KCET 2007 Physics Question Paper with Answer and Solution

(A) Total mass of the shell $M = 20 \,kg$.
Given the ratio of masses $m_1 : m_2 = 2 : 3$.
Thus,$m_1 = (2/5) \times 20 = 8 \,kg$ and $m_2 = (3/5) \times 20 = 12 \,kg$.
According to the law of conservation of linear momentum,the initial momentum is zero,so the final momentum must also be zero: $m_1 v_1 + m_2 v_2 = 0$.
Taking the smaller fragment velocity $v_1 = 6 \,ms^{-1}$,we have $8 \times 6 + 12 \times v_2 = 0$.
$48 + 12 v_2 = 0 \implies v_2 = -4 \,ms^{-1}$.
The magnitude of the velocity of the larger fragment is $4 \,ms^{-1}$.
Kinetic energy of the larger fragment $K_2 = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} \times 12 \times (4)^2 = 6 \times 16 = 96 \,J$.

(B) The acceleration due to gravity at the surface of the earth is $g = \frac{GM}{R^2}$.
At a height $h$ above the surface,the acceleration due to gravity $g'$ is given by $g' = \frac{GM}{(R+h)^2}$.
Dividing the two expressions,we get $\frac{g'}{g} = \left(\frac{R}{R+h}\right)^2$.
Given that $g' = \frac{g}{2}$,we substitute this into the equation:
$\frac{1}{2} = \left(\frac{R}{R+h}\right)^2$.
Taking the square root on both sides:
$\frac{1}{\sqrt{2}} = \frac{R}{R+h}$.
Rearranging the terms:
$R+h = R\sqrt{2}$.
$h = R(\sqrt{2}-1)$.

(B) According to the equation of continuity, the volume flow rate remains constant throughout the system.
$A_{1} v_{1} = A_{2} v_{2}$
Here, $A_{1} = 8 \,cm^{2} = 8 \times 10^{-4} \,m^{2}$.
The speed inside the tube is $v_{1} = 0.15 \,m \,min^{-1} = \frac{0.15}{60} \,m \,s^{-1} = 0.0025 \,m \,s^{-1}$.
The total area of the $40$ holes is $A_{2} = 40 \times 10^{-8} \,m^{2}$.
Substituting these values into the continuity equation:
$(8 \times 10^{-4}) \times (0.0025) = (40 \times 10^{-8}) \times v_{2}$
$v_{2} = \frac{8 \times 10^{-4} \times 0.0025}{40 \times 10^{-8}}$
$v_{2} = \frac{2 \times 10^{-6}}{40 \times 10^{-8}} = \frac{200}{40} = 5 \,m \,s^{-1}$.
Thus, the speed with which the liquid is ejected is $5 \,m \,s^{-1}$.

(A) The rise or fall of liquids in narrow tubes or porous materials is known as capillarity.
Plant fibres act as a network of fine capillary tubes.
Therefore,water rises in plant fibres due to the phenomenon of capillarity.

(A) The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$.
The formula for horizontal range is $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$.
According to the problem,$H = \frac{R}{2}$.
Substituting the formulas: $\frac{u^2 \sin^2 \theta}{2g} = \frac{1}{2} \left( \frac{2u^2 \sin \theta \cos \theta}{g} \right)$.
Simplifying the equation: $\frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin \theta \cos \theta}{g}$.
Dividing both sides by $\frac{u^2 \sin \theta}{g}$ (assuming $\sin \theta \neq 0$): $\frac{\sin \theta}{2} = \cos \theta$.
Therefore,$\tan \theta = 2$,which gives $\theta = \tan^{-1}(2)$.

(A) Let the temperature of the junction $C$ be $\theta$.
Since the rods are in series,the rate of heat flow through both rods must be equal in the steady state.
The rate of heat flow is given by $H = \frac{KA(\Delta T)}{L}$.
Since the rods are identical,$A$ and $L$ are the same for both.
Thus,$K_1(100 - \theta) = K_2(\theta - 25)$.
Rearranging,we get $\frac{K_1}{K_2} = \frac{\theta - 25}{100 - \theta}$.
Given $\frac{K_1}{K_2} = \frac{2}{3}$,we have $\frac{2}{3} = \frac{\theta - 25}{100 - \theta}$.
Cross-multiplying gives $2(100 - \theta) = 3(\theta - 25)$.
$200 - 2\theta = 3\theta - 75$.
$5\theta = 275$.
$\theta = 55^{\circ}C$.

(C) According to Wien's displacement law,the product of the wavelength of maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ is constant:
$\lambda_{m1} T_1 = \lambda_{m2} T_2$
Given:
$\lambda_{m1} = 500 \,nm$,$T_1 = 6000 \,K$
$\lambda_{m2} = 400 \,nm$,$T_2 = ?$
Substituting the values into the equation:
$500 \,nm \times 6000 \,K = 400 \,nm \times T_2$
$T_2 = \frac{500 \times 6000}{400}$
$T_2 = 5 \times 1500$
$T_2 = 7500 \,K$
Therefore,the temperature of the star is $7500 \,K$.

(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink in Kelvin.
Given: $T_1 = 127^{\circ} C = 127 + 273 = 400 \ K$ and $T_2 = 27^{\circ} C = 27 + 273 = 300 \ K$.
Substituting the values: $\eta = 1 - \frac{300}{400} = 1 - 0.75 = 0.25$ or $\frac{1}{4}$.
Efficiency is also defined as $\eta = \frac{W}{Q_1}$,where $W$ is the work done and $Q_1$ is the heat supplied.
Given $Q_1 = 40 \ kJ$,we have $\frac{1}{4} = \frac{W}{40 \ kJ}$.
Therefore,$W = \frac{40}{4} \ kJ = 10 \ kJ$.

(B) The equation for an adiabatic process is given by $P V^{\gamma} = \text{constant}$.
Given that the cube of the pressure is inversely proportional to the fourth power of the volume, we have $P^3 \propto V^{-4}$, which implies $P^3 V^4 = \text{constant}$.
Taking the cube root of both sides, we get $(P^3 V^4)^{1/3} = \text{constant}^{1/3}$, which simplifies to $P V^{4/3} = \text{constant}$.
Comparing this with the standard adiabatic equation $P V^{\gamma} = \text{constant}$, we identify $\gamma = 4/3$.
Thus, the ratio of specific heats is $\gamma = 1.33$.

(B) At constant pressure,the heat required is given by $Q_p = n C_p \Delta T$.
Given $n = 2 \ moles$,$\Delta T = 35^{\circ} C - 25^{\circ} C = 10 \ K$,and $Q_p = 310 \ J$.
$310 = 2 \times C_p \times 10 \Rightarrow C_p = \frac{310}{20} = 15.5 \ J \ mol^{-1} K^{-1}$.
Using the relation $C_p - C_V = R$,where $R \approx 8.3 \ J \ mol^{-1} K^{-1}$:
$C_V = C_p - R = 15.5 - 8.3 = 7.2 \ J \ mol^{-1} K^{-1}$.
At constant volume,the heat required is $Q_V = n C_V \Delta T$.
$Q_V = 2 \times 7.2 \times 10 = 144 \ J$.

(A) Impulse is defined as the product of force and time interval.
Impulse = Force $\times$ Time
Dimensional formula of Force = $[MLT^{-2}]$
Dimensional formula of Time = $[T]$
Therefore,the dimensional formula of impulse = $[MLT^{-2}] \times [T] = [MLT^{-1}]$.

(A) The equation of a simple harmonic wave is given by $y = a \sin \frac{2 \pi}{\lambda} (vt - x)$.
By differentiating the displacement $y$ with respect to time $t$,we obtain the particle velocity:
$v_p = \frac{dy}{dt} = a \cdot \frac{2 \pi v}{\lambda} \cos \frac{2 \pi}{\lambda} (vt - x)$.
The maximum particle velocity $(v_{p, \max})$ occurs when the cosine term is $1$:
$v_{p, \max} = \frac{2 \pi v a}{\lambda}$.
According to the problem,the maximum particle velocity is half the wave velocity $(v)$:
$v_{p, \max} = \frac{v}{2}$.
Equating the two expressions for $v_{p, \max}$:
$\frac{v}{2} = \frac{2 \pi v a}{\lambda}$.
Solving for the amplitude $a$:
$a = \frac{\lambda}{4 \pi}$.

(C) The problem involves two steps of the Doppler effect.
First,the wall acts as an observer receiving sound from the moving engine (source). The frequency received by the wall is $f_1 = f \left( \frac{v}{v - v_s} \right)$,where $f = 1.2 \ kHz$,$v = 350 \ m/s$,and $v_s = 50 \ m/s$.
$f_1 = 1.2 \left( \frac{350}{350 - 50} \right) = 1.2 \left( \frac{350}{300} \right) = 1.4 \ kHz$.
Second,the wall acts as a stationary source reflecting this frequency $f_1$ to the driver (observer) moving towards the wall with velocity $v_o = 50 \ m/s$.
The frequency heard by the driver is $f' = f_1 \left( \frac{v + v_o}{v} \right)$.
$f' = 1.4 \left( \frac{350 + 50}{350} \right) = 1.4 \left( \frac{400}{350} \right) = 1.4 \times \frac{8}{7} = 1.6 \ kHz$.

(A) For an open organ pipe of length $l$,the fundamental frequency is given by $f = \frac{v}{2l}$,where $v$ is the speed of sound in air.
When the tube is placed vertically in water such that half its length is submerged,it acts as a closed organ pipe of length $l' = \frac{l}{2}$.
The fundamental frequency of a closed organ pipe is given by $f' = \frac{v}{4l'}$.
Substituting $l' = \frac{l}{2}$ into the expression for $f'$,we get:
$f' = \frac{v}{4(l/2)} = \frac{v}{2l}$.
Comparing this with the initial frequency $f = \frac{v}{2l}$,we find that $f' = f$.

(C) The velocity of sound in an ideal gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$,where $\gamma$ is the adiabatic index,$R$ is the gas constant,$T$ is the temperature,and $M$ is the molar mass.
Since the temperature $T$ is the same for both gases,the ratio of the velocities is $\frac{v_{H_2}}{v_{He}} = \sqrt{\frac{\gamma_{H_2} M_{He}}{\gamma_{He} M_{H_2}}}$.
For hydrogen $(H_2)$,$M_{H_2} = 2 \times 10^{-3} \ kg/mol$ and $\gamma_{H_2} = \frac{7}{5}$.
For helium $(He)$,$M_{He} = 4 \times 10^{-3} \ kg/mol$ and $\gamma_{He} = \frac{5}{3}$.
Substituting these values: $\frac{v_{H_2}}{v_{He}} = \sqrt{\frac{(7/5) \times 4}{(5/3) \times 2}} = \sqrt{\frac{7}{5} \times \frac{3}{5} \times \frac{4}{2}} = \sqrt{\frac{7 \times 3 \times 2}{5 \times 5}} = \sqrt{\frac{42}{25}} = \frac{\sqrt{42}}{5}$.

(B) The phase angle $\theta$ between current $I$ and voltage $V$ in an $LCR$ series circuit is given by the formula:
$\tan \theta = \frac{X_L - X_C}{R}$
First,calculate the inductive reactance $X_L$:
$X_L = 2 \pi f L = 2 \pi \times 50 \times \left( \frac{200}{\pi} \times 10^{-3} \right) = 20 \, \Omega$
Next,calculate the capacitive reactance $X_C$:
$X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \times 50 \times (10^{-3} / \pi)} = \frac{1}{0.1} = 10 \, \Omega$
Given resistance $R = 10 \, \Omega$.
Substituting these values into the phase angle formula:
$\tan \theta = \frac{20 - 10}{10} = \frac{10}{10} = 1$
Since $\tan \theta = 1$,we have $\theta = \tan^{-1}(1) = \frac{\pi}{4}$ radians.
Thus,the phase angle of the circuit is $\frac{\pi}{4}$.

(C) Efficiency of a transformer is defined as the ratio of output power to input power: $\eta = \frac{P_{out}}{P_{in}}$.
Given, $\eta = 88 \% = 0.88$ and $P_{out} = 880 \,W$.
Substituting these values: $0.88 = \frac{880}{P_{in}}$.
Therefore, $P_{in} = \frac{880}{0.88} = 1000 \,W$.
The input power is also given by $P_{in} = V_{in} \times I_{in}$, where $V_{in} = 2200 \,V$.
Thus, $I_{in} = \frac{P_{in}}{V_{in}} = \frac{1000}{2200} \,A$.
$I_{in} = \frac{10}{22} \,A \approx 0.4545 \,A$.
Rounding to two decimal places, the input current is $0.45 \,A$.

(A) The number of spectral lines emitted when an electron transitions from energy level $n$ to lower levels is given by $N = \frac{n(n-1)}{2}$.
For the first case,$N = 3$:
$3 = \frac{n_{1}(n_{1}-1)}{2} \Rightarrow n_{1}^2 - n_{1} - 6 = 0 \Rightarrow (n_{1}-3)(n_{1}+2) = 0$.
Since $n_{1} > 0$,we have $n_{1} = 3$.
For the second case,$N = 6$:
$6 = \frac{n_{2}(n_{2}-1)}{2} \Rightarrow n_{2}^2 - n_{2} - 12 = 0 \Rightarrow (n_{2}-4)(n_{2}+3) = 0$.
Since $n_{2} > 0$,we have $n_{2} = 4$.
The orbital speed of an electron in the $n$-th orbit is given by $v_n = \frac{Ze^2}{2\varepsilon_0 hn}$,which implies $v \propto \frac{1}{n}$.
Therefore,the ratio of the speeds is $\frac{v_1}{v_2} = \frac{n_2}{n_1} = \frac{4}{3}$.

(D) In the Raman effect, when light is scattered by molecules, the scattered light contains frequencies different from the incident frequency.
Stokes' lines are the spectral lines observed in the scattered light that have a frequency lower than the incident (original) frequency.
Since frequency $(f)$ and wavelength $(\lambda)$ are inversely related by the equation $c = f\lambda$, a lower frequency corresponds to a longer (greater) wavelength.
Therefore, Stokes' lines have a wavelength greater than that of the original line.

(A) The ionization energy of a hydrogen-like ion is given by the formula $E = R c h Z^2$,where $R$ is the Rydberg constant,$c$ is the speed of light,$h$ is Planck's constant,and $Z$ is the atomic number.
For the lithium ion $Li^{2+}$,the atomic number $Z = 3$.
Substituting the value of $Z$ into the formula:
$E = R c h (3)^2 = 9 R c h$.
Therefore,the ionization energy of $Li^{2+}$ is $9 h c R$.

(A) The original capacitance of the air-filled parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$.
When a dielectric slab of constant $K$ is introduced to fill one-fourth of the area,the capacitor can be considered as two capacitors in parallel.
One capacitor has air as the dielectric with area $\frac{3A}{4}$ and plate separation $d$. Its capacitance is $C_1 = \frac{\varepsilon_0 (3A/4)}{d} = \frac{3}{4} \frac{\varepsilon_0 A}{d} = \frac{3C}{4}$.
The other capacitor has a dielectric $K$ with area $\frac{A}{4}$ and plate separation $d$. Its capacitance is $C_2 = \frac{K \varepsilon_0 (A/4)}{d} = \frac{K}{4} \frac{\varepsilon_0 A}{d} = \frac{KC}{4}$.
Since these two capacitors are in parallel,the equivalent capacitance is $C_{net} = C_1 + C_2$.
$C_{net} = \frac{3C}{4} + \frac{KC}{4} = \frac{C}{4}(K+3)$.

(D) The loss of energy when two capacitors are connected in parallel is given by the formula: $\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$.
Given: $C_1 = C_2 = 5 \mu F = 5 \times 10^{-6} \ F$,$V_1 = 2 \ kV = 2000 \ V$,and $V_2 = 1 \ kV = 1000 \ V$.
Substituting the values:
$\Delta U = \frac{1}{2} \times \frac{(5 \times 10^{-6}) \times (5 \times 10^{-6})}{5 \times 10^{-6} + 5 \times 10^{-6}} \times (2000 - 1000)^2$
$\Delta U = \frac{1}{2} \times \frac{25 \times 10^{-12}}{10 \times 10^{-6}} \times (1000)^2$
$\Delta U = \frac{1}{2} \times 2.5 \times 10^{-6} \times 10^6$
$\Delta U = 1.25 \ J$.

(B) In the given circuit, the two $2 \Omega$ resistors are connected in parallel.
Their equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{2} + \frac{1}{2} = 1 \Omega^{-1}$, so $R_p = 1 \Omega$.
An ideal ammeter has zero resistance and is connected in parallel with the $4 \Omega$ resistor. This effectively short-circuits the $4 \Omega$ resistor, meaning no current flows through it.
The total resistance of the circuit is the sum of the internal resistance of the battery $(1 \Omega)$, the series resistor $(2 \Omega)$, and the parallel combination $(1 \Omega)$.
$R_{net} = 1 \Omega + 2 \Omega + 1 \Omega = 4 \Omega$.
The total current $I$ flowing from the battery is $I = \frac{V}{R_{net}} = \frac{4 \text{ V}}{4 \Omega} = 1 \text{ A}$.
Since the ammeter is in parallel with the $4 \Omega$ resistor, all the current $I$ passes through the ideal ammeter.

(B) The formula for drift velocity is given by $v_{d} = \frac{I}{n e A}$.
Here, $I = 5 \,A$, $n = 5 \times 10^{26} \,m^{-3}$, $e = 1.6 \times 10^{-19} \,C$, and $A = 4 \times 10^{-6} \,m^{2}$.
Substituting the values:
$v_{d} = \frac{5}{(5 \times 10^{26}) \times (1.6 \times 10^{-19}) \times (4 \times 10^{-6})}$
$v_{d} = \frac{5}{5 \times 1.6 \times 4 \times 10^{26-19-6}}$
$v_{d} = \frac{1}{6.4 \times 10^{1}}$
$v_{d} = \frac{1}{64} = 0.015625 \,ms^{-1} = 1.56 \times 10^{-2} \,ms^{-1}$.

(B) The resistance of a bulb is given by $R = \frac{V^2}{P}$.
For the first bulb: $R_1 = \frac{220^2}{25} = 1936 \ \Omega$.
For the second bulb: $R_2 = \frac{220^2}{100} = 484 \ \Omega$.
Since the bulbs are connected in series,the total resistance is $R_{net} = R_1 + R_2 = 1936 + 484 = 2420 \ \Omega$.
The current flowing through the circuit is $I = \frac{V_{supply}}{R_{net}} = \frac{440}{2420} = \frac{2}{11} \ A$.
The potential difference across the $25 \ W$ bulb is $V_1 = I \times R_1 = \frac{2}{11} \times 1936 = 352 \ V$.
The potential difference across the $100 \ W$ bulb is $V_2 = I \times R_2 = \frac{2}{11} \times 484 = 88 \ V$.
Since the potential difference across the $25 \ W$ bulb $(352 \ V)$ exceeds its rated voltage $(220 \ V)$,the $25 \ W$ bulb will fuse.

(A) The balanced condition for a Wheatstone bridge is $\frac{P}{R} = \frac{Q}{S}$.
Given values: $P=10 \Omega, Q=20 \Omega, R=15 \Omega, S=30 \Omega$.
Checking the ratio: $\frac{P}{R} = \frac{10}{15} = \frac{2}{3}$ and $\frac{Q}{S} = \frac{20}{30} = \frac{2}{3}$.
Since $\frac{P}{R} = \frac{Q}{S}$,the bridge is balanced,and no current flows through the galvanometer.
Now,the circuit simplifies to two parallel branches: one with $(P+R)$ and the other with $(Q+S)$.
Resistance of the first branch,$R_1 = P + R = 10 + 15 = 25 \Omega$.
Resistance of the second branch,$R_2 = Q + S = 20 + 30 = 50 \Omega$.
Since $R_1$ and $R_2$ are in parallel,the equivalent resistance $R_{eq}$ is given by:
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{25} + \frac{1}{50} = \frac{2+1}{50} = \frac{3}{50} \Omega^{-1}$.
Therefore,$R_{eq} = \frac{50}{3} \Omega$.
The current $I$ from the battery of voltage $V = 6 \text{ V}$ is:
$I = \frac{V}{R_{eq}} = \frac{6}{50/3} = \frac{6 \times 3}{50} = \frac{18}{50} = 0.36 \text{ A}$.

(B) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K = qV$ is the kinetic energy.
Substituting the values: $h = 6.63 \times 10^{-34} \ J \cdot s$,$m = 1.67 \times 10^{-27} \ kg$,$q = 1.6 \times 10^{-19} \ C$,and $V = 1000 \ V$.
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.6 \times 10^{-19} \times 1000}}$
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{5.344 \times 10^{-43}}} = \frac{6.63 \times 10^{-34}}{7.31 \times 10^{-22}} \approx 0.9 \times 10^{-12} \ m$.

(D) According to Einstein's photoelectric equation:
$E_{k} = h\nu - \Phi$
where $E_{k}$ is the maximum kinetic energy,$h$ is Planck's constant,$\nu$ is the frequency of incident light,and $\Phi = h\nu_{0}$ is the work function of the metal surface ($\nu_{0}$ is the threshold frequency).
This equation is of the form $y = mx + c$,where:
$y = E_{k}$
$x = \nu$
$m = h$ (slope)
$c = -\Phi$ (y-intercept)
$1$. For $\nu < \nu_{0}$,the energy of the incident photon is less than the work function,so no photoelectric emission occurs,and $E_{k} = 0$.
$2$. For $\nu = \nu_{0}$,$E_{k} = 0$.
$3$. For $\nu > \nu_{0}$,$E_{k}$ increases linearly with frequency $\nu$.
Graph $D$ correctly shows that $E_{k}$ is zero until the threshold frequency $\nu_{0}$ is reached,after which it increases linearly.

(B) The threshold energy (work function) is given by $\Phi = h \nu_0$. Given $h = 6.6 \times 10^{-34} \ J \cdot s$ and $1 \ eV = 1.6 \times 10^{-19} \ J$.
For metal $A$:
$\Phi_A = h \nu_A = (6.6 \times 10^{-34}) \times (1.8 \times 10^{14}) \ J = 11.88 \times 10^{-20} \ J$.
Converting to $eV$: $\Phi_A = \frac{11.88 \times 10^{-20}}{1.6 \times 10^{-19}} \ eV = 0.7425 \ eV$.
For metal $B$:
$\Phi_B = h \nu_B = (6.6 \times 10^{-34}) \times (2.2 \times 10^{14}) \ J = 14.52 \times 10^{-20} \ J$.
Converting to $eV$: $\Phi_B = \frac{14.52 \times 10^{-20}}{1.6 \times 10^{-19}} \ eV = 0.9075 \ eV$.
The energy of incident photons is $E = 0.825 \ eV$.
Since $E > \Phi_A$ $(0.825 \ eV > 0.7425 \ eV)$ and $E < \Phi_B$ $(0.825 \ eV < 0.9075 \ eV)$,photoelectrons are emitted only from metal $A$.

(A) The classical electromagnetic theory of light treats light as a continuous wave. According to this theory,the energy of the wave depends on its intensity (amplitude). However,experimental observations of the photoelectric effect show that the emission of electrons depends on the frequency of the incident light,not its intensity. Furthermore,the emission is instantaneous,which contradicts the classical prediction of a time lag for energy absorption. Therefore,the electromagnetic theory of light failed to explain the photoelectric effect.

(C) The resistance of the bulb is given by $R = \frac{V^2}{P} = \frac{100^2}{50} = 200 \, \Omega$.
For the bulb to operate at its rated power, the current $I$ must be $I = \frac{P}{V} = \frac{50}{100} = 0.5 \, A$.
When connected to a $200 \, V$ $AC$ source, the total impedance $Z$ of the circuit is $Z = \frac{V_{source}}{I} = \frac{200}{0.5} = 400 \, \Omega$.
The impedance of an $RL$ circuit is $Z = \sqrt{R^2 + X_L^2}$, where $X_L = 2\pi fL$.
Substituting the values: $400 = \sqrt{200^2 + X_L^2}$.
$160000 = 40000 + X_L^2 \implies X_L^2 = 120000$.
$X_L = \sqrt{120000} = 200\sqrt{3} \, \Omega$.
Since $X_L = 2\pi fL$, we have $200\sqrt{3} = 2 \times \pi \times 50 \times L$.
$L = \frac{200\sqrt{3}}{100\pi} = \frac{2\sqrt{3}}{\pi} \approx \frac{2 \times 1.732}{3.14} \approx 1.1 \, H$.

(C) The magnitude of the induced emf in a coil is given by the formula: $|e| = L \left| \frac{dI}{dt} \right|$.
Given:
Initial current $I_1 = 4 \,A$.
Final current $I_2 = 0 \,A$.
Change in current $dI = I_2 - I_1 = 0 - 4 = -4 \,A$.
Time interval $dt = 0.1 \,s$.
Induced emf $|e| = 100 \,V$.
Substituting these values into the formula:
$100 = L \times \frac{|-4|}{0.1}$.
$100 = L \times \frac{4}{0.1}$.
$100 = L \times 40$.
$L = \frac{100}{40} = 2.5 \,H$.
Therefore,the self-inductance of the coil is $2.5 \,H$.

(B) In a vacuum,all components of the electromagnetic spectrum travel with the same speed,which is the speed of light,denoted by $c$.
This value is approximately $3 \times 10^{8} \ m/s$.
While their frequencies and wavelengths differ,their velocity remains constant in a vacuum.

(D) The force of repulsion between two charges is given by Coulomb's law: $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$.
Given: $F = 10 \text{ mg wt} = 10 \times 10^{-3} \text{ kg} \times 10 \text{ ms}^{-2} = 0.1 \text{ N}$.
Distance $r = 0.6 \text{ m}$.
Since the charges are identical,let $q_{1} = q_{2} = q$.
Substituting the values: $0.1 = (9 \times 10^{9}) \times \frac{q^{2}}{(0.6)^{2}}$.
$q^{2} = \frac{0.1 \times 0.36}{9 \times 10^{9}} = \frac{0.036}{9 \times 10^{9}} = 0.004 \times 10^{-9} = 4 \times 10^{-12} \text{ C}^{2}$.
Taking the square root: $q = \sqrt{4 \times 10^{-12}} = 2 \times 10^{-6} \text{ C} = 2 \mu\text{C}$.

(A) The electric field intensity $\vec{E}$ is related to the electric potential $V$ by the relation $\vec{E} = -\nabla V$.
Since the potential $V$ depends only on $x$,the electric field is given by $E_x = -\frac{dV}{dx}$.
Given $V = 3x^2 + 5$.
Calculating the derivative: $\frac{dV}{dx} = \frac{d}{dx}(3x^2 + 5) = 6x$.
Therefore,$E_x = -6x$.
At the point $(-2, 1, 0)$,the $x$-coordinate is $-2$.
Substituting $x = -2$ into the expression for $E_x$:
$E_x = -6(-2) = +12 \ Vm^{-1}$.
Thus,the intensity of the electric field at $(-2, 1, 0)$ is $+12 \ Vm^{-1}$.

(C) Let the radius of each small drop be $r$ and its charge be $q$. The potential of a small drop is $V' = \frac{kq}{r}$.
When $8$ drops combine to form a large drop of radius $R$ and charge $Q$, the volume remains constant: $\frac{4}{3} \pi R^3 = 8 \times \frac{4}{3} \pi r^3$, which gives $R = 2r$.
The total charge of the large drop is $Q = 8q$.
The potential of the large drop is $V = \frac{kQ}{R} = \frac{k(8q)}{2r} = 4 \left( \frac{kq}{r} \right) = 4V'$.
Given $V = 20 \,V$, we have $20 = 4V'$.
Therefore, $V' = \frac{20}{4} = 5 \,V$.

(D) The force per unit length between two parallel conductors carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi r}$. The force is attractive if currents are in the same direction.
Force on $C$ due to $A$ $(F_{AC})$:
$F_{AC} = \frac{\mu_0 I_A I_C L}{2 \pi r_{AC}} = \frac{2 \times 10^{-7} \times 2 \times 3 \times 1}{0.05} = \frac{12 \times 10^{-7}}{0.05} = 2.4 \times 10^{-5} \, N$ (towards $A$, attractive).
Force on $C$ due to $B$ $(F_{BC})$:
$F_{BC} = \frac{\mu_0 I_B I_C L}{2 \pi r_{BC}} = \frac{2 \times 10^{-7} \times 4 \times 3 \times 1}{0.08} = \frac{24 \times 10^{-7}}{0.08} = 3.0 \times 10^{-5} \, N$ (towards $B$, attractive).
Since $F_{BC} > F_{AC}$, the net force is $F_{net} = F_{BC} - F_{AC} = (3.0 - 2.4) \times 10^{-5} \, N = 0.6 \times 10^{-5} \, N$ towards $B$.

(B) The magnetic field $B$ at the centre of a circular coil with $N$ turns,radius $r$,and current $I$ is given by $B = \frac{\mu_{0} N I}{2 r}$.
Initially,for $N=1$ turn,the magnetic field is $B_{0} = \frac{\mu_{0} I}{2 r}$.
When the coil is rewound to have $N' = 3$ turns using the same length of wire,the new radius $r'$ becomes $r' = \frac{r}{3}$.
The new magnetic field $B'$ at the centre is $B' = \frac{\mu_{0} N' I}{2 r'} = \frac{\mu_{0} (3) I}{2 (r/3)}$.
Simplifying this,we get $B' = \frac{9 \mu_{0} I}{2 r} = 9 B_{0}$.

(B) For a charged particle moving in a perpendicular magnetic field,the magnetic force provides the centripetal force: $\frac{mv^2}{r} = Bqv$.
This simplifies to the radius formula: $r = \frac{mv}{Bq} = \frac{p}{Bq}$,where $p$ is the momentum.
Since the kinetic energy $E$ is the same for both,we use the relation $p = \sqrt{2mE}$.
Substituting this into the radius formula: $r = \frac{\sqrt{2mE}}{Bq}$.
For a proton $(p)$ and a deuteron $(d)$,the charges are equal $(q_p = q_d = e)$ and the kinetic energies are equal $(E_p = E_d = E)$.
The ratio of the radii is $\frac{r_p}{r_d} = \frac{\sqrt{2m_p E} / (Be)}{\sqrt{2m_d E} / (Be)} = \sqrt{\frac{m_p}{m_d}}$.
Given that the mass of a deuteron is twice the mass of a proton $(m_d = 2m_p)$,we have $\frac{r_p}{r_d} = \sqrt{\frac{m_p}{2m_p}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

(A) When a charged particle moves perpendicular to a magnetic field,the magnetic Lorentz force provides the necessary centripetal force for circular motion.
$F_m = F_c$
$Bqv = \frac{mv^2}{r}$
Since the angular velocity $\omega = \frac{v}{r}$,we can write $Bq = m\omega$.
Substituting $\omega = 2\pi f$,we get $Bq = m(2\pi f)$.
Solving for frequency $f$,we obtain $f = \frac{Bq}{2\pi m}$.

(A) The radius of a nucleus is given by $R = R_{0} A^{1/3}$,where $R_{0} \approx 1.2 \times 10^{-15} \text{ m}$.
The volume of a nucleus $(V)$ is given by the formula for the volume of a sphere: $V = \frac{4}{3} \pi R^{3}$.
Substituting the expression for $R$ into the volume formula:
$V = \frac{4}{3} \pi (R_{0} A^{1/3})^{3}$
$V = \frac{4}{3} \pi R_{0}^{3} A$
Since $\frac{4}{3}$,$\pi$,and $R_{0}^{3}$ are constants,it follows that $V \propto A$.
Therefore,the volume of a nucleus is directly proportional to the mass number $A$.

(D) Leptons are a class of elementary particles that do not participate in strong nuclear interactions. They interact primarily through electromagnetic and weak nuclear forces.
An electron is a fundamental particle belonging to the lepton family.

(B) For a radioactive element to form its own isotope,the atomic number $(Z)$ must remain the same,while the mass number $(A)$ changes.
An $\alpha$-particle emission decreases the atomic number by $2$ and the mass number by $4$ $(_{2}He^{4})$.
$A$ $\beta$-particle emission increases the atomic number by $1$ and keeps the mass number unchanged $(_{-1}\beta^{0})$.
To keep the atomic number $Z$ constant after $3$ disintegrations,we need to balance the change in $Z$: $\Delta Z = (n_{\alpha} \times -2) + (n_{\beta} \times 1) = 0$.
Given $n_{\alpha} + n_{\beta} = 3$,substituting $n_{\alpha} = 1$ gives $(-2) + (2) = 0$.
Therefore,the emission of $1$ $\alpha$-particle and $2$ $\beta$-particles results in the same atomic number,forming an isotope.

(A) The law of radioactive decay is given by the formula: $N = N_{0} \left(\frac{1}{2}\right)^{t/T}$.
Here,the initial number of nuclei $N_{0} = 10000$,the half-life $T = 20 \text{ days}$,and the time elapsed $t = 10 \text{ days}$.
Substituting these values into the formula:
$N = 10000 \times \left(\frac{1}{2}\right)^{10/20}$
$N = 10000 \times \left(\frac{1}{2}\right)^{1/2}$
$N = \frac{10000}{\sqrt{2}}$
Using $\sqrt{2} \approx 1.414$:
$N = \frac{10000}{1.414} \approx 7072.13$
Rounding to the nearest whole number,we get $N \approx 7070$.

(A) The equivalent focal length $(F)$ of two thin lenses separated by a distance $d$ is given by the formula:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$
Given $f_1 = 0.2 \,m$,$f_2 = 0.2 \,m$,and $d = 0.5 \,m$.
Substituting these values into the formula:
$\frac{1}{F} = \frac{1}{0.2} + \frac{1}{0.2} - \frac{0.5}{(0.2)(0.2)}$
$\frac{1}{F} = 5 + 5 - \frac{0.5}{0.04}$
$\frac{1}{F} = 10 - 12.5$
$\frac{1}{F} = -2.5$
Therefore,$F = -\frac{1}{2.5} = -0.4 \,m$.

(C) The condition for the two headlights to be just resolved is given by the Rayleigh criterion: $\theta = 1.22 \frac{\lambda}{D}$, where $\theta$ is the angular resolution, $\lambda$ is the wavelength of light, and $D$ is the diameter of the pupil.
Also, $\theta = \frac{d}{x}$, where $d$ is the separation between the headlights and $x$ is the distance of the jeep from the observer.
Equating the two expressions: $\frac{d}{x} = 1.22 \frac{\lambda}{D} \Rightarrow x = \frac{d \times D}{1.22 \times \lambda}$.
Given: $d = 1.2 \,m$, $D = 2 \,mm = 2 \times 10^{-3} \,m$, $\lambda = 5896 \,\text{Å} = 5896 \times 10^{-10} \,m$.
Substituting the values: $x = \frac{1.2 \times 2 \times 10^{-3}}{1.22 \times 5896 \times 10^{-10}} \approx 3336 \,m$.
Converting to kilometers: $x \approx 3.34 \,km$.

(B) ray of light incident normally on one face of a right-angled isosceles prism passes undeviated into the prism.
When it reaches the hypotenuse,the angle of incidence $i$ is $45^{\circ}$.
Since the ray grazes the hypotenuse,the angle of refraction is $90^{\circ}$.
Thus,the angle of incidence is equal to the critical angle $C$,so $C = 45^{\circ}$.
The refractive index $\mu$ is given by $\mu = \frac{1}{\sin C}$.
Substituting the value of $C$,we get $\mu = \frac{1}{\sin 45^{\circ}} = \frac{1}{1/\sqrt{2}} = \sqrt{2}$.
Therefore,$\mu \approx 1.414$.

(C) The dispersive power $\omega$ of a prism is given by $\omega = \frac{\delta_{B} - \delta_{R}}{\delta_{y}}$,where $\delta_{B}$ and $\delta_{R}$ are the deviations for blue and red rays,and $\delta_{y}$ is the mean deviation,calculated as $\delta_{y} = \frac{\delta_{B} + \delta_{R}}{2}$.
For the first prism:
$\delta_{B1} = 12^{\circ}$,$\delta_{R1} = 8^{\circ}$.
$\delta_{y1} = \frac{12^{\circ} + 8^{\circ}}{2} = 10^{\circ}$.
$\omega_{1} = \frac{12^{\circ} - 8^{\circ}}{10^{\circ}} = \frac{4}{10} = \frac{2}{5}$.
For the second prism:
$\delta_{B2} = 14^{\circ}$,$\delta_{R2} = 10^{\circ}$.
$\delta_{y2} = \frac{14^{\circ} + 10^{\circ}}{2} = 12^{\circ}$.
$\omega_{2} = \frac{14^{\circ} - 10^{\circ}}{12^{\circ}} = \frac{4}{12} = \frac{1}{3}$.
The ratio of dispersive powers is $\frac{\omega_{1}}{\omega_{2}} = \frac{2/5}{1/3} = \frac{2}{5} \times 3 = \frac{6}{5}$.

(B) The refractive index of glass with respect to air is $\mu = 1.5$.
The critical angle $C$ is given by $\sin C = \frac{1}{\mu} = \frac{1}{1.5} = \frac{2}{3} \approx 0.667$.
$C = \sin^{-1}(0.667) \approx 41.8^{\circ} \approx 42^{\circ}$.
Since the angle of incidence $i = 50^{\circ}$ is greater than the critical angle $C = 42^{\circ}$,the light ray undergoes total internal reflection.
In total internal reflection,the angle of reflection $r$ is equal to the angle of incidence $i$,so $r = 50^{\circ}$.
The deviation $\delta$ is the angle between the original path of the incident ray and the reflected ray.
From the geometry of the reflection,the deviation is $\delta = 180^{\circ} - (i + r) = 180^{\circ} - (50^{\circ} + 50^{\circ}) = 180^{\circ} - 100^{\circ} = 80^{\circ}$.

(B) The apparent depth of an object in a medium is given by the formula: $\text{Apparent depth} = \frac{\text{Real depth}}{\text{Refractive index}}$.
The vessel has a total height of $2d$. It is half-filled with two immiscible liquids,so the real depth of each liquid is $d$.
For the first liquid with refractive index $\mu_1 = \sqrt{2}$,the apparent depth $x_1$ is:
$x_1 = \frac{d}{\sqrt{2}}$
For the second liquid with refractive index $\mu_2 = n$,the apparent depth $x_2$ is:
$x_2 = \frac{d}{n}$
The total apparent depth of the bottom of the vessel is the sum of the apparent depths of the two layers:
$\text{Total apparent depth} = x_1 + x_2 = \frac{d}{\sqrt{2}} + \frac{d}{n}$
Taking the common denominator:
$\text{Total apparent depth} = \frac{dn + d\sqrt{2}}{n\sqrt{2}} = \frac{d(n + \sqrt{2})}{n\sqrt{2}}$

(B) The reverse bias resistance $(R)$ of a diode is defined as the ratio of the change in voltage $(\Delta V)$ to the change in current $(\Delta I)$.
Given:
Change in voltage, $\Delta V = 1 \, V$
Change in current, $\Delta I = 0.5 \, \mu A = 0.5 \times 10^{-6} \, A$
Using the formula:
$R = \frac{\Delta V}{\Delta I}$
$R = \frac{1}{0.5 \times 10^{-6}} \, \Omega$
$R = \frac{1}{0.5} \times 10^{6} \, \Omega$
$R = 2 \times 10^{6} \, \Omega$
Therefore, the reverse bias resistance is $2 \times 10^{6} \, \Omega$.

(D) The given truth table shows that the output $Y$ is $1$ when either $A$ or $B$ or both are $0$,and the output $Y$ is $0$ only when both inputs $A$ and $B$ are $1$.
This behavior corresponds to the Boolean expression $Y = \overline{A \cdot B}$.
This is the characteristic truth table of a $NAND$ gate,which is an $AND$ gate followed by a $NOT$ gate.

(A) In a $p$-type semiconductor,the density of mobile holes is significantly higher than the density of conduction electrons due to the addition of trivalent impurity atoms.
Therefore,the majority charge carriers are holes,and the minority charge carriers are conduction (free) electrons.

(D) $LASER$ action relies on the following fundamental processes:
$(i)$ Population inversion: Achieving a state where more atoms are in an excited state than in the ground state.
(ii) Stimulated emission: An incident photon triggers an excited atom to emit a second photon identical in phase,frequency,and direction.
(iii) Amplification: The process of increasing the intensity of light by repeating stimulated emission within an optical cavity.
Therefore,all the given options are correct.

(C) The condition for the $n^{th}$ secondary maximum in a single slit diffraction pattern is given by $x_n = \frac{(2n+1) \lambda D}{2a}$,where $n$ is the order of the secondary maximum.
For the second secondary maximum $(n=2)$ of red light $(\lambda_1 = 6500 Å)$: $x_2 = \frac{(2(2)+1) \lambda_1 D}{2a} = \frac{5 \lambda_1 D}{2a}$.
For the third secondary maximum $(n=3)$ of an unknown wavelength $(\lambda_2)$: $x_3 = \frac{(2(3)+1) \lambda_2 D}{2a} = \frac{7 \lambda_2 D}{2a}$.
Since the maxima coincide,$x_2 = x_3$,which implies $\frac{5 \lambda_1 D}{2a} = \frac{7 \lambda_2 D}{2a}$.
Simplifying,we get $5 \lambda_1 = 7 \lambda_2$.
Substituting $\lambda_1 = 6500 Å$,we have $5 \times 6500 = 7 \times \lambda_2$.
$\lambda_2 = \frac{32500}{7} Å \approx 4642.8 Å$.

(B) According to Brewster's Law,when the reflected ray is completely polarized,the angle of incidence is equal to the polarizing angle $(\theta_p)$.
Given,$\theta_p = 60^{\circ}$.
The refractive index of glass $(\mu_g)$ is given by $\mu_g = \tan \theta_p$.
$\mu_g = \tan 60^{\circ} = \sqrt{3}$.
We know that the refractive index is the ratio of the speed of light in vacuum $(c)$ to the speed of light in the medium $(v_g)$:
$\mu_g = \frac{c}{v_g}$.
Substituting the values,$\sqrt{3} = \frac{3 \times 10^8 \text{ m/s}}{v_g}$.
$v_g = \frac{3 \times 10^8}{\sqrt{3}} \text{ m/s} = \sqrt{3} \times 10^8 \text{ m/s}$.

(D) For solution $A$: $L_{1} = 20 \ cm$,$\theta_{1} = +38^{\circ}$. Let concentration be $C_{1}$. The specific rotation is $\alpha_{1} = \frac{\theta_{1}}{L_{1} C_{1}} = \frac{38^{\circ}}{20 C_{1}}$.
For solution $B$: $L_{2} = 30 \ cm$,$\theta_{2} = -24^{\circ}$ (left-handed). Let concentration be $C_{2}$. The specific rotation is $\alpha_{2} = \frac{\theta_{2}}{L_{2} C_{2}} = \frac{-24^{\circ}}{30 C_{2}}$.
In the mixture,the volume ratio is $1:2$. Thus,the new concentrations are $C_{1}' = \frac{C_{1}}{3}$ and $C_{2}' = \frac{2C_{2}}{3}$.
The total optical rotation $\theta$ for a path length $l = 30 \ cm$ is given by $\theta = (\alpha_{1} C_{1}' + \alpha_{2} C_{2}') l$.
Substituting the values: $\theta = \left( \frac{38^{\circ}}{20 C_{1}} \cdot \frac{C_{1}}{3} + \frac{-24^{\circ}}{30 C_{2}} \cdot \frac{2 C_{2}}{3} \right) \times 30$.
$\theta = \left( \frac{38^{\circ}}{60} - \frac{48^{\circ}}{90} \right) \times 30 = \left( \frac{19^{\circ}}{30} - \frac{16^{\circ}}{30} \right) \times 30 = 19^{\circ} - 16^{\circ} = +3^{\circ}$.
Since the result is positive,it is a right-handed rotation of $3^{\circ}$.

(D) For coherent sources,the intensity at the central maximum is given by $I_{0} = (\sqrt{I_{1}} + \sqrt{I_{2}})^2$. Since both sources have the same amplitude $A$,their individual intensities are equal,say $I_{1} = I_{2} = I$.
Thus,$I_{0} = (\sqrt{I} + \sqrt{I})^2 = (2\sqrt{I})^2 = 4I$.
This implies that the intensity of each individual source is $I = \frac{I_{0}}{4}$.
When the sources are incoherent,the interference term averages to zero over time. Therefore,the resultant intensity is simply the sum of the individual intensities: $I_{res} = I_{1} + I_{2} = I + I = 2I$.
Substituting $I = \frac{I_{0}}{4}$ into the expression,we get $I_{res} = 2 \times (\frac{I_{0}}{4}) = \frac{I_{0}}{2}$.

(A) In Young's double slit experiment,the half angular width $\theta$ of the central maximum is given by the relation $\sin \theta = \frac{\lambda}{d}$.
Given wavelength $\lambda = 589 \ nm = 589 \times 10^{-9} \ m$.
Given slit separation $d = 0.589 \ mm = 0.589 \times 10^{-3} \ m$.
Substituting the values:
$\sin \theta = \frac{589 \times 10^{-9}}{0.589 \times 10^{-3}}$
$\sin \theta = \frac{589 \times 10^{-9}}{589 \times 10^{-6}} = 10^{-3} = 0.001$.
Therefore,$\theta = \sin^{-1}(0.001)$.

(B) The current $I$ through a tangent galvanometer is given by $I = \frac{2r B_H}{\mu_0 N} \tan \theta$, where $r$ is the radius, $N$ is the number of turns, and $\theta$ is the deflection.
Since the galvanometers are connected in parallel to a cell of emf $V$, the potential difference across each is the same $(V_A = V_B = V)$.
Given $V = IR$, we have $I = V/R$. Since $R$ is the same for both $(8 \Omega)$, the currents $I_A$ and $I_B$ are equal.
Thus, $\frac{2 r_A B_H}{\mu_0 N_A} \tan \theta_A = \frac{2 r_B B_H}{\mu_0 N_B} \tan \theta_B$.
Simplifying, we get $\frac{r_A \tan \theta_A}{N_A} = \frac{r_B \tan \theta_B}{N_B}$.
Substituting the given values: $r_A = 8 \text{ cm}$, $r_B = 16 \text{ cm}$, $N_A = 2$, $\theta_A = 30^{\circ}$, $\theta_B = 60^{\circ}$.
$\frac{8 \tan 30^{\circ}}{2} = \frac{16 \tan 60^{\circ}}{N_B}$.
$4 \times \frac{1}{\sqrt{3}} = \frac{16 \times \sqrt{3}}{N_B}$.
$N_B = \frac{16 \times \sqrt{3} \times \sqrt{3}}{4} = \frac{16 \times 3}{4} = 12$ turns.