KCET 2024 Physics Question Paper with Answer and Solution

(A) The acceleration due to gravity at a height $h$ above the Earth's surface is given by the formula: $g_h = \frac{g}{(1 + \frac{h}{R})^2}$.
Given that the height $h = \frac{R}{2}$,where $R$ is the radius of the Earth.
Substituting the value of $h$ into the formula:
$g_h = \frac{g}{(1 + \frac{R/2}{R})^2} = \frac{g}{(1 + \frac{1}{2})^2} = \frac{g}{(\frac{3}{2})^2}$.
$g_h = \frac{g}{9/4} = \frac{4}{9}g$.
Using $g = 9.8 \ m/s^2$:
$g_h = \frac{4}{9} \times 9.8 \approx 4.355 \ m/s^2 \approx 4.4 \ m/s^2$.

(A) The ratio of molar specific heats $(\gamma = C_p / C_V)$ is given by the formula $\gamma = 1 + \frac{2}{f}$,where $f$ is the degree of freedom of the gas molecule.
Oxygen $(O_2)$ is a diatomic gas at room temperature.
For a diatomic gas,the number of degrees of freedom is $f = 5$ ($3$ translational and $2$ rotational).
Substituting the value of $f$ into the formula:
$\gamma = 1 + \frac{2}{5} = 1 + 0.4 = 1.4$.

(B) The force pulling the block down the inclined plane is $F = mg \sin 30^{\circ} = mg \times 0.5 = 0.5 mg$.
The maximum static frictional force (limiting friction) is $f_{s,max} = \mu_s R = \mu_s mg \cos 30^{\circ}$.
Given $\mu_s = 0.6$,we have $f_{s,max} = 0.6 \times mg \times \frac{\sqrt{3}}{2} = 0.3 \times 1.732 \times mg = 0.5196 mg$.
Since the pulling force $F = 0.5 mg$ is less than the maximum static frictional force $f_{s,max} = 0.5196 mg$,the block will not move.
Therefore,the acceleration of the block is zero.

(D) The resultant $R$ of two vectors $A$ and $B$ lies in the range $|A-B| \leq R \leq |A+B|$.
For option $A$: $|1-2| \leq R \leq |1+2| \implies 1 \leq R \leq 3$. Since $3$ is in the range,the resultant can be $3$.
For option $B$: $|2-5| \leq R \leq |2+5| \implies 3 \leq R \leq 7$. Since $3$ is in the range,the resultant can be $3$.
For option $C$: $|3-6| \leq R \leq |3+6| \implies 3 \leq R \leq 9$. Since $3$ is in the range,the resultant can be $3$.
For option $D$: $|4-8| \leq R \leq |4+8| \implies 4 \leq R \leq 12$. The range is $[4, 12]$. Since $3$ is not in this range,the resultant can never be $3$ units.

(C) Given: Flow rate $Q = 0.314 \,m^3/s$ and radius $r = 10 \,cm = 0.1 \,m$.
Using the equation of continuity, $Q = A \times v$, where $A = \pi r^2$ is the cross-sectional area.
Substituting the values: $0.314 = \pi \times (0.1)^2 \times v$.
Taking $\pi \approx 3.14$, we get $0.314 = 3.14 \times 0.01 \times v$.
$0.314 = 0.0314 \times v$.
$v = \frac{0.314}{0.0314} = 10 \,m/s$.

(B) Consider an element of length $dx$ at a distance $x$ from the free end of the wire.
The weight of the portion of the wire below this element is $dw = (A \cdot x \cdot \rho) g$,where $A$ is the cross-sectional area.
The stress at this section is $\sigma = \frac{dw}{A} = \rho g x$.
The strain is $\frac{d(\Delta l)}{dx} = \frac{\sigma}{Y} = \frac{\rho g x}{Y}$.
Integrating from $x = 0$ to $x = L$ to find the total elongation $\Delta L$:
$\Delta L = \int_{0}^{L} \frac{\rho g x}{Y} dx = \frac{\rho g}{Y} \left[ \frac{x^2}{2} \right]_{0}^{L} = \frac{\rho g L^2}{2 Y}$.

(A) Given,diameter,$d = 80 \ m$.
Therefore,radius,$r = d/2 = 40 \ m$.
Distance travelled after completion of $3/4$ revolution is given by:
Distance $= (3/4) \times (2 \pi r) = (3/2) \times \pi \times 40 = 60 \pi \ m$.
Displacement is the shortest distance between the initial point $A$ and the final point $B$. Since the athlete covers $3/4$ of the circle,the angle between the initial and final position vectors is $90^{\circ}$.
Using the Pythagorean theorem for the right-angled triangle formed by the two radii:
Displacement $= \sqrt{r^2 + r^2} = r \sqrt{2} = 40 \sqrt{2} \ m$.

(B) We know that for a particle executing $SHM$,the velocity $v$ and acceleration $a$ are given by the equations:
$v = \omega \sqrt{A^2 - y^2}$
$a = -\omega^2 y$
At the mean position,the displacement $y = 0$.
Substituting $y = 0$ into the velocity equation:
$v = \omega \sqrt{A^2 - 0} = \omega A$
Thus,the velocity is maximum $(v_{\max} = \omega A)$.
Substituting $y = 0$ into the acceleration equation:
$a = -\omega^2 (0) = 0$
Thus,the acceleration is zero.

(D) Given,mass of the body,$m_1 = 1 \,kg$.
Mass of the pulley,$m_2 = 2 \,kg$.
According to the law of conservation of energy,the potential energy lost by the mass is equal to the sum of the kinetic energy of the mass and the rotational kinetic energy of the pulley.
$m_1 g h = \frac{1}{2} m_1 v^2 + \frac{1}{2} I \omega^2$
Since the pulley is a disc,its moment of inertia is $I = \frac{1}{2} m_2 R^2$. Also,$\omega = \frac{v}{R}$.
Substituting these values:
$m_1 g h = \frac{1}{2} m_1 v^2 + \frac{1}{2} \left( \frac{1}{2} m_2 R^2 \right) \left( \frac{v}{R} \right)^2$
$m_1 g h = \frac{1}{2} m_1 v^2 + \frac{1}{4} m_2 v^2$
Substituting the given values $(m_1 = 1 \,kg, m_2 = 2 \,kg, g = 10 \,ms^{-2}, h = 1.6 \,m)$:
$1 \times 10 \times 1.6 = \frac{1}{2} \times 1 \times v^2 + \frac{1}{4} \times 2 \times v^2$
$16 = 0.5 v^2 + 0.5 v^2$
$16 = v^2$
$v = 4 \,ms^{-1}$

(C) The total heat energy $Q$ required to convert the solid cube at $\theta_0$ to vapour at $\theta_2$ is the sum of heat required for each phase change and temperature change step:
$1$. Heat to raise temperature of solid from $\theta_0$ to $\theta_1$: $Q_1 = m s_1(\theta_1 - \theta_0)$
$2$. Heat for fusion at $\theta_1$: $Q_2 = m L_f$
$3$. Heat to raise temperature of liquid from $\theta_1$ to $\theta_2$: $Q_3 = m s_2(\theta_2 - \theta_1)$
$4$. Heat for vaporisation at $\theta_2$: $Q_4 = m L_v$
Total heat $Q = Q_1 + Q_2 + Q_3 + Q_4 = m s_1(\theta_1 - \theta_0) + m L_f + m s_2(\theta_2 - \theta_1) + m L_v$.

(D) The work done by the gas in a cyclic process is equal to the area enclosed by the cycle on the $p-V$ diagram.
Since the cycle $MNOM$ is traversed in a clockwise direction, the work done is positive.
The area of the triangle $MNO$ is given by:
$W = \text{Area of } \triangle MNO = \frac{1}{2} \times \text{base} \times \text{height}$
$W = \frac{1}{2} \times (ON) \times (OM)$
From the graph, the base $ON = 3V_0 - V_0 = 2V_0$ and the height $OM = 3p_0 - p_0 = 2p_0$.
Substituting these values:
$W = \frac{1}{2} \times (2V_0) \times (2p_0)$
$W = 2p_0 V_0$
Thus, the work done by the gas is $2p_0 V_0$.

(C) The activity of a radioactive substance is defined as the rate of decay,which is the number of disintegrations per unit time.
Mathematically,Activity $A = -\frac{dN}{dt}$.
Since $N$ (number of nuclei) is a dimensionless quantity and $t$ represents time,the dimensions of activity are $[T^{-1}]$.
Therefore,the dimensional formula is $[M^0 L^0 T^{-1}]$.

(A) The motor-cyclist moves towards a cliff,so the cliff acts as a stationary source of reflected sound.
Speed of the motor-cyclist,$v_o = 18 \ km/h = 18 \times \frac{5}{18} \ m/s = 5 \ m/s$.
Frequency of the source horn,$f = 325 \ Hz$.
Speed of sound in air,$v = 330 \ m/s$.
The frequency of the reflected sound heard by the motor-cyclist is given by the Doppler effect formula for a moving observer and a stationary source:
$f' = f \left( \frac{v + v_o}{v} \right) = 325 \left( \frac{330 + 5}{330} \right) = 325 \left( \frac{335}{330} \right) \approx 329.92 \ Hz$.
The number of beats heard is the difference between the reflected frequency and the original frequency:
$\Delta f = f' - f = 329.92 - 325 = 4.92 \ Hz \approx 5 \ Hz$.

(D) To determine the direction of the angular velocity vector $\vec{\omega}$,we use the right-hand thumb rule.
According to this rule,if we curl the fingers of our right hand in the direction of rotation of the fan,the thumb points in the direction of the angular velocity vector.
In the given figure,the fan is rotating in a clockwise direction when viewed from above.
Since the axle is along the $Z$-axis,the rotation occurs in the $XY$-plane.
Applying the right-hand rule,the thumb points downwards,which is in the negative $Z$-direction.
Therefore,the direction of angular velocity is along $-\hat{k}$.

(C) The area under the $P-t$ graph is equal to the work done, which is equal to the change in kinetic energy $(\Delta K)$.
Given mass $m = 500 \text{ g} = 0.5 \text{ kg}$.
The area of the trapezoid $OABC$ is given by:
$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{height})$
$\text{Area} = \frac{1}{2} \times (2 + 8) \times 5 = \frac{1}{2} \times 10 \times 5 = 25 \text{ J}$.
Since the particle starts from rest, the initial kinetic energy is $0$. Thus, the final kinetic energy at $t = 5 \text{ s}$ is $K = 25 \text{ J}$.
We know that kinetic energy $K = \frac{p^2}{2m}$, where $p$ is the momentum.
$25 = \frac{p^2}{2 \times 0.5}$
$25 = \frac{p^2}{1}$
$p^2 = 25$
$p = 5 \text{ kg m/s} = 5 \text{ N-s}$.

(C) The resonance frequency $f_0$ in an $L-C-R$ series circuit is given by the formula:
$f_0 = \frac{1}{2 \pi \sqrt{L C}}$
From this expression,we can see that the resonance frequency is inversely proportional to the square root of the capacitance:
$f_0 \propto \frac{1}{\sqrt{C}}$
As the value of capacitance $C$ increases,the resonance frequency $f_0$ decreases. This relationship represents a hyperbolic curve,which corresponds to the graph shown in option $C$.

(C) In a series $L-C-R$ circuit,the circuit is capacitive when the capacitive reactance $X_C$ is greater than the inductive reactance $X_L$ (i.e.,$X_C > X_L$).
From the given graph,the intersection point $B$ represents the resonance frequency where $X_L = X_C$.
For frequencies less than the resonance frequency (i.e.,to the left of point $B$),the curve for $X_C$ lies above the curve for $X_L$,meaning $X_C > X_L$.
Among the given options,point $A$ lies to the left of point $B$,where $X_C > X_L$. Therefore,the circuit is capacitive at point $A$.

(D) For better tuning of an $L-C-R$ circuit used in communication,the quality factor $Q$ must be high.
The formula for the quality factor is $Q = \frac{1}{R} \sqrt{\frac{L}{C}}$.
To maximize $Q$,we need a small resistance $R$,a large inductance $L$,and a small capacitance $C$.
Comparing the given options:
$A: R=20, L=1.5, C=35$
$B: R=25, L=2.5, C=45$
$C: R=25, L=1.5, C=45$
$D: R=15, L=3.5, C=30$
Option $D$ has the minimum resistance $(15 \Omega)$,the maximum inductance $(3.5 \text{ H})$,and the minimum capacitance $(30 \mu\text{F})$.
Therefore,the combination in option $D$ provides the highest quality factor,resulting in better tuning.

(D) The distance of closest approach $r_0$ is determined by equating the initial kinetic energy of the alpha particle to the electrostatic potential energy at the point of closest approach:
$\frac{1}{2} m v^2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$
From this equation,we can see that $v^2 \propto \frac{1}{r_0}$,which implies $r_0 \propto \frac{1}{v^2}$.
Given that the initial distance of closest approach is $d$ for velocity $v$,let the new distance be $d'$ for velocity $2v$.
$\frac{d'}{d} = \frac{v^2}{(2v)^2} = \frac{v^2}{4v^2} = \frac{1}{4}$
Therefore,$d' = \frac{d}{4}$.

(D) The radius of the $n^{th}$ orbit of a hydrogen atom is given by $r_n \propto n^2$.
Since the area $A$ of the orbit is $A = \pi r_n^2$,we have $A_n \propto (n^2)^2 = n^4$.
For the ground state,$n_1 = 1$.
For the first excited state,$n_2 = 2$.
The ratio of the area of the first excited state $(A_2)$ to the ground state $(A_1)$ is:
$\frac{A_2}{A_1} = \left(\frac{n_2}{n_1}\right)^4 = \left(\frac{2}{1}\right)^4 = 16$.
Therefore,the ratio is $16: 1$.

(D) The volume of a nucleus is given by $V = \frac{4}{3} \pi R^3$ and its surface area is $S = 4 \pi R^2$.
The ratio of volume to surface area is $\frac{V}{S} = \frac{\frac{4}{3} \pi R^3}{4 \pi R^2} = \frac{R}{3}$.
The radius of a nucleus is given by $R = R_0 A^{1/3}$,where $A$ is the mass number.
For $Al^{27}$,$A = 27$. Thus,$R = R_0 (27)^{1/3} = 3 R_0$.
Substituting this into the ratio: $\frac{V}{S} = \frac{3 R_0}{3} = R_0$.
Given $R_0 = 1.2 \times 10^{-15} \ m$,the ratio is $1.2 \times 10^{-15} \ m$.

(A) Given: Capacitance $C = 5 \mu \text{F} = 5 \times 10^{-6} \text{ F}$.
Potential difference $V = 4 \text{ V}$.
Rate of change of potential difference $\frac{dV}{dt} = 0.6 \text{ Vs}^{-1}$.
The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.
To find the time rate at which energy is stored,we differentiate $U$ with respect to time $t$:
$\frac{dU}{dt} = \frac{d}{dt} (\frac{1}{2}CV^2) = \frac{1}{2}C \cdot 2V \cdot \frac{dV}{dt} = CV \frac{dV}{dt}$.
Substituting the given values:
$\frac{dU}{dt} = (5 \times 10^{-6} \text{ F}) \times (4 \text{ V}) \times (0.6 \text{ Vs}^{-1})$.
$\frac{dU}{dt} = 20 \times 0.6 \times 10^{-6} \text{ W} = 12 \times 10^{-6} \text{ W} = 12 \mu \text{W}$.

(C) We know that Ohm's law is given by $V = I R$.
Substituting $V = E \cdot l$ and $R = \rho \cdot \frac{l}{A}$,where $l$ is the length and $A$ is the cross-sectional area of the conductor,we get:
$E \cdot l = I \cdot \rho \cdot \frac{l}{A}$
Dividing both sides by $l$,we get:
$E = \left( \frac{I}{A} \right) \rho$
Since current density $J = \frac{I}{A}$,we substitute this into the equation:
$E = J \rho$ or $E = \rho J$.

(A) The total charge $Q$ flowing through the cross-section of the conductor is equal to the area under the $I-t$ graph.
From the graph,for $0 \leq t \leq 10 \ s$,the current increases linearly from $100 \ mA$ to $300 \ mA$. The area is a trapezoid: $A_1 = \frac{(100 + 300) \times 10^{-3} \ A}{2} \times 10 \ s = 2 \ C$.
For $10 \ s \leq t \leq 20 \ s$,the current is constant at $300 \ mA$. The area is a rectangle: $A_2 = 300 \times 10^{-3} \ A \times (20 - 10) \ s = 3 \ C$.
Total charge $Q = A_1 + A_2 = 2 \ C + 3 \ C = 5 \ C$.
Using the quantization of charge,$Q = ne$,where $e = 1.6 \times 10^{-19} \ C$ is the charge of an electron.
$n = \frac{Q}{e} = \frac{5}{1.6 \times 10^{-19}} = 3.125 \times 10^{19}$.

(C) First, calculate the resistance of the bulb using the formula $R = \frac{V^2}{P}$.
$R = \frac{120 \times 120}{60} = 240 \, \Omega$.
Next, calculate the rated current of the bulb using $I = \frac{P}{V}$.
$I = \frac{60}{120} = 0.5 \, A$.
For the bulb to glow properly, it must draw the same current $I = 0.5 \, A$ when connected to the $220 \, V$ source.
The total resistance $R_{total}$ required in the circuit is $R_{total} = \frac{V_{source}}{I} = \frac{220}{0.5} = 440 \, \Omega$.
Since the bulb and the additional resistor are in series, $R_{total} = R + R_{series}$.
Therefore, $R_{series} = R_{total} - R = 440 \, \Omega - 240 \, \Omega = 200 \, \Omega$.

(A) From the $I-V$ graph,the slope represents $1/R$.
For temperature $t_1 = 100^{\circ} C$,the angle is $45^{\circ}$,so $1/R_1 = \tan 45^{\circ} = 1$,which gives $R_1 = 1 \Omega$.
For temperature $t_2 = 400^{\circ} C$,the angle is $30^{\circ}$,so $1/R_2 = \tan 30^{\circ} = 1/\sqrt{3}$,which gives $R_2 = \sqrt{3} \Omega \approx 1.732 \Omega$.
The temperature coefficient of resistance $\alpha$ is given by the formula:
$\alpha = \frac{R_2 - R_1}{R_1 t_2 - R_2 t_1} = \frac{\sqrt{3} - 1}{1 \times 400 - \sqrt{3} \times 100} = \frac{0.732}{400 - 173.2} = \frac{0.732}{226.8} \approx 3.22 \times 10^{-3} /^{\circ} C$.
Rounding to the nearest given option,the value is approximately $3 \times 10^{-3} /^{\circ} C$.

(C) Standard resistance,$S=4 \ \Omega$.
At $t_1=0^{\circ} C$,the balancing length is $l_1=50 \ cm$.
The resistance of the coil $X$ is given by $R_1 = \frac{l_1}{100-l_1} \times S$.
$R_1 = \frac{50}{100-50} \times 4 = \frac{50}{50} \times 4 = 4 \ \Omega$.
At $t_2=100^{\circ} C$,the balancing length is $l_2=60 \ cm$.
The resistance of the coil $X$ is $R_2 = \frac{l_2}{100-l_2} \times S$.
$R_2 = \frac{60}{100-60} \times 4 = \frac{60}{40} \times 4 = 6 \ \Omega$.
The temperature coefficient of resistance $\alpha$ is given by $\alpha = \frac{R_2 - R_1}{R_1(t_2 - t_1)}$.
$\alpha = \frac{6 - 4}{4(100 - 0)} = \frac{2}{400} = 0.005^{\circ} C^{-1}$.

(D) The circuit consists of two networks connected in series.
Each network forms a balanced Wheatstone bridge.
For the first network,the equivalent resistance $R_1$ is calculated by simplifying the parallel combinations of resistors,which results in $R_1 = R$.
For the second network,the equivalent resistance $R_2$ is calculated similarly,which results in $R_2 = 2R$.
The total equivalent resistance of the circuit is $R_{AB} = R_1 + R_2 = R + 2R = 3R$.
Using Ohm's law,the current $I$ is given by $I = V_0 / R_{AB} = V_0 / 3R$.

(A) Given,the original wavelength $\lambda = 600 \ nm = 600 \times 10^{-9} \ m$.
The observed wavelength $\lambda' = 601 \ nm = 601 \times 10^{-9} \ m$.
The change in wavelength is $\Delta \lambda = \lambda' - \lambda = (601 - 600) \times 10^{-9} \ m = 1 \times 10^{-9} \ m$.
According to the Doppler effect for light,the relation is $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$,where $v$ is the speed of the galaxy and $c$ is the speed of light $(c = 3 \times 10^8 \ m/s)$.
Therefore,$v = c \cdot \frac{\Delta \lambda}{\lambda} = (3 \times 10^8 \ m/s) \cdot \frac{1 \times 10^{-9} \ m}{600 \times 10^{-9} \ m}$.
$v = 3 \times 10^8 \cdot \frac{1}{600} = \frac{3 \times 10^8}{600} = 0.5 \times 10^6 \ m/s$.
$v = 500,000 \ m/s = 500 \ km/s$.

(C) Given,energy of incident light is $E$ and work function is $\phi_0 = \frac{E}{3}$.
According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by:
$K_{\max} = E - \phi_0$
$K_{\max} = E - \frac{E}{3} = \frac{2E}{3}$.
Since photoelectrons are emitted with a range of energies from $0$ to $K_{\max}$ due to collisions within the metal,the kinetic energy $K$ of the photoelectrons lies in the range $0 \leq K \leq \frac{2E}{3}$.

(C) Given,work function $\phi_0 = 2.4 \ eV$.
Photoemission occurs only if the incident photon energy $E \ge \phi_0$,which implies the incident wavelength $\lambda \le \lambda_0$.
The threshold wavelength $\lambda_0$ is calculated as:
$\lambda_0 = \frac{1240 \ eV \cdot nm}{\phi_0} = \frac{1240}{2.4} \approx 516.7 \ nm$.
Photoemission will not take place if the incident wavelength $\lambda > \lambda_0$.
Comparing the given options:
$A) 200 \ nm < 516.7 \ nm$ (Emission occurs)
$B) 300 \ nm < 516.7 \ nm$ (Emission occurs)
$C) 700 \ nm > 516.7 \ nm$ (No emission)
$D) 400 \ nm < 516.7 \ nm$ (Emission occurs)
Therefore,for $700 \ nm$,photoemission does not take place.

(A) Given,$B = 2 \text{ mT} = 2 \times 10^{-3} \text{ T}$.
Radius $R = \frac{10}{\sqrt{\pi}} \text{ cm} = \frac{10}{\sqrt{\pi}} \times 10^{-2} \text{ m} = \frac{10^{-1}}{\sqrt{\pi}} \text{ m}$.
The area of the circular cross-section $S_3$ is $A = \pi R^2 = \pi \times \left(\frac{10^{-1}}{\sqrt{\pi}}\right)^2 = \pi \times \frac{10^{-2}}{\pi} = 10^{-2} \text{ m}^2$.
For the hemisphere $S_1$,the magnetic field lines enter the surface. The angle between the area vector (outward normal) and the magnetic field vector is $180^{\circ}$.
Flux $\Phi_{S_1} = B A \cos(180^{\circ}) = (2 \times 10^{-3} \text{ T}) \times (10^{-2} \text{ m}^2) \times (-1) = -2 \times 10^{-5} \text{ Wb} = -20 \times 10^{-6} \text{ Wb} = -20 \mu \text{ Wb}$.
Since the total magnetic flux through a closed surface is zero (Gauss's Law for magnetism),the flux leaving the cone $S_2$ must be equal in magnitude to the flux entering the hemisphere $S_1$.
Therefore,$\Phi_{S_1} + \Phi_{S_2} = 0 \implies \Phi_{S_2} = -\Phi_{S_1} = +20 \mu \text{ Wb}$.

(C) The magnetic field $B$ produced by a long straight current-carrying conductor at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.
As the ring falls towards the conductor,the distance $r$ decreases,causing the magnetic field $B$ passing through the ring to increase.
According to the right-hand rule,the magnetic field lines from the conductor point out of the plane of the paper in the region where the ring is located.
Since the magnetic flux linked with the ring is increasing in the outward direction,according to Lenz's law,the induced current in the ring will create a magnetic field pointing into the plane of the paper to oppose this increase.
Using the right-hand grip rule,a magnetic field directed into the plane of the paper corresponds to a clockwise induced current in the ring.

(B) Given: Induced current $I = 2 \text{ A}$,Resistance $R = 10 \text{ } \Omega$,Time interval $\Delta t = 1 \text{ ms} = 10^{-3} \text{ s}$.
According to Faraday's law of induction,the magnitude of induced emf is $|\varepsilon| = \frac{\Delta \phi}{\Delta t}$.
Since $|\varepsilon| = I R$,we have $I R = \frac{\Delta \phi}{\Delta t}$.
Therefore,the change in magnetic flux is $\Delta \phi = I R \Delta t$.
Substituting the values: $\Delta \phi = 2 \times 10 \times 10^{-3} \text{ Wb}$.
$\Delta \phi = 20 \times 10^{-3} \text{ Wb} = 2 \times 10^{-2} \text{ Wb}$.

(A) The magnetic field $B$ produced by an infinitely long wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.
The magnetic flux $\phi$ through the square loop is $\phi = \int B \cdot dA = \int_{x}^{x+a} \frac{\mu_0 I}{2\pi r} (a \, dr) = \frac{\mu_0 I a}{2\pi} \ln\left(\frac{x+a}{x}\right)$.
The mutual inductance $M$ is defined as $M = \frac{\phi}{I} = \frac{\mu_0 a}{2\pi} \ln\left(1 + \frac{a}{x}\right)$.
Since the loop is moving at a constant speed $v$,the distance $x$ increases with time $t$ as $x = x_0 + vt$.
Substituting this into the expression for $M$,we get $M(t) = \frac{\mu_0 a}{2\pi} \ln\left(1 + \frac{a}{x_0 + vt}\right)$.
As $t$ increases,the term $\frac{a}{x_0 + vt}$ decreases,and therefore $\ln(1 + \frac{a}{x_0 + vt})$ decreases.
This corresponds to a curve that starts at a maximum value and decreases asymptotically towards zero as $t \to \infty$,which matches the graph shown in option $A$.

(A) The intensity $I$ is defined as the power per unit area,so the power $P = I A$.
The momentum $p$ of a photon is given by $p = E / c$,where $E$ is the energy.
For a perfectly reflecting surface,the change in momentum $\Delta p$ for a photon is $p_{final} - p_{initial} = (-p) - (p) = -2p = -2E / c$.
The force $F$ is the rate of change of momentum,$F = \frac{dp}{dt} = \frac{2}{c} \frac{dE}{dt}$.
Since the power $P = \frac{dE}{dt} = I A$,the force exerted on the surface is $F = \frac{2 I A}{c}$.

(C) The charge on the body is given by $q = -3.2 \mu C = -3.2 \times 10^{-6} \ C$.
According to the quantization of charge,$q = ne$,where $n$ is the number of electrons and $e$ is the elementary charge $(e = -1.6 \times 10^{-19} \ C)$.
Therefore,the number of excess electrons is given by $n = \frac{q}{e}$.
$n = \frac{-3.2 \times 10^{-6} \ C}{-1.6 \times 10^{-19} \ C}$.
$n = 2 \times 10^{13}$.

(B) According to Coulomb's law and Newton's third law of motion,the electrostatic force exerted by one point charge on another is equal in magnitude and opposite in direction.
Let $\vec{F}_{AB}$ be the force on $A$ due to $B$ $(F_1)$ and $\vec{F}_{BA}$ be the force on $B$ due to $A$ $(F_2)$.
According to Newton's third law,$\vec{F}_{AB} = -\vec{F}_{BA}$.
Therefore,$F_1 = -F_2$.

(B) stationary charge produces an electric field in the space surrounding it.
When a charge is in motion,it constitutes an electric current.
An electric current produces a magnetic field in the space surrounding it.
Therefore,a moving electron produces both an electric field and a magnetic field.

(B) The volume of the cube is $V = (1 \ cm)^3 = 10^{-6} \ m^3$.
The number density of molecules is $n = \frac{100}{10^{-6}} = 10^8 \ m^{-3}$.
The polarization $P$ is given by $P = n \cdot p$,where $p$ is the induced dipole moment.
$P = 10^8 \times 0.2 \times 10^{-6} = 0.02 \ C \cdot m^{-2}$.
The relation between polarization $P$,electric susceptibility $\chi_e$,and electric field $E$ is $P = \epsilon_0 \chi_e E$. However,in many contexts,the susceptibility $\chi_e$ is defined via $P = \chi_e E$.
Using $P = \chi_e E$,we get $\chi_e = \frac{P}{E} = \frac{0.02}{4} = 0.005$.
Re-evaluating based on the provided solution logic: $\chi_e = \frac{n \cdot p}{E} = \frac{10^8 \times 0.2 \times 10^{-6}}{4} = \frac{20}{4} = 5$.

(B) Given,electric field $\vec{E} = 3 \times 10^5 \hat{j} \text{ NC}^{-1}$.
Area of the rectangle $A = 10 \text{ cm} \times 30 \text{ cm} = 0.1 \text{ m} \times 0.3 \text{ m} = 3 \times 10^{-2} \text{ m}^2$.
Since the plane of the rectangle is parallel to the $ZX$-plane,its area vector $\vec{A}$ is directed along the $Y$-axis (normal to the $ZX$-plane).
Therefore,$\vec{A} = 3 \times 10^{-2} \hat{j} \text{ m}^2$.
The electric flux $\phi$ is given by the dot product of the electric field and the area vector:
$\phi = \vec{E} \cdot \vec{A} = (3 \times 10^5 \hat{j}) \cdot (3 \times 10^{-2} \hat{j})$
$\phi = 9 \times 10^3 \text{ Vm}$.

(A) An electric dipole consists of two equal and opposite charges,$+q$ and $-q$,separated by a small distance.
Therefore,the net charge $q_{net}$ enclosed by the spherical surface is $q + (-q) = 0$.
According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{net}}{\varepsilon_0}$.
Substituting the value of $q_{net}$,we get $\phi = \frac{0}{\varepsilon_0} = 0$.

(C) In electrostatic equilibrium,the electric field inside a conductor is zero. According to Gauss's law,any excess charge placed on a conductor must reside entirely on its surface to minimize mutual repulsion.
The electric field just outside the surface of a charged conductor is given by $E = \frac{\sigma}{\varepsilon_0}$,where $\sigma$ is the surface charge density.
The electric potential inside a charged conductor is constant and equal to the potential at its surface,not zero.
The net electric field is always perpendicular (normal) to the surface of the conductor,not tangential.
Therefore,the correct statement is that any excess charge resides on the surface of the conductor.

(A) Case-$I$: Number of turns,$N_1 = 9$. The magnetic field at the centre is given by $B_1 = \frac{\mu_0 N_1 I}{2 R} = \frac{9 \mu_0 I}{2 R}$.
Case-$II$: Number of turns,$N_2 = 3$. Let the new radius be $R'$. Since the total length of the wire remains constant,$N_1 (2 \pi R) = N_2 (2 \pi R')$.
Substituting the values: $9 (2 \pi R) = 3 (2 \pi R') \Rightarrow R' = 3 R$.
The new magnetic field is $B_2 = \frac{\mu_0 N_2 I}{2 R'} = \frac{\mu_0 \times 3 \times I}{2 \times 3 R} = \frac{\mu_0 I}{2 R}$.
Comparing $B_2$ with $B_1$: $\frac{B_2}{B_1} = \frac{\mu_0 I / 2 R}{9 \mu_0 I / 2 R} = \frac{1}{9}$.
Therefore,$B_2 = \frac{B_1}{9}$.

(D) The magnetic field at the centre of a circular coil is given by $B_{\text{centre}} = \frac{\mu_0 I}{2R}$.
The magnetic field at a distance $x$ on the axis of the coil is given by $B_{\text{axis}} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
According to the problem,$B_{\text{centre}} = 64 \times B_{\text{axis}}$.
Substituting the formulas,we get $\frac{\mu_0 I}{2R} = 64 \times \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
Canceling common terms,we have $\frac{1}{R} = \frac{64 R^2}{(R^2 + x^2)^{3/2}}$.
This simplifies to $(R^2 + x^2)^{3/2} = 64 R^3$.
Taking the cube root of both sides,$(R^2 + x^2)^{1/2} = 4R$.
Squaring both sides,$R^2 + x^2 = 16R^2$.
Therefore,$x^2 = 15R^2$,which gives $x = R\sqrt{15}$.

(D) The time period of a charged particle in a uniform magnetic field is given by $T = \frac{2 \pi m}{q B}$.
Given $q/m = \pi \text{ C kg}^{-1}$ and $B = 2 \text{ T}$,we have:
$T = \frac{2 \pi}{\pi \times 2} = 1 \text{ s}$.
The particle is projected along the $X$-axis. After time $t = \frac{1}{12} \text{ s}$,the angle of deviation $\theta$ is given by:
$\theta = \frac{t}{T} \times 360^{\circ} = \frac{1/12}{1} \times 360^{\circ} = 30^{\circ}$.
Since the magnetic field is in the $-\hat{k}$ direction,the particle moves in the $XY$-plane and deflects towards the positive $Y$-axis.
The velocity vector $\vec{v}$ at time $t$ is given by:
$\vec{v} = v_0 (\cos \theta \hat{i} + \sin \theta \hat{j})$
$\vec{v} = 10 (\cos 30^{\circ} \hat{i} + \sin 30^{\circ} \hat{j})$
$\vec{v} = 10 (\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}) = 5(\sqrt{3} \hat{i} + \hat{j}) \text{ ms}^{-1}$.

(C) Magnetic hysteresis is a phenomenon where the magnetization of a material lags behind the applied external magnetic field.
This behavior is characteristic of ferromagnetic materials,such as iron $(Fe)$,nickel $(Ni)$,and cobalt $(Co)$.
In these materials,the formation and movement of magnetic domains result in energy loss during a complete cycle of magnetization and demagnetization,which is represented by the hysteresis loop.
Paramagnetic and diamagnetic materials do not exhibit this property.

(C) For paramagnetic materials,the magnetic susceptibility $\chi$ is inversely proportional to the absolute temperature $T$,according to Curie's Law: $\chi \propto \frac{1}{T}$.
Given: $\chi_1 = 1.2 \times 10^{-5}$ at $T_1 = 300 \ K$.
We need to find $\chi_2$ at $T_2 = 200 \ K$.
Using the relation $\frac{\chi_2}{\chi_1} = \frac{T_1}{T_2}$,we get:
$\chi_2 = \chi_1 \times \frac{T_1}{T_2}$
$\chi_2 = 1.2 \times 10^{-5} \times \frac{300}{200}$
$\chi_2 = 1.2 \times 10^{-5} \times 1.5$
$\chi_2 = 1.8 \times 10^{-5}$.

(D) The energy released in a nuclear reaction is given by the difference in the total binding energy of the products and the reactants.
$Q = (BE_{\text{products}}) - (BE_{\text{reactants}})$
Given:
$BE(^{235}_{92}U) = 1800 \ MeV$
$BE(^{144}_{56}Ba) = 1200 \ MeV$
$BE(^{89}_{36}Kr) = 780 \ MeV$
Total $BE$ of reactants $= 1800 \ MeV$
Total $BE$ of products $= 1200 + 780 = 1980 \ MeV$
Energy released $Q = 1980 - 1800 = 180 \ MeV$
Since this energy is carried by $3$ neutrons,the average kinetic energy per neutron is:
$KE_{\text{avg}} = \frac{180 \ MeV}{3} = 60 \ MeV$

(C) The activity of a radioactive sample is given by $R = R_0 e^{-\lambda t}$.
Taking the natural logarithm on both sides,we get $\ln R = \ln R_0 - \lambda t$.
Comparing this with the equation of a straight line $y = mx + c$,the slope of the graph is $m = -\lambda$.
From the given graph,at $t = 0$,$\ln R_0 = 2$,which implies $R_0 = e^2 = 7.5$.
The slope of the graph is $\lambda = -\frac{\Delta(\ln R)}{\Delta t} = -\frac{1 - 2}{10 \times 10^3 - 0} = \frac{1}{10^4} = 10^{-4} \text{ s}^{-1}$.
We know that the activity $R_0 = \lambda N_0$,where $N_0$ is the number of undecayed nuclei at $t = 0$.
Therefore,$N_0 = \frac{R_0}{\lambda} = \frac{7.5}{10^{-4}} = 7.5 \times 10^4 = 75000$.

(D) From the diagram,the lens can be considered as a combination of two plano-convex lenses.
For the left half,the focal length $f_1$ is given by:
$\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{0.5}{R}$
For the right half,the focal length $f_2$ is given by:
$\frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (1.2 - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right) = \frac{0.2}{R}$
The combined power of the lens is $P = P_1 + P_2$,so $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$.
$\frac{1}{f} = \frac{0.5}{R} + \frac{0.2}{R} = \frac{0.7}{14} = \frac{0.7}{14} = 0.05 \ cm^{-1}$.
Thus,$f = \frac{1}{0.05} = 20 \ cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $u = -40 \ cm$ and $f = 20 \ cm$:
$\frac{1}{20} = \frac{1}{v} - \frac{1}{-40}$
$\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{2-1}{40} = \frac{1}{40}$
Therefore,$v = 40 \ cm$.

(A) Given: $n_1 = \frac{4}{3}$,$n_2 = \frac{3}{2}$,$R = 10 \text{ cm}$.
The object distance $u = -2R = -20 \text{ cm}$ (using sign convention).
The refraction formula at a spherical surface is given by:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Substituting the values:
$\frac{3/2}{v} - \frac{4/3}{-20} = \frac{3/2 - 4/3}{10}$
$\frac{3}{2v} + \frac{4}{60} = \frac{1/6}{10}$
$\frac{3}{2v} + \frac{1}{15} = \frac{1}{60}$
$\frac{3}{2v} = \frac{1}{60} - \frac{1}{15} = \frac{1 - 4}{60} = -\frac{3}{60} = -\frac{1}{20}$
$\frac{3}{2v} = -\frac{1}{20}$
$2v = -60 \implies v = -30 \text{ cm}$.
Since $v$ is negative,the image is formed on the same side as the object (in the rarer medium) at a distance of $30 \text{ cm}$ from the pole $P$.

(D) An astronomical telescope consists of two lenses: an objective lens and an eyepiece.
The objective lens forms a real,inverted,and diminished image of a distant object at its focal plane.
This image acts as an object for the eyepiece,which acts as a simple magnifier.
The eyepiece is adjusted so that the image lies within its focal length,resulting in a virtual,inverted,and magnified final image relative to the original distant object.

(C) Given,the prism is equilateral,so the angle of the prism $A = 60^{\circ}$.
Given that the angle of minimum deviation $\delta_m = A = 60^{\circ}$.
The refractive index $\mu$ of the prism is given by the formula:
$\mu = \frac{\sin \left(\frac{A + \delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
Substituting the values:
$\mu = \frac{\sin \left(\frac{60^{\circ} + 60^{\circ}}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)} = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.
We know that the refractive index $\mu$ is related to the speed of light in vacuum $c$ and the speed of light in the medium $v$ as $\mu = \frac{c}{v}$.
Therefore,$v = \frac{c}{\mu} = \frac{3 \times 10^8 \ m/s}{\sqrt{3}} = \sqrt{3} \times 10^8 \ m/s$.

(D) From the circuit diagram,the total current $I$ flowing through the series resistor $R_S$ is given by the voltage drop across it divided by its resistance:
$I = \frac{V_0 - V_Z}{R_S}$
According to Kirchhoff's current law at the junction point,the total current $I$ splits into the Zener diode current $I_Z$ and the load current $I_L$:
$I = I_Z + I_L$
Rearranging this equation to solve for the Zener diode current $I_Z$:
$I_Z = I - I_L$
Substituting the expression for $I$:
$I_Z = \left(\frac{V_0 - V_Z}{R_S}\right) - I_L$

(B) Given:
Battery emf,$V = 5.7 \ V$
Barrier potential of the diode,$V_B = 0.7 \ V$
Series resistance,$R_S = 5 \ k\Omega = 5 \times 10^3 \ \Omega$
In a forward-biased $p-n$ junction diode,the effective voltage across the resistor is the difference between the battery emf and the barrier potential.
Effective voltage,$V_{eff} = V - V_B = 5.7 \ V - 0.7 \ V = 5.0 \ V$
Using Ohm's law,the current $I$ in the circuit is:
$I = \frac{V_{eff}}{R_S} = \frac{5.0 \ V}{5 \times 10^3 \ \Omega} = 1 \times 10^{-3} \ A$
$I = 1 \ mA$

(B) In an unbiased $p-n$ junction diode,the diffusion of charge carriers across the junction creates a region depleted of mobile charge carriers.
This region is known as the depletion region.
It consists of immobile ionized donor and acceptor atoms (ions) fixed in the crystal lattice.
Since the mobile charge carriers (free electrons and holes) have diffused away or recombined,this region contains neither free electrons nor holes.

(B) In conductors,there is no forbidden energy gap between the conduction band and the valence band.
For materials like copper,which is a metal (conductor),the valence band and the conduction band overlap.
This overlapping allows electrons to move freely,which is why copper is a good conductor of electricity.
Silicon,germanium,and carbon (in diamond form) are semiconductors or insulators,where a forbidden energy gap exists.

(B) When unpolarised light of intensity $I_0$ passes through the first polaroid,the intensity of the transmitted light is $I_1 = \frac{I_0}{2}$.
The angle between the pass axes of the first and second polaroid is $\theta_{12} = 30^{\circ}$. According to Malus' Law,the intensity of light transmitted through the second polaroid is $I_2 = I_1 \cos^2(30^{\circ}) = \frac{I_0}{2} \times (\frac{\sqrt{3}}{2})^2 = \frac{I_0}{2} \times \frac{3}{4} = \frac{3 I_0}{8}$.
The angle between the pass axes of the second and third polaroid is $\theta_{23} = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
According to Malus' Law,the intensity of light transmitted through the third polaroid is $I_3 = I_2 \cos^2(60^{\circ}) = \frac{3 I_0}{8} \times (\frac{1}{2})^2 = \frac{3 I_0}{8} \times \frac{1}{4} = \frac{3 I_0}{32}$.

(D) In Young's double slit experiment, the fringe width is given by $\beta = \frac{D \lambda}{d}$.
Since the experimental setup ($D$ and $d$) remains the same, $\beta \propto \lambda$.
According to the de Broglie hypothesis, the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.
For the same speed $v$, the wavelength is inversely proportional to the mass $m$ of the particle, i.e., $\lambda \propto \frac{1}{m}$.
Since the mass of a proton $(m_p)$ is much greater than the mass of an electron $(m_e)$, we have $m_p > m_e$.
Therefore, the wavelength of the proton beam $(\lambda_2)$ will be smaller than the wavelength of the electron beam $(\lambda_1)$, i.e., $\lambda_2 < \lambda_1$.
Since $\beta \propto \lambda$, it follows that $\beta_2 < \beta_1$, or $\beta_1 > \beta_2$.