TS EAMCET 2017 Physics Question Paper with Answer and Solution

(C) In an elastic collision between two bodies of equal mass where one is initially at rest,the conservation of linear momentum and kinetic energy leads to the equation: $\vec{v}_1 = \vec{v}_1' + \vec{v}_2'$.
Since the collision is elastic and masses are equal,the conservation of kinetic energy gives: $v_1^2 = (v_1')^2 + (v_2')^2$.
Comparing these two equations with the vector relation $\vec{v}_1^2 = (\vec{v}_1' + \vec{v}_2')^2 = (v_1')^2 + (v_2')^2 + 2\vec{v}_1' \cdot \vec{v}_2'$,we find that $2\vec{v}_1' \cdot \vec{v}_2' = 0$.
This implies that the dot product of the final velocity vectors is zero,meaning the angle between the two balls after the collision is $90^{\circ}$ (provided the collision is not head-on).

(B) The particle moves in a circle with uniform speed $v$. Let the initial velocity be $\vec{v}_i = v \hat{i}$.
At the diametrically opposite point,the velocity is $\vec{v}_f = -v \hat{i}$.
The change in momentum is $\Delta \vec{p} = m \vec{v}_f - m \vec{v}_i = M(-v \hat{i}) - M(v \hat{i}) = -2 M v \hat{i}$.
The magnitude of the change in momentum is $|\Delta \vec{p}| = 2 M v$.
Since the speed $v$ is uniform,the kinetic energy $K = \frac{1}{2} M v^2$ remains constant throughout the motion.
Therefore,the change in kinetic energy is $0$.

(D) $SONAR$ stands for Sound Navigation and Ranging. It operates by emitting ultrasonic pulses that travel through water,hit an object,and reflect back to the receiver. By measuring the time taken for the echo to return,the distance to the object can be calculated. Therefore,the principle used is the reflection of ultrasonic waves.

(A) According to the law of conservation of angular momentum at the perihelion and aphelion points:
$m v_1 r_1 = m v_2 r_2$
$\Rightarrow v_2 = \frac{v_1 r_1}{r_2}$ $(i)$
From the law of conservation of total mechanical energy:
$-\frac{G M m}{r_1} + \frac{1}{2} m v_1^2 = -\frac{G M m}{r_2} + \frac{1}{2} m v_2^2$
Substituting $v_2$ from $(i)$:
$-\frac{G M}{r_1} + \frac{1}{2} v_1^2 = -\frac{G M}{r_2} + \frac{1}{2} \left(\frac{v_1 r_1}{r_2}\right)^2$
$\frac{1}{2} v_1^2 \left(1 - \frac{r_1^2}{r_2^2}\right) = G M \left(\frac{1}{r_1} - \frac{1}{r_2}\right)$
$\frac{1}{2} v_1^2 \left(\frac{r_2^2 - r_1^2}{r_2^2}\right) = G M \left(\frac{r_2 - r_1}{r_1 r_2}\right)$
$\frac{1}{2} v_1^2 \frac{(r_2 - r_1)(r_2 + r_1)}{r_2^2} = G M \frac{(r_2 - r_1)}{r_1 r_2}$
$v_1^2 = \frac{2 G M r_2}{r_1(r_1 + r_2)}$
$v_1 = \sqrt{\frac{2 G M r_2}{r_1(r_1 + r_2)}}$
Angular momentum $L = m v_1 r_1 = m \sqrt{\frac{2 G M r_2}{r_1(r_1 + r_2)}} \cdot r_1 = m \sqrt{\frac{2 G M r_1 r_2}{r_1 + r_2}}$

(A) According to the law of conservation of energy,the potential energy of the object at the top of the ramp is converted into kinetic energy at the bottom of the ramp.
Potential energy at height $h$ = $mgh$.
When the object moves on the flat track,the work done by the frictional force $f = \mu N = \mu mg$ stops the object over a distance $d$.
The work done by friction is $W = f \times d = \mu mgd$.
Equating the initial potential energy to the work done by friction:
$mgh = \mu mgd$
Dividing both sides by $mg$,we get:
$d = \frac{h}{\mu}$

(D) To keep the box in static equilibrium,the downward force along the incline $(mg \sin \theta)$ must be balanced by the maximum static friction force $(f_{max})$.
The normal force $N$ acting on the box is the sum of the component of weight perpendicular to the incline and the applied force $F$:
$N = mg \cos \theta + F$
The maximum static friction force is given by:
$f_{max} = \mu N = \mu(mg \cos \theta + F)$
For equilibrium,$mg \sin \theta \leq f_{max}$,so the minimum force $F$ required is when $mg \sin \theta = \mu(mg \cos \theta + F)$.
Rearranging for $F$:
$F = \frac{mg \sin \theta}{\mu} - mg \cos \theta = mg \left( \frac{\sin \theta}{\mu} - \cos \theta \right)$
Given $m = 1 \ kg$,$g = 10 \ m/s^2$,$\theta = 30^{\circ}$,and $\mu = 0.2$:
$F = 1 \times 10 \left( \frac{\sin 30^{\circ}}{0.2} - \cos 30^{\circ} \right)$
$F = 10 \left( \frac{0.5}{0.2} - \frac{\sqrt{3}}{2} \right) = 10 \left( 2.5 - 0.866 \right)$
$F = 10 \times 1.634 = 16.34 \ N$
Thus,the minimum magnitude of $F$ is $\geq 16.3 \ N$.

(B) Given:
Percentage error in length,$\frac{\Delta L}{L} \times 100 = 3 \%$
Percentage error in force,$\frac{\Delta F}{F} \times 100 = 4 \%$
We know that pressure $P$ is defined as force per unit area:
$P = \frac{F}{A}$
Since the plate is square,the area $A = L^2$.
Therefore,$P = \frac{F}{L^2}$.
The relative error in pressure is given by:
$\frac{\Delta P}{P} = \frac{\Delta F}{F} + 2 \frac{\Delta L}{L}$
To find the percentage error,multiply by $100$:
$\frac{\Delta P}{P} \times 100 = \left( \frac{\Delta F}{F} \times 100 \right) + 2 \left( \frac{\Delta L}{L} \times 100 \right)$
Substituting the given values:
$\frac{\Delta P}{P} \times 100 = 4 \% + 2(3 \%) = 4 \% + 6 \% = 10 \%$
Thus,the permissible error in the calculation of pressure is $10 \%$.

(D) Given: Diameter of pipe $d = 5 \,mm = 5 \times 10^{-3} \,m$. Density of gasoline $\rho = 720 \,kg/m^3$. Viscosity of gasoline $\eta = 6 \times 10^{-3} \,Poise = 6 \times 10^{-4} \,Pa \cdot s$. The critical Reynolds number for pipe flow is $R_e = 2000$. The formula for critical velocity $v_c$ is $v_c = \frac{R_e \cdot \eta}{\rho \cdot d}$. Substituting the values: $v_c = \frac{2000 \times 6 \times 10^{-4}}{720 \times 5 \times 10^{-3}} = \frac{1.2}{3.6} = \frac{1}{3} \approx 0.33 \,m/s$. Thus,the flow becomes turbulent when the velocity is $ > 0.33 \,m/s$.

(D) Given:
Young's modulus $(Y)$ = $2 \times 10^{11} \,N/m^2$
Elastic limit (Stress,$\sigma$) = $1 \times 10^8 \,N/m^2$
Original length $(L)$ = $1 \,m$
We know that Young's modulus is defined as the ratio of stress to strain:
$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\sigma}{\Delta L / L}$
Rearranging the formula to find the maximum elongation $(\Delta L)$:
$\Delta L = \frac{\sigma \times L}{Y}$
Substituting the given values:
$\Delta L = \frac{1 \times 10^8 \,N/m^2 \times 1 \,m}{2 \times 10^{11} \,N/m^2}$
$\Delta L = 0.5 \times 10^{-3} \,m$
Converting meters to millimeters $(1 \,m = 1000 \,mm)$:
$\Delta L = 0.5 \,mm$

(A) At the maximum height, the final velocity of the object is $v = 0 \,m/s$.
Let the time taken to reach the maximum height be $t_{max}$.
We know that the velocity of an object at time $t$ before reaching the maximum height is equal to the velocity it gains in time $t$ while falling from the maximum height (due to symmetry).
Here, $t = 0.5 \,s$ and the acceleration due to gravity is $g = 9.8 \,m/s^2$.
Using the first equation of motion for the downward journey starting from rest at the maximum height:
$v = u + gt$
$v = 0 + (9.8 \,m/s^2) \times (0.5 \,s)$
$v = 4.9 \,m/s$.

(A) Given deceleration $\omega = -\frac{dv}{dt} = a \sqrt{v}$.
$1$. To find time $t$:
$\frac{dv}{dt} = -a \sqrt{v} \implies \int_{v_0}^{0} v^{-1/2} dv = \int_{0}^{t} -a dt$
$[2 \sqrt{v}]_{v_0}^{0} = -at \implies -2 \sqrt{v_0} = -at \implies t = \frac{2 \sqrt{v_0}}{a}$.
$2$. To find distance $s$:
Using $\omega = v \frac{dv}{ds} = a \sqrt{v} \implies v \frac{dv}{ds} = -a \sqrt{v} \implies \sqrt{v} dv = -a ds$.
Integrating both sides: $\int_{v_0}^{0} v^{1/2} dv = \int_{0}^{s} -a ds$
$[\frac{2}{3} v^{3/2}]_{v_0}^{0} = -as \implies -\frac{2}{3} v_0^{3/2} = -as \implies s = \frac{2 v_0^{3/2}}{3 a}$.

(D) Given,
Initial speed of the ball $(u) = 20 \,m/s$
Angle of projection $(\theta) = 30^{\circ}$
Acceleration due to gravity $(g) = 10 \,m/s^2$
The formula for the maximum height $(H)$ reached by a projectile is given by:
$H = \frac{u^2 \sin^2 \theta}{2g}$
Substituting the given values into the formula:
$H = \frac{(20)^2 \times (\sin 30^{\circ})^2}{2 \times 10}$
$H = \frac{400 \times (1/2)^2}{20}$
$H = \frac{400 \times (1/4)}{20}$
$H = \frac{100}{20}$
$H = 5 \,m$
Therefore,the maximum height reached by the ball is $5 \,m$.

(A) Given the parametric equations of motion:
$x = a \cos t$
$y = a \sin t$
$z = t$
First,consider the projection of the motion on the $xy$-plane:
$x^2 + y^2 = (a \cos t)^2 + (a \sin t)^2 = a^2(\cos^2 t + \sin^2 t) = a^2$
This represents a circle of radius $a$ in the $xy$-plane.
Simultaneously,the particle moves along the $z$-axis with a constant velocity since $z = t$,which implies $\frac{dz}{dt} = 1$.
Since the particle moves in a circular path in the $xy$-plane while simultaneously advancing linearly along the $z$-axis,the resulting trajectory is a helix.

(B) Given:
Width of the river $(w) = 200 \ m$
Velocity of the river $(v_r) = 2 \ m/s$
Velocity of the swimmer with respect to the river $(v_{sr}) = 1 \ m/s$
To cross the river in the shortest time,the swimmer should swim perpendicular to the river flow.
The time taken to cross the river is $t = \frac{w}{v_{sr}} = \frac{200 \ m}{1 \ m/s} = 200 \ s$.
During this time,the swimmer is carried downstream by the river flow.
The distance drifted downstream $(d)$ is given by $d = v_r \times t$.
$d = 2 \ m/s \times 200 \ s = 400 \ m$.
Therefore,the swimmer reaches the opposite bank at a distance of $400 \ m$ from the point directly opposite to the starting point.

(D) Given: Length of simple pendulum $L = 1 \,m$,acceleration of elevator $a = 2 \,m/s^2$,and acceleration due to gravity $g = 10 \,m/s^2$.
When an elevator moves upward with an acceleration $a$,the effective acceleration due to gravity $g_{eff}$ is given by $g_{eff} = g + a$.
Substituting the values,$g_{eff} = 10 + 2 = 12 \,m/s^2$.
The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.
Substituting the values,$T = 2\pi \sqrt{\frac{1}{12}} = 2\pi \frac{1}{\sqrt{4 \times 3}} = 2\pi \frac{1}{2\sqrt{3}} = \frac{\pi}{\sqrt{3}} \,s$.

(D) Given: Area $A = 1.0 \,m^2$, Thickness $l = 3 \,cm = 0.03 \,m$, Temperature difference $\Delta \theta = 30^{\circ} C - 0^{\circ} C = 30^{\circ} C$, Thermal conductivity $K = 0.03 \,W/mK$, Latent heat $L = 3.00 \times 10^5 \,J/kg$, Time $t = 24 \times 3600 \,s = 86400 \,s$.
Using the heat conduction formula: $Q = \frac{K A \Delta \theta t}{l}$.
Since $Q = m L$, we have $m = \frac{K A \Delta \theta t}{L l}$.
Substituting the values: $m = \frac{0.03 \times 1.0 \times 30 \times 86400}{3.00 \times 10^5 \times 0.03}$.
$m = \frac{0.9 \times 86400}{9000} = \frac{77760}{9000} = 8.64 \,kg$.

$(A)$ Given: Coefficient of linear expansion of steel, $\alpha = 11 \times 10^{-6} /{ }^{\circ} C$.
Required accuracy $\Delta l = 0.06 \,mm = 6 \times 10^{-5} \,m$ for a length $l = 1 \,m$.
We know the formula for linear expansion: $\Delta l = l \alpha \Delta t$.
Rearranging for the temperature change: $\Delta t = \frac{\Delta l}{l \alpha}$.
Substituting the values: $\Delta t = \frac{6 \times 10^{-5}}{1 \times 11 \times 10^{-6}} = \frac{60}{11} \approx 5.45^{\circ} C$.
Since the scale is accurate at $25^{\circ} C$, the allowed temperature range is $25^{\circ} C \pm 5.45^{\circ} C$.
This gives a range from $19.55^{\circ} C$ to $30.45^{\circ} C$.
Rounding to the nearest integer values provided in the options, the range is approximately $19^{\circ} C$ to $31^{\circ} C$.
Thus, option $A$ is correct.

(C) The change in internal energy $(\Delta U)$ of an ideal gas is given by the formula $\Delta U = n C_v \Delta T$.
Given:
Number of moles $(n) = 2000 = 2 \times 10^3 \text{ mol}$.
Initial temperature $(T_i) = 34^{\circ} C$,Final temperature $(T_f) = 24^{\circ} C$.
Temperature change $(\Delta T) = 24 - 34 = -10 \text{ K}$.
For air,$\gamma = 1.4$. The molar heat capacity at constant volume is $C_v = \frac{R}{\gamma - 1} = \frac{R}{1.4 - 1} = \frac{R}{0.4}$.
Substituting the values:
$\Delta U = n \left( \frac{R}{0.4} \right) \Delta T$
$\Delta U = (2 \times 10^3) \times \left( \frac{8.314}{0.4} \right) \times (-10)$
$\Delta U = - \frac{2 \times 8.314 \times 10^4}{0.4}$
$\Delta U = - \frac{16.628 \times 10^4}{0.4} = -41.57 \times 10^4 \text{ J} \approx -4.2 \times 10^5 \text{ J}$.

(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ is the sink temperature.
Given $\eta_1 = \frac{1}{6}$, we have $\frac{1}{6} = 1 - \frac{T_2}{T_1}$, which implies $\frac{T_2}{T_1} = \frac{5}{6}$ or $T_2 = \frac{5}{6} T_1$ (Eq. $1$).
When the sink temperature is reduced by $62^{\circ} C$, the new sink temperature is $T_2' = T_2 - 62$. The new efficiency is $\eta_2 = 2 \times \eta_1 = 2 \times \frac{1}{6} = \frac{1}{3}$.
Thus, $\frac{1}{3} = 1 - \frac{T_2 - 62}{T_1}$.
Substituting $T_2 = \frac{5}{6} T_1$ into the equation: $\frac{1}{3} = 1 - \frac{\frac{5}{6} T_1 - 62}{T_1} = 1 - \frac{5}{6} + \frac{62}{T_1} = \frac{1}{6} + \frac{62}{T_1}$.
Rearranging gives $\frac{1}{3} - \frac{1}{6} = \frac{62}{T_1}$, so $\frac{1}{6} = \frac{62}{T_1}$, which means $T_1 = 372 \,K$.
Using Eq. $1$, $T_2 = \frac{5}{6} \times 372 = 310 \,K$.

(C) The frequency of a sound wave is a characteristic property of the source that produces it.
When a wave travels from one medium to another (e.g.,from air to water),its speed and wavelength change,but its frequency remains constant.
Therefore,the frequency of the wave at the bottom of the reservoir remains $v \text{ Hz}$.

(A) Given:
$F_{1} = (\hat{i} + 2\hat{j} + 3\hat{k}) \text{ N}$
$F_{2} = (4\hat{i} - 5\hat{j} - 2\hat{k}) \text{ N}$
$r_{1} = (20\hat{i} + 15\hat{j}) \text{ cm} = (0.2\hat{i} + 0.15\hat{j}) \text{ m}$
$r_{2} = (7\hat{k}) \text{ cm} = (0.07\hat{k}) \text{ m}$
Total force $F = F_{1} + F_{2} = (1+4)\hat{i} + (2-5)\hat{j} + (3-2)\hat{k} = (5\hat{i} - 3\hat{j} + \hat{k}) \text{ N}$.
Displacement $s = r_{2} - r_{1} = (0\hat{i} + 0\hat{j} + 0.07\hat{k}) - (0.2\hat{i} + 0.15\hat{j} + 0\hat{k}) = (-0.2\hat{i} - 0.15\hat{j} + 0.07\hat{k}) \text{ m}$.
Work done $W = F \cdot s = (5\hat{i} - 3\hat{j} + \hat{k}) \cdot (-0.2\hat{i} - 0.15\hat{j} + 0.07\hat{k})$.
$W = (5 \times -0.2) + (-3 \times -0.15) + (1 \times 0.07)$.
$W = -1.0 + 0.45 + 0.07 = -0.48 \text{ J}$.

(D) Given parameters:
$E = 10 \, V$, $\omega = 200 \, rad/s$, $R = 50 \, \Omega$, $L = 400 \, mH = 0.4 \, H$, $C = 200 \, \mu F = 200 \times 10^{-6} \, F$.
First, calculate the inductive reactance $(X_L)$ and capacitive reactance $(X_C)$:
$X_L = \omega L = 200 \times 0.4 = 80 \, \Omega$.
$X_C = \frac{1}{\omega C} = \frac{1}{200 \times 200 \times 10^{-6}} = \frac{1}{0.04} = 25 \, \Omega$.
Now, calculate the impedance $(Z)$ of the $LCR$ circuit:
$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{50^2 + (80 - 25)^2} = \sqrt{2500 + 55^2} = \sqrt{2500 + 3025} = \sqrt{5525} \approx 74.33 \, \Omega$.
The rms current $(I)$ in the circuit is:
$I = \frac{E}{Z} = \frac{10}{74.33} \approx 0.1345 \, A$.
The rms voltage across the inductor $(V_L)$ is given by:
$V_L = I \times X_L = 0.1345 \times 80 \approx 10.76 \, V$.
Rounding to one decimal place, we get $10.8 \, V$.

(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
For the ground state $(n_1 = 1)$, the energy is $E_1 = -13.6 \text{ eV}$.
For the excited state $(n_2 = 3)$, the energy is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \text{ eV}$.
The energy required to excite the atom from $n=1$ to $n=3$ is $\Delta E = E_3 - E_1$.
$\Delta E = -1.51 - (-13.6) = 12.09 \text{ eV}$.
Rounding this value, we get $\Delta E \approx 12.1 \text{ eV}$.

(B) The $4 \mu F$ capacitor is in series with the parallel combination of the $2 \mu F$ and $3 \mu F$ capacitors.
When a charge of $+80 \mu C$ is placed on the upper plate of the $4 \mu F$ capacitor,an equal and opposite charge of $-80 \mu C$ is induced on its lower plate.
This charge of $+80 \mu C$ is then distributed between the upper plates of the $2 \mu F$ and $3 \mu F$ capacitors,which are connected in parallel.
Since the capacitors are in parallel,the charge $q$ distributes in the ratio of their capacitances:
$q_1 = \left( \frac{C_1}{C_1 + C_2} \right) Q_{total}$
For the $3 \mu F$ capacitor:
$q = \left( \frac{3 \mu F}{3 \mu F + 2 \mu F} \right) \times 80 \mu C$
$q = \left( \frac{3}{5} \right) \times 80 \mu C = 3 \times 16 \mu C = 48 \mu C$.

(B) Given:
Frequency of message signal $(f_m) = 1 \text{ kHz} = 1 \times 10^3 \text{ Hz}$.
Peak voltage of message signal $(E_m) = 5 \text{ V}$.
Carrier frequency $(f_c) = 1 \text{ MHz} = 1 \times 10^6 \text{ Hz}$.
Peak voltage of carrier $(E_c) = 15 \text{ V}$.
The equation of an amplitude modulated wave is given by:
$e(t) = E_c [1 + \mu \sin(\omega_m t)] \sin(\omega_c t)$
where $\mu = \frac{E_m}{E_c}$ is the modulation index.
$\mu = \frac{5}{15} = \frac{1}{3}$.
Substituting the values:
$e(t) = 15 [1 + \frac{1}{3} \sin(2 \pi f_m t)] \sin(2 \pi f_c t)$
$e(t) = 15 [1 + \frac{1}{3} \sin(2 \pi \times 10^3 t)] \sin(2 \pi \times 10^6 t)$.

(D) The circuit consists of two parallel branches, each containing three ideal batteries in series.
Each branch has an equivalent emf of $E_{eq} = 3 \times 20 \, V = 60 \, V$.
Since the two branches are connected in parallel, the total equivalent emf of the battery combination remains $E_{total} = 60 \, V$.
As the batteries are ideal, they have no internal resistance.
Therefore, the current $i$ flowing through the external resistance $R = 4 \, \Omega$ is given by Ohm's law:
$i = \frac{E_{total}}{R} = \frac{60 \, V}{4 \, \Omega} = 15 \, A$.

(C) Copper is a metallic conductor, and its resistance decreases as the temperature decreases because the frequency of collisions between electrons and lattice ions decreases.
Germanium is a semiconductor. In semiconductors, the number of charge carriers (electrons and holes) decreases exponentially as the temperature decreases, which leads to a significant increase in resistance.
Therefore, when both are cooled from room temperature to $80 \text{ K}$, the resistance of copper decreases, and the resistance of germanium increases.

(A) Given:
$t_1 = 30^{\circ} C$,$R_1 = 3.1 \Omega$
$t_2 = 100^{\circ} C$,$R_2 = 4.5 \Omega$
We know that the resistance at temperature $t$ is given by $R_t = R_0(1 + \alpha \Delta t)$.
Alternatively,the temperature coefficient of resistance $\alpha$ is defined as:
$\alpha = \frac{R_2 - R_1}{R_1(t_2 - t_1) + R_2(t_1 - t_2)}$ is not the standard form,so we use $\alpha = \frac{R_2 - R_1}{R_1(t_2 - t_1)}$ if $R_1$ is at $0^{\circ} C$,but here we use the general formula:
$R_2 = R_1[1 + \alpha(t_2 - t_1)]$
$\alpha = \frac{R_2 - R_1}{R_1(t_2 - t_1)}$
$\alpha = \frac{4.5 - 3.1}{3.1(100 - 30)}$
$\alpha = \frac{1.4}{3.1 \times 70}$
$\alpha = \frac{1.4}{217} \approx 0.00645^{\circ} C^{-1}$
Wait,recalculating using the provided formula in the prompt: $\alpha = \frac{R_2 - R_1}{R_1 t_2 - R_2 t_1} = \frac{1.4}{310 - 135} = \frac{1.4}{175} = 0.008^{\circ} C^{-1}$.
Thus,the correct option is $A$.

(D) We know that the de-Broglie wavelength $\lambda$ is related to kinetic energy $K$ by the formula: $\lambda = \frac{h}{\sqrt{2mK}}$.
This implies that $K \propto \frac{1}{\lambda^2}$.
Let $K_1$ be the initial kinetic energy at $\lambda_1 = 1 \,nm$ and $K_2$ be the final kinetic energy at $\lambda_2 = 0.5 \,nm$.
Then, $\frac{K_2}{K_1} = \left( \frac{\lambda_1}{\lambda_2} \right)^2 = \left( \frac{1}{0.5} \right)^2 = 2^2 = 4$.
So, $K_2 = 4K_1$.
The energy that should be added to the electron is $\Delta K = K_2 - K_1 = 4K_1 - K_1 = 3K_1$.
Therefore, the energy to be added is three-times the initial energy.

(A) The solenoid is connected to a $DC$ source with a constant $emf$ $(V)$.
The current in the solenoid is given by $I = V/R$,where $R$ is the resistance of the solenoid wire.
When the iron core is pulled out,the self-inductance $(L)$ of the solenoid changes,but the resistance $(R)$ of the wire remains constant.
Since the $DC$ source provides a constant $emf$ and the resistance of the circuit does not change,the steady-state current $I$ remains unchanged.
Therefore,the current will remain the same.

(B) Given: Number of turns in the coil $= n$. Resistance of the coil $= R$. Resistance of the galvanometer $= 4R$. Total resistance of the circuit $R_{total} = R + 4R = 5R$.
According to Faraday's Law of electromagnetic induction,the induced electromotive force (emf) is given by $e = -n \frac{\Delta \phi}{\Delta t}$.
Here,the change in flux is $\Delta \phi = \phi_2 - \phi_1$ and the time taken is $\Delta t = t$.
Therefore,$e = -n \frac{(\phi_2 - \phi_1)}{t}$.
The induced current $i$ is given by $i = \frac{e}{R_{total}}$.
Substituting the values,we get $i = \frac{-n(\phi_2 - \phi_1)}{5Rt}$.

(A) Given:
Number of turns in the circular coil $(N) = 100$
Area $(A) = 2 \times 10^{-2} \,m^2$
Magnetic field $(B) = 0.01 \,T$
Frequency of rotation $(f) = 50 \,Hz$
The maximum induced electromotive force (emf) in a rotating coil is given by the formula:
$e_{max} = N B A \omega$
Since angular frequency $\omega = 2 \pi f$, we have:
$e_{max} = N B A (2 \pi f)$
Substituting the given values:
$e_{max} = 100 \times 0.01 \times (2 \times 10^{-2}) \times 2 \times 3.14159 \times 50$
$e_{max} = 1 \times 0.02 \times 314.159$
$e_{max} = 0.02 \times 314.159 = 6.283 \,V$
Thus, the maximum emf produced is approximately $6.28 \,V$.

(C) The motional electromotive force $(EMF)$ induced in a conductor moving through a magnetic field is given by the formula $e = \int (\vec{v} \times \vec{B}) \cdot d\vec{l}$.
For a metal ball of radius $r$,the effective length $l$ between the points of maximum potential difference is the diameter of the ball,which is $l = 2r$.
The induced $EMF$ is given by $e = B v l \sin \alpha$,where $\alpha$ is the angle between the velocity vector $\vec{v}$ and the magnetic field vector $\vec{B}$.
Substituting $l = 2r$ into the formula,we get $e = B v (2r) \sin \alpha$.
Therefore,the maximum potential difference is $2r|\vec{B}||\vec{v}| \sin \alpha$.

(C) According to Gauss's law,the electric field $E$ at a distance $r$ from a point charge $q$ is given by $\oint E \cdot ds = \frac{q_{enclosed}}{\varepsilon_0}$.
For a Gaussian surface of radius $r$ (where $r < R_1$) inside the cavity,the only charge enclosed is the point charge $q$ placed at the center.
Thus,$E(4 \pi r^2) = \frac{q}{\varepsilon_0}$.
Solving for $E$,we get $E = \frac{q}{4 \pi \varepsilon_0 r^2}$.
The charge $Q$ on the conducting shell does not contribute to the electric field inside the cavity because the field due to a spherically symmetric shell of charge is zero inside it.

(C) According to Oersted's experiment,when an electric current flows through a conductor,it creates a magnetic field in the space surrounding it. While a stationary charge produces only an electric field,a moving charge (or current) produces both an electric field (due to its charge) and a magnetic field (due to its motion). However,in the context of a current-carrying wire where the net charge is zero,the primary effect observed in its neighbourhood is the magnetic field.

(D) The nuclear decay reaction is given by: ${ }_{94}^{239} Pu \rightarrow { }_{92}^{235} U + { }_{Z}^{A} X$.
Applying the law of conservation of mass number: $239 = 235 + A \Rightarrow A = 4$.
Applying the law of conservation of atomic number: $94 = 92 + Z \Rightarrow Z = 2$.
$A$ particle with mass number $4$ and atomic number $2$ is an alpha particle $({ }_{2}^{4} He)$.
Therefore,an alpha particle is emitted during this decay.

(A) Given: Distance of light source $(u) = -1.5 \ m = -\frac{3}{2} \ m$. Radius of curvature of the convex mirror $(R) = +1 \ m$. Focal length $(f) = \frac{R}{2} = +0.5 \ m$.
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
Substituting the values: $\frac{1}{0.5} = \frac{1}{v} + \frac{1}{-1.5} \Rightarrow 2 = \frac{1}{v} - \frac{2}{3}$.
Solving for $v$: $\frac{1}{v} = 2 + \frac{2}{3} = \frac{6+2}{3} = \frac{8}{3}$.
Therefore,$v = \frac{3}{8} = 0.375 \ m$.
Since $v$ is positive,the image is formed behind the mirror,making it virtual and upright.
Magnification $(m) = -\frac{v}{u} = -\frac{0.375}{-1.5} = \frac{0.375}{1.5} = 0.25$.

(A) According to Snell's Law,the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the refractive indices of the two media:
$\frac{\sin \alpha_1}{\sin \alpha_2} = \frac{\mu_2}{\mu_1}$
We also know that the refractive index $\mu$ is inversely proportional to the wavelength $\lambda$ in the medium,given by $\mu = \frac{c}{v} = \frac{\lambda_0}{\lambda}$,where $\lambda_0$ is the wavelength in vacuum.
Therefore,$\frac{\mu_2}{\mu_1} = \frac{\lambda_1}{\lambda_2}$.
Equating the two expressions,we get:
$\frac{\sin \alpha_1}{\sin \alpha_2} = \frac{\lambda_1}{\lambda_2}$
Rearranging for $\lambda_2$,we find:
$\lambda_2 = \lambda_1 \frac{\sin \alpha_2}{\sin \alpha_1}$

(C) In an $n$-type semiconductor,the majority charge carriers are electrons,not holes. Therefore,the statement that the majority carriers in $n$-type semiconductors are holes is incorrect. The other statements are true: intrinsic semiconductors have a negative temperature coefficient of resistance,trivalent doping creates $p$-type material,and a $p-n$ junction functions as a diode.

(A) Let $I_0$ be the intensity of the incident light.
Intensity reflected from the upper surface,$I_1 = 25 \% \text{ of } I_0 = \frac{I_0}{4}$.
Intensity transmitted into the glass plate = $I_0 - \frac{I_0}{4} = \frac{3I_0}{4}$.
Intensity reflected from the lower surface,$I_2 = 50 \% \text{ of } \frac{3I_0}{4} = \frac{1}{2} \times \frac{3I_0}{4} = \frac{3I_0}{8}$.
The ratio of maximum to minimum intensity is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2} = \left( \frac{\sqrt{\frac{I_0}{4}} + \sqrt{\frac{3I_0}{8}}}{\sqrt{\frac{I_0}{4}} - \sqrt{\frac{3I_0}{8}}} \right)^2 = \left( \frac{\frac{1}{2} + \sqrt{\frac{3}{8}}}{\frac{1}{2} - \sqrt{\frac{3}{8}}} \right)^2$.