer cylindrical shell.
Since the system is in a steady state and there is no heat loss across the cylindrical surface,the total heat flow $Q_{\text{total}}$ is the sum of heat flows through the two parts: $Q_{\text{total}} = Q_{1} + Q_{2}$.
The rate of heat flow is given by $Q = \frac{KA \Delta \theta}{L}$.
For the inner cylinder,area $A_{1} = \pi R^{2}$.
For the outer shell,area $A_{2} = \pi (2R)^{2} - \pi R^{2} = 3\pi R^{2}$.
The total area is $A = A_{1} + A_{2} = 4\pi R^{2}$.
Equating the heat flow: $\frac{K(4\pi R^{2}) \Delta \theta}{L} = \frac{K_{1}(\pi R^{2}) \Delta \theta}{L} + \frac{K_{2}(3\pi R^{2}) \Delta \theta}{L}$.
Canceling common terms $\frac{\pi R^{2} \Delta \theta}{L}$ from both sides:
$4K = K_{1} + 3K_{2}$.
Therefore,the effective thermal conductivity is $K = \frac{K_{1} + 3K_{2}}{4}$.

(C) According to Wien's displacement law,the product of the wavelength of maximum emission and the absolute temperature is constant: $\lambda T = b$.
Therefore,$\lambda_A T_A = \lambda_B T_B$.
This implies the ratio of temperatures is inversely proportional to the ratio of wavelengths: $\frac{T_A}{T_B} = \frac{\lambda_B}{\lambda_A}$.
Given $\lambda_A = 360 \ nm$ and $\lambda_B = 480 \ nm$.
Substituting the values: $\frac{T_A}{T_B} = \frac{480}{360} = \frac{4}{3}$.
Thus,the ratio of the surface temperatures of $A$ and $B$ is $4: 3$.

(D) The efficiency of a Carnot heat engine is given by the formula $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink.
For option $A$: $\eta_A = 1 - \frac{400}{600} = 1 - \frac{2}{3} = \frac{1}{3} \approx 0.333$.
For option $B$: $\eta_B = 1 - \frac{200}{400} = 1 - \frac{1}{2} = \frac{1}{2} = 0.500$.
For option $C$: $\eta_C = 1 - \frac{300}{500} = 1 - \frac{3}{5} = \frac{2}{5} = 0.400$.
For option $D$: $\eta_D = 1 - \frac{100}{300} = 1 - \frac{1}{3} = \frac{2}{3} \approx 0.667$.
Comparing the values,$\eta_D$ is the maximum efficiency.
Therefore,the correct combination is $T_1 = 300 \ K$ and $T_2 = 100 \ K$.

(B) $1$. Thermal conductivity $(K)$: From $Q = \frac{KA(T_2 - T_1)t}{d}$,we get $[K] = [M^{1} L^{1} T^{-3} K^{-1}]$. This is correct.
$2$. Potential $(V)$: $V = \frac{W}{q}$. $[V] = \frac{[M^{1} L^{2} T^{-2}]}{[A^{1} T^{1}]} = [M^{1} L^{2} T^{-3} A^{-1}]$. The given option $B$ states $[M^{1} L^{2} T^{3} A^{-1}]$,which is incorrect.
$3$. Permeability of free space $(\mu_{0})$: From $F = \frac{\mu_{0} I_1 I_2 L}{2\pi d}$,we get $[\mu_{0}] = [M^{1} L^{1} T^{-2} A^{-2}]$. This is correct.
Therefore,option $B$ is the incorrect statement.

(B) Given equation: $y=0.05 \sin \pi(2 t-0.02 x)$
Expanding the equation: $y=0.05 \sin (2 \pi t - 0.02 \pi x)$
Comparing this with the standard wave equation $y=a \sin (\omega t - k x)$:
Angular frequency $\omega = 2 \pi \ rad/s$
Wave number $k = 0.02 \pi \ m^{-1}$
The minimum distance between two particles in phase is the wavelength $\lambda$.
Since $k = \frac{2 \pi}{\lambda}$,we have $\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{0.02 \pi} = \frac{2}{0.02} = 100 \ m$.
The wave velocity $v$ is given by $v = \frac{\omega}{k}$.
$v = \frac{2 \pi}{0.02 \pi} = \frac{2}{0.02} = 100 \ m/s$.

(B) The person is moving towards a wall with a tuning fork. The frequency of the sound heard directly from the fork is $f = 338 \,Hz$.
The frequency of the sound reflected from the wall, $f'$, is perceived by the moving observer. Since the observer is moving towards the wall, the reflected sound acts as a source at rest and the observer is moving towards it.
The formula for the frequency of the reflected sound heard by the observer is $f' = f \left( \frac{v + v_0}{v - v_s} \right)$.
Here, the source (the wall reflecting the sound) is at rest, so $v_s = 0$. The observer is moving towards the wall with $v_0 = 2 \,ms^{-1}$.
Thus, $f' = f \left( \frac{v + v_0}{v} \right) = 338 \left( \frac{340 + 2}{340} \right) = 338 \left( \frac{342}{340} \right) = 338 \times 1.00588 \approx 340 \,Hz$.
The beat frequency is the difference between the reflected frequency and the original frequency: $\Delta f = f' - f = 340 - 338 = 2 \,Hz$.
Wait, let's re-evaluate: The observer is moving towards the wall. The sound reaches the wall and reflects back. The frequency of the reflected sound as heard by the observer is $f' = f \left( \frac{v + v_0}{v - v_0} \right)$ is incorrect because the source is the wall. The correct approach is: The wall receives sound of frequency $f_{wall} = f \left( \frac{v}{v - v_0} \right)$. The wall reflects this frequency, and the observer moving towards the wall receives $f' = f_{wall} \left( \frac{v + v_0}{v} \right) = f \left( \frac{v + v_0}{v - v_0} \right)$.
Calculating: $f' = 338 \left( \frac{340 + 2}{340 - 2} \right) = 338 \left( \frac{342}{338} \right) = 342 \,Hz$.
Beat frequency = $f' - f = 342 - 338 = 4 \,Hz$.

(A) For an open pipe of length $L_o$,the frequency of the $n^{th}$ harmonic is $f_n = \frac{n v}{2 L_o}$. The second overtone corresponds to the $3^{rd}$ harmonic $(n=3)$,so $f_{o} = \frac{3 v}{2 L_o}$.
For a closed pipe of length $L_c$,the frequency of the $n^{th}$ harmonic is $f_n = \frac{(2n-1) v}{4 L_c}$. The first overtone corresponds to the $3^{rd}$ harmonic $(n=2)$,so $f_{c} = \frac{3 v}{4 L_c}$.
Given that $f_o = f_c$,we have $\frac{3 v}{2 L_o} = \frac{3 v}{4 L_c}$.
Simplifying this,we get $\frac{1}{2 L_o} = \frac{1}{4 L_c}$,which implies $\frac{L_o}{L_c} = \frac{4}{2} = \frac{2}{1}$.

(C) According to the law of conservation of linear momentum,the magnitude of momentum of the gun $(p_g)$ and the bullet $(p_b)$ must be equal after firing,i.e.,$p_g = p_b = p$.
The kinetic energy $K$ of an object of mass $m$ with momentum $p$ is given by the relation $K = \frac{p^2}{2m}$.
Since the momentum $p$ is the same for both the gun and the bullet,we have $K \propto \frac{1}{m}$.
Because the mass of the gun $(M)$ is much greater than the mass of the bullet $(m)$,the kinetic energy of the gun $(K_g = \frac{p^2}{2M})$ will be significantly less than the kinetic energy of the bullet $(K_b = \frac{p^2}{2m})$.
Therefore,the kinetic energy of the gun is less than $K$.

(A) According to the work-energy theorem,the work done is equal to the change in kinetic energy.
Work done to accelerate from $0$ to $v$ is $W_1 = \frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2$.
Work done to accelerate from $v$ to $2v$ is $W_2 = \Delta K = \frac{1}{2}m(2v)^2 - \frac{1}{2}mv^2$.
$W_2 = \frac{1}{2}m(4v^2) - \frac{1}{2}mv^2 = 2mv^2 - 0.5mv^2 = 1.5mv^2$.
Comparing $W_2$ with $W_1$:
$W_2 = 3 \times (\frac{1}{2}mv^2) = 3W_1$.
Therefore,the work done is three times the work done in accelerating it from rest to $v$.

(A) The initial impedance is $Z = \sqrt{R^2 + X_c^2}$,where $X_c = \frac{1}{\omega C}$.
The initial current is $I = \frac{V}{Z}$.
When the frequency changes to $\omega' = \frac{\omega}{3}$,the new capacitive reactance is $X_c' = \frac{1}{\omega' C} = \frac{1}{(\omega/3) C} = 3X_c$.
The new impedance is $Z' = \sqrt{R^2 + (X_c')^2} = \sqrt{R^2 + (3X_c)^2}$.
Given that the new current $I' = \frac{I}{2}$,we have $\frac{V}{Z'} = \frac{1}{2} \frac{V}{Z}$,which implies $Z' = 2Z$.
Squaring both sides,$(Z')^2 = 4Z^2$,so $R^2 + 9X_c^2 = 4(R^2 + X_c^2)$.
$R^2 + 9X_c^2 = 4R^2 + 4X_c^2$.
$5X_c^2 = 3R^2$.
$\frac{X_c^2}{R^2} = \frac{3}{5} = 0.6$.
Therefore,the ratio $\frac{X_c}{R} = \sqrt{0.6}$.

(A) Initially, the potential difference across each element is $V_R = V_L = V_C = 20 \, V$. Since $V_L = V_C$, the circuit is at resonance, meaning the impedance $Z = R$. The source voltage is $V = V_R = 20 \, V$.
When the resistance $R$ is doubled to $2R$, the new impedance becomes $Z' = \sqrt{(2R)^2 + (X_L - X_C)^2}$. Since $X_L = X_C$ at resonance, $Z' = 2R$.
The new current in the circuit is $I' = V / Z' = V / (2R) = I / 2$, where $I$ is the original current.
The new potential difference across the resistor is $V_R' = I' \times (2R) = (I/2) \times (2R) = IR = 20 \, V$.
The new potential difference across the inductor is $V_L' = I' X_L = (I/2) X_L = V_L / 2 = 20 / 2 = 10 \, V$.
The new potential difference across the capacitor is $V_C' = I' X_C = (I/2) X_C = V_C / 2 = 20 / 2 = 10 \, V$.
Thus, the new potential differences are $20 \, V, 10 \, V, 10 \, V$.

(B) Given: Turns ratio $\frac{n_{s}}{n_{p}} = 3$,Efficiency $\eta = 0.75$,Primary current $I_{p} = 2 \,A$,Primary voltage $V_{p} = 100 \,V$.
For a transformer,the voltage ratio is equal to the turns ratio: $\frac{V_{s}}{V_{p}} = \frac{n_{s}}{n_{p}} = 3$.
Therefore,$V_{s} = 3 \times V_{p} = 3 \times 100 \,V = 300 \,V$.
Efficiency $\eta$ is defined as the ratio of output power to input power: $\eta = \frac{V_{s} I_{s}}{V_{p} I_{p}}$.
Substituting the values: $0.75 = \frac{300 \times I_{s}}{100 \times 2}$.
$0.75 = \frac{300 \times I_{s}}{200} = 1.5 \times I_{s}$.
$I_{s} = \frac{0.75}{1.5} = 0.5 \,A$.
Thus,the secondary voltage is $300 \,V$ and the secondary current is $0.5 \,A$.

(D) The line emission spectrum is produced by an excited substance in the atomic state,such as a mercury vapour lamp.
Line absorption spectrum is produced when electromagnetic radiations pass through a medium; therefore,the sunlight spectrum is a line absorption spectrum.
The band spectrum is used to study the molecular structure.
Since all the statements $A$,$B$,and $C$ are correct,the correct option is $D$.

(B) Quarks are elementary particles that combine to form composite particles known as hadrons.
Protons and neutrons are baryons,which are made of three quarks.
$\pi$-mesons (pions) are mesons,which are made of a quark and an antiquark.
$A$ positron is an antiparticle of the electron,which is a lepton.
Leptons are fundamental particles and are not composed of quarks.
Therefore,the positron is not made of quarks.

(A) When the electron de-excites from $3E$ to $E$,the energy difference is $\Delta E_1 = 3E - E = 2E$.
The energy of the emitted photon is given by $\frac{hc}{\lambda} = 2E$ ... $(i)$
When the electron de-excites from $\frac{5E}{3}$ to $E$,the energy difference is $\Delta E_2 = \frac{5E}{3} - E = \frac{2E}{3}$.
The energy of the emitted photon is $\frac{hc}{\lambda'} = \frac{2E}{3}$ ... (ii)
Dividing equation $(i)$ by equation (ii):
$\frac{hc/\lambda}{hc/\lambda'} = \frac{2E}{2E/3}$
$\frac{\lambda'}{\lambda} = \frac{2E \times 3}{2E} = 3$
$\lambda' = 3\lambda$

(C) In the given arrangement,plate $Q$ is common to two capacitors connected in parallel,as both plates $P$ and $R$ are grounded (at $0 \,V$).
Let $V$ be the potential of plate $Q$.
The capacitance of the first capacitor (between $P$ and $Q$) is $C_1 = \frac{\varepsilon_0 A}{d}$.
The capacitance of the second capacitor (between $Q$ and $R$) is $C_2 = \frac{\varepsilon_0 A}{2d}$.
Since they are in parallel,the effective capacitance is $C_{\text{eff}} = C_1 + C_2 = \frac{\varepsilon_0 A}{d} + \frac{\varepsilon_0 A}{2d} = \frac{3 \varepsilon_0 A}{2d}$.
Given $q = 8.85 \times 10^{-8} \,C$,$A = 2.0 \,m^2$,$d = 2 \times 10^{-3} \,m$,and $\varepsilon_0 = 8.85 \times 10^{-12} \,F/m$.
Using $q = C_{\text{eff}} V$,we have $V = \frac{q}{C_{\text{eff}}} = \frac{q \cdot 2d}{3 \varepsilon_0 A}$.
Substituting the values: $V = \frac{8.85 \times 10^{-8} \times 2 \times (2 \times 10^{-3})}{3 \times (8.85 \times 10^{-12}) \times 2.0} = \frac{8.85 \times 4 \times 10^{-11}}{3 \times 8.85 \times 2 \times 10^{-12}} = \frac{4 \times 10}{3 \times 2} = \frac{20}{3} \approx 6.67 \,V$.

(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.
Let the initial charge be $q_1 = q$ and the final charge be $q_2 = q + 2 \text{ C}$.
The initial energy is $U_1 = \frac{q^2}{2C}$ and the final energy is $U_2 = \frac{(q+2)^2}{2C}$.
Given that the energy increases by $21 \%$,we have $U_2 = U_1 + 0.21 U_1 = 1.21 U_1$.
Substituting the expressions for $U_1$ and $U_2$:
$\frac{(q+2)^2}{2C} = 1.21 \times \frac{q^2}{2C}$.
Canceling $\frac{1}{2C}$ from both sides,we get $(q+2)^2 = 1.21 q^2$.
Taking the square root of both sides: $q + 2 = 1.1 q$.
Rearranging the terms: $1.1 q - q = 2$,which gives $0.1 q = 2$.
Therefore,$q = \frac{2}{0.1} = 20 \text{ C}$.

(B) From the given circuit diagram, the two $2 \, \mu F$ capacitors in the top branch are in series. Their equivalent capacitance is $C_{1} = \frac{2 \times 2}{2+2} = 1 \, \mu F$.
Similarly, the two $2 \, \mu F$ capacitors in the right and bottom branches are in series. Their equivalent capacitance is $C_{2} = \frac{2 \times 2}{2+2} = 1 \, \mu F$.
Now, the circuit simplifies to a parallel combination of $C \, \mu F$ and $C_{2} = 1 \, \mu F$, which is in series with $C_{1} = 1 \, \mu F$.
The effective capacitance $C_{\text{eff}}$ is given by:
$C_{\text{eff}} = \frac{(C + 1) \times 1}{(C + 1) + 1} = \frac{C + 1}{C + 2} \, \mu F$.
Given, charge $q = 0.75 \, mC = 0.75 \times 10^{-3} \, C$ and potential $V = 10^{3} \, V$.
The effective capacitance is $C_{\text{eff}} = \frac{q}{V} = \frac{0.75 \times 10^{-3}}{10^{3}} = 0.75 \times 10^{-6} \, F = 0.75 \, \mu F$.
Equating the two expressions for $C_{\text{eff}}$:
$0.75 = \frac{C + 1}{C + 2}$
$\frac{3}{4} = \frac{C + 1}{C + 2}$
$3(C + 2) = 4(C + 1)$
$3C + 6 = 4C + 4$
$C = 2 \, \mu F$.

(A) Let the resistances of the three conductors be $R_1, R_2$ and $R_3$. When connected individually across a battery of voltage $V$, the currents are $I_1 = 1 \,A, I_2 = 2 \,A$ and $I_3 = 3 \,A$.
Using Ohm's law, $V = I R$, we have:
$R_1 = \frac{V}{1} = V$
$R_2 = \frac{V}{2}$
$R_3 = \frac{V}{3}$
When connected in series, the equivalent resistance $R_{eq}$ is:
$R_{eq} = R_1 + R_2 + R_3 = V + \frac{V}{2} + \frac{V}{3} = V \left( 1 + \frac{1}{2} + \frac{1}{3} \right) = V \left( \frac{6 + 3 + 2}{6} \right) = \frac{11V}{6}$
The current $I$ drawn from the same battery in series is:
$I = \frac{V}{R_{eq}} = \frac{V}{11V/6} = \frac{6}{11} \,A$.

(B) The current $I$ splits into $I_1$ and $I_2$ through the parallel resistors $R$ and $2R$. Using the current divider rule:
$I_1 = I \times \frac{2R}{R + 2R} = \frac{2I}{3}$
$I_2 = I \times \frac{R}{R + 2R} = \frac{I}{3}$
The heat produced in a resistor is given by $H = I^2 R t$. Assuming time $t$ is the same for all:
$H_1 = I_1^2 R = \left(\frac{2I}{3}\right)^2 R = \frac{4I^2 R}{9}$
$H_2 = I_2^2 (2R) = \left(\frac{I}{3}\right)^2 (2R) = \frac{2I^2 R}{9}$
For the resistor $1.5R$,the total current $I$ flows through it:
$H_3 = I^2 (1.5R) = 1.5 I^2 R = \frac{13.5 I^2 R}{9}$
The ratio $H_1 : H_2 : H_3 = \frac{4}{9} : \frac{2}{9} : \frac{13.5}{9} = 4 : 2 : 13.5 = 8 : 4 : 27$.

(D) For a tangent galvanometer,the current $I$ is given by $I = K \tan \theta$,where $K = \frac{2r B_{H}}{n \mu_{0}}$.
Since the galvanometers are connected in parallel,the voltage $V$ across them is the same. Thus,$I_1 R_1 = I_2 R_2$,which implies $\frac{I_1}{I_2} = \frac{R_2}{R_1} = \frac{3}{1}$.
Using the formula $I = K \tan \theta$,we have $\frac{I_1}{I_2} = \frac{n_2}{n_1} \cdot \frac{\tan \theta_1}{\tan \theta_2}$.
Substituting the given values: $\frac{3}{1} = \frac{n_2}{n_1} \cdot \frac{\tan 30^{\circ}}{\tan 60^{\circ}}$.
$\frac{3}{1} = \frac{n_2}{n_1} \cdot \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{n_2}{n_1} \cdot \frac{1}{3}$.
Therefore,$\frac{n_1}{n_2} = \frac{1}{3 \times 3} = \frac{1}{9}$.
Wait,re-evaluating: $\frac{3}{1} = \frac{n_2}{n_1} \cdot \frac{1}{3} \implies \frac{n_2}{n_1} = 9 \implies \frac{n_1}{n_2} = \frac{1}{9}$.
Given the options,let's re-check the ratio: $\frac{R_2}{R_1} = 3$. $\frac{I_1}{I_2} = 3$. $\frac{n_2}{n_1} = \frac{I_1 \tan \theta_2}{I_2 \tan \theta_1} = 3 \cdot \frac{\sqrt{3}}{1/\sqrt{3}} = 3 \cdot 3 = 9$. The ratio $n_1:n_2$ is $1:9$. Since $1:9$ is not an option,let's re-verify the resistance ratio logic. If $R_1:R_2 = 1:3$,then $I_1:I_2 = 3:1$. The correct ratio is $1:9$.

(A) Let $R_{1} = R_{2} = R$. The current through $R_{2}$ is $(I - 1.5) \ A$.
Applying Kirchhoff's voltage law $(KVL)$ in the loop containing $E, R_{1}$,and $R_{2}$:
$E - I R_{1} - (I - 1.5) R_{2} = 0$
Since $R_{1} = R_{2} = R$,we have $E = I R + (I - 1.5) R = R(2I - 1.5) \quad ... (i)$
Applying $KVL$ in the outer loop containing $E, R_{1}$,and $R'$:
$E - I R_{1} - 1.5 R' = 0$
$E = I R + 1.5 R' \quad ... (ii)$
From the provided circuit diagram,the voltage across $R_{2}$ is the same as the voltage across $R'$,so $V_{R_{2}} = V_{R'}$.
$(I - 1.5) R = 1.5 R'$
$R' = \frac{(I - 1.5) R}{1.5}$
Substituting $R'$ into equation $(ii)$:
$E = I R + 1.5 \left[ \frac{(I - 1.5) R}{1.5} \right] = I R + (I - 1.5) R = R(2I - 1.5)$
This confirms the consistency. Given the options,let's test $R = 60 \ \Omega$ and $E = 180 \ V$:
$180 = 60(2I - 1.5) \Rightarrow 3 = 2I - 1.5 \Rightarrow 2I = 4.5 \Rightarrow I = 2.25 \ A$.
Then $I - 1.5 = 2.25 - 1.5 = 0.75 \ A$.
Voltage across $R_{2} = 0.75 \times 60 = 45 \ V$.
Voltage across $R' = 1.5 \times R' = 45 \ V \Rightarrow R' = 30 \ \Omega$.
This is a valid physical circuit. Thus,$E = 180 \ V$ and $R_{1} = 60 \ \Omega$ is the correct pair.

(D) Given,the ratio of masses is $m_{1}: m_{2}: m_{3} = 1: 3: 5$ and the ratio of lengths is $l_{1}: l_{2}: l_{3} = 5: 3: 1$.
We know that the electrical resistance $R$ is given by $R = \rho \frac{l}{A}$.
Since density $d = \frac{m}{V} = \frac{m}{Al}$,we have $A = \frac{m}{dl}$.
Substituting $A$ in the resistance formula,we get $R = \rho \frac{l}{(m/dl)} = \rho d \frac{l^{2}}{m}$.
Since $\rho$ and $d$ are constant for copper wires,$R \propto \frac{l^{2}}{m}$.
Therefore,the ratio of resistances is $R_{1}: R_{2}: R_{3} = \frac{l_{1}^{2}}{m_{1}}: \frac{l_{2}^{2}}{m_{2}}: \frac{l_{3}^{2}}{m_{3}}$.
Substituting the given values: $R_{1}: R_{2}: R_{3} = \frac{5^{2}}{1}: \frac{3^{2}}{3}: \frac{1^{2}}{5} = \frac{25}{1}: \frac{9}{3}: \frac{1}{5} = 25: 3: 0.2$.
To simplify the ratio,multiply by $5$: $125: 15: 1$.

(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2 m E_{k}}}$.
Since $h$ and $m$ are constants,we have $\lambda \propto \frac{1}{\sqrt{E_{k}}}$.
Let the initial kinetic energy be $E_{k1} = E$ and the final kinetic energy be $E_{k2}$.
Given $\lambda_1 = 1 \ nm$ and $\lambda_2 = 0.5 \ nm$,we have $\frac{\lambda_1}{\lambda_2} = \frac{1}{0.5} = 2$.
Using the relation $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{E_{k2}}{E_{k1}}}$,we get $2 = \sqrt{\frac{E_{k2}}{E}}$,which implies $\frac{E_{k2}}{E} = 4$,so $E_{k2} = 4E$.
The additional energy required is $\Delta E = E_{k2} - E_{k1} = 4E - E = 3E$.
Thus,the additional energy is $3$ times the initial kinetic energy.

(D) The energy of a photon emitted during a transition from $n_i$ to $n_f$ is given by $E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$.
For the transition $n=3$ to $n=2$, the energy is $E_{3-2} = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{5}{36} \right) \approx 1.89 \text{ eV}$.
Since photoelectrons are just emitted, the work function of the metal is $\Phi = 1.89 \text{ eV}$.
For the photoelectric effect to occur, the energy of the incident photon must be greater than or equal to the work function $(E \ge \Phi)$.
Let's calculate the energy for the given options:
$A$: $n=2$ to $n=1$: $E = 13.6 (1 - 1/4) = 10.2 \text{ eV} > 1.89 \text{ eV}$ (Possible).
$B$: $n=3$ to $n=1$: $E = 13.6 (1 - 1/9) = 12.09 \text{ eV} > 1.89 \text{ eV}$ (Possible).
$C$: $n=5$ to $n=2$: $E = 13.6 (1/4 - 1/25) = 13.6 (21/100) = 2.856 \text{ eV} > 1.89 \text{ eV}$ (Possible).
$D$: $n=4$ to $n=3$: $E = 13.6 (1/9 - 1/16) = 13.6 (7/144) \approx 0.66 \text{ eV} < 1.89 \text{ eV}$ (Not possible).
Therefore, the photoelectric effect is not possible for the transition from $n=4$ to $n=3$.

(C) Given: Maximum velocity $v = 1.8 \times 10^{6} \ m/s$ and specific charge $\frac{e}{m} = 1.8 \times 10^{11} \ C/kg$.
The kinetic energy of the fastest photoelectron is equal to the work done by the stopping potential $V_{0}$,given by the equation: $e V_{0} = \frac{1}{2} m v^{2}$.
Dividing both sides by $m$,we get: $V_{0} \left(\frac{e}{m}\right) = \frac{v^{2}}{2}$.
Substituting the given values:
$V_{0} \times (1.8 \times 10^{11}) = \frac{(1.8 \times 10^{6})^{2}}{2}$.
$V_{0} \times 1.8 \times 10^{11} = \frac{3.24 \times 10^{12}}{2}$.
$V_{0} \times 1.8 \times 10^{11} = 1.62 \times 10^{12}$.
$V_{0} = \frac{1.62 \times 10^{12}}{1.8 \times 10^{11}} = 0.9 \times 10 = 9 \ V$.

$(A)$ Given: Number of turns $n = 100$, Area $A = 0.1 \,m \times 0.05 \,m = 0.005 \,m^{2}$.
Initial magnetic field $B_{1} = 0.1 \,T$, final magnetic field $B_{2} = 0.05 \,T$, and time interval $dt = 0.05 \,s$.
The magnetic flux $\phi$ is given by $\phi = nBA \cos \theta$. Since the coil is perpendicular to the field, the angle $\theta = 0^{\circ}$, so $\cos 0^{\circ} = 1$.
The induced e.m.f. $e$ is given by Faraday's Law: $e = \left| -\frac{d\phi}{dt} \right| = nA \frac{|dB|}{dt}$.
Substituting the values: $e = 100 \times 0.005 \times \frac{(0.1 - 0.05)}{0.05}$.
$e = 0.5 \times \frac{0.05}{0.05} = 0.5 \,V$.

(C) According to Lenz's law,the induced current in coil $B$ will oppose the cause that produces it.
As the current in coil $A$ increases,the magnetic flux linked with coil $B$ increases.
To oppose this increase in magnetic flux,an induced current flows in coil $B$ in a direction opposite to the current in coil $A$.
Since the currents in the adjacent sides of the two coils flow in opposite directions,they exert a repulsive force on each other.
Therefore,coil $B$ is repelled.

(B) Given that the oil drop is at rest,the gravitational force acting downwards must be balanced by the electric force acting upwards.
$qE = mg$
Since the electric field $E$ between two plates is given by $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the distance between the plates:
$q \left(\frac{V}{d}\right) = mg$
Rearranging to solve for the charge $q$:
$q = \frac{mgd}{V}$
Given values:
$m = 10^{-6} \,kg$
$g = 10 \,ms^{-2}$
$d = 1 \,mm = 10^{-3} \,m$
$V = 500 \,V$
Substituting these values into the equation:
$q = \frac{10^{-6} \times 10 \times 10^{-3}}{500}$
$q = \frac{10^{-8}}{500} = \frac{10^{-8}}{5 \times 10^2} = 0.2 \times 10^{-10} \,C = 2 \times 10^{-11} \,C$

(D) In a uniform electric field, the electric potential $V$ decreases in the direction of the electric field lines. The potential at any point $(x, y)$ in a uniform electric field $\vec{E} = E\hat{i}$ is given by $V = -Ex + \text{constant}$.
Points on the same vertical line (perpendicular to the field lines) have the same potential because they have the same $x$-coordinate.
Looking at the diagram:
$1$. Points $A$ and $B$ are at the same vertical position relative to the field lines, but they are not on the same vertical line. However, the question asks for the relationship between potentials. Let's analyze the $x$-coordinates.
$2$. Point $C$ is at the leftmost position, so it has the smallest $x$-coordinate, meaning $V_{C}$ is the highest.
$3$. Point $D$ is at the rightmost position, so it has the largest $x$-coordinate, meaning $V_{D}$ is the lowest.
$4$. Points $A$ and $B$ have the same $x$-coordinate, so $V_{A} = V_{B}$.
$5$. Since $C$ is to the left of $D$, $V_{C} > V_{D}$.
Thus, the correct relationship is $V_{A} = V_{B}$ and $V_{C} > V_{D}$.

(C) When two metal spheres are joined by a wire,charge flows from the sphere at higher potential to the sphere at lower potential until they reach the same potential.
Let the final charges be $q_1'$ and $q_2'$. Since the potential $V$ is equal,$V_1 = V_2$.
$\frac{k q_1'}{r_1} = \frac{k q_2'}{r_2} \implies \frac{q_1'}{q_2'} = \frac{r_1}{r_2} = \frac{0.01}{0.02} = \frac{1}{2}$.
The total charge is conserved: $q_1' + q_2' = 15 \ mC + 45 \ mC = 60 \ mC$.
Using the ratio,$q_1' = \left( \frac{1}{1+2} \right) \times 60 \ mC = \frac{1}{3} \times 60 \ mC = 20 \ mC$.
Since $20 \ mC = 20 \times 10^{-3} \ C$,the final charge on the first sphere is $20 \times 10^{-3} \ C$.

(A) The electric potential at the centre of a charged spherical shell of radius $R$ with charge $q$ is $V = \frac{q}{4 \pi \varepsilon_{0} R}$.
For the two concentric spheres,the total potential $V$ at the common centre is the sum of potentials due to each sphere:
$V = V_{1} + V_{2} = \frac{q_{1}}{4 \pi \varepsilon_{0} R} + \frac{q_{2}}{4 \pi \varepsilon_{0} r}$.
Given that the surface charge densities are equal,$\sigma = \frac{q_{1}}{4 \pi R^{2}} = \frac{q_{2}}{4 \pi r^{2}}$.
This implies $q_{1} = 4 \pi R^{2} \sigma$ and $q_{2} = 4 \pi r^{2} \sigma$.
Substituting these values into the potential equation:
$V = \frac{4 \pi R^{2} \sigma}{4 \pi \varepsilon_{0} R} + \frac{4 \pi r^{2} \sigma}{4 \pi \varepsilon_{0} r}$.
$V = \frac{R \sigma}{\varepsilon_{0}} + \frac{r \sigma}{\varepsilon_{0}} = \frac{\sigma}{\varepsilon_{0}}(R + r)$.

(D) The magnetic field at the centre $O$ due to a circular loop of radius $R$ carrying current $I_{1}$ is given by: $B_{1} = \frac{\mu_{0} I_{1}}{2 R}$.
The magnetic field at the centre $O$ due to a long straight conductor carrying current $I_{2}$ at a perpendicular distance $d = OA + AB = R + R = 2R$ is given by: $B_{2} = \frac{\mu_{0} I_{2}}{2 \pi d} = \frac{\mu_{0} I_{2}}{2 \pi (2R)} = \frac{\mu_{0} I_{2}}{4 \pi R}$.
Given that the net magnetic field at $O$ is zero,the magnitudes of the magnetic fields produced by the loop and the straight conductor must be equal: $B_{1} = B_{2}$.
Substituting the expressions: $\frac{\mu_{0} I_{1}}{2 R} = \frac{\mu_{0} I_{2}}{4 \pi R}$.
Simplifying the equation: $\frac{I_{1}}{2} = \frac{I_{2}}{4 \pi}$.
Therefore,the ratio of the currents is: $\frac{I_{1}}{I_{2}} = \frac{2}{4 \pi} = \frac{1}{2 \pi}$.

(A) The radius $r$ of the circular path of a charged particle in a magnetic field $B$ is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mE}}{qB}$.
Given that the kinetic energy $E$ and the magnetic field $B$ are the same for both particles,we have $r \propto \frac{\sqrt{m}}{q}$.
For a proton,mass $m_P = m$ and charge $q_P = q$.
For a helium nucleus (alpha particle),mass $m_{He} = 4m$ and charge $q_{He} = 2q$.
Therefore,the ratio of their radii is $\frac{r_P}{r_{He}} = \frac{\sqrt{m_P}/q_P}{\sqrt{m_{He}}/q_{He}} = \sqrt{\frac{m}{4m}} \times \frac{2q}{q} = \frac{1}{2} \times 2 = 1$.
Thus,the ratio is $1: 1$.

(C) When a charged particle passes undeflected through mutually perpendicular electric and magnetic fields,the electric force $(F_e)$ must balance the magnetic force $(F_m)$.
$F_e = F_m$
$qE = qvB \sin \theta$
Since the fields are mutually perpendicular,$\theta = 90^{\circ}$,so $\sin 90^{\circ} = 1$.
Thus,$E = vB$.
Given:
Velocity $v = 2 \times 10^{3} \ ms^{-1}$
Magnetic field $B = 1.5 \ T$
$E = (2 \times 10^{3}) \times 1.5 = 3 \times 10^{3} \ V/m$ (or $NC^{-1}$).
Therefore,the magnitude of the electric field is $3 \times 10^{3} \ NC^{-1}$.

(B) The energy required to remove the first electron from a neutral helium atom $(He)$ is given as $24.6 eV$.
After the first electron is removed,the remaining ion is $He^+$,which is a hydrogen-like species with atomic number $Z = 2$.
The energy required to remove the second electron from the ground state of $He^+$ is given by the formula $E = Z^2 \times 13.6 eV$.
Substituting $Z = 2$,we get $E = (2)^2 \times 13.6 eV = 4 \times 13.6 eV = 54.4 eV$.
The total energy required to remove both electrons is the sum of the energy to remove the first and the second electron.
Total Energy $= 24.6 eV + 54.4 eV = 79 eV$.

(C) Given:
Binding energy per nucleon for ${ }_{1} H^{2} = 1.1 \ MeV$.
Binding energy per nucleon for ${ }_{2} He^{4} = 7 \ MeV$.
The nuclear fusion reaction is: ${ }_{1} H^{2} + { }_{1} H^{2} \longrightarrow { }_{2} He^{4} + Q$.
Total binding energy of reactants (two deuterons) = $2 \times (2 \times 1.1 \ MeV) = 4.4 \ MeV$.
Total binding energy of the product (one helium nucleus) = $4 \times 7 \ MeV = 28 \ MeV$.
The energy released $Q$ is the difference between the total binding energy of the product and the reactants:
$Q = BE_{\text{product}} - BE_{\text{reactants}}$
$Q = 28 \ MeV - 4.4 \ MeV = 23.6 \ MeV$.

(D) Statement $I$: During $\beta^-$-decay,a neutron transforms into a proton,an electron,and an antineutrino. Thus,it is accompanied by an antineutrino (or neutrino in $\beta^+$-decay). This statement is correct.
Statement $II$: Nuclear forces act between nucleons (protons and neutrons) regardless of their charge. Thus,nuclear force is charge independent. This statement is correct.
Statement $III$: Nuclear fusion of hydrogen into helium is the primary source of energy in stars. This statement is correct.
Therefore,all three statements are correct.

(B) Let the recoil speed of the daughter nucleus be $v^{\prime}$.
According to the law of conservation of linear momentum,the initial momentum of the nucleus is zero.
Therefore,the final momentum of the system must also be zero.
Let the mass of the $\alpha$-particle be $4$ units and the mass of the daughter nucleus be $(A-4)$ units.
$0 = (A-4) v^{\prime} + 4 v$
$(A-4) v^{\prime} = -4 v$
$v^{\prime} = -\frac{4 v}{A-4}$
The magnitude of the recoil speed is $\frac{4 v}{A-4}$.

(C) The number of radioactive nuclei decaying in a given time interval is proportional to the number of nuclei present.
Let $N_0$ be the initial number of nuclei.
In the first $2 \,s$, $100$ particles are emitted, so the remaining nuclei are $N_0 - 100$.
In the next $2 \,s$, $50$ particles are emitted.
Since the number of decays is halved in equal time intervals, the half-life $T_{1/2}$ is $2 \,s$.
The mean life $T_m$ is related to the half-life by the formula $T_m = \frac{T_{1/2}}{\ln(2)} = \frac{T_{1/2}}{0.693}$.
Substituting $T_{1/2} = 2 \,s$, we get $T_m = \frac{2}{0.693} \,s$.

(D) Let us analyze each statement:
$(i)$ In forward biased condition,the depletion layer width decreases,allowing current to flow; thus,a diode conducts. This is correct.
(ii) The packing fraction is defined as $f = (M - A) / A$. $A$ smaller or negative packing fraction indicates higher stability of the nucleus. This is correct.
(iii) Binding energy is the energy required to disassemble a nucleus into its constituent protons and neutrons,which is equivalent to the mass defect $\Delta m$ via Einstein's equation $E = \Delta m c^2$. This is correct.
(iv) Spontaneous fission is a rare radioactive decay process that occurs only in very heavy nuclei (e.g.,$U-238$,$Cf-252$). It is not a general property of all radioactive elements. Therefore,the statement that radioactive elements (implying all) can undergo spontaneous fission is incorrect.

(D) The activity $A$ is given by $A = \lambda N = \frac{0.693}{T_{1/2}} N$.
Since the initial number of atoms $N_0$ for $A$ and $B$ are equal,the initial activity $A_0$ is inversely proportional to the half-life $T_{1/2}$.
Therefore,$\frac{A_0(A)}{A_0(B)} = \frac{T_{1/2}(B)}{T_{1/2}(A)} = \frac{2 \text{ days}}{1 \text{ day}} = 2$.
Given $A_0(A) + A_0(B) = 1200 \text{ dis/min}$.
Substituting $A_0(A) = 2 A_0(B)$,we get $2 A_0(B) + A_0(B) = 1200$,which implies $3 A_0(B) = 1200$,so $A_0(B) = 400 \text{ dis/min}$ and $A_0(A) = 800 \text{ dis/min}$.
After $t = 4 \text{ days}$,the activity of $A$ is $A(A) = \frac{A_0(A)}{2^{t/T_{1/2}(A)}} = \frac{800}{2^{4/1}} = \frac{800}{16} = 50 \text{ dis/min}$.
The activity of $B$ is $A(B) = \frac{A_0(B)}{2^{t/T_{1/2}(B)}} = \frac{400}{2^{4/2}} = \frac{400}{4} = 100 \text{ dis/min}$.
The total activity after $4 \text{ days}$ is $50 + 100 = 150 \text{ dis/min}$.

(A) For total internal reflection to occur, the light must travel from a denser medium to a rarer medium, and the angle of incidence $\theta$ must be greater than the critical angle $C$.
Given speeds of light in media $M_{1}$ and $M_{2}$ are $v_{1} = 1.5 \times 10^{8} \text{ m/s}$ and $v_{2} = 2 \times 10^{8} \text{ m/s}$.
Since $v_{1} < v_{2}$, medium $M_{1}$ is denser than $M_{2}$.
The critical angle $C$ is given by $\sin C = \frac{v_{1}}{v_{2}}$.
Substituting the values: $\sin C = \frac{1.5 \times 10^{8}}{2 \times 10^{8}} = \frac{1.5}{2} = \frac{3}{4}$.
Therefore, $C = \sin^{-1}\left(\frac{3}{4}\right)$.
For total internal reflection, the angle of incidence $\theta$ must satisfy $\theta > C$.
Hence, $\theta > \sin^{-1}\left(\frac{3}{4}\right)$.

(D) Given: Ratio of radii of curvature $\frac{R_{1}}{R_{2}} = \frac{1}{2}$, focal length $f = 6 \,cm$, and refractive index $\mu = 1.5$.
Let $R_{1} = R$ and $R_{2} = 2R$.
For a converging lens, the lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_{1}} + \frac{1}{R_{2}} \right)$.
Substituting the values: $\frac{1}{6} = (1.5 - 1) \left( \frac{1}{R} + \frac{1}{2R} \right)$.
$\frac{1}{6} = 0.5 \left( \frac{2 + 1}{2R} \right) = 0.5 \left( \frac{3}{2R} \right) = \frac{1.5}{2R} = \frac{3}{4R}$.
Solving for $R$: $4R = 18$, so $R = 4.5 \,cm$.
Therefore, $R_{1} = 4.5 \,cm$ and $R_{2} = 2 \times 4.5 = 9 \,cm$.

(B) Given,magnification $m = 2$ and the image is real. For a real image,the distance between the object and the image is $D = |u| + |v| = 0.72 \ m$.
Since $m = \frac{|v|}{|u|} = 2$,we have $|v| = 2|u|$.
Substituting this into the distance equation: $|u| + 2|u| = 0.72 \ m \Rightarrow 3|u| = 0.72 \ m \Rightarrow |u| = 0.24 \ m$ and $|v| = 0.48 \ m$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $v = 0.48 \ m$ and $u = -0.24 \ m$:
$\frac{1}{f} = \frac{1}{0.48} - \frac{1}{-0.24} = \frac{1 + 2}{0.48} = \frac{3}{0.48} \Rightarrow f = 0.16 \ m$.
When the object is moved $0.04 \ m$ towards the lens,the new object distance is $u' = -(0.24 - 0.04) = -0.20 \ m$.
Using the lens formula again: $\frac{1}{v'} - \frac{1}{-0.20} = \frac{1}{0.16} \Rightarrow \frac{1}{v'} = \frac{1}{0.16} - \frac{1}{0.20} = \frac{5 - 4}{0.80} = \frac{1}{0.80} \Rightarrow v' = 0.80 \ m$.
The new magnification is $m' = \frac{v'}{u'} = \frac{0.80}{-(-0.20)} = 4$.

(C) Let $n_1$ be the refractive index of the medium outside the first surface, $n_2$ be the refractive index of the prism, and $n_3$ be the refractive index of the medium outside the second surface.
For a ray to bend towards the normal, it must travel from a rarer to a denser medium $(n_{incident} < n_{refracted})$.
For a ray to bend away from the normal, it must travel from a denser to a rarer medium $(n_{incident} > n_{refracted})$.
In case $(a)$, the ray bends towards the normal at the first surface $(n_1 < n_2)$ and grazes the second surface, implying total internal reflection or critical angle condition $(n_2 > n_3)$. Thus, $n_2 > n_1$ and $n_2 > n_3$ is correct.
In case $(b)$, the ray enters undeviated at the first surface $(n_1 = n_2)$ and bends away from the normal at the second surface $(n_2 > n_3)$. This is correct.
In case $(c)$, the ray bends towards the normal at the first surface $(n_1 < n_2)$ and bends away from the normal at the second surface $(n_2 > n_3)$. The statement says $n_2 < n_1$ and $n_2 = n_3$, which contradicts the observed bending. Thus, $(c)$ is the wrong statement.

(C) The lateral shift is given by $L_{s} = t \frac{\sin(i-r)}{\cos r}$. As the angle of incidence $i$ increases,the lateral shift $L_{s}$ increases. Thus,statement $A$ is correct.
As the refractive index $\mu$ increases,the angle of refraction $r$ decreases,which leads to an increase in the lateral shift $L_{s}$. Thus,statement $B$ is correct.
The normal shift is given by $L_{N} = t(1 - \frac{1}{\mu})$. As the refractive index $\mu$ increases,the term $\frac{1}{\mu}$ decreases,which means $(1 - \frac{1}{\mu})$ increases. Therefore,the normal shift $L_{N}$ increases as the refractive index increases. Thus,statement $C$ is incorrect.
Both $L_{s}$ and $L_{N}$ are directly proportional to the thickness $t$ of the medium. Thus,statement $D$ is correct.
Therefore,the wrong statement is $C$.

(A) In a $p-n$ junction diode,when a reverse bias is applied,the external electric field is in the same direction as the internal electric field of the depletion region. This causes the majority charge carriers to move away from the junction,thereby increasing the width of the depletion region. Conversely,in forward bias,the external field opposes the internal field,which reduces the width of the depletion region.

(D) To use a transistor as an amplifier,the input circuit (emitter-base junction) must be forward biased to allow current flow,and the output circuit (collector-base junction) must be reverse biased to provide high resistance and voltage gain.
Therefore,the emitter-base junction is forward biased,and the collector-base junction is reverse biased.

(D) Let the inputs of the $OR$ gate be $A$ and $B$. The output of the $OR$ gate is $Y = A + B$.
This output $Y$ is connected to both inputs of a $NAND$ gate. Let the inputs of the $NAND$ gate be $X_1$ and $X_2$,where $X_1 = X_2 = Y = A + B$.
The output of a $NAND$ gate with inputs $X_1$ and $X_2$ is given by $\overline{X_1 \cdot X_2}$.
Substituting $X_1 = X_2 = A + B$,the final output $Y^{\prime}$ is $\overline{(A + B) \cdot (A + B)}$.
Using the Boolean identity $X \cdot X = X$,we get $Y^{\prime} = \overline{A + B}$.
The expression $\overline{A + B}$ represents the Boolean operation of a $NOR$ gate.
Therefore,the combination acts as a $NOR$ gate.

(C) $p$-type semiconductor is formed by doping an intrinsic semiconductor with a trivalent impurity atom (Group $13$ element).
Gallium $(Ga)$ is a trivalent element.
Therefore,when Germanium $(Ge)$ is doped with Gallium,it creates a $p$-type semiconductor.
Bismuth $(Bi)$,Antimony $(Sb)$,and Phosphorus $(P)$ are pentavalent elements (Group $15$),which produce $n$-type semiconductors.

(A) The optical rotation $\theta$ produced by a sugar solution is given by the formula: $\theta = S \cdot l \cdot C$, where $S$ is the specific rotation, $l$ is the length of the tube, and $C$ is the concentration of the solution.
In the first case, $\theta = S \cdot l \cdot C$, where $l = 0.2 \,m$ and $C = \frac{m}{V}$, where $m$ is the mass of sugar and $V$ is the volume of the tube $(V = \pi R^2 l)$.
Thus, $\theta = S \cdot l \cdot \frac{m}{\pi R^2 l} = \frac{S \cdot m}{\pi R^2}$.
In the second case, the same amount of sugar $m$ is placed in a tube of length $l_1 = 0.3 \,m$ with the same radius $R$. The volume of the new tube is $V_1 = \pi R^2 l_1 = \pi R^2 (0.3)$.
The concentration of the solution in the second case is $C_1 = \frac{m}{V_1} = \frac{m}{\pi R^2 (0.3)}$.
The new rotation $\theta_1$ is given by $\theta_1 = S \cdot l_1 \cdot C_1 = S \cdot (0.3) \cdot \frac{m}{\pi R^2 (0.3)} = \frac{S \cdot m}{\pi R^2}$.
Comparing the two expressions, we find $\theta_1 = \theta$.

(C) The wave theory of light,proposed by Huygens,successfully explains the phenomena of interference,diffraction,and the fact that the velocity of light in a denser medium is less than in a rarer medium.
However,the wave theory does not explain the scattering of light,which is better described by the particle nature of light or quantum theory.
Therefore,phenomena $(2)$,$(3)$,and $(4)$ support the wave theory of light.

(C) For a thin film,the condition for constructive interference (maxima) in reflected light is $2 \mu t \cos r = (n + 1/2) \lambda_1$,where $n$ is an integer.
For destructive interference (minima),the condition is $2 \mu t \cos r = m \lambda_2$,where $m$ is an integer.
Given $\lambda_1 = 600 \ nm$ and $\lambda_2 = 450 \ nm$. Assuming normal incidence $(\cos r = 1)$:
$2 \mu t = (n + 1/2) \lambda_1 = (2n + 1) \frac{\lambda_1}{2} = (2n + 1) \times 300 \ nm$
$2 \mu t = m \lambda_2 = m \times 450 \ nm$
Equating the two: $(2n + 1) \times 300 = m \times 450 \implies (2n + 1) \times 2 = 3m \implies 4n + 2 = 3m$.
For the smallest thickness with no minimum in between,we test $n=1$: $4(1) + 2 = 6 = 3m \implies m = 2$.
Substituting $n=1$ into the maxima condition:
$2 \times 1.5 \times t = (1 + 0.5) \times 600 \ nm$
$3t = 1.5 \times 600 \ nm = 900 \ nm$
$t = 300 \ nm = 3 \times 10^{-7} \ m$.
Thus,the thickness is $3$ units.

(A) The fringe width $\beta$ in a Young's double-slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance to the screen,and $d$ is the slit separation.
This implies that $\beta \propto \lambda$.
According to the photoelectric effect,the energy of a photon is $E = \frac{hc}{\lambda}$. For the photoelectric effect to occur,the energy of the incident photon must be greater than or equal to the work function of the metal $(\Phi)$,i.e.,$\frac{hc}{\lambda} \geq \Phi$.
This means that a smaller wavelength corresponds to higher energy.
Since $\lambda_{1}$ can produce the photoelectric effect and $\lambda_{2}$ cannot,it implies that $\lambda_{1}$ has higher energy than $\lambda_{2}$,which means $\lambda_{1} < \lambda_{2}$.
Since $\beta \propto \lambda$,it follows that $\beta_{1} < \beta_{2}$.