KCET 2008 Physics Question Paper with Answer and Solution

(A) Let $T$ be the tension in the spring balance. The forces acting on the mass $M$ are the tension $T$ along the string,the gravitational force $mg$ acting vertically downwards,and the horizontal force $F$ applied by the string.
In equilibrium,the vertical component of the tension $T$ must balance the weight of the mass.
$T \cos \theta = mg$
Given $m = 10 \ kg$,$\theta = 60^{\circ}$,and $g$ is the acceleration due to gravity.
The reading of the spring balance corresponds to the tension $T$ in $kg-wt$.
$T = \frac{mg}{\cos 60^{\circ}}$
Since $1 \ kg-wt = 1 \ kg \times g$,the reading in $kg-wt$ is simply $T/g = m / \cos 60^{\circ}$.
$T_{reading} = \frac{10}{\cos 60^{\circ}} = \frac{10}{1/2} = 20 \ kg-wt$.

(C) The distance travelled by a body in the $n^{\text{th}}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.
For the first second $(n=1)$: $5 = u + \frac{a}{2}(2(1) - 1) \implies 5 = u + \frac{a}{2}$ (Eq. $i$).
For the third second $(n=3)$: $2 = u + \frac{a}{2}(2(3) - 1) \implies 2 = u + \frac{5a}{2}$ (Eq. $ii$).
Subtracting Eq. $i$ from Eq. $ii$: $(2 - 5) = (u + \frac{5a}{2}) - (u + \frac{a}{2}) \implies -3 = 2a \implies a = -1.5 \,m/s^2$.
The negative sign indicates deceleration.
The magnitude of the force is $F = |m \times a| = 4 \,kg \times 1.5 \,m/s^2 = 6 \,N$.

(B) The upthrust on a body is given by the weight of the liquid displaced. The ratio of upthrusts is equal to the ratio of the densities (or specific gravities) of the liquids.
$\frac{\text{Upthrust in liquid}}{\text{Upthrust in water}} = \frac{\text{Specific gravity of liquid}}{\text{Specific gravity of water}}$
Upthrust in water $= 50 \,g - 40 \,g = 10 \,g$.
Let the upthrust in the liquid be $x$.
$\frac{x}{10 \,g} = \frac{1.5}{1}$
$x = 15 \,g$.
The weight of the body in the liquid is given by: $\text{Weight in air} - \text{Upthrust in liquid}$.
Weight $= 50 \,g - 15 \,g = 35 \,g$.

(D) Let the body be projected with initial velocity $u$. The equation of motion for a height $h$ is given by $h = ut - \frac{1}{2}gt^2$.
Rearranging this,we get a quadratic equation in $t$: $\frac{1}{2}gt^2 - ut + h = 0$.
The roots of this equation are $t_1$ and $t_2$,which represent the times at which the body reaches height $h$ during ascent and descent.
From the properties of quadratic equations,the sum of the roots is $t_1 + t_2 = \frac{-(-u)}{\frac{1}{2}g} = \frac{2u}{g}$.
Solving for $u$,we get $u = \frac{g(t_1 + t_2)}{2}$.

(B) The time period of a simple pendulum at rest is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
When the lift accelerates upwards with an acceleration $a$,the effective acceleration due to gravity becomes $g_{eff} = g + a$.
The new time period is $T' = 2\pi \sqrt{\frac{l}{g+a}}$.
Given that $T' = T/2$,we have $\frac{T}{2} = 2\pi \sqrt{\frac{l}{g+a}}$.
Dividing the expression for $T'$ by $T$,we get $\frac{1}{2} = \sqrt{\frac{g}{g+a}}$.
Squaring both sides,$\frac{1}{4} = \frac{g}{g+a}$.
This implies $g + a = 4g$,so $a = 3g$.

(D) Let the temperature at the junction be $\theta$. In steady state,the rate of heat flow through the copper rod must be equal to the rate of heat flow through the steel rod.
$H_{Cu} = H_{steel}$
$\frac{K_{Cu} A (100 - \theta)}{L_{Cu}} = \frac{K_{steel} A (\theta - 0)}{L_{steel}}$
Given that $K_{Cu} = 9 K_{steel}$,$L_{Cu} = 18 \text{ cm}$,and $L_{steel} = 6 \text{ cm}$.
Substituting the values:
$\frac{9 K_{steel} (100 - \theta)}{18} = \frac{K_{steel} (\theta - 0)}{6}$
$\frac{100 - \theta}{2} = \theta$
$100 - \theta = 2\theta$
$3\theta = 100$
$\theta = \frac{100}{3} \approx 33.3^{\circ} C$
Thus,the temperature at the junction is approximately $33^{\circ} C$.

(A) According to Stefan-Boltzmann law,the power radiated by a black body is proportional to the fourth power of its absolute temperature: $E \propto T^{4}$.
Given:
Initial temperature $T_{1} = 27^{\circ} C = 27 + 273 = 300 \ K$.
Final temperature $T_{2} = 627^{\circ} C = 627 + 273 = 900 \ K$.
Initial energy emitted $E_{1} = 0.5 \ J/s$.
Using the ratio formula: $\frac{E_{2}}{E_{1}} = \left(\frac{T_{2}}{T_{1}}\right)^{4}$.
Substituting the values: $\frac{E_{2}}{0.5} = \left(\frac{900}{300}\right)^{4}$.
$\frac{E_{2}}{0.5} = (3)^{4} = 81$.
$E_{2} = 81 \times 0.5 = 40.5 \ J/s$.

(B) The net work done in a cyclic process is equal to the area enclosed by the $p-V$ diagram.
For the given triangle $ABC$,the base is $(3 V_{1} - V_{1}) = 2 V_{1}$ and the height is $(4 p_{1} - p_{1}) = 3 p_{1}$.
The magnitude of the area is $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2 V_{1}) \times (3 p_{1}) = 3 p_{1} V_{1}$.
Since the cycle $A \rightarrow B \rightarrow C \rightarrow A$ is traversed in an anticlockwise direction,the work done by the gas is negative.
Therefore,the net work done is $W_{\text{cycle}} = -3 p_{1} V_{1}$.

(D) In an isothermal process,the temperature of the system remains constant. Since the internal energy of an ideal gas is a function of temperature only $(U = f(T))$,the internal energy remains constant in an isothermal process. Any heat supplied to the system is entirely converted into work done by the system,and vice versa.

(C) According to Mayer's relation for an ideal gas:
$C_{p} - C_{V} = R$ $(i)$
where $C_{p}$ is the molar specific heat at constant pressure and $C_{V}$ is the molar specific heat at constant volume.
By definition,the ratio of specific heats is:
$\gamma = \frac{C_{p}}{C_{V}}$
This implies $C_{p} = \gamma C_{V}$ (ii)
Substituting equation (ii) into equation $(i)$:
$\gamma C_{V} - C_{V} = R$
$C_{V}(\gamma - 1) = R$
Therefore,$C_{V} = \frac{R}{\gamma - 1}$

(A) According to Newton's law of gravitation,the force $F$ between two masses $M_1$ and $M_2$ separated by a distance $R$ is given by:
$F = \frac{G M_1 M_2}{R^2}$
Rearranging the formula to solve for the universal gravitational constant $G$:
$G = \frac{F R^2}{M_1 M_2}$
Substituting the dimensional formulas for force $[F] = [MLT^{-2}]$,distance $[R] = [L]$,and mass $[M] = [M]$:
$[G] = \frac{[MLT^{-2}] [L^2]}{[M] [M]} = \frac{[ML^3 T^{-2}]}{[M^2]} = [M^{-1} L^3 T^{-2}]$

(A) The frequency of a vibrating string is given by the formula: $v = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tension,and $m$ is the linear mass density.
From this,we can see that $v \propto \frac{\sqrt{T}}{l}$.
Let the initial state be $v_1 = 200 \ Hz$,$l_1 = l$,and $T_1 = T$. Let the final state be $v_2 = 300 \ Hz$,$l_2 = 2l$,and $T_2 = T'$.
Taking the ratio: $\frac{v_2}{v_1} = \frac{\sqrt{T'}/l_2}{\sqrt{T}/l_1} = \sqrt{\frac{T'}{T}} \cdot \frac{l_1}{l_2}$.
Substituting the values: $\frac{300}{200} = \sqrt{\frac{T'}{T}} \cdot \frac{l}{2l}$.
$1.5 = \sqrt{\frac{T'}{T}} \cdot 0.5$.
$\sqrt{\frac{T'}{T}} = \frac{1.5}{0.5} = 3$.
Squaring both sides,we get $\frac{T'}{T} = 3^2 = 9$.
Therefore,the ratio of the new tension to the original tension is $9: 1$.

(D) The formula for the apparent frequency $v'$ heard by an observer moving towards a stationary source is given by:
$v' = v \left( \frac{v + v_o}{v} \right)$
Given that the apparent frequency $v'$ is double the actual frequency $v$,we have $v' = 2v$.
Substituting this into the equation:
$2v = v \left( \frac{v + v_o}{v} \right)$
$2 = \frac{v + v_o}{v}$
$2v = v + v_o$
$v_o = v$
Therefore,the observer should approach the source with a velocity equal to the velocity of sound,$v$.

(A) The sound intensity level $L$ in decibels $(dB)$ is given by the formula $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity of the sound and $I_0$ is the reference intensity.
For a $60 \ dB$ sound: $60 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) \Rightarrow 6 = \log_{10} \left( \frac{I_1}{I_0} \right) \Rightarrow \frac{I_1}{I_0} = 10^6$.
For a $30 \ dB$ sound: $30 = 10 \log_{10} \left( \frac{I_2}{I_0} \right) \Rightarrow 3 = \log_{10} \left( \frac{I_2}{I_0} \right) \Rightarrow \frac{I_2}{I_0} = 10^3$.
To find how many times more intense the $60 \ dB$ sound is compared to the $30 \ dB$ sound,we calculate the ratio $\frac{I_1}{I_2}$:
$\frac{I_1}{I_2} = \frac{I_1 / I_0}{I_2 / I_0} = \frac{10^6}{10^3} = 10^3 = 1000$.
Thus,a $60 \ dB$ sound is $1000$ times more intense than a $30 \ dB$ sound.

(D) The standard equation of a progressive wave is $y = A \sin(2\pi ft - kx)$.
Comparing the given equation $y = 6 \sin 2\pi(2t - 0.1x)$ with the standard form,we get the wave number $k = 2\pi \times 0.1 = 0.2\pi \ rad/mm$.
The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = k \Delta x$.
Given the separation between particles $\Delta x = 2 \ mm$.
Substituting the values,$\Delta \phi = (0.2\pi) \times 2 = 0.4\pi \ rad$.
To convert radians to degrees,multiply by $\frac{180^{\circ}}{\pi}$:
$\Delta \phi = 0.4\pi \times \frac{180^{\circ}}{\pi} = 0.4 \times 180^{\circ} = 72^{\circ}$.

(C) Let $I$ be the current flowing through the secondary coil.
Input power in the primary coil is $P_{in} = V \times i = 220 \,V \times 5 \,A = 1100 \,W$.
Since $50 \%$ of the power is lost, the output power $P_{out}$ is $50 \%$ of the input power.
$P_{out} = 0.50 \times P_{in} = 0.50 \times 1100 \,W = 550 \,W$.
The output power is also given by $P_{out} = V^{\prime} \times I$, where $V^{\prime} = 2200 \,V$.
Therefore, $2200 \,V \times I = 550 \,W$.
$I = \frac{550}{2200} \,A = 0.25 \,A$.

(C) The wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n^2} - \frac{1}{m^2} \right)$.
For the minimum wavelength (shortest wavelength),the transition occurs from $m = \infty$ to the ground state $n$ of the series.
For the Lyman series,$n = 1$ and $m = \infty$:
$\frac{1}{\lambda_{L}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \implies \lambda_{L} = \frac{1}{R}$.
For the Balmer series,$n = 2$ and $m = \infty$:
$\frac{1}{\lambda_{B}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4} \implies \lambda_{B} = \frac{4}{R}$.
The ratio of the minimum wavelengths is $\frac{\lambda_{L}}{\lambda_{B}} = \frac{1/R}{4/R} = \frac{1}{4} = 0.25$.

(C) The de-Broglie wavelength $\lambda$ is given by the relation $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mKE}}$.
From this,the kinetic energy $KE$ can be expressed as $KE = \frac{h^2}{2m\lambda^2}$.
Since the de-Broglie wavelength $\lambda$ is the same for both the electron and the proton,we have $KE \propto \frac{1}{m}$.
Because the mass of an electron $(m_e \approx 9.11 \times 10^{-31} \ kg)$ is much smaller than the mass of a proton $(m_p \approx 1.67 \times 10^{-27} \ kg)$,the kinetic energy of the electron will be greater than that of the proton.

(D) In a series $LCR$ circuit at resonance,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$ $(X_L = X_C)$.
At resonance,the phase difference $\phi$ between voltage and current is $0^{\circ}$.
$1$. The power factor is $\cos \phi = \cos 0^{\circ} = 1$. Thus,option $D$ is incorrect.
$2$. Average power $P_{avg} = V_{rms} I_{rms} \cos \phi = V_{rms} I_{rms} (1) = V_{rms} I_{rms}$,which is equal to the apparent power. Thus,option $B$ is true.
$3$. Peak energy stored in the capacitor is $\frac{1}{2} C V_0^2$ and in the inductor is $\frac{1}{2} L I_0^2$. At resonance,these values are equal. Thus,option $A$ is true.
$4$. Wattless current is $I_{rms} \sin \phi$. Since $\phi = 0^{\circ}$,$\sin 0^{\circ} = 0$,so wattless current is zero. Thus,option $C$ is true.
Therefore,the statement that is not true is that the power factor is zero.

(C) In a series $LCR$ circuit,the total alternating emf $(V)$ is given by the phasor sum of the individual voltages across the resistor $(V_R)$,inductor $(V_L)$,and capacitor $(V_C)$.
The formula is:
$V = \sqrt{V_R^2 + (V_L - V_C)^2}$
Given values:
$V_R = 80 \ V$
$V_L = 40 \ V$
$V_C = 100 \ V$
Substituting these values into the formula:
$V = \sqrt{(80)^2 + (40 - 100)^2}$
$V = \sqrt{6400 + (-60)^2}$
$V = \sqrt{6400 + 3600}$
$V = \sqrt{10000}$
$V = 100 \ V$
Therefore,the value of the alternating emf is $100 \ V$.

(D) The solar spectrum consists of a continuous spectrum of light emitted by the hot interior of the Sun,which is crossed by numerous dark lines known as Fraunhofer lines. These dark lines are produced because the cooler gases in the Sun's outer atmosphere (the photosphere and chromosphere) absorb specific wavelengths of light from the continuous spectrum. Therefore,the solar spectrum is classified as a line absorption spectrum.

(D) $X$-rays are high-energy electromagnetic radiations produced by transitions between inner electron shells (like $K$-shell to $L$-shell) in atoms with high atomic numbers $(Z)$.
In a hydrogen atom,there is only one electron and the energy difference between its energy levels is very small (in the order of $eV$).
$X$-ray emission requires energy transitions in the order of $keV$.
Therefore,a hydrogen atom cannot emit $X$-rays because the energy levels are too close to each other to produce such high-energy photons.

(D) To achieve a potential difference of $600 \ V$ using capacitors rated at $200 \ V$ each,we must connect $n$ capacitors in series such that $n \times 200 \ V = 600 \ V$. Thus,$n = 3$.
The equivalent capacitance of these $3$ capacitors in series is given by $\frac{1}{C_s} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$,so $C_s = 2 \ \mu F$.
To achieve a total capacitance of $18 \ \mu F$,we need to connect $m$ such series branches in parallel,where $m \times C_s = 18 \ \mu F$.
Thus,$m \times 2 \ \mu F = 18 \ \mu F$,which gives $m = 9$.
The total number of capacitors required is $n \times m = 3 \times 9 = 27$.

(C) The $6 \mu F$ and $3 \mu F$ capacitors are connected in series.
Therefore,the equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2}$.
Thus,$C_1 = 2 \mu F$.
This equivalent capacitor $C_1$ is in parallel with the $2 \mu F$ capacitor.
Therefore,the total equivalent capacitance of the system is $C_{eq} = C_1 + 2 \mu F = 2 \mu F + 2 \mu F = 4 \mu F$.
The total energy $U$ stored in the system is given by $U = \frac{1}{2} C_{eq} V^2$.
Given $V = 2 \text{ V}$,we have $U = \frac{1}{2} \times 4 \mu F \times (2 \text{ V})^2 = 2 \times 4 = 8 \mu J$.

(C) The circuit is a Wheatstone bridge-like structure. Due to symmetry,the potential at the center point is the same as the potential at the midpoint of the diagonals. However,a simpler way to solve this is by using the symmetry of the circuit.
Let the points be $A, B, C, D$. The resistance between $A$ and $B$ is to be found.
By symmetry,the circuit can be simplified. The resistors connected to the center point can be treated as parallel branches.
Alternatively,using the symmetry of the circuit,the equivalent resistance $R_{eq}$ between $A$ and $B$ is given by the parallel combination of the paths.
For this specific symmetric circuit with all resistors equal to $R = 15 \Omega$,the equivalent resistance between any two adjacent corners is $R_{eq} = \frac{2}{3}R$.
$R_{eq} = \frac{2}{3} \times 15 \Omega = 10 \Omega$.

(B) To find the potential difference between $A$ and $B$,we apply Kirchhoff's Voltage Law $(KVL)$ along the path from $A$ to $B$.
Starting from point $A$,the current $I = 2 \ A$ flows through the circuit.
The potential at $A$ is $V_A$.
Moving through the $6 \ \Omega$ resistor in the direction of current,the potential drop is $I \times R = 2 \times 6 = 12 \ V$.
Moving through the $12 \ V$ battery from positive to negative terminal,the potential drop is $12 \ V$.
Moving through the $9 \ \Omega$ resistor,the potential drop is $2 \times 9 = 18 \ V$.
Moving through the $4 \ V$ battery from negative to positive terminal,the potential gain is $4 \ V$.
Moving through the $5 \ \Omega$ resistor,the potential drop is $2 \times 5 = 10 \ V$.
Equating this to the potential at $B$ $(V_B)$:
$V_A - (2 \times 6) - 12 - (2 \times 9) + 4 - (2 \times 5) = V_B$
$V_A - 12 - 12 - 18 + 4 - 10 = V_B$
$V_A - 48 = V_B$
$V_A - V_B = 48 \ V$

(D) The drift velocity $v_d$ is given by the formula $v_d = \frac{eE\tau}{m}$,where $e$ is the charge of the electron,$E$ is the electric field,$\tau$ is the relaxation time,and $m$ is the mass of the electron.
When the temperature of the metal wire increases,the atoms in the metal lattice vibrate with greater amplitude.
This leads to more frequent collisions between the electrons and the lattice ions,which decreases the relaxation time $\tau$.
Since $v_d \propto \tau$,the drift velocity $v_d$ decreases.
Additionally,the thermal velocity of the electrons is proportional to the square root of the absolute temperature $(v_{th} \propto \sqrt{T})$,so as the temperature increases,the thermal velocity of the electrons increases.

(A) The frequency of revolution of the electron is given by $f = \frac{v}{2 \pi r}$.
Substituting the given values: $f = \frac{2.2 \times 10^{6}}{2 \pi (5 \times 10^{-11})} \approx 7.0 \times 10^{15} \, Hz$.
The current $i$ associated with the motion of the electron is $i = qf$, where $q$ is the charge of the electron $(1.6 \times 10^{-19} \, C)$.
$i = (1.6 \times 10^{-19} \, C) \times (7.0 \times 10^{15} \, Hz) = 11.2 \times 10^{-4} \, A$.
Converting to milliamperes: $i = 1.12 \times 10^{-3} \, A = 1.12 \, mA$.

(A) The principle of a tangent galvanometer is given by $I = K \tan \theta$,where $K = \frac{2rB_H}{\mu_0 N}$.
Since the galvanometers are identical except for the number of turns $N$ and are connected in series,the current $I$ flowing through them is the same.
Thus,$I = \frac{2rB_H}{\mu_0 N} \tan \theta$,which implies $N \tan \theta = \text{constant}$.
Therefore,$N_A \tan \theta_A = N_B \tan \theta_B$.
The ratio of the number of turns is $\frac{N_A}{N_B} = \frac{\tan \theta_B}{\tan \theta_A}$.
Given $\theta_A = 60^{\circ}$ and $\theta_B = 30^{\circ}$,we have $\frac{N_A}{N_B} = \frac{\tan 30^{\circ}}{\tan 60^{\circ}}$.
$\frac{N_A}{N_B} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3}$.
Thus,the ratio is $1: 3$.

(B) Let the initial current be $I$ and the galvanometer resistance be $G$. The deflection is proportional to the current, so $I = k \times 100$, where $k$ is a constant.
When a shunt $S = 1 \ \Omega$ is connected in parallel, the new current through the galvanometer $I_g$ is given by the current divider rule: $I_g = I \left( \frac{S}{S+G} \right)$.
The new deflection is $1$ division, so $I_g = k \times 1$.
Substituting the values: $k = (k \times 100) \left( \frac{1}{1+G} \right)$.
Dividing both sides by $k$: $1 = \frac{100}{1+G}$.
$1+G = 100$.
$G = 99 \ \Omega$.

(A) According to Einstein's photoelectric equation: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)$.
For the first case: $3 eV_{s} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)$ --- $(i)$
For the second case: $eV_{s} = hc \left( \frac{1}{2 \lambda} - \frac{1}{\lambda_{0}} \right)$ --- (ii)
Dividing equation $(i)$ by equation (ii):
$\frac{3 eV_{s}}{eV_{s}} = \frac{hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)}{hc \left( \frac{1}{2 \lambda} - \frac{1}{\lambda_{0}} \right)}$
$3 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_{0}}}{\frac{1}{2 \lambda} - \frac{1}{\lambda_{0}}}$
$3 \left( \frac{1}{2 \lambda} - \frac{1}{\lambda_{0}} \right) = \frac{1}{\lambda} - \frac{1}{\lambda_{0}}$
$\frac{3}{2 \lambda} - \frac{3}{\lambda_{0}} = \frac{1}{\lambda} - \frac{1}{\lambda_{0}}$
$\frac{3}{2 \lambda} - \frac{1}{\lambda} = \frac{3}{\lambda_{0}} - \frac{1}{\lambda_{0}}$
$\frac{1}{2 \lambda} = \frac{2}{\lambda_{0}}$
$\lambda_{0} = 4 \lambda$.

(C) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ of emitted electrons is given by $K_{max} = h\nu - \Phi_0$, where $h$ is Planck's constant, $\nu$ is the frequency of incident light, and $\Phi_0$ is the work function of the metal.
Since $\nu = c / \lambda$, the kinetic energy depends on the frequency $\nu$ and the wavelength $\lambda$ of the incident light, as well as the work function $\Phi_0$ of the material.
The intensity of incident light determines the number of photons striking the surface per unit time, which affects the number of emitted electrons (photoelectric current), but it does not affect the maximum kinetic energy of individual emitted electrons.

(A) When two circular loops carrying current in the same direction are moved further apart,the magnetic field produced by one coil at the location of the other decreases.
This results in a decrease in the magnetic flux linked with each coil.
According to Lenz's law,the system opposes this change in flux by inducing an electromotive force $(EMF)$ that acts to increase the current in the coils.
Therefore,the electric current will increase in both coils to maintain the magnetic flux.

(C) The speed of light in vacuum $(c)$ is related to the fundamental electromagnetic constants of free space,namely the permeability of free space $(\mu_{0})$ and the permittivity of free space $(\varepsilon_{0})$.
According to Maxwell's equations,the speed of electromagnetic waves in vacuum is given by the formula:
$c = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$
Thus,the correct expression is $\sqrt{\frac{1}{\mu_{0} \varepsilon_{0}}}$.

(C) The initial electrostatic force between the spheres is given by Coulomb's law: $F_{\text{initial}} = \frac{1}{4 \pi \varepsilon_0} \frac{|q_1 q_2|}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{|12 \times (-8)|}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{96}{r^2}$.
When the spheres are brought into contact,the total charge is redistributed equally because the spheres are identical: $q_{\text{new}} = \frac{q_1 + q_2}{2} = \frac{12 + (-8)}{2} = \frac{4}{2} = 2 \mu C$.
After contact,the new electrostatic force is: $F_{\text{final}} = \frac{1}{4 \pi \varepsilon_0} \frac{|q_{\text{new}} q_{\text{new}}|}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{2 \times 2}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{4}{r^2}$.
The ratio of the magnitudes of the forces is: $\frac{|F_{\text{initial}}|}{|F_{\text{final}}|} = \frac{96/r^2}{4/r^2} = \frac{96}{4} = 24$.
Thus,the ratio is $24: 1$.

(A) The potential $V_{1}$ of the smaller sphere (radius $r$,charge $q$) is due to its own charge and the potential due to the outer sphere (radius $R$,charge $Q$).
$V_{1} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R}$
The potential $V_{2}$ of the bigger sphere (radius $R$,charge $Q$) is due to its own charge and the potential due to the inner sphere (which acts as a point charge at the center for points outside it).
$V_{2} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}$
The potential difference between the spheres is $V = V_{1} - V_{2}$.
$V = \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} \right) - \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R} \right)$
$V = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r} + \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} - \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} - \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}$
$V = \frac{1}{4 \pi \varepsilon_{0}} \left( \frac{q}{r} - \frac{q}{R} \right)$

(A) The electrostatic potential energy $U$ of a system of three charges $q_1, q_2, q_3$ separated by distances $r_{12}, r_{23}, r_{31}$ is given by $U = \frac{1}{4\pi\epsilon_0} [\frac{q_1q_2}{r_{12}} + \frac{q_2q_3}{r_{23}} + \frac{q_3q_1}{r_{31}}]$.
Here,the charges are $q_1 = +q$,$q_2 = +q$,and $q_3 = Q$. The distance between each pair is $a$.
Substituting these values into the formula,we get:
$U = \frac{K}{a} [q \cdot q + q \cdot Q + Q \cdot q] = 0$
Since $K/a \neq 0$,the term inside the bracket must be zero:
$q^2 + 2qQ = 0$
$2qQ = -q^2$
$Q = -\frac{q^2}{2q} = -\frac{q}{2}$

(D) The force on the horizontal segments $PQ$ and $RS$ are equal in magnitude but opposite in direction,so their net force is zero.
The force between two parallel current-carrying conductors is given by $F = \frac{\mu_0 I_1 I_2 l}{2 \pi r}$,where $r$ is the distance between the conductors.
For segment $PS$ (at distance $r_1 = 2 \text{ cm} = 0.02 \text{ m}$):
$F_{PS} = \frac{2 \times 10^{-7} \times 20 \times 20 \times 0.15}{0.02} = 6 \times 10^{-4} \text{ N}$ (Attractive force towards the wire).
For segment $QR$ (at distance $r_2 = 2 \text{ cm} + 10 \text{ cm} = 12 \text{ cm} = 0.12 \text{ m}$):
$F_{QR} = \frac{2 \times 10^{-7} \times 20 \times 20 \times 0.15}{0.12} = 1 \times 10^{-4} \text{ N}$ (Repulsive force away from the wire).
The net force is $F_{\text{net}} = F_{PS} - F_{QR} = 6 \times 10^{-4} - 1 \times 10^{-4} = 5 \times 10^{-4} \text{ N}$.

(C) The magnetic field at the centre of a circular coil of radius $r$ carrying current $I$ is given by:
$B_{c} = \frac{\mu_{0} I}{2r}$
The magnetic field on the axis of the circular coil at a distance $x$ from the centre is given by:
$B_{a} = \frac{\mu_{0} I r^{2}}{2(r^{2} + x^{2})^{3/2}}$
Given that the distance $x = r$,we substitute this into the expression for $B_{a}$:
$B_{a} = \frac{\mu_{0} I r^{2}}{2(r^{2} + r^{2})^{3/2}} = \frac{\mu_{0} I r^{2}}{2(2r^{2})^{3/2}} = \frac{\mu_{0} I r^{2}}{2(2^{3/2} r^{3})} = \frac{\mu_{0} I}{2 \cdot 2\sqrt{2} \cdot r} = \frac{\mu_{0} I}{4\sqrt{2} r}$
Now,calculating the ratio $B_{c} : B_{a}$:
$\frac{B_{c}}{B_{a}} = \frac{\frac{\mu_{0} I}{2r}}{\frac{\mu_{0} I}{4\sqrt{2} r}} = \frac{4\sqrt{2} r}{2r} = 2\sqrt{2}$
Therefore,the ratio $B_{c} : B_{a}$ is $2\sqrt{2} : 1$.

(A) The conductor consists of three parts: two semi-infinite straight wires and a circular arc of $270^\circ$ (or $\frac{3\pi}{2}$ radians).
$1$. For the straight wire $(A)$,the point $O$ lies on its axis,so the magnetic field $B_A = 0$.
$2$. For the circular arc $(B)$,the magnetic field at the centre is $B_B = \frac{\mu_0 I \theta}{4 \pi r} = \frac{\mu_0 I (3\pi/2)}{4 \pi r} = \frac{3 \mu_0 I}{8 r}$. The direction is inwards (using the right-hand rule).
$3$. For the straight wire $(C)$,the point $O$ is at a perpendicular distance $r$ from the wire. The magnetic field due to a semi-infinite wire is $B_C = \frac{\mu_0 I}{4 \pi r}$. The direction is also inwards.
$4$. The total magnetic field $B = B_A + B_B + B_C = 0 + \frac{3 \mu_0 I}{8 r} + \frac{\mu_0 I}{4 \pi r}$.
Factoring out $\frac{\mu_0 I}{4 \pi r}$,we get $B = \frac{\mu_0 I}{4 \pi r} \left( \frac{3\pi}{2} + 1 \right)$.

(B) The energy released per fission is $E = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
Given power output $P = 1.6 \text{ MW} = 1.6 \times 10^6 \text{ W}$.
The rate of fission $R$ is given by the formula $R = P / E$.
Substituting the values: $R = (1.6 \times 10^6) / (3.2 \times 10^{-11})$.
$R = 0.5 \times 10^{17} = 5 \times 10^{16} \text{ fissions/s}$.

(C) Let the number of $\alpha$ particles emitted be $x$ and the number of $\beta$ particles emitted be $y$.
The nuclear decay reaction is represented as:
${ }_{92} U^{235} \longrightarrow x({ }_{2} \alpha^{4}) + y({ }_{-1} \beta^{0}) + { }_{82} Pb^{203}$
Equating the mass numbers on both sides:
$235 = 4x + 203$
$4x = 235 - 203 = 32$
$x = 8$
Equating the atomic numbers on both sides:
$92 = 2x - y + 82$
Substituting $x = 8$ into the equation:
$92 = 2(8) - y + 82$
$92 = 16 - y + 82$
$92 = 98 - y$
$y = 98 - 92 = 6$
Therefore,$8$ $\alpha$ particles and $6$ $\beta$ particles are emitted.

(C) The nuclear force $F_{n}$ is a short-range force that acts effectively only within the range of approximately $10^{-15} \text{ m}$ (or $1 \text{ fm}$).
Beyond this distance,the nuclear force decreases exponentially and becomes negligible.
The given separation is $40 \text{ Å} = 40 \times 10^{-10} \text{ m} = 4 \times 10^{-9} \text{ m}$.
Since $4 \times 10^{-9} \text{ m} \gg 10^{-15} \text{ m}$,the nuclear force $F_{n}$ is essentially zero at this distance.
However,the electrostatic force $F_{e}$ follows the inverse-square law $(F_{e} \propto 1/r^2)$ and remains significant even at this distance.
Therefore,$F_{n} \ll F_{e}$.

(C) The nuclear radius $R$ is given by the formula $R = R_0 A^{1/3}$,where $A$ is the mass number and $R_0$ is a constant.
Given mass numbers are $A_1 = 216$ and $A_2 = 125$.
The ratio of the radii is $\frac{R_1}{R_2} = \frac{R_0 (A_1)^{1/3}}{R_0 (A_2)^{1/3}}$.
Substituting the values,$\frac{R_1}{R_2} = \frac{(216)^{1/3}}{(125)^{1/3}}$.
Since $216 = 6^3$ and $125 = 5^3$,we have $\frac{R_1}{R_2} = \frac{6}{5}$ or $6:5$.

(C) Baryons are a family of subatomic particles made of three quarks.
Among all baryons,the proton is the only stable particle that does not decay into other particles.
While the free neutron decays into a proton,an electron,and an antineutrino with a mean life of approximately $880 \ s$,the proton is considered stable with a lifetime exceeding $10^{34}$ years.
Therefore,the proton is the most stable particle in the Baryon group.

(A) The activity $R$ of a radioactive substance is given by $R = \lambda N$,where $N$ is the number of radioactive nuclei present at time $t$. Since the initial masses are the same,the initial number of nuclei $N_0$ is the same for both substances.
The activity at time $t$ is given by $R = R_0 \left( \frac{1}{2} \right)^{t / T_{1/2}}$,where $T_{1/2}$ is the half-life.
For the first substance: $T_{1/2, 1} = 1 \ yr$,$t = 4 \ yr$.
$R_1 = R_0 \left( \frac{1}{2} \right)^{4/1} = R_0 \left( \frac{1}{2} \right)^4$.
For the second substance: $T_{1/2, 2} = 2 \ yr$,$t = 4 \ yr$.
$R_2 = R_0 \left( \frac{1}{2} \right)^{4/2} = R_0 \left( \frac{1}{2} \right)^2$.
The ratio of their activities is:
$\frac{R_1}{R_2} = \frac{R_0 (1/2)^4}{R_0 (1/2)^2} = \left( \frac{1}{2} \right)^{4-2} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Thus,the ratio is $1: 4$.

(B) The equivalent focal length $F$ of two thin lenses separated by a distance $d$ is given by the formula:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$
Here,$f_1$ is the focal length of the convex lens (positive) and $f_2$ is the focal length of the concave lens (negative).
When the lenses are put in contact,the separation distance $d = 0$.
Substituting $d = 0$ into the formula,we get:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
Since $f_1 > 0$ and $f_2 < 0$,let $f_1 = f$ and $f_2 = -f'$ (where $f, f' > 0$).
$\frac{1}{F} = \frac{1}{f} - \frac{1}{f'}$
As the lenses are brought from a separated state to contact,the term $-\frac{d}{f_1 f_2}$ (which was negative since $f_1 f_2 < 0$) is removed. Effectively,the power of the combination changes such that the focal length increases compared to the separated state.

(C) The lens is composed of $3$ different materials,each having a different refractive index.
Since the lens is formed by $3$ distinct parts,each part acts as an independent lens with its own focal length.
When a point object is placed on the principal axis,light rays passing through each of the $3$ different sections will be refracted differently.
Consequently,each section will form an image of the object at a position determined by its specific focal length.
Therefore,$3$ distinct images will be formed,one by each section of the lens.

(A) The rays are incident normally on the first surface of the prism,so they pass undeviated and strike the hypotenuse at an angle of incidence $i = 45^{\circ}$.
For total internal reflection to occur at the second surface,the angle of incidence must be greater than the critical angle $(i > C)$.
Given critical angles are $C_{red} = 46^{\circ}$,$C_{green} = 44^{\circ}$,and $C_{blue} = 43^{\circ}$.
For red light: $i = 45^{\circ} < C_{red} = 46^{\circ}$. Thus,red light will emerge from the prism.
For green light: $i = 45^{\circ} > C_{green} = 44^{\circ}$. Thus,green light will undergo total internal reflection.
For blue light: $i = 45^{\circ} > C_{blue} = 43^{\circ}$. Thus,blue light will undergo total internal reflection.
Since green and blue light undergo total internal reflection while red light emerges,the arrangement separates red color from blue and green colors.

(C) According to the law of reflection,the angle of incidence $i$ is equal to the angle of reflection $r$,so $i = r$.
Given that the reflected and refracted rays are mutually perpendicular,the sum of the angle of reflection $r$,the angle between them $(90^{\circ})$,and the angle of refraction $r'$ is $180^{\circ}$.
Thus,$r + 90^{\circ} + r' = 180^{\circ}$.
Since $r = i$,we have $i + 90^{\circ} + r' = 180^{\circ}$,which implies $r' = 90^{\circ} - i$.
Applying Snell's Law at the interface: $\mu_r \sin i = \mu_d \sin r'$.
Substituting $r' = 90^{\circ} - i$,we get $\mu_r \sin i = \mu_d \sin(90^{\circ} - i) = \mu_d \cos i$.
Therefore,$\frac{\mu_d}{\mu_r} = \frac{\sin i}{\cos i} = \tan i$.
For the critical angle $C$,we know that $\sin C = \frac{\mu_r}{\mu_d}$.
Thus,$\sin C = \frac{1}{\tan i} = \cot i$.
Hence,the critical angle is $C = \sin^{-1}(\cot i)$.

(B) When an observer is in a denser medium (water,refractive index $n$) and the object (bird) is in a rarer medium (air,refractive index $1$),the apparent height of the object is increased.
For an observer in a medium of refractive index $n$,the apparent height $h'$ of an object at a real height $h$ is given by $h' = n \times h$.
Here,the bird is at a height $y$ above the water surface.
Therefore,the apparent height of the bird as seen by the fish from the water surface is $h' = n \times y$.
The fish is at a depth $x$ below the water surface.
Thus,the total distance of the bird as estimated by the fish is the sum of the depth of the fish and the apparent height of the bird: $D = x + ny$.

(B) In an unbiased $p-n$ junction,a depletion region is formed at the junction due to the diffusion of charge carriers.
This creates a built-in potential barrier.
According to the potential-distance graph for a $p-n$ junction,the potential in the $n$-region is higher than the potential in the $p$-region.
Therefore,the potential at $p$ is less than that at $n$.

(C) The given circuit consists of an $OR$ gate followed by an $AND$ gate. The Boolean expression for the output $Y$ is given by $Y = (A + B) \cdot C$.
For the output to be $Y = 1$,both inputs to the $AND$ gate must be $1$. Therefore,$(A + B) = 1$ and $C = 1$.
Checking the options:
For option $C$: $A = 1, B = 0, C = 1$.
Substituting these values: $Y = (1 + 0) \cdot 1 = 1 \cdot 1 = 1$.
Thus,the inputs $A = 1, B = 0, C = 1$ result in an output $Y = 1$.

(B) For the first diffraction minimum,the condition is given by $a \sin \theta = n\lambda$. For $n=1$,$a \sin \theta = \lambda$.
Given $\theta = 30^{\circ}$,we have $a = \frac{\lambda}{\sin 30^{\circ}} = 2\lambda$.
For the first secondary maximum,the condition is $a \sin \theta' = \frac{3\lambda}{2}$.
Substituting the value of $a = 2\lambda$ into the equation:
$2\lambda \sin \theta' = \frac{3\lambda}{2}$
$\sin \theta' = \frac{3\lambda}{2 \times 2\lambda} = \frac{3}{4}$.
Therefore,$\theta' = \sin^{-1}\left(\frac{3}{4}\right)$.

(D) The degree of diffraction is directly proportional to the wavelength of the incident wave.
Since diffraction occurs when the size of the obstacle or aperture is comparable to the wavelength of the wave,waves with longer wavelengths exhibit more pronounced diffraction effects.
Among the given options,radiowaves have the longest wavelength.
Therefore,radiowaves undergo maximum diffraction.

(B) The blue colour of sea water is primarily due to the scattering of sunlight by water molecules. When sunlight enters the water,the shorter wavelengths (blue light) are scattered more effectively by the water molecules compared to longer wavelengths,which gives the sea its characteristic blue appearance.

(A) When an unpolarised light beam of intensity $I_{0}$ passes through a polaroid,it becomes plane-polarised.
According to the properties of polaroids,the intensity of the emergent plane-polarised light is exactly half of the intensity of the incident unpolarised light.
Therefore,the intensity of the emergent light is $I = \frac{I_{0}}{2}$.

(B) dichroic crystal is a material that exhibits the property of dichroism,which is the selective absorption of light based on its polarization direction.
Some crystals,such as tourmaline,have the property of strongly absorbing light with vibrations perpendicular to a specific direction (called the transmission axis) while transmitting light with vibrations parallel to it.
This selective absorption of light is known as dichroism.
Therefore,tourmaline is a well-known example of a dichroic crystal.

(D) In Young's double slit experiment,the interference pattern is formed by the superposition of light waves from two coherent sources. If a third slit is introduced between the two slits,it acts as an additional source of light. This additional light contributes to the background illumination on the screen,which does not participate in the interference pattern. As a result,the intensity of the dark fringes increases,while the intensity of the bright fringes remains relatively unchanged. This leads to a decrease in the visibility or contrast between the bright and dark fringes.

(A) The fringe width $\beta$ in a Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance of the screen from the slits, and $d$ is the slit separation.
When the screen is moved by $\Delta D = 5 \times 10^{-2} \, m$, the change in fringe width is $\Delta \beta = 3 \times 10^{-5} \, m$.
From the formula, the change in fringe width is $\Delta \beta = \frac{\lambda}{d} \Delta D$.
Rearranging for $\lambda$, we get $\lambda = \frac{\Delta \beta \cdot d}{\Delta D}$.
Substituting the given values:
$\lambda = \frac{(3 \times 10^{-5} \, m) \times (10^{-3} \, m)}{5 \times 10^{-2} \, m}$.
$\lambda = \frac{3 \times 10^{-8}}{5 \times 10^{-2}} \, m = 0.6 \times 10^{-6} \, m = 6 \times 10^{-7} \, m$.
Converting to $\text{Å}$ $(1 \, \text{Å} = 10^{-10} \, m)$:
$\lambda = 6000 \times 10^{-10} \, m = 6000 \, \text{Å}$.