KCET 2020 Physics Question Paper with Answer and Solution

(A) The particle is moving in a uniform circular motion on a smooth horizontal table.
In this motion,the only horizontal force acting on the particle towards the center of the circle is the tension $T$ in the string.
According to Newton's second law of motion,the net force acting on a body in circular motion must be equal to the centripetal force required for that motion.
The centripetal force $F_c$ is given by the formula $F_c = \frac{m v^{2}}{l}$.
Since the tension $T$ is the only force providing this centripetal force,we have:
Net force (directed towards the centre) $= T = \frac{m v^{2}}{l}$.
Therefore,the net force on the particle directed towards the centre is $T$.

(D) Given,mass,$m = (0.3 \pm 0.003) \text{ g}$
Radius,$r = (0.5 \pm 0.005) \text{ mm}$
Length,$l = (6 \pm 0.06) \text{ cm}$
Density of cylinder,$\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{m}{\pi r^2 l}$
The fractional error in $\rho$ is given by:
$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$
Percentage error in $\rho$ is:
$\frac{\Delta \rho}{\rho} \times 100 = \left( \frac{\Delta m}{m} \times 100 \right) + 2 \left( \frac{\Delta r}{r} \times 100 \right) + \left( \frac{\Delta l}{l} \times 100 \right)$
Substituting the values:
$\frac{\Delta \rho}{\rho} \times 100 = \left( \frac{0.003}{0.3} \times 100 \right) + 2 \left( \frac{0.005}{0.5} \times 100 \right) + \left( \frac{0.06}{6} \times 100 \right)$
$\frac{\Delta \rho}{\rho} \times 100 = 1\% + 2(1\%) + 1\% = 4\%$

(A) According to the Law of Floatation,for an object floating in a liquid,the weight of the object is equal to the weight of the liquid displaced.
Let $V$ be the total volume of the iceberg and $V_{sub}$ be the volume submerged in water.
Weight of the iceberg = $V \cdot \rho_{i} \cdot g$
Weight of the water displaced = $V_{sub} \cdot \rho_{w} \cdot g$
Equating the two: $V \cdot \rho_{i} \cdot g = V_{sub} \cdot \rho_{w} \cdot g$
The fraction of the volume submerged is $\frac{V_{sub}}{V} = \frac{\rho_{i}}{\rho_{w}}$.
Given $\rho_{i} = 0.917 \ g \ cm^{-3}$ and $\rho_{w} = 1.00 \ g \ cm^{-3}$.
Therefore,$\frac{V_{sub}}{V} = \frac{0.917}{1.00} = 0.917$.

(A) When a soap bubble is charged,the charge distributes uniformly over its surface.
Due to the electrostatic repulsion between the like charges on the surface,an outward pressure is exerted on the bubble.
This additional outward pressure acts in the same direction as the pressure due to the air inside the bubble.
Consequently,the bubble expands to balance these forces,leading to an increase in its radius.

(C) For a perfect rigid body,the strain is always zero for any applied stress.
Young's modulus $(Y)$ is defined as the ratio of normal stress to longitudinal strain:
$Y = \frac{\text{Normal stress}}{\text{Longitudinal strain}}$
Since the strain for a perfectly rigid body is $0$,the denominator becomes $0$.
Therefore,$Y = \frac{\text{Stress}}{0} = \infty$.
Thus,the Young's modulus of a perfect rigid body is infinity.

(B) The value of acceleration due to gravity at height $h = 10 \,km$ is given by $g_h = g(1 - \frac{2h}{R_e}) = x$,where $R_e$ is the radius of the Earth.
At a depth $d$ below the surface,the acceleration due to gravity is given by $g_d = g(1 - \frac{d}{R_e}) = x$.
Equating the two expressions for $x$:
$g(1 - \frac{2h}{R_e}) = g(1 - \frac{d}{R_e})$
$1 - \frac{2h}{R_e} = 1 - \frac{d}{R_e}$
$\frac{2h}{R_e} = \frac{d}{R_e}$
$d = 2h$.
Given $h = 10 \,km$,we get $d = 2 \times 10 \,km = 20 \,km$.

(C) Let $h$ be the height of the escalator.
Speed of the girl walking on the stationary escalator is $v_{g} = \frac{h}{20}$.
Speed of the moving escalator is $v_{e} = \frac{h}{30}$.
When the girl walks on the moving escalator, her effective speed is the sum of her walking speed and the escalator's speed: $v_{total} = v_{g} + v_{e}$.
$v_{total} = \frac{h}{20} + \frac{h}{30} = h \left( \frac{3+2}{60} \right) = \frac{5h}{60} = \frac{h}{12}$.
The time taken to cover the distance $h$ is $t = \frac{h}{v_{total}} = \frac{h}{h/12} = 12 \,s$.

(B) Let the body start from rest $(u = 0)$ and move with a constant acceleration $a$.
At any time $t$,the velocity $v$ of the body is given by the first equation of motion: $v = u + at = 0 + at = at$.
The force $F$ acting on the body is given by Newton's second law: $F = ma$.
Power $P$ delivered to the body is defined as the product of force and velocity: $P = F \cdot v$.
Substituting the expressions for $F$ and $v$: $P = (ma) \cdot (at) = ma^2t$.
Since mass $m$ and acceleration $a$ are constants,the term $ma^2$ is constant.
Therefore,$P \propto t$.

(D) The velocity of rain is $\vec{v}_r = -12 \hat{j} \ ms^{-1}$.
The velocity of the woman is $\vec{v}_w = -12 \hat{i} \ ms^{-1}$ (since she is moving from east to west).
The relative velocity of rain with respect to the woman is $\vec{v}_{rw} = \vec{v}_r - \vec{v}_w = -12 \hat{j} - (-12 \hat{i}) = 12 \hat{i} - 12 \hat{j} \ ms^{-1}$.
To protect herself from rain,the woman must hold her umbrella in the direction of the relative velocity of the rain,which is opposite to $\vec{v}_{rw}$.
The angle $\theta$ with the vertical is given by $\tan \theta = \frac{|v_w|}{|v_r|} = \frac{12}{12} = 1$ .
Therefore,$\theta = \tan^{-1}(1) = 45^{\circ}$.
Since the woman is moving towards the west,the relative velocity vector points towards the west,so she must hold the umbrella at an angle of $45^{\circ}$ towards the west to counteract the rain.

(A) Given: Mass of the plate,$M = 2 \, kg$.
Moment of inertia about the $x$-axis,$I_{x} = 0.2 \, kg \cdot m^{2}$.
Moment of inertia about the $y$-axis,$I_{y} = 0.3 \, kg \cdot m^{2}$.
According to the perpendicular axis theorem,the moment of inertia $I_{z}$ of a planar body about an axis perpendicular to its plane (passing through the origin $O$) is given by:
$I_{z} = I_{x} + I_{y}$
$I_{z} = 0.2 + 0.3 = 0.5 \, kg \cdot m^{2}$.
We know that the moment of inertia is also related to the radius of gyration $k$ by the formula:
$I = M k^{2}$
Substituting the values:
$0.5 = 2 \cdot k^{2}$
$k^{2} = \frac{0.5}{2} = 0.25 \, m^{2}$
$k = \sqrt{0.25} = 0.5 \, m$.
Converting to centimeters:
$k = 0.5 \times 100 \, cm = 50 \, cm$.
Thus,the radius of gyration is $50 \, cm$.

(A) Initial angular velocity of the wheel,$\omega_{0} = 0 \ rad/s$.
Final angular velocity,$\omega = 10 \ rad/s$.
Time taken,$t = 5 \ s$.
Using the first equation of rotational motion:
$\omega = \omega_{0} + \alpha t$
$10 = 0 + \alpha \times 5$
$\alpha = \frac{10}{5} = 2 \ rad/s^{2}$.
Now,using the second equation of rotational motion to find the total angle $\theta$:
$\theta = \omega_{0} t + \frac{1}{2} \alpha t^{2}$
$\theta = 0 \times 5 + \frac{1}{2} \times 2 \times (5)^{2}$
$\theta = 0 + 25 = 25 \ rad$.

(C) According to Newton's law of cooling,the rate of cooling is proportional to the surface area of the body $(dQ/dt \propto A)$.
For a given mass and material,the volume is constant. Among a sphere,a cube,and a thin circular plate,the thin circular plate has the largest surface area,while the sphere has the smallest surface area.
Since the rate of cooling is directly proportional to the surface area,the body with the largest surface area will lose heat the fastest.
Therefore,the plate will cool the fastest and the sphere will cool the slowest.

(C) According to the first law of thermodynamics,the heat supplied $Q$ is equal to the change in internal energy $\Delta U$ plus the work done $W$: $Q = \Delta U + W$.
For a process at constant pressure,the heat supplied is $Q = n C_p \Delta T$.
The change in internal energy is $\Delta U = n C_V \Delta T$.
The work done is $W = Q - \Delta U = n C_p \Delta T - n C_V \Delta T = n R \Delta T$.
The fraction of heat energy converted into work is $\frac{W}{Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p}$.
For a monoatomic ideal gas,$C_p = \frac{5}{2} R$.
Therefore,the fraction is $\frac{W}{Q} = \frac{R}{\frac{5}{2} R} = \frac{2}{5}$.

(A) For an adiabatic process,the system does work on the surroundings at the expense of its internal energy.
Since the internal energy of an ideal gas depends only on its temperature $(U = n C_v T)$,a decrease in internal energy implies a decrease in temperature $(T)$.
According to the ideal gas equation,$p V = n R T$.
Since $n$ and $R$ are constants and the temperature $T$ decreases during adiabatic expansion,the product $p V$ must also decrease.

(D) According to the Doppler effect,when a source of sound moves with a constant speed $v_s$ towards a stationary observer,the observed frequency $n'$ is given by $n' = n \left( \frac{v}{v - v_s} \right)$,where $v$ is the speed of sound in air. Since $v - v_s < v$,it follows that $n' > n$. This frequency remains constant as long as the train approaches the observer at a constant speed.
When the train passes the observer and moves away,the source is now moving away from the stationary observer. The observed frequency $n''$ is given by $n'' = n \left( \frac{v}{v + v_s} \right)$. Since $v + v_s > v$,it follows that $n'' < n$. This frequency also remains constant as long as the train moves away at a constant speed.
Therefore,the frequency is higher than the actual frequency $n$ before passing the observer and drops to a value lower than $n$ after passing the observer. This step-like change is correctly represented by the graph in option $D$.

(C) Mass of the tray, $m = 12 \,kg$.
Time period, $T = 1.5 \,s$.
Let $k$ be the spring constant of each spring.
Since the springs are connected in parallel, the effective spring constant is $k_{\text{net}} = k + k = 2k$.
The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{k_{\text{net}}}}$.
Substituting the values, we get $1.5 = 2\pi \sqrt{\frac{12}{2k}}$.
$1.5 = 2\pi \sqrt{\frac{6}{k}}$.
Squaring both sides, we have $(1.5)^2 = (2\pi)^2 \cdot \frac{6}{k}$.
$2.25 = 4\pi^2 \cdot \frac{6}{k}$.
$k = \frac{24\pi^2}{2.25} \approx \frac{24 \times 9.87}{2.25} \approx 105.28 \,Nm^{-1}$.
Thus, the spring constant of each spring is approximately $105 \,Nm^{-1}$.

(D) The magnetic field at the center of a circular loop is $B = \frac{\mu_0 i}{2r}$.
The magnetic moment of the loop is $M = i \pi r^2$.
The ratio is $x = \frac{B}{M} = \frac{\mu_0 i / 2r}{i \pi r^2} = \frac{\mu_0}{2 \pi r^3}$.
When the current $i$ becomes $2i$ and the radius $r$ becomes $2r$,the new magnetic field $B'$ is $B' = \frac{\mu_0 (2i)}{2(2r)} = \frac{\mu_0 i}{2r} = B$.
The new magnetic moment $M'$ is $M' = (2i) \pi (2r)^2 = (2i) \pi (4r^2) = 8(i \pi r^2) = 8M$.
The new ratio $x'$ is $x' = \frac{B'}{M'} = \frac{B}{8M} = \frac{1}{8} \left( \frac{B}{M} \right) = \frac{x}{8}$.

(C) Given,peak voltages are $V_{C}=30 \,V$,$V_{L}=110 \,V$,and $V_{R}=60 \,V$.
The peak voltage $(V_{0})$ across a series $L-C-R$ circuit is given by the formula:
$V_{0} = \sqrt{V_{R}^{2} + (V_{L} - V_{C})^{2}}$
Substituting the given values:
$V_{0} = \sqrt{(60)^{2} + (110 - 30)^{2}}$
$V_{0} = \sqrt{60^{2} + 80^{2}}$
$V_{0} = \sqrt{3600 + 6400} = \sqrt{10000} = 100 \,V$
The rms value of the applied voltage is related to the peak voltage by $V_{\text{rms}} = \frac{V_{0}}{\sqrt{2}}$.
$V_{\text{rms}} = \frac{100}{\sqrt{2}} \approx 70.7 \,V$.

(C) Given, $L = 0.5 \, mH = 0.5 \times 10^{-3} \, H$ and $C = 20 \, \mu F = 20 \times 10^{-6} \, F$.
The resonant frequency of an $L-C$ circuit is given by the formula:
$f = \frac{1}{2 \pi \sqrt{L C}}$
Substituting the values:
$f = \frac{1}{2 \pi \sqrt{0.5 \times 10^{-3} \times 20 \times 10^{-6}}}$
$f = \frac{1}{2 \pi \sqrt{10 \times 10^{-9}}} = \frac{1}{2 \pi \sqrt{10^{-8}}}$
$f = \frac{1}{2 \pi \times 10^{-4}} = \frac{10^4}{2 \pi} \approx \frac{10000}{6.283} \approx 1592.3 \, Hz$
Thus, the resonant frequency is approximately $1592 \, Hz$.

(B) Given,power factor $= \frac{1}{\sqrt{3}}$.
Inductive reactance,$X_{L} = 2 \Omega$.
The power factor in an $R-L$ circuit is given by:
$\text{Power factor} = \frac{R}{Z} = \frac{R}{\sqrt{R^{2} + X_{L}^{2}}}$
Substituting the given values:
$\frac{1}{\sqrt{3}} = \frac{R}{\sqrt{R^{2} + 2^{2}}} = \frac{R}{\sqrt{R^{2} + 4}}$
Squaring both sides:
$\frac{1}{3} = \frac{R^{2}}{R^{2} + 4}$
$R^{2} + 4 = 3R^{2}$
$2R^{2} = 4$
$R^{2} = 2$
$R = \sqrt{2} \Omega$.

(B) According to Rutherford's $\alpha$-particle scattering experiment,the following observations were made:
$(i)$ Most of the $\alpha$-particles passed through the gold foil undeflected,corresponding to path $A-A^{\prime}$.
(ii) $A$ small fraction of $\alpha$-particles were scattered by small angles,corresponding to path $C-C^{\prime}$.
(iii) $A$ very small fraction of $\alpha$-particles were scattered by large angles (or reflected back),corresponding to path $B-B^{\prime}$.
Since the number of particles scattered decreases as the scattering angle increases,the number of particles $n$ follows the order $n_{A^{\prime}} > n_{C^{\prime}} > n_{B^{\prime}}$.
Therefore,the number of $\alpha$-particles is maximum in $A^{\prime}$ and minimum in $B^{\prime}$.

(B) The angular momentum of an electron in an orbit of radius $r$ is given by Bohr's quantization condition:
$mvr = \frac{nh}{2\pi}$
For the ground state $(n=1)$,the momentum $p = mv$ is:
$p = \frac{h}{2\pi r}$
The de-Broglie wavelength $\lambda$ is defined as:
$\lambda = \frac{h}{p}$
Substituting the value of $p$ from the angular momentum equation:
$\lambda = \frac{h}{(h / 2\pi r)} = 2\pi r$
In the ground state of a hydrogen atom,the Bohr radius $r = 0.53 \text{ Å}$.
Therefore,$\lambda = 2 \times 3.14 \times 0.53 \text{ Å} \approx 3.33 \text{ Å}$.
Thus,the correct option is $3.3 \text{ Å}$.

(C) The time period of revolution $T$ of an electron in the $n^{th}$ orbit is defined as the ratio of the circumference of the orbit to the orbital velocity of the electron: $T = \frac{2 \pi r_{n}}{v_{n}}$.
According to Bohr's theory,the radius of the $n^{th}$ orbit is $r_{n} \propto n^{2}$ and the velocity of the electron in the $n^{th}$ orbit is $v_{n} \propto \frac{1}{n}$.
Substituting these proportionalities into the formula for the time period:
$T_{n} \propto \frac{r_{n}}{v_{n}} \propto \frac{n^{2}}{1/n} = n^{3}$.
Thus,the time period of revolution is proportional to $n^{3}$.

(C) Let the capacitance of each identical capacitor be $C$.
When connected in parallel,the equivalent capacitance is $C_{p} = C + C = 2C$.
When connected in series,the equivalent capacitance is $\frac{1}{C_{s}} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$,which gives $C_{s} = \frac{C}{2}$.
According to the problem,the difference between these equivalent capacitances is $6 \mu F$:
$C_{p} - C_{s} = 6 \mu F$
$2C - \frac{C}{2} = 6 \mu F$
$\frac{3C}{2} = 6 \mu F$
$3C = 12 \mu F$
$C = 4 \mu F$.

(D) The given resistance is $R = 0.2 \ k\Omega \pm 10 \%$.
Converting this to ohms,we get $R = 200 \ \Omega \pm 10 \%$.
This can be written as $R = 20 \times 10^1 \ \Omega \pm 10 \%$.
According to the standard carbon resistor colour code:
The digit $2$ corresponds to the colour red.
The digit $0$ corresponds to the colour black.
The multiplier $10^1$ corresponds to the colour brown.
The tolerance of $10 \%$ corresponds to the colour silver.
Therefore,the colour code is red,black,brown,and silver.

(A) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length of the conductor in the direction of current flow,and $A$ is the cross-sectional area perpendicular to the current flow.
From the relation $R \propto \frac{L}{A}$,we see that for a fixed volume of the material,the resistance is proportional to the square of the length $(L^2)$ or inversely proportional to the square of the area $(A^2)$.
Alternatively,for a fixed rod,the resistance is maximum when the current flows through the smallest cross-sectional area $A$,which results in the longest effective length $L$.
The three possible cross-sectional areas are:
$1$. $A_1 = 10 \text{ cm} \times 1 \text{ cm} = 10 \text{ cm}^2$
$2$. $A_2 = 10 \text{ cm} \times 0.5 \text{ cm} = 5 \text{ cm}^2$
$3$. $A_3 = 1 \text{ cm} \times 0.5 \text{ cm} = 0.5 \text{ cm}^2$
The resistance is maximum when the cross-sectional area $A$ is minimum. The minimum area is $0.5 \text{ cm}^2$,which corresponds to the faces of dimensions $1 \text{ cm} \times 0.5 \text{ cm}$.

(A) Let the total current entering the cubical network at point $A$ be $6I$. Due to the symmetry of the cube, the current splits equally into three branches at $A$, each carrying $2I$.
At the next set of nodes, these currents split again.
Applying Kirchhoff's voltage law to a path from $A$ to $B$ (e.g., $A \rightarrow \text{node} \rightarrow \text{node} \rightarrow B$), the potential drop $V$ across the network is the sum of potential drops across the resistors in any path.
For a path consisting of three edges, the potential drop is $V = (2I \times R) + (I \times R) + (2I \times R) = 5IR$.
Given $R = 1 \Omega$, we have $V = 5I$.
The equivalent resistance $R_{eq}$ is given by $R_{eq} = \frac{V}{I_{total}} = \frac{5I}{6I} = \frac{5}{6} \Omega$.

(B) Given:
Emf of the battery, $E = 12 \, V$
Internal resistance, $r = 2 \times 10^{-2} \, \Omega = 0.02 \, \Omega$
Current drawn by the starter motor, $I = 80 \, A$
When the starter is $ON$, the battery supplies current to the motor. The terminal voltage $V$ is given by the formula:
$V = E - Ir$
Substituting the given values:
$V = 12 - (80 \times 0.02)$
$V = 12 - 1.6$
$V = 10.4 \, V$
Therefore, the terminal voltage when the starter is $ON$ is $10.4 \, V$.

(D) The $I-V$ characteristic of a copper wire of length $L$ and area of cross-section $A$ is shown in the figure.
Since the $I-V$ characteristic is a straight line,it obeys Ohm's law.
The slope of the $I-V$ graph is given by $\text{slope} = \tan \theta = \frac{I}{V}$.
From Ohm's law,$V = IR$,so $\frac{I}{V} = \frac{1}{R}$.
Since $R = \rho \frac{L}{A}$,we have $\text{slope} = \frac{1}{\rho \frac{L}{A}} = \frac{A}{\rho L}$.
$(i)$ If the length $L$ is increased,the slope $\frac{A}{\rho L}$ decreases.
(ii) If the area $A$ is increased,the slope $\frac{A}{\rho L}$ increases.
(iii) If a steel wire of the same dimensions is used,since $\rho_{\text{steel}} > \rho_{\text{copper}}$,the slope $\frac{A}{\rho L}$ decreases.
(iv) If the experiment is performed at a higher temperature,the resistivity $\rho$ of copper increases,so the slope $\frac{A}{\rho L}$ decreases.
Therefore,the correct statement is that the slope becomes less if the length of the wire is increased.

(A) Given,length of potentiometer wire $l = 5 \,m$.
Emf of the primary battery $E = 10 \,V$.
The potential gradient $K$ across the potentiometer wire is given by $K = \frac{E}{l} = \frac{10 \,V}{5 \,m} = 2 \,V/m$.
When the secondary cell is connected,the balancing length is $l_1 = 200 \,cm = 2 \,m$.
The emf of the secondary cell $E_s$ is given by $E_s = K \times l_1$.
Substituting the values,$E_s = 2 \,V/m \times 2 \,m = 4 \,V$.

(C) The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $K.E. = eV$.
Since the electron starts from rest, its kinetic energy is $\frac{1}{2}mv^2$.
Equating the two, we get $\frac{1}{2}mv^2 = eV$.
Solving for velocity $v$, we have $v = \sqrt{\frac{2eV}{m}}$.
Substituting the values: $e = 1.6 \times 10^{-19} \,C$, $m = 9.1 \times 10^{-31} \,kg$, and $V = 1200 \,V$.
$v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 1200}{9.1 \times 10^{-31}}} = \sqrt{\frac{3.84 \times 10^{-16}}{9.1 \times 10^{-31}}} = \sqrt{0.42198 \times 10^{15}} = \sqrt{42.198 \times 10^{13}} \approx 2.05 \times 10^{7} \,ms^{-1}$.
Rounding this value, we get $v \approx 2.1 \times 10^{7} \,ms^{-1}$.

(C) Given: Intensity $I = 20 \,W/cm^2 = 20 \times 10^4 \,W/m^2$.
Area $A = 25 \,cm \times 15 \,cm = 375 \,cm^2 = 375 \times 10^{-4} \,m^2$.
Time $t = 1 \,s$.
The energy $E$ incident on the surface per second is $E = I \times A \times t = 20 \times 10^4 \times 375 \times 10^{-4} \times 1 = 7500 \,J$.
For a perfectly reflecting surface,the momentum $p$ imparted by light is given by $p = \frac{2E}{c}$,where $c = 3 \times 10^8 \,m/s$ is the speed of light.
$p = \frac{2 \times 7500}{3 \times 10^8} = \frac{15000}{3 \times 10^8} = 5000 \times 10^{-8} = 5 \times 10^{-5} \,kg \cdot m/s$.

(B) Given,angular momentum of electron in $H$-atom $= \frac{3h}{2\pi} \dots (i)$
From Bohr's postulate,angular momentum $= \frac{nh}{2\pi} \dots (ii)$
Comparing equations $(i)$ and $(ii)$,we get $n = 3$.
The kinetic energy $(KE)$ of an electron in a hydrogen atom is given by the formula:
$KE = \frac{13.6 \times Z^2}{n^2} \text{ eV}$
For hydrogen atom,atomic number $Z = 1$. Substituting $Z = 1$ and $n = 3$:
$KE = \frac{13.6 \times 1^2}{3^2} \text{ eV}$
$KE = \frac{13.6}{9} \text{ eV}$
$KE = 1.51 \text{ eV}$

(A) From the graph,we observe that the stopping potential $V_{0}$ is the same for curves $1$ and $2$. However,for curve $3$,the stopping potential is greater in magnitude than that for curves $1$ and $2$.
We know that the stopping potential is given by the relation: $e V_{0} = E_{\max} = h \gamma - \phi_{0}$,where $\phi_{0}$ is the work function of the metal.
Since $V_{0}$ is the same for curves $1$ and $2$,it implies that the incident frequencies are equal,i.e.,$\gamma_{1} = \gamma_{2}$.
For curve $3$,the stopping potential is higher,which means $\gamma_{3} > \gamma_{1} = \gamma_{2}$.
Regarding the saturation current,it depends on the intensity of the incident light. Curves $2$ and $3$ reach the same saturation current,meaning $I_{2} = I_{3}$,while curve $1$ has a lower saturation current,so $I_{1} < I_{2} = I_{3}$.
Therefore,comparing the options,$\gamma_{1} = \gamma_{2}$ and $I_{1} \neq I_{2}$ is the correct statement.

(B) Given: Inductance $L = 0.2 \,H$,initial current $I_1 = 5 \,A$,final current $I_2 = 2 \,A$,and time interval $\Delta t = 0.5 \,s$.
The change in current is $\Delta I = I_1 - I_2 = 5 \,A - 2 \,A = 3 \,A$.
The magnitude of the average induced emf $|e|$ in the coil is given by the formula:
$|e| = L \left| \frac{\Delta I}{\Delta t} \right| = 0.2 \,H \times \frac{3 \,A}{0.5 \,s} = 0.2 \times 6 \,V = 1.2 \,V$.

(B) Given: Length of the rod $l = 2 \, m$, speed $v = 5 \, ms^{-1}$, magnetic field $B = 0.04 \, T$, and resistance $R = 3 \, \Omega$.
The motional electromotive force $(EMF)$ induced in the rod is given by:
$\varepsilon = B l v$
$\varepsilon = 0.04 \, T \times 2 \, m \times 5 \, ms^{-1} = 0.4 \, V$
The current $I$ induced in the rod is given by Ohm's law:
$I = \frac{\varepsilon}{R}$
$I = \frac{0.4 \, V}{3 \, \Omega} = 0.1333... \, A$
$I \approx 0.133 \, A = 133 \, mA$.

(B) The nuclear force is a short-range force, typically effective within a range of a few $fm$ $(1 \, fm = 10^{-15} \, m)$.
Whereas, the electromagnetic force is a long-range force that follows the inverse-square law.
Given the separation is $10 \, nm = 10 \times 10^{-9} \, m = 10^{-8} \, m$.
Since $10^{-8} \, m$ is much larger than the range of the nuclear force $(\, 10^{-15} \, m)$, the nuclear force $F_{n}$ is effectively zero at this distance.
Therefore, the electromagnetic force $F_{e}$ is significantly greater than the nuclear force $F_{n}$, i.e., $F_{e} \gg F_{n}$.

(B) The torque $\tau$ on an electric dipole placed in a uniform electric field $E$ is given by $\tau = p E \sin \theta$,where $p$ is the electric dipole moment and $\theta$ is the angle between $p$ and $E$.
For small angles $\theta$,we can approximate $\sin \theta \approx \theta$.
Thus,the torque becomes $\tau = p E \theta$.
Since the torque is also given by $\tau = I \alpha$,where $I$ is the moment of inertia and $\alpha$ is the angular acceleration,we have $I \alpha = p E \theta$.
Rearranging for angular acceleration,we get $\alpha = \frac{p E}{I} \theta$.
This equation is of the form $\alpha = -\omega^2 \theta$,which represents simple harmonic motion.
Comparing the terms,we find $\omega^2 = \frac{p E}{I}$,so $\omega = \sqrt{\frac{p E}{I}}$.
The time period $T$ is given by $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{I}{p E}}$.

(A) The density of electric field lines is directly proportional to the magnitude of the electric field.
Given that the separation of field lines on the left (at point $B$) is twice the separation of those on the right (at point $A$),the electric field at $B$ is half the electric field at $A$.
$E_B = \frac{E_A}{2} = \frac{40 \ Vm^{-1}}{2} = 20 \ Vm^{-1}$.
The force $F$ on a charge $q$ in an electric field $E$ is given by $F = qE$.
Given $q = 20 \ \mu C = 20 \times 10^{-6} \ C$.
$F = (20 \times 10^{-6} \ C) \times (20 \ Vm^{-1}) = 400 \times 10^{-6} \ N = 4 \times 10^{-4} \ N$.

(C) Given,linear charge density of the infinitely long wire,$\lambda = \frac{1}{4} \times 10^{-2} \text{ C/m} = 0.25 \times 10^{-2} \text{ C/m} = 2.5 \times 10^{-3} \text{ C/m}$.
Distance from the wire,$r = 20 \text{ cm} = 0.2 \text{ m}$.
The electric field $E$ due to an infinitely long straight wire is given by the formula: $E = \frac{\lambda}{2 \pi \varepsilon_0 r} = \frac{2k\lambda}{r}$,where $k = \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$.
Substituting the values:
$E = \frac{2 \times (9 \times 10^9) \times (2.5 \times 10^{-3})}{0.2}$
$E = \frac{18 \times 10^9 \times 2.5 \times 10^{-3}}{0.2}$
$E = \frac{45 \times 10^6}{0.2} = 225 \times 10^6 = 2.25 \times 10^8 \text{ N/C}$.
Note: Based on the provided options and the calculation,the correct magnitude is $2.25 \times 10^8 \text{ N/C}$. Given the options provided in the prompt,option $C$ is the intended answer assuming a typo in the exponent of the charge density in the original question.

(B) According to Gauss's Law,the total electric flux through a closed surface is $\frac{q_{enclosed}}{\varepsilon_{0}}$.
When a charge $q$ is placed at one corner of a cube,it is shared equally by $8$ such identical cubes to enclose the charge completely.
Therefore,the flux through one cube is $\Phi_{cube} = \frac{q}{8 \varepsilon_{0}}$.
$A$ cube has $6$ faces. The charge $q$ is located at a corner. The three faces meeting at this corner (in this case,faces $ADHE$,$ABFE$,and $ABCD$ if the charge is at $A$) contain the charge $q$ on their plane. For these faces,the electric field lines are parallel to the surface,meaning the angle between the area vector and the electric field is $90^{\circ}$. Thus,the flux through these $3$ faces is $0$.
The remaining $3$ faces share the total flux $\Phi_{cube}$ equally due to symmetry.
Therefore,the flux through each of the remaining $3$ faces is $\Phi_{face} = \frac{\Phi_{cube}}{3} = \frac{q / 8 \varepsilon_{0}}{3} = \frac{q}{24 \varepsilon_{0}}$.
Thus,the electric flux through the face $ABCD$ is $\frac{q}{24 \varepsilon_{0}}$.

(B) According to the given figure,the line $A B$ is perpendicular to the direction of the electric field lines. Hence,the surface passing through line $A B$ and perpendicular to the electric field lines behaves like an equipotential surface,therefore
$V_{A}=V_{B} \dots (i)$
Electric field and electric potential are related as
$E=-\frac{d V}{d x} \Rightarrow V=-\int E d x$
This indicates that the electric potential decreases in the direction of the electric field,i.e.
$V_{B}>V_{C} \dots (ii)$
From Eqs. $(i)$ and $(ii)$,we have
$V_{A}=V_{B}>V_{C}$

(C) The total magnetic field at point $O$ is the vector sum of the magnetic fields produced by the three segments: the straight wire $AB$,the circular arc $BC$,and the straight wire $CD$.
$1$. For the straight wire $AB$,the point $O$ lies on the axis of the wire. Therefore,the magnetic field due to segment $AB$ is $B_{AB} = 0$.
$2$. For the straight wire $CD$,the point $O$ is at a perpendicular distance $r$ from the wire. The wire extends from $C$ to infinity. The magnetic field due to a semi-infinite wire is given by $B_{CD} = \frac{\mu_0 I}{4 \pi r}$.
$3$. For the circular arc $BC$,the angle subtended at the center $O$ is $\theta = 270^\circ = \frac{3\pi}{2}$ radians. The magnetic field due to a circular arc is $B_{arc} = \frac{\mu_0 I}{2r} \cdot \frac{\theta}{2\pi} = \frac{\mu_0 I}{2r} \cdot \frac{3\pi/2}{2\pi} = \frac{3\mu_0 I}{8r}$.
Since the directions of the magnetic fields from the arc and the semi-infinite wire are the same (into the page),the net magnetic field is:
$B_{net} = B_{AB} + B_{arc} + B_{CD} = 0 + \frac{3\mu_0 I}{8r} + \frac{\mu_0 I}{4\pi r} = \frac{3}{8} \frac{\mu_0 I}{r} + \frac{\mu_0 I}{4\pi r}$.

(A) Let the two wires be located at $x = -d$ and $x = +d$ on the $xx'$ axis. The currents are flowing out of the plane of the paper.
Using the right-hand rule, the magnetic field due to a wire carrying current $i$ at a distance $r$ is $B = \frac{\mu_0 i}{2\pi r}$.
For the wire at $x = -d$, the field is $B_1 = \frac{\mu_0 i}{2\pi (x+d)}$ in the $(-\hat{j})$ direction for $x > -d$.
For the wire at $x = +d$, the field is $B_2 = \frac{\mu_0 i}{2\pi (d-x)}$ in the $\hat{j}$ direction for $x < d$.
In the region between the wires $(-d < x < d)$, the net magnetic field is $B_{net} = B_2 - B_1 = \frac{\mu_0 i}{2\pi} \left[ \frac{1}{d-x} - \frac{1}{d+x} \right]$.
At the midpoint $x = 0$, $B_{net} = 0$. As $x$ approaches $d$, $B_{net} \to \infty$ in the $\hat{j}$ direction. As $x$ approaches $-d$, $B_{net} \to -\infty$ in the $(-\hat{j})$ direction.
Outside the wires, the fields add up in magnitude. The correct graphical representation showing the field passing through zero at the midpoint and having opposite signs on either side is represented by option $A$.

(A) According to the Biot-Savart law,the magnetic field $dB$ at a point due to a current element $i d l$ is given by the expression:
$dB = \frac{\mu_{0}}{4 \pi} \frac{i (d l \times r)}{r^{3}}$
Here,$i$ is the current,$d l$ is the length element vector,$r$ is the position vector of the point where the field is to be calculated relative to the current element,and $\mu_{0}$ is the permeability of free space.
Thus,the correct expression is $\frac{\mu_{0} i}{4 \pi} \frac{d l \times r}{r^{3}}$.

(C) For a long cylindrical wire of radius $R$ carrying a uniform current $I$:
$1$. Inside the wire $(r < R)$,the magnetic field $B_{\text{in}}$ is given by $B_{\text{in}} = \frac{\mu_0 I r}{2 \pi R^2}$. Since $\mu_0, I, R$ are constants,we have $B_{\text{in}} \propto r$. This represents a straight line passing through the origin.
$2$. Outside the wire $(r > R)$,the magnetic field $B_{\text{out}}$ is given by $B_{\text{out}} = \frac{\mu_0 I}{2 \pi r}$. Since $\mu_0, I$ are constants,we have $B_{\text{out}} \propto \frac{1}{r}$. This represents a rectangular hyperbola.
$3$. At the surface $(r = R)$,the magnetic field is maximum,$B_{\text{max}} = \frac{\mu_0 I}{2 \pi R}$.
Combining these,the graph shows a linear increase for $r < R$ and a hyperbolic decrease for $r > R$,which corresponds to Graph $C$.

(C) When a charged particle is accelerated in a cyclotron,the radius $r$ of its path is given by $r = \frac{\sqrt{2Km}}{Bq}$.
Squaring both sides,we get $r^2 = \frac{2Km}{B^2q^2}$,which implies $K = \frac{B^2q^2r^2}{2m}$.
Since $B$ and $r$ are the same for all particles at the exit,$K \propto \frac{q^2}{m}$.
For protons $\left({ }_{1}^{1} H\right)$,deuterons $\left({ }_{1}^{2} H\right)$,and $\alpha$-particles $\left({ }_{2}^{4} He\right)$:
Charge ratio: $q_p : q_d : q_{\alpha} = 1 : 1 : 2$.
Mass ratio: $m_p : m_d : m_{\alpha} = 1 : 2 : 4$.
Calculating the ratio of kinetic energies $K \propto \frac{q^2}{m}$:
$K_p \propto \frac{1^2}{1} = 1$.
$K_d \propto \frac{1^2}{2} = 0.5$.
$K_{\alpha} \propto \frac{2^2}{4} = 1$.
Comparing the values,the minimum kinetic energy is gained by deuterons.

(B) Given: $M_{1} = 8 \text{ Am}^{-1}$,$B_{1} = 0.6 \text{ T}$,$T_{1} = 4 \text{ K}$,$B_{2} = 0.2 \text{ T}$,and $T_{2} = 16 \text{ K}$.
According to Curie's law for paramagnetic materials,the magnetization $M$ is given by $M = \frac{C B}{T}$,where $C$ is the Curie constant.
Therefore,$M_{1} = \frac{C B_{1}}{T_{1}}$ and $M_{2} = \frac{C B_{2}}{T_{2}}$.
Taking the ratio: $\frac{M_{2}}{M_{1}} = \frac{B_{2}}{B_{1}} \times \frac{T_{1}}{T_{2}}$.
Substituting the values: $\frac{M_{2}}{8} = \frac{0.2}{0.6} \times \frac{4}{16}$.
$\frac{M_{2}}{8} = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$.
$M_{2} = \frac{8}{12} = \frac{2}{3} \text{ Am}^{-1}$.

(D) permanent magnet is made of ferromagnetic material. In a permanent magnet,the magnetic domains are aligned in a specific direction due to the manufacturing process (like cooling in an external magnetic field). Therefore,at room temperature,the domains are all perfectly aligned to produce a net magnetic moment.

(C) The process of $\beta^{-}$-decay is represented by the following nuclear reaction:
${ }_{Z}^{A} X \longrightarrow{ }_{Z+1}^{A} Y + e^{-} + \bar{\nu} + Q$
In this process,a neutron inside the nucleus transforms into a proton,emitting an electron $(e^{-})$ and an anti-neutrino $(\bar{\nu})$.
Therefore,the correct statement is that a neutron in the nucleus decays,emitting an electron.

(C) Given,half-life,$T_{1/2} = 15$ years.
Time,$t = 30$ years.
Number of half-lives,$n = \frac{t}{T_{1/2}} = \frac{30}{15} = 2$.
The fraction of nuclei remaining undecayed after $n$ half-lives is given by $\frac{N}{N_0} = (\frac{1}{2})^n$.
Substituting $n = 2$,we get $\frac{N}{N_0} = (\frac{1}{2})^2 = \frac{1}{4} = 0.25$.
The fraction of the element that has decayed is $1 - \frac{N}{N_0} = 1 - 0.25 = 0.75$.

(A) Given, distance between poles $(d) = 3.14 \, m$.
Resolving power $(\theta) = 1 \, min = (1/60)^{\circ} = (1/60) \times (\pi/180) \, radians$.
Let the maximum distance from which the two poles can be identified distinctly be $x$.
Using the formula for small angles, $\theta = d/x$ (where $\theta$ is in radians).
$\theta = 1 \, min = (1/60) \times (\pi/180) \, rad$.
Substituting the values: $(1/60) \times (\pi/180) = 3.14 / x$.
Since $\pi \approx 3.14$, we have $(1/60) \times (3.14/180) = 3.14 / x$.
$1 / (60 \times 180) = 1 / x$.
$x = 60 \times 180 = 10800 \, m$.
$x = 10.8 \, km$.

(C) From the lens formula,$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Differentiating with respect to time $t$,we get $0 = -\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt}$.
Rearranging,we find the velocity of the image $v_i = \frac{dv}{dt} = \left(\frac{v^2}{u^2}\right) \frac{du}{dt} = m^2 v_o$,where $m = \frac{v}{u}$ is the magnification and $v_o$ is the object speed.
Since $m = \frac{f}{f+u}$,the magnification $m$ changes as the object moves towards the focus $(u \to -f)$.
Differentiating the velocity $v_i$ with respect to time to find acceleration $a_i = \frac{dv_i}{dt} = \frac{d}{dt}(m^2 v_o) = 2m v_o \frac{dm}{dt}$.
As the object approaches the focus,$m$ changes non-linearly,making $\frac{dm}{dt}$ variable. Thus,the image moves away from the lens with a non-uniform acceleration.

(D) Since the beam of light is converging at point $P$,it acts as a virtual object for the concave lens. The distance of point $P$ from the lens position is $u = +12 \,cm$ (as it is in the direction of incident light).
The focal length of the concave lens is $f = -16 \,cm$.
Using the lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Substituting the values:
$\frac{1}{v} - \frac{1}{12} = \frac{1}{-16}$
$\frac{1}{v} = \frac{1}{12} - \frac{1}{16}$
$\frac{1}{v} = \frac{4 - 3}{48} = \frac{1}{48}$
Therefore,$v = 48 \,cm$.
Since the beam converges at a distance $x$ from the lens,$x = 48 \,cm$.

(D) Given,angle of prism $= A$.
Refractive index of prism,$\mu = \cot \frac{A}{2}$.
We know the formula for refractive index in terms of minimum deviation $\delta_m$ is:
$\mu = \frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
Substituting the given value of $\mu$:
$\cot \left(\frac{A}{2}\right) = \frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
Using $\cot \theta = \frac{\cos \theta}{\sin \theta}$:
$\frac{\cos (A/2)}{\sin (A/2)} = \frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin (A/2)}$
$\cos (A/2) = \sin \left(\frac{A+\delta_m}{2}\right)$
Since $\cos \theta = \sin (90^{\circ} - \theta)$,we have:
$\sin (90^{\circ} - A/2) = \sin \left(\frac{A+\delta_m}{2}\right)$
Equating the angles:
$90^{\circ} - \frac{A}{2} = \frac{A+\delta_m}{2}$
$180^{\circ} - A = A + \delta_m$
$\delta_m = 180^{\circ} - 2A$.

(D) The circuit acts as a half-wave rectifier with a capacitor filter.
When the $AC$ supply is connected, the $p-n$ junction diode conducts only during the positive half-cycle (forward biased condition).
During this cycle, the capacitor charges to the peak value of the input $AC$ voltage.
Once charged, the capacitor maintains this peak voltage as there is no discharge path provided in the circuit.
The potential difference $V$ across the capacitor is equal to the peak value of the input $AC$ voltage $(V_0)$.
Given that the $RMS$ voltage $V_{\text{rms}} = 220 \,V$, the peak voltage is calculated as:
$V = V_0 = V_{\text{rms}} \times \sqrt{2}$
$V = 220 \times \sqrt{2} \,V = 220 \sqrt{2} \,V$.

(B) The circuit consists of two cross-coupled $NAND$ gates, forming an $S-R$ latch.
Let the inputs be $S=1$ and $R=0$.
The output $P$ is given by $P = \overline{1 \cdot Q} = \overline{Q}$.
The output $Q$ is given by $Q = \overline{0 \cdot P} = \overline{0} = 1$.
Since $Q=1$, substituting this into the equation for $P$ gives $P = \overline{1} = 0$.
Thus, the stable state of the circuit is $P=0$ and $Q=1$.

(B) At room temperature,some of the valence electrons acquire thermal energy greater than $E_{g}$ (energy gap between the conduction and valence band) and become free.
When a valence electron is shifted to the conduction band,it leaves behind a vacancy in the valence band.
This vacancy is known as a hole and it acts as a charge carrier with a positive charge.

(C) When unpolarised light of intensity $I_{0}$ passes through the first polaroid $P_{1}$,the intensity of the transmitted light becomes $I_{1} = \frac{I_{0}}{2}$.
Given $I_{0} = 128 \ Wm^{-2}$,so $I_{1} = \frac{128}{2} = 64 \ Wm^{-2}$.
According to Malus's law,when light of intensity $I_{1}$ passes through a polaroid whose pass axis makes an angle $\theta$ with the incident light's polarization direction,the transmitted intensity is $I = I_{1} \cos^{2} \theta$.
For $P_{2}$,the angle between the pass axis of $P_{1}$ and $P_{2}$ is $\theta_{1} = 45^{\circ}$.
Intensity after $P_{2}$ is $I_{2} = I_{1} \cos^{2} 45^{\circ} = 64 \times (\frac{1}{\sqrt{2}})^{2} = 64 \times \frac{1}{2} = 32 \ Wm^{-2}$.
For $P_{3}$,the angle between the pass axis of $P_{2}$ and $P_{3}$ is $\theta_{2} = 45^{\circ}$.
Intensity after $P_{3}$ is $I_{3} = I_{2} \cos^{2} 45^{\circ} = 32 \times (\frac{1}{\sqrt{2}})^{2} = 32 \times \frac{1}{2} = 16 \ Wm^{-2}$.

(D) Given: Distance between slits and screen $D = 1.2 \, m$, distance between slits $d = 2.4 \, mm = 2.4 \times 10^{-3} \, m$, thickness of mica sheet $t = 1 \, \mu m = 1 \times 10^{-6} \, m$, and refractive index $\mu = 1.5$.
When a transparent sheet is introduced in the path of one of the interfering beams, the shift in the position of the central bright fringe is given by the formula:
$y = (\mu - 1) t \frac{D}{d}$
Substituting the given values:
$y = (1.5 - 1) \times (1 \times 10^{-6} \, m) \times \frac{1.2 \, m}{2.4 \times 10^{-3} \, m}$
$y = 0.5 \times 10^{-6} \times \frac{1.2}{2.4 \times 10^{-3}}$
$y = 0.5 \times 10^{-6} \times 0.5 \times 10^{3}$
$y = 0.25 \times 10^{-3} \, m = 0.25 \, mm$
Thus, the shift in the position of the central bright fringe is $0.25 \, mm$.