AP EAMCET 2015 Physics Question Paper with Answer and Solution

(B) Let the mass of the body be $m$ and the mass of the earth be $M$. The escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$.
The initial velocity of the body is $v = \frac{v_e}{2} = \frac{1}{2} \sqrt{\frac{2GM}{R}}$.
Using the law of conservation of mechanical energy between the surface of the earth and the maximum height $h$ (where final velocity is $0$):
$K_i + U_i = K_f + U_f$
$\frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$
Substituting $v^2 = \frac{1}{4} \cdot \frac{2GM}{R} = \frac{GM}{2R}$:
$\frac{1}{2}m \left( \frac{GM}{2R} \right) - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$\frac{GMm}{4R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$-\frac{3GMm}{4R} = - \frac{GMm}{R+h}$
$\frac{3}{4R} = \frac{1}{R+h}$
$3(R+h) = 4R$
$3R + 3h = 4R$
$3h = R$
$h = R/3$

(B) Let the mass of both balls be $m$. Let the initial velocities be $u_P = v$ and $u_Q = -10 \ ms^{-1}$.
After the collision,the final velocity of ball $P$ is $v_P = 0$ and let the final velocity of ball $Q$ be $v_Q$.
By the law of conservation of linear momentum:
$m(v) + m(-10) = m(0) + m(v_Q)$
$v - 10 = v_Q$
Now,using the coefficient of restitution formula $e = \frac{\text{velocity of separation}}{\text{velocity of approach}}$:
$e = \frac{v_Q - v_P}{u_P - u_Q}$
Given $e = 0.6$,$v_P = 0$,$u_P = v$,and $u_Q = -10$:
$0.6 = \frac{v_Q - 0}{v - (-10)}$
$0.6 = \frac{v_Q}{v + 10}$
Substitute $v_Q = v - 10$ into the equation:
$0.6 = \frac{v - 10}{v + 10}$
$0.6(v + 10) = v - 10$
$0.6v + 6 = v - 10$
$16 = 0.4v$
$v = \frac{16}{0.4} = 40 \ ms^{-1}$

(A) Using the principle of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.
At the surface: $E_i = K_i + U_i = \frac{1}{2}mv^2 - \frac{GMm}{R}$
At maximum height $h$: $E_f = K_f + U_f = 0 - \frac{GMm}{R+h}$
Since $E_i = E_f$,we have $\frac{1}{2}mv^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$
Given $v = \frac{v_e}{2} = \frac{1}{2}\sqrt{\frac{2GM}{R}}$,so $v^2 = \frac{GM}{2R}$.
Substituting $v^2$ into the energy equation:
$\frac{1}{2}m(\frac{GM}{2R}) - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$\frac{GMm}{4R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$-\frac{3GMm}{4R} = - \frac{GMm}{R+h}$
$\frac{3}{4R} = \frac{1}{R+h} \Rightarrow 3(R+h) = 4R \Rightarrow 3R + 3h = 4R \Rightarrow 3h = R \Rightarrow h = \frac{R}{3}$

(C) The orbital velocity of a satellite is given by $v_0 = \sqrt{\frac{GM}{R}}$.
The escape velocity of a satellite is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Therefore,the relationship between escape velocity and orbital velocity is $v_e = \sqrt{2} v_0$.
The additional velocity required to escape is $\Delta v = v_e - v_0$.
Substituting the values: $\Delta v = \sqrt{2} v_0 - v_0 = v_0(\sqrt{2} - 1)$.
Given $v_0 = 10 \ km/s$ and $\sqrt{2} \approx 1.414$,we get:
$\Delta v = 10 \times (1.414 - 1) = 10 \times 0.414 = 4.14 \ km/s$.

(D) Helium is a monoatomic gas. For an ideal gas,the relationship between molar specific heat at constant pressure $(C_p)$ and molar specific heat at constant volume $(C_V)$ is given by Mayer's relation: $C_p - C_V = R$.
Given: $C_V = 12.6 \ J \ mol^{-1} \ K^{-1}$ and $R = 8.314 \ J \ mol^{-1} \ K^{-1}$.
Substituting these values into the equation: $C_p = C_V + R$.
$C_p = 12.6 + 8.314 = 20.914 \ J \ mol^{-1} \ K^{-1}$.
Rounding this value to the nearest integer,we get $C_p \approx 21 \ J \ mol^{-1} \ K^{-1}$.

(A) The average kinetic energy of a gas molecule is given by the formula $KE_{av} = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature of the gas.
Since the argon and chlorine gases are in the same flask and in thermal equilibrium,they are at the same temperature $T = 27^{\circ} C = 300 \ K$.
Because the average kinetic energy per molecule depends only on the temperature $T$ and is independent of the mass or the nature of the gas molecules,the ratio of the average kinetic energies of the two gases is $1:1$.

(C) The force $F$ required to just move the body is equal to the limiting static friction force $f_s$.
$f_s = \mu_s N = \mu_s mg$
Given $\mu_s = 0.8$, $m = 4 \,kg$, and $g = 10 \,ms^{-2}$.
$F = 0.8 \times 4 \times 10 = 32 \,N$.
Once the body starts moving, the kinetic friction force $f_k$ acts on it.
$f_k = \mu_k N = \mu_k mg$
Given $\mu_k = 0.6$.
$f_k = 0.6 \times 4 \times 10 = 24 \,N$.
The net force acting on the body is $F_{net} = F - f_k$.
$F_{net} = 32 - 24 = 8 \,N$.
Using Newton's second law, $F_{net} = ma$.
$8 = 4 \times a$.
$a = 2 \,ms^{-2}$.

(B) Given force vector $\vec{F} = (2 \hat{i} + \hat{j} - \hat{k}) \text{ N}$.
Initial velocity $\vec{u} = 0 \text{ ms}^{-1}$.
Final velocity $\vec{v} = (4 \hat{i} + 2 \hat{j} - 2 \hat{k}) \text{ ms}^{-1}$.
Time $t = 20 \text{ s}$.
Using Newton's Second Law, $\vec{F} = m \vec{a} = m \left( \frac{\vec{v} - \vec{u}}{t} \right)$.
Rearranging for mass: $m = \frac{\vec{F} \cdot t}{\vec{v} - \vec{u}}$.
Since $\vec{v} - \vec{u} = (4 \hat{i} + 2 \hat{j} - 2 \hat{k}) = 2(2 \hat{i} + \hat{j} - \hat{k}) = 2 \vec{F}$.
Substituting the values: $m = \frac{\vec{F} \cdot 20}{2 \vec{F}} = \frac{20}{2} = 10 \text{ kg}$.

(C) The pressure $p$ on a circular plate is given by the formula $p = \frac{F}{A}$,where $F$ is the force and $A$ is the area of the plate. Since the plate is circular,$A = \pi R^2$,where $R$ is the radius.
Therefore,$p = \frac{F}{\pi R^2}$.
Using the rules of propagation of errors,the relative error in $p$ is given by:
$\frac{\Delta p}{p} = \frac{\Delta F}{F} + 2 \frac{\Delta R}{R}$.
To find the percentage error,we multiply by $100$:
$\frac{\Delta p}{p} \times 100 = \left( \frac{\Delta F}{F} \times 100 \right) + 2 \left( \frac{\Delta R}{R} \times 100 \right)$.
Given that the percentage error in force $\frac{\Delta F}{F} \times 100 = 5 \%$ and the percentage error in radius $\frac{\Delta R}{R} \times 100 = 3 \%$,we substitute these values:
$\text{Percentage error in } p = 5 \% + 2(3 \%) = 5 \% + 6 \% = 11 \%$.
Thus,the percentage error in the measurement of pressure is $11 \%$.

(B) The pressure at depth $h$ is given by $\Delta P = h \rho g$.
Bulk modulus $B$ is defined as $B = \frac{\Delta P}{\Delta V / V}$.
Therefore,the fractional compression is $\frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{h \rho g}{B}$.
Given: $h = 3000 \,m$,$\rho = 1000 \,kg/m^3$,$g = 9.8 \,m/s^2$,and $B = 2.2 \times 10^9 \,N/m^2$.
Substituting these values:
$\frac{\Delta V}{V} = \frac{3000 \times 1000 \times 9.8}{2.2 \times 10^9} = \frac{2.94 \times 10^7}{2.2 \times 10^9} = 1.336 \times 10^{-2} \approx 1.34 \times 10^{-2}$.

(C) According to Hooke's Law,the extension in a wire is proportional to the applied force: $F = k(l - l_0)$,where $l_0$ is the original length and $k$ is the force constant.
For the first case: $F_1 = k(l_1 - l_0)$ -- $(1)$
For the second case: $F_2 = k(l_2 - l_0)$ -- $(2)$
Dividing $(1)$ by $(2)$,we get: $\frac{F_1}{F_2} = \frac{l_1 - l_0}{l_2 - l_0}$
Cross-multiplying: $F_1(l_2 - l_0) = F_2(l_1 - l_0)$
$F_1 l_2 - F_1 l_0 = F_2 l_1 - F_2 l_0$
$F_2 l_0 - F_1 l_0 = F_2 l_1 - F_1 l_2$
$l_0(F_2 - F_1) = F_2 l_1 - F_1 l_2$
$l_0 = \frac{F_2 l_1 - F_1 l_2}{F_2 - F_1}$

(D) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin 2\theta}{g}$.
Let $R$ be the distance to the target.
For $\theta_1 = 15^{\circ}$,the range is $R_1 = R - 10$. So,$R - 10 = \frac{u^2 \sin 30^{\circ}}{g} = \frac{u^2}{2g}$.
For $\theta_2 = 45^{\circ}$,the range is $R_2 = R + 15$. So,$R + 15 = \frac{u^2 \sin 90^{\circ}}{g} = \frac{u^2}{g}$.
Dividing the two equations: $\frac{R - 10}{R + 15} = \frac{u^2/2g}{u^2/g} = \frac{1}{2}$.
$2R - 20 = R + 15 \Rightarrow R = 35 \ m$.
Now,$R + 15 = \frac{u^2}{g} \Rightarrow 35 + 15 = \frac{u^2}{g} \Rightarrow \frac{u^2}{g} = 50 \ m$.
To hit the target,the range must be $R = 35 \ m$. Thus,$35 = 50 \sin 2\theta$.
$\sin 2\theta = \frac{35}{50} = \frac{7}{10}$.
$\theta = \frac{1}{2} \sin^{-1}\left(\frac{7}{10}\right)$.

(D) Let $\vec{v}_m$ be the velocity of the man,$\vec{v}_r$ be the velocity of the rain,and $\vec{v}_{rm}$ be the velocity of the rain with respect to the man.
Given,$\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$.
When the man is running,the rain appears to fall vertically,meaning $\vec{v}_{rm}$ is vertical.
When the man stops,the rain falls at $60^{\circ}$ with the horizontal,meaning $\vec{v}_r$ makes an angle of $60^{\circ}$ with the horizontal.
From the vector triangle,the angle between $\vec{v}_r$ and the vertical is $90^{\circ} - 60^{\circ} = 30^{\circ}$.
In the right-angled triangle formed by $\vec{v}_m$,$\vec{v}_r$,and $\vec{v}_{rm}$:
$\tan(30^{\circ}) = \frac{|\vec{v}_m|}{|\vec{v}_{rm}|}$
$\frac{1}{\sqrt{3}} = \frac{5}{|\vec{v}_{rm}|}$
$|\vec{v}_{rm}| = 5\sqrt{3} \text{ km/h}$.

(A) The kinetic energy $(KE)$ of a simple harmonic oscillator is given by $KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$.
The potential energy $(PE)$ is given by $PE = \frac{1}{2} m \omega^2 x^2$.
Here,$A = 10 \ cm$ and $x = 4 \ cm$.
The ratio of kinetic energy to potential energy is $\frac{KE}{PE} = \frac{\frac{1}{2} m \omega^2 (A^2 - x^2)}{\frac{1}{2} m \omega^2 x^2} = \frac{A^2 - x^2}{x^2}$.
Substituting the values: $\frac{KE}{PE} = \frac{10^2 - 4^2}{4^2} = \frac{100 - 16}{16} = \frac{84}{16}$.
Calculating the result: $\frac{84}{16} = 5.25$.

(C) According to the Stefan-Boltzmann law,the net rate of energy emission $E$ by a black body at temperature $T$ in surroundings at temperature $T_S$ is given by $E = \sigma A (T^4 - T_S^4)$.
Given $T_1 = 600 \ K$,$T_2 = 900 \ K$,and $T_S = 300 \ K$.
The ratio of energies is $\frac{E_1}{E_2} = \frac{T_1^4 - T_S^4}{T_2^4 - T_S^4}$.
Substituting the values: $\frac{E_1}{E_2} = \frac{(600)^4 - (300)^4}{(900)^4 - (300)^4}$.
Factoring out $(300)^4$: $\frac{E_1}{E_2} = \frac{(300)^4 [2^4 - 1^4]}{(300)^4 [3^4 - 1^4]}$.
$\frac{E_1}{E_2} = \frac{16 - 1}{81 - 1} = \frac{15}{80} = \frac{3}{16}$.

(B) For a heat engine,efficiency is defined as $\eta = \frac{W}{Q_H} = 1 - \frac{Q_L}{Q_H}$.
For a refrigerator,the coefficient of performance is defined as $\alpha = \frac{Q_L}{W} = \frac{Q_L}{Q_H - Q_L}$.
Taking the reciprocal,$\frac{1}{\alpha} = \frac{Q_H - Q_L}{Q_L} = \frac{Q_H}{Q_L} - 1$.
Thus,$\frac{Q_H}{Q_L} = 1 + \frac{1}{\alpha} = \frac{\alpha + 1}{\alpha}$.
Therefore,$\frac{Q_L}{Q_H} = \frac{\alpha}{1 + \alpha}$.
Substituting this into the efficiency formula: $\eta = 1 - \frac{Q_L}{Q_H} = 1 - \frac{\alpha}{1 + \alpha} = \frac{1 + \alpha - \alpha}{1 + \alpha} = \frac{1}{1 + \alpha}$.

(B) From the first law of thermodynamics, we have the equation $dQ = dU + dW$.
In an adiabatic process, there is no exchange of heat with the surroundings, so $dQ = 0$.
Substituting this into the equation, we get $0 = dU + dW$, which implies $dU = -dW$.
Given that the gas does $4.5 \,J$ of external work, $dW = 4.5 \,J$.
Therefore, $dU = -4.5 \,J$.
Since the change in internal energy $dU$ is negative, the internal energy decreases by $4.5 \,J$.

(B) The given equation of the transverse wave is $y=2 \sin (30 t-40 x)$.
Comparing this with the standard wave equation $y=A \sin (\omega t-k x)$,we get the angular frequency $\omega = 30 \ rad/s$ and the wave number $k = 40 \ m^{-1}$.
The velocity of propagation $v$ is given by the formula $v = \frac{\omega}{k}$.
Substituting the values,we get $v = \frac{30}{40} = 0.75 \ ms^{-1}$.

(A) The fundamental frequency of a closed pipe is given by $f = \frac{v}{4l}$,where $v$ is the speed of sound in the gas and $l$ is the length of the pipe.
Since $v = \sqrt{\frac{\gamma RT}{M}}$,we have $f = \frac{1}{4l} \sqrt{\frac{\gamma RT}{M}}$.
Given that the fundamental frequency $f$ and temperature $T$ are the same for both pipes,we have $l \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass of the gas.
For oxygen $(O_2)$,$M_1 = 32 \ g/mol$. For hydrogen $(H_2)$,$M_2 = 2 \ g/mol$.
The ratio of their lengths is $\frac{l_1}{l_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
Thus,the ratio of their lengths is $1:4$.

(C) The power $P$ is defined as the work done $W$ per unit time $t$,given by $P = \frac{W}{t}$.
Here,the work done is against gravity to lift the man and the object: $W = (m + M)gh$,where $m$ is the mass of the object and $M$ is the mass of the man.
Given: $P = 2000 \ W$,$M = 50 \ kg$,$h = 20 \ m$,$t = 10 \ s$,and $g = 10 \ m/s^2$.
Substituting the values into the power formula: $P = \frac{(m + M)gh}{t}$.
$2000 = \frac{(m + 50) \times 10 \times 20}{10}$.
$2000 = (m + 50) \times 20$.
Dividing both sides by $20$: $100 = m + 50$.
Therefore,$m = 100 - 50 = 50 \ kg$.

(A) The power dissipated in an $AC$ circuit is given by the formula:
$P = V_{rms} I_{rms} \cos \phi$
Given:
$V_0 = 50 \ V$
$I_0 = 50 \ mA = 50 \times 10^{-3} \ A$
Phase difference $\phi = \frac{\pi}{3}$
The $RMS$ values are $V_{rms} = \frac{V_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}}$.
Substituting these into the power formula:
$P = \frac{V_0}{\sqrt{2}} \times \frac{I_0}{\sqrt{2}} \times \cos \phi = \frac{V_0 I_0}{2} \cos \phi$
$P = \frac{50 \times 50 \times 10^{-3}}{2} \times \cos(\frac{\pi}{3})$
Since $\cos(\frac{\pi}{3}) = 0.5$:
$P = \frac{2500 \times 10^{-3}}{2} \times 0.5 = 1.25 \times 0.5 = 0.625 \ W$

(B) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the longest wavelength line,the transition occurs between adjacent energy levels,i.e.,$n_2 = n_1 + 1$.
For the Balmer series,$n_1 = 2$,so $n_2 = 3$. The longest wavelength $\lambda_{BL}$ is:
$\frac{1}{\lambda_{BL}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) \implies \lambda_{BL} = \frac{36}{5R} \quad ... (A)$
For the Paschen series,$n_1 = 3$,so $n_2 = 4$. The longest wavelength $\lambda_{PL}$ is:
$\frac{1}{\lambda_{PL}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{7}{144} \right) \implies \lambda_{PL} = \frac{144}{7R} \quad ... (B)$
Taking the ratio of the longest wavelengths:
$\frac{\lambda_{BL}}{\lambda_{PL}} = \frac{36}{5R} \times \frac{7R}{144} = \frac{36 \times 7}{5 \times 144} = \frac{7}{5 \times 4} = \frac{7}{20}$.

(C) The energy stored in a capacitor is given by $E = \frac{1}{2} C V^2$.
Since the battery remains connected,the potential difference $V$ across the plates remains constant.
The capacitance of a parallel plate capacitor is $C = \frac{A \epsilon_0}{d}$,which implies $C \propto \frac{1}{d}$.
If the distance $d$ is doubled $(d' = 2d)$,the new capacitance $C'$ becomes $C' = \frac{C}{2}$.
The new energy $E'$ is given by $E' = \frac{1}{2} C' V^2$.
Substituting $C' = \frac{C}{2}$,we get $E' = \frac{1}{2} (\frac{C}{2}) V^2 = \frac{1}{2} (\frac{1}{2} C V^2) = \frac{E}{2}$.

(C) The modulation index $m$ is defined as the ratio of the difference and the sum of the maximum and minimum amplitudes of the modulated wave.
Formula: $m = \frac{E_{\max} - E_{\min}}{E_{\max} + E_{\min}}$
Given: $E_{\max} = 16 \,V$ and $E_{\min} = 4 \,V$
Substituting the values into the formula:
$m = \frac{16 - 4}{16 + 4}$
$m = \frac{12}{20}$
$m = \frac{3}{5} = 0.6$
Therefore, the modulation index is $0.6$.

(B) The drift velocity $v_d$ is given by the formula $v_d = \frac{eE\tau}{m}$,where $E = \frac{V}{l}$.
Substituting $E$,we get $v_d = \frac{eV\tau}{ml}$.
Since the potential difference $V$ is constant,$v_d \propto \frac{1}{l}$.
If the length $l$ is increased to $4l$,the new drift velocity $v_{d'}$ becomes $v_{d'} = \frac{eV\tau}{m(4l)} = \frac{v_d}{4}$.
Therefore,the drift velocity decreases by $4$ times.

(B) Let the initial resistances be $R_1 = 2 \ \Omega$ and $R_2 = 3 \ \Omega$. Let the initial balancing length be $l_1$.
Using the meter bridge principle: $\frac{R_1}{R_2} = \frac{l_1}{100 - l_1} \Rightarrow \frac{2}{3} = \frac{l_1}{100 - l_1} \Rightarrow 200 - 2l_1 = 3l_1 \Rightarrow 5l_1 = 200 \Rightarrow l_1 = 40 \ cm$.
When a shunt $S$ is added in parallel to $3 \ \Omega$,the new resistance $R_2'$ becomes $\frac{3S}{3+S}$.
The new balancing length $l_2$ is shifted by $22.5 \ cm$. Since $R_2$ decreases,the balancing point shifts towards the $3 \ \Omega$ side,so $l_2 = 40 + 22.5 = 62.5 \ cm$.
Using the new balance condition: $\frac{R_1}{R_2'} = \frac{l_2}{100 - l_2} \Rightarrow \frac{2}{\frac{3S}{3+S}} = \frac{62.5}{100 - 62.5} = \frac{62.5}{37.5} = \frac{5}{3}$.
$\frac{2(3+S)}{3S} = \frac{5}{3} \Rightarrow \frac{6+2S}{S} = 5 \Rightarrow 6 + 2S = 5S \Rightarrow 3S = 6 \Rightarrow S = 2 \ \Omega$.

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the velocity of the particle.
Since both the electron and the proton are moving with the same velocity $v$,the wavelength is inversely proportional to the mass of the particle: $\lambda \propto \frac{1}{m}$.
Therefore,the ratio of the de-Broglie wavelength of the electron $(\lambda_e)$ to that of the proton $(\lambda_p)$ is given by $\frac{\lambda_e}{\lambda_p} = \frac{m_p}{m_e}$.
Thus,the ratio is $m_p : m_e$.

(C) The energy flux $I = 9 \ Wcm^{-2} = 9 \times 10^4 \ Wm^{-2}$.
The area $A = 20 \ cm^2 = 20 \times 10^{-4} \ m^2$.
The time $t = 1 \ hour = 3600 \ s$.
The speed of light $c = 3 \times 10^8 \ ms^{-1}$.
For a non-reflecting surface, the momentum $p$ delivered by radiation is given by $p = \frac{U}{c} = \frac{IAt}{c}$.
Substituting the values:
$p = \frac{(9 \times 10^4) \times (20 \times 10^{-4}) \times 3600}{3 \times 10^8}$.
$p = \frac{180 \times 3600}{3 \times 10^8} = \frac{648000}{3 \times 10^8} = 216000 \times 10^{-8} \ kgms^{-1}$.
$p = 2.16 \times 10^{-3} \ kgms^{-1}$.

(A) The formula for induced emf $(e)$ in an inductor is given by $e = -L \frac{di}{dt}$.
Given:
Inductance $L = 30 \ mH = 30 \times 10^{-3} \ H$.
Initial current $i_1 = 6 \ A$.
Final current $i_2 = 2 \ A$.
Time interval $dt = 2 \ s$.
Change in current $di = i_2 - i_1 = 2 \ A - 6 \ A = -4 \ A$.
Substituting the values:
$e = - (30 \times 10^{-3} \ H) \times \left( \frac{-4 \ A}{2 \ s} \right)$.
$e = - (30 \times 10^{-3}) \times (-2) = 60 \times 10^{-3} \ V = 0.06 \ V$.
The magnitude of the induced emf is $0.06 \ V$.

(A) The relationship between the dielectric constant $K$ and the electric susceptibility $\chi_{e}$ is given by the formula:
$K = 1 + \frac{\chi_{e}}{\varepsilon_0}$
Given that $K = \frac{4}{3}$,we substitute this value into the equation:
$\frac{4}{3} = 1 + \frac{\chi_{e}}{\varepsilon_0}$
Subtracting $1$ from both sides:
$\frac{\chi_{e}}{\varepsilon_0} = \frac{4}{3} - 1$
$\frac{\chi_{e}}{\varepsilon_0} = \frac{1}{3}$
Therefore,the electric susceptibility is:
$\chi_{e} = \frac{\varepsilon_0}{3}$

(D) According to Coulomb's Law,the force between two point charges is given by $F = k \frac{|q_1 q_2|}{r^2}$.
Initially,$q_1 = +8 \mu C$ and $q_2 = +12 \mu C$,and the force $F_1 = 48 \ N$.
When an additional charge of $-10 \mu C$ is added to each,the new charges are:
$q_1' = 8 \mu C - 10 \mu C = -2 \mu C$
$q_2' = 12 \mu C - 10 \mu C = +2 \mu C$
Since the distance $r$ remains unchanged,the ratio of the forces is:
$\frac{F_2}{F_1} = \frac{|q_1' q_2'|}{|q_1 q_2|} = \frac{|(-2) \times 2|}{|8 \times 12|} = \frac{4}{96} = \frac{1}{24}$
Therefore,$F_2 = \frac{F_1}{24} = \frac{48 \ N}{24} = 2 \ N$.
Since the charges have opposite signs (one is negative and one is positive),the force is attractive.

(B) The force experienced by an electron of charge $e$ in an electric field $E$ is given by $F = eE$.
According to Newton's second law,the acceleration $a$ of the electron is $a = \frac{F}{m_e} = \frac{eE}{m_e}$.
Since the electron is released from rest,its initial velocity $u = 0$.
Using the kinematic equation for distance $S$ travelled in time $t$:
$S = ut + \frac{1}{2}at^2$
$S = 0 \cdot t + \frac{1}{2} \left( \frac{eE}{m_e} \right) t^2$
$S = \frac{eEt^2}{2m_e}$.

(C) The magnetic field $B$ at a distance $r$ from a long straight current-carrying conductor is given by $B = \frac{\mu_0 I}{2 \pi r}$.
Let the two conductors be placed at a distance $d = 10 \text{ cm} = 0.1 \text{ m}$.
At the midpoint,the distance from each conductor is $r = \frac{d}{2} = 0.05 \text{ m}$.
For the first conductor,the magnetic field $B_1$ at the midpoint is directed into the plane (using the right-hand thumb rule).
For the second conductor,the magnetic field $B_2$ at the midpoint is directed out of the plane.
Since the currents are equal $(I_1 = I_2 = 3 \text{ A})$ and the distances are equal,the magnitudes of the magnetic fields are equal: $B_1 = B_2 = \frac{\mu_0 I}{2 \pi r}$.
Since the fields are equal in magnitude and opposite in direction,the net magnetic field $B_{net} = B_1 - B_2 = 0$.

(B) For an electron to travel in a straight line in a crossed field (where electric and magnetic fields are perpendicular), the net Lorentz force must be zero.
$F_{net} = F_e + F_m = 0$
$eE = evB$
Where $e$ is the electronic charge, $E$ is the electric field intensity, $v$ is the velocity, and $B$ is the magnetic field induction.
$v = \frac{E}{B}$
Given $E = 20 \times 10^3 \,V/m$ and $B = 2.0 \,T$.
$v = \frac{20 \times 10^3}{2.0} = 10 \times 10^3 \,m/s$.

(C) The magnetic flux $\phi$ is given by the formula $\phi = B \cdot A$.
Since $B = \mu H$ and $\mu = \mu_0(1 + \chi_m)$,we have $\phi = \mu_0(1 + \chi_m) H A$.
Given:
Area $A = 0.25 \ cm^2 = 0.25 \times 10^{-4} \ m^2$.
Magnetic field strength $H = 1000 \ Am^{-1}$.
Susceptibility $\chi_m = 313$.
Permeability of free space $\mu_0 = 4 \pi \times 10^{-7} \ Hm^{-1}$.
Substituting the values:
$\phi = (4 \pi \times 10^{-7}) \times (1 + 313) \times 1000 \times (0.25 \times 10^{-4})$
$\phi = (4 \pi \times 10^{-7}) \times 314 \times 10^3 \times 0.25 \times 10^{-4}$
$\phi = 4 \pi \times 314 \times 0.25 \times 10^{-8}$
$\phi = 314 \pi \times 10^{-8} \approx 986.45 \times 10^{-8} \approx 9.87 \times 10^{-6} \ Wb$.

(D) The given reaction is $n \rightarrow p + e^{-} + \bar{\nu}$.
This represents the $\beta^{-}$ decay process,where a neutron transforms into a proton,an electron,and an antineutrino.
In $\beta^{-}$ decay,the conservation of lepton number requires the emission of an antineutrino $(\bar{\nu})$ to balance the electron (lepton number $+1$) produced in the reaction.
Therefore,$x$ stands for antineutrino.

(A) The refraction formula at a spherical surface is given by:
$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$
Given values:
$\mu_1 = 1$ (air),$\mu_2 = 1.5$ (glass)
$u = -100 \ cm$ (distance of object from the surface,taken as negative by sign convention)
$v = +100 \ cm$ (distance of image from the surface,taken as positive as it is formed inside the glass)
Substituting these values into the formula:
$\frac{1.5}{100} - \frac{1}{-100} = \frac{1.5 - 1}{R}$
$\frac{1.5}{100} + \frac{1}{100} = \frac{0.5}{R}$
$\frac{2.5}{100} = \frac{0.5}{R}$
$R = \frac{0.5 \times 100}{2.5} = \frac{50}{2.5} = 20 \ cm$
Thus,the radius of curvature is $20 \ cm$.

(D) The circuit consists of an $AND$ gate followed by a $NAND$ gate. The inputs to the $AND$ gate are $A$ and $B$,so its output is $A \cdot B$.
The input $C$ passes through a $NOT$ gate,so its output is $\bar{C}$.
These two signals are the inputs to the $NAND$ gate,which produces the final output $Y$.
Thus,$Y = \overline{(A \cdot B) \cdot \bar{C}}$.
Using De Morgan's law,$Y = \overline{A \cdot B} + \overline{\bar{C}} = \bar{A} + \bar{B} + C$.
For the output $Y$ to be zero $(Y=0)$,we must have $\bar{A} + \bar{B} + C = 0$.
This requires $\bar{A} = 0$,$\bar{B} = 0$,and $C = 0$ simultaneously.
Therefore,$A = 1$,$B = 1$,and $C = 0$.

(C) Given the ratio of intensities of two coherent sources is $\frac{I_1}{I_2} = \frac{9}{4}$.
Let $I_1 = 9k$ and $I_2 = 4k$,where $k$ is a constant.
The ratio of maximum intensity to minimum intensity in an interference pattern is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \left(\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}}\right)^2$
Substituting the values:
$\frac{I_{\max}}{I_{\min}} = \left(\frac{\sqrt{9k} + \sqrt{4k}}{\sqrt{9k} - \sqrt{4k}}\right)^2$
$\frac{I_{\max}}{I_{\min}} = \left(\frac{3\sqrt{k} + 2\sqrt{k}}{3\sqrt{k} - 2\sqrt{k}}\right)^2$
$\frac{I_{\max}}{I_{\min}} = \left(\frac{5\sqrt{k}}{\sqrt{k}}\right)^2 = (5)^2 = 25$
Thus,the ratio is $25:1$.