TS EAMCET 2024 Physics Question Paper with Answer and Solution

(D) For an object rolling down an inclined plane of length $l$ and angle $\theta$,the acceleration is given by $a = \frac{g \sin \theta}{1 + K^2/R^2}$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$ with $u=0$ and $s=l$,we get $l = \frac{1}{2} \left( \frac{g \sin \theta}{1 + K^2/R^2} \right) t^2$.
Thus,$t = \sqrt{\frac{2l(1 + K^2/R^2)}{g \sin \theta}}$,which implies $t \propto \sqrt{1 + K^2/R^2}$.
For a hollow cylinder,the radius of gyration $K^2 = R^2$,so $K^2/R^2 = 1$. Thus,$t_1 \propto \sqrt{1 + 1} = \sqrt{2}$.
For a solid cylinder,the radius of gyration $K^2 = R^2/2$,so $K^2/R^2 = 1/2$. Thus,$t_2 \propto \sqrt{1 + 1/2} = \sqrt{3/2}$.
Taking the ratio,$\frac{t_2}{t_1} = \frac{\sqrt{3/2}}{\sqrt{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.
Given $t_1 = 2 \ s$,we have $t_2 = \frac{\sqrt{3}}{2} \times 2 = \sqrt{3} \approx 1.732 \ s$.

(B) For equal masses $m$,the final temperature $T_{mix}$ of a mixture of two liquids with specific heats $C_1, C_2$ and initial temperatures $T_1, T_2$ is given by $T_{mix} = \frac{T_1 C_1 + T_2 C_2}{C_1 + C_2}$.
For liquids $A$ and $B$:
$20 = \frac{15 C_A + 24 C_B}{C_A + C_B} \Rightarrow 20 C_A + 20 C_B = 15 C_A + 24 C_B \Rightarrow 5 C_A = 4 C_B \Rightarrow C_A = \frac{4}{5} C_B$.
For liquids $B$ and $C$:
$26 = \frac{24 C_B + 30 C_C}{C_B + C_C} \Rightarrow 26 C_B + 26 C_C = 24 C_B + 30 C_C \Rightarrow 2 C_B = 4 C_C \Rightarrow C_B = 2 C_C$.
Expressing all in terms of $C_C$:
$C_B = 2 C_C$
$C_A = \frac{4}{5} (2 C_C) = \frac{8}{5} C_C$
Thus,$C_A : C_B : C_C = \frac{8}{5} C_C : 2 C_C : C_C = 8 : 10 : 5$.

(B) Given: $m_{\text{ice}} = 54 \ g$,$T_{\text{ice}} = -20^{\circ} C$,$m_{\text{steam}} = 25 \ g$,$T_{\text{steam}} = 100^{\circ} C$.
Heat required to bring ice to $0^{\circ} C$: $Q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T = 54 \times 2.1 \times 20 = 2268 \ J$.
Heat required to melt the ice at $0^{\circ} C$: $Q_2 = m_{\text{ice}} \cdot L_f = 54 \times 334 = 18036 \ J$.
Total heat required to convert ice at $-20^{\circ} C$ to water at $0^{\circ} C$: $Q_{\text{total}} = Q_1 + Q_2 = 2268 + 18036 = 20304 \ J$.
Heat released by condensing $25 \ g$ of steam at $100^{\circ} C$ to water at $100^{\circ} C$: $Q_{\text{condense}} = m_{\text{steam}} \cdot L_v = 25 \times 2260 = 56500 \ J$.
Since $Q_{\text{condense}} > Q_{\text{total}}$,all ice melts and the water temperature rises to $100^{\circ} C$.
Heat required to raise the temperature of $54 \ g$ of water from $0^{\circ} C$ to $100^{\circ} C$: $Q_3 = m_{\text{ice}} \cdot c_{\text{water}} \cdot \Delta T = 54 \times 4.2 \times 100 = 22680 \ J$.
Remaining heat available for condensation: $Q_{\text{rem}} = Q_{\text{condense}} - (Q_{\text{total}} + Q_3) = 56500 - (20304 + 22680) = 56500 - 42984 = 13516 \ J$.
Mass of steam condensed by this remaining heat: $m_{\text{condensed}} = Q_{\text{rem}} / L_v = 13516 / 2260 \approx 6 \ g$.
Final mass of water = $54 \ g$ (melted ice) + $(25 - 6) \ g$ (condensed steam) = $73 \ g$.
Final mass of steam = $6 \ g$ (remaining steam).
Thus,the mixture contains $73 \ g$ of water and $6 \ g$ of steam at $100^{\circ} C$.

(C) By the principle of calorimetry,Heat lost by water $=$ Heat gained by ice.
Let $T$ be the final equilibrium temperature.
Heat lost by $74 \ g$ of water $= m_w c_w (T_i - T) = 74 \times 1 \times (70 - T)$.
Heat gained by $37 \ g$ of ice to melt at $0^{\circ} C = m_i L_f = 37 \times 80$.
Heat gained by $37 \ g$ of melted water to reach temperature $T = m_i c_w (T - 0) = 37 \times 1 \times T$.
Equating the heat lost and gained:
$74(70 - T) = 37 \times 80 + 37T$.
Dividing by $37$:
$2(70 - T) = 80 + T$.
$140 - 2T = 80 + T$.
$3T = 60$.
$T = 20^{\circ} C$.

(A) For water,$m_1 = 360 \ g$,$T_1 = 40^{\circ} C$,$T_f = 100^{\circ} C$.
Heat required by water to reach $100^{\circ} C$ is $H_1 = m_1 C_w (T_f - T_1) = 360 \times 1 \times (100 - 40) = 21600 \ cal$.
At equilibrium,heat released by steam condensing at $100^{\circ} C$ equals the heat required by water.
Let $m'$ be the mass of steam that condenses. $m' \times L = H_1 \implies m' \times 540 = 21600 \implies m' = 40 \ g$.
Remaining mass of steam $m_s = 60 - 40 = 20 \ g$.
Total mass of water $m_w = 360 + 40 = 400 \ g$.
The ratio of masses of steam to water is $m_s : m_w = 20 : 400 = 1 : 20$.

(C) According to the principle of calorimetry,Heat lost by hot body = Heat gained by cold body.
Let the final temperature be $T$.
Heat lost by $74 \ g$ of water at $70^{\circ} C$ to reach temperature $T$ is $Q_{lost} = m_w c_w (70 - T) = 74 \times 1 \times (70 - T)$.
Heat gained by $37 \ g$ of ice at $0^{\circ} C$ to melt and reach temperature $T$ is $Q_{gained} = m_i L_f + m_i c_w (T - 0) = 37 \times 80 + 37 \times 1 \times T$.
Equating the two: $74(70 - T) = 2960 + 37T$.
$5180 - 74T = 2960 + 37T$.
$111T = 2220$.
$T = 20^{\circ} C$.

(B) Given: Thickness $\Delta x = 5 \ mm = 5 \times 10^{-3} \ m$,Area $A = 5 \ cm^2 = 5 \times 10^{-4} \ m^2$,Temperature difference $\Delta T = 14^{\circ} C$,Heat flow rate $Q = 42 \ J/s = 42 \ W$.
The formula for heat conduction in steady state is $Q = \frac{K A \Delta T}{\Delta x}$.
Rearranging for thermal conductivity $K$: $K = \frac{Q \Delta x}{A \Delta T}$.
Substituting the values: $K = \frac{42 \times 5 \times 10^{-3}}{5 \times 10^{-4} \times 14}$.
Simplifying the expression: $K = \frac{42 \times 10^{-3}}{10^{-4} \times 14} = \frac{3 \times 10^{-3}}{10^{-4}} = 3 \times 10 = 30 \ W \ m^{-1} \ K^{-1}$.

(C) According to Stefan-Boltzmann law,the energy radiated per second $E$ is proportional to the surface area $A$ of the body,assuming temperature $T$ and emissivity $e$ remain constant.
$E = e \sigma A T^4 \implies E \propto A$
For a solid sphere of radius $r$,the surface area is $A_{sphere} = 4 \pi r^2$.
When the sphere is cut into two hemispheres,each hemisphere has a curved surface area of $2 \pi r^2$ and a flat circular base of area $\pi r^2$.
Therefore,the total surface area of one hemisphere is $A_{hemisphere} = 2 \pi r^2 + \pi r^2 = 3 \pi r^2$.
The ratio of energy radiated by one hemisphere to the whole sphere is $\frac{E_{hemisphere}}{E_{sphere}} = \frac{A_{hemisphere}}{A_{sphere}} = \frac{3 \pi r^2}{4 \pi r^2} = \frac{3}{4}$.

(D) The rate of heat flow $H$ through a cylindrical rod is given by $H = \frac{kA \Delta \theta}{\ell}$,where $k$ is thermal conductivity,$A = \pi r^2$ is the cross-sectional area,$\Delta \theta$ is the temperature difference,and $\ell$ is the length.
Since the material is the same,$k_A = k_B$.
Given: $\frac{\Delta \theta_A}{\Delta \theta_B} = \frac{2}{3}$,$\frac{H_A}{H_B} = \frac{5}{9}$,and $\frac{r_A}{r_B} = \frac{1}{2}$.
Using the formula $\frac{H_A}{H_B} = \left( \frac{r_A}{r_B} \right)^2 \left( \frac{\Delta \theta_A}{\Delta \theta_B} \right) \left( \frac{\ell_B}{\ell_A} \right)$:
$\frac{5}{9} = \left( \frac{1}{2} \right)^2 \left( \frac{2}{3} \right) \left( \frac{\ell_B}{\ell_A} \right)$
$\frac{5}{9} = \left( \frac{1}{4} \right) \left( \frac{2}{3} \right) \left( \frac{\ell_B}{\ell_A} \right)$
$\frac{5}{9} = \frac{2}{12} \left( \frac{\ell_B}{\ell_A} \right) = \frac{1}{6} \left( \frac{\ell_B}{\ell_A} \right)$
$\frac{\ell_B}{\ell_A} = \frac{5}{9} \times 6 = \frac{30}{9} = \frac{10}{3}$
Therefore,$\frac{\ell_A}{\ell_B} = \frac{3}{10}$.

(C) The relationship between Fahrenheit $(F)$ and Celsius $(C)$ temperature scales is given by the formula: $F = \frac{9}{5} C + 32$.
According to the problem,the temperature on the Fahrenheit scale is twice the temperature on the Celsius scale,so $F = 2C$,which implies $C = \frac{F}{2}$.
Substituting this into the formula: $F = \frac{9}{5} \left( \frac{F}{2} \right) + 32$.
$F = \frac{9F}{10} + 32$.
Multiplying the entire equation by $10$: $10F = 9F + 320$.
$10F - 9F = 320$.
Therefore,$F = 320^{\circ} F$.

(C) The formula for the total internal energy $U$ of an ideal gas is given by $U = n \frac{f}{2} R T$.
For a diatomic gas,the degrees of freedom $f = 5$.
Given: number of moles $n = 4$,temperature $T = 27^{\circ} C = 27 + 273 = 300 \ K$,and $R = 8.31 \ J \ mol^{-1} \ K^{-1}$.
Substituting these values into the formula:
$U = 4 \times \frac{5}{2} \times 8.31 \times 300$
$U = 2 \times 5 \times 8.31 \times 300$
$U = 10 \times 2493 = 24930 \ J$
$U = 24.93 \ kJ$.

(D) At constant pressure,$W = P \Delta V = nR \Delta T$,which implies $W \propto \Delta T$.
Given $W_1 = W$ and $W_2 = 2W$,we have $\frac{(\Delta T)_2}{(\Delta T)_1} = \frac{W_2}{W_1} = 2$,so $(\Delta T)_2 = 2(\Delta T)_1$.
For a monatomic gas,the degrees of freedom $f_1 = 3$. The heat supplied is $Q_1 = \Delta U_1 + W_1 = \frac{f_1}{2} nR(\Delta T)_1 + nR(\Delta T)_1 = (\frac{3}{2} + 1) nR(\Delta T)_1 = \frac{5}{2} nR(\Delta T)_1$.
For a diatomic gas,the degrees of freedom $f_2 = 5$. The heat supplied is $Q_2 = \Delta U_2 + W_2 = \frac{f_2}{2} nR(\Delta T)_2 + 2W = \frac{5}{2} nR(2(\Delta T)_1) + 2nR(\Delta T)_1 = (5 + 2) nR(\Delta T)_1 = 7nR(\Delta T)_1$.
Therefore,the ratio is $\frac{Q_1}{Q_2} = \frac{\frac{5}{2} nR(\Delta T)_1}{7nR(\Delta T)_1} = \frac{5}{14}$.
Thus,$Q_1 : Q_2 = 5 : 14$.

(A) The efficiency of a reversible heat engine is given by $\eta = 1 - \frac{T_2}{T_1} = 0.5$.
From this,we find $\frac{T_2}{T_1} = 1 - 0.5 = 0.5$.
The coefficient of performance $(COP)$ of a refrigerator working between the same temperatures is given by $\beta = \frac{T_2}{T_1 - T_2}$.
Dividing the numerator and denominator by $T_1$,we get $\beta = \frac{T_2/T_1}{1 - T_2/T_1}$.
Substituting the value $\frac{T_2}{T_1} = 0.5$,we get $\beta = \frac{0.5}{1 - 0.5} = \frac{0.5}{0.5} = 1$.

(D) For a diatomic gas at constant pressure:
$1$. The work done is $dW = P dV = nR dT$.
$2$. The change in internal energy is $dU = n C_v dT$. For a diatomic gas,the molar heat capacity at constant volume is $C_v = \frac{5}{2} R$,so $dU = n \left( \frac{5}{2} R \right) dT$.
$3$. The heat absorbed is $dQ = n C_p dT$. For a diatomic gas,the molar heat capacity at constant pressure is $C_p = \frac{7}{2} R$,so $dQ = n \left( \frac{7}{2} R \right) dT$.
$4$. The ratio $dW : dU : dQ$ is:
$dW : dU : dQ = nR dT : n \left( \frac{5}{2} R \right) dT : n \left( \frac{7}{2} R \right) dT$
Dividing by $nR dT$,we get:
$1 : \frac{5}{2} : \frac{7}{2}$
Multiplying by $2$ to simplify:
$2 : 5 : 7$.

(D) Given,the ratio of specific heat capacities $\gamma = 1.5 = 3/2$.
For an adiabatic process,$P_i V_i^\gamma = P_f V_f^\gamma$. Given $V_f = 2V_i$ and $P_f = P_1$,we have $P_{i,ad} V_i^{1.5} = P_1 (2V_i)^{1.5}$.
Thus,$P_{i,ad} = P_1 (2)^{1.5} = P_1 (2\sqrt{2})$.
For an isothermal process,$P_i V_i = P_f V_f$. Given $V_f = 2V_i$ and $P_f = P_2$,we have $P_{i,iso} V_i = P_2 (2V_i)$.
Thus,$P_{i,iso} = 2P_2$.
Given $P_1 = P_2$,the ratio of initial pressures is $\frac{P_{i,ad}}{P_{i,iso}} = \frac{P_1 (2\sqrt{2})}{2P_1} = \frac{2\sqrt{2}}{2} = \sqrt{2} : 1$.

(A) The given equation is $m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + kx = 0$.
According to the principle of homogeneity of dimensions,each term in the equation must have the same dimensions.
Thus,the dimensions of $m \frac{d^2 x}{dt^2}$,$b \frac{dx}{dt}$,and $kx$ are equal.
Equating the dimensions of $b \frac{dx}{dt}$ and $kx$:
$[b] [v] = [k] [x] \implies [b] = [k] [x] / [v] = [k] [x] / ([x] / [t]) = [k] [t]$.
Therefore,$[b] = [k] [T]$.
Now,consider the expression $\frac{b}{\sqrt{km}}$.
Substituting $[b] = [k] [T]$ into the expression:
$\frac{[b]}{\sqrt{[k][m]}} = \frac{[k][T]}{\sqrt{[k][m]}}$.
Since $k = F/x = ma/x$,the dimensions of $k$ are $[M T^{-2}]$.
Also,the angular frequency of a damped oscillator is related to $\omega = \sqrt{k/m}$,so $\sqrt{k/m}$ has dimensions of $[T^{-1}]$.
Alternatively,using $[b] = [k] [T]$:
$\frac{b}{\sqrt{km}} = \frac{k T}{\sqrt{km}} = \sqrt{\frac{k}{m}} T = [T^{-1}] [T] = [M^0 L^0 T^0]$.

(B) Reductionism is the scientific approach that attempts to explain a complex system by breaking it down into its fundamental,simpler components and understanding the interactions between them. By studying these constituent parts,one can derive the macroscopic properties of the larger system.

(A) Physical laws are based on conservation principles. Conserved quantities can be scalars (like energy,mass,or charge) or vectors (like linear momentum or angular momentum). Therefore,the statement that all conserved quantities are necessarily scalars is incorrect.

(D) The speed of a transverse wave in a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since $\mu = \text{Area} \times \text{density} = (\pi r^2) \rho$,the speed becomes $v = \sqrt{\frac{T}{\pi r^2 \rho}}$.
Given that the tension $T$ and radius $r$ are the same for both strings,we have $v \propto \frac{1}{\sqrt{\rho}}$.
Let $\rho_A = \rho$. Then $\rho_B = \rho + 0.02\rho = 1.02\rho$.
The speed in string '$B$' is $v_B = \sqrt{\frac{T}{\pi r^2 (1.02\rho)}} = \frac{1}{\sqrt{1.02}} \sqrt{\frac{T}{\pi r^2 \rho}}$.
Therefore,$v_B = \frac{v}{\sqrt{1.02}}$.

(C) Given frequencies are $f_1 = n-1$,$f_2 = n$,and $f_3 = n+1$.
Beats are produced due to the difference in frequencies of the sound sources.
The number of beats produced per second is determined by the difference between the maximum and minimum frequencies present in the system.
$\text{Beats produced} = f_{\text{max}} - f_{\text{min}} = (n+1) - (n-1) = 2 \text{ beats/sec}$.
Since all three sources are vibrating together,the interference pattern results in a beat frequency of $2 \text{ Hz}$.
Therefore,the number of beats produced is $2$ and the number of beats heard is also $2$.
Thus,the correct option is $C$.

(D) Given equations are $y_1 = 5 \sin(400 \pi t)$ and $y_2 = 8 \sin(408 \pi t)$.
Comparing these with the standard equation $y = A \sin(\omega t)$,we get angular frequencies $\omega_1 = 400 \pi \text{ rad/s}$ and $\omega_2 = 408 \pi \text{ rad/s}$.
The frequencies $f_1$ and $f_2$ are given by $f = \frac{\omega}{2 \pi}$.
$f_1 = \frac{400 \pi}{2 \pi} = 200 \text{ Hz}$ and $f_2 = \frac{408 \pi}{2 \pi} = 204 \text{ Hz}$.
The beat frequency is $|f_2 - f_1| = |204 - 200| = 4 \text{ beats per second}$.
To find the number of beats per minute,we multiply by $60$:
$\text{Beats per minute} = 4 \times 60 = 240 \text{ beats/min}$.

(D) The speed of the car is $v_s = 54 \ km/h = 54 \times \frac{5}{18} = 15 \ m/s$.
The person is standing between the car and the wall.
The frequency of the sound received directly from the car (source moving towards stationary observer) is $f_1 = f \left( \frac{v}{v - v_s} \right)$.
The frequency of the sound reflected from the wall is equivalent to the sound coming from an image of the car moving towards the observer at the same speed $v_s$. Thus,$f_2 = f \left( \frac{v}{v - v_s} \right)$.
Since $f_1 = f_2$,the difference in frequencies is $f_2 - f_1 = 0 \ Hz$.

(A) Using Doppler's effect formula:
$1$) Frequency when the train approaches: $f_{\text{app}} = f_0 \times \frac{v}{v - v_s}$
$2$) Frequency when the train recedes: $f_{\text{rec}} = f_0 \times \frac{v}{v + v_s}$
Given: $v = 330 \text{ m/s}$,$v_s = 108 \text{ km/h} = 108 \times \frac{5}{18} = 30 \text{ m/s}$.
The difference in frequency is $\Delta f = f_{\text{app}} - f_{\text{rec}} = f_0 \left( \frac{v}{v - v_s} - \frac{v}{v + v_s} \right)$.
$\Delta f = f_0 \left( \frac{v(v + v_s) - v(v - v_s)}{v^2 - v_s^2} \right) = f_0 \left( \frac{2 v v_s}{v^2 - v_s^2} \right)$.
Percentage change = $\frac{\Delta f}{f_0} \times 100 = \left( \frac{2 v v_s}{v^2 - v_s^2} \right) \times 100$.
Substituting the values: $\frac{2 \times 330 \times 30}{330^2 - 30^2} \times 100 = \frac{19800}{108900 - 900} \times 100 = \frac{19800}{108000} \times 100 = \frac{11}{60} \times 100 \approx 18.33 \%$.

(B) Let $v$ be the speed of sound and $f = 102 \ Hz$ be the frequency of the source.
Since the source is at rest $(v_s = 0)$,the frequency $f'$ heard by an observer moving away from the source with speed $v_o$ is given by the Doppler effect formula: $f' = f \left( \frac{v - v_o}{v} \right)$.
Given that both observers move away from the source with speed $v_o = 10 \% \text{ of } v = \frac{v}{10}$,the frequency heard by observer $1$ is:
$f_1 = f \left( \frac{v - v/10}{v} \right) = f \left( \frac{0.9v}{v} \right) = 0.9f$.
Similarly,the frequency heard by observer $2$ is:
$f_2 = f \left( \frac{v - v/10}{v} \right) = f \left( \frac{0.9v}{v} \right) = 0.9f$.
Therefore,the ratio of the frequencies heard by the observers is:
$\frac{f_1}{f_2} = \frac{0.9f}{0.9f} = 1: 1$.

(A) Let $v$ be the speed of sound and $l$ be the length of the pipes.
For an open organ pipe,the fundamental frequency is $f_{o} = \frac{v}{2l}$.
For a closed organ pipe,the fundamental frequency is $f_{c} = \frac{v}{4l}$.
Given that the difference between these fundamental frequencies is $100 \ Hz$:
$f_{o} - f_{c} = 100 \ Hz \implies \frac{v}{2l} - \frac{v}{4l} = 100 \ Hz \implies \frac{v}{4l} = 100 \ Hz$.
The second harmonic of the open pipe is $f_{2,o} = 2 \times f_{o} = 2 \times \frac{v}{2l} = \frac{v}{l}$.
The third harmonic of the closed pipe is $f_{3,c} = 3 \times f_{c} = 3 \times \frac{v}{4l} = \frac{3v}{4l}$.
The difference between these frequencies is:
$f_{2,o} - f_{3,c} = \frac{v}{l} - \frac{3v}{4l} = \frac{4v - 3v}{4l} = \frac{v}{4l}$.
Since $\frac{v}{4l} = 100 \ Hz$,the required difference is $100 \ Hz$.

(C) Given: Path difference $\Delta x = 50 \ cm = 0.5 \ m$,Phase difference $\Delta \phi = 1.8 \pi$,Speed of sound $v = 340 \ ms^{-1}$.
We know the relation between path difference and phase difference is $\Delta x = \frac{\lambda}{2 \pi} \cdot \Delta \phi$.
Substituting the values: $0.5 = \frac{\lambda}{2 \pi} \times 1.8 \pi$.
$0.5 = \lambda \times 0.9 \Rightarrow \lambda = \frac{0.5}{0.9} = \frac{5}{9} \ m$.
The frequency $f$ is given by $f = \frac{v}{\lambda}$.
$f = \frac{340}{5/9} = \frac{340 \times 9}{5} = 68 \times 9 = 612 \ Hz$.

(B) Given: Radius $r = 9 \ km = 9000 \ m$,Speed $v = 540 \ kmh^{-1} = 540 \times \frac{5}{18} = 150 \ ms^{-1}$,Acceleration due to gravity $g = 10 \ ms^{-2}$.
For a banked aircraft in a horizontal loop,the banking angle $\theta$ is given by the relation: $\tan \theta = \frac{v^2}{rg}$.
Substituting the values: $\tan \theta = \frac{150 \times 150}{9000 \times 10} = \frac{22500}{90000} = \frac{1}{4}$.
Therefore,$\theta = \tan^{-1}(0.25) = \cot^{-1}(4)$.

(D) Let the body slide up to a distance '$s$' before coming to rest.
The height reached is $h = s \sin \theta$.
According to the work-energy theorem,the total work done by all forces equals the change in kinetic energy:
$W_{friction} + W_{gravity} = K_f - K_i$
$-\mu mg \cos \theta \cdot s - mg \sin \theta \cdot s = 0 - E$
$\mu mgs \cos \theta + mgs \sin \theta = E$
$s(mg(\mu \cos \theta + \sin \theta)) = E$
$s = \frac{E}{mg(\mu \cos \theta + \sin \theta)}$
The work done against friction is $W_{against\ friction} = |W_{friction}| = \mu mg \cos \theta \cdot s$.
Substituting the value of '$s$':
$W_{against\ friction} = \mu mg \cos \theta \cdot \left( \frac{E}{mg(\mu \cos \theta + \sin \theta)} \right)$
$W_{against\ friction} = \frac{\mu E \cos \theta}{\sin \theta + \mu \cos \theta}$

(C) Given: Mass $m = 4 \ kg$. Velocity vector $\vec{v} = (2 \hat{i} - 4 \hat{j} - \hat{k}) \ ms^{-1}$.
First,calculate the magnitude of the velocity $v = |\vec{v}| = \sqrt{(2)^2 + (-4)^2 + (-1)^2}$.
$v = \sqrt{4 + 16 + 1} = \sqrt{21} \ ms^{-1}$.
The kinetic energy $K.E.$ is given by the formula $K.E. = \frac{1}{2} mv^2$.
Substituting the values: $K.E. = \frac{1}{2} \times 4 \times (\sqrt{21})^2$.
$K.E. = 2 \times 21 = 42 \ J$.

(B) Let the mass of the body be $m$ and the initial velocity be $u$. At maximum height $h$,the total energy is $E = mgh$.
At a height $y = 0.4h$ from the ground,the potential energy is $PE = mgy = mg(0.4h) = 0.4mgh$.
According to the law of conservation of energy,the total energy remains constant.
Therefore,$KE + PE = E$.
$KE = E - PE = mgh - 0.4mgh = 0.6mgh$.
The ratio of kinetic energy to potential energy is $\frac{KE}{PE} = \frac{0.6mgh}{0.4mgh} = \frac{0.6}{0.4} = \frac{3}{2}$.
Thus,the ratio is $3: 2$.

(A) The time period of oscillation of a current-carrying coil in an external magnetic field is given by $T = 2\pi \sqrt{\frac{I_{moment}}{MB}}$,where $I_{moment}$ is the moment of inertia and $M$ is the magnetic dipole moment.
Since the magnetic dipole moment $M = NIA$,where $N$ is the number of turns,$I$ is the current,and $A$ is the area,we have $M \propto I$.
Therefore,$T \propto \frac{1}{\sqrt{M}} \propto \frac{1}{\sqrt{I}}$.
Given $I_1 = 2.5 \ A$ and $I_2 = 10 \ A$,we have:
$\frac{T_2}{T_1} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{2.5}{10}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,$T_2 = \frac{T_1}{2} = \frac{T}{2}$.

(D) The torque $\tau$ on a bar magnet in a uniform magnetic field $B$ is given by the formula $\tau = MB \sin \theta$,where $M$ is the magnetic dipole moment and $\theta$ is the angle between the magnetic moment and the magnetic field.
Initial torque $\tau_1 = MB \sin 30^{\circ} = MB \times 0.5$.
Final torque $\tau_2 = MB \sin 45^{\circ} = MB \times \frac{1}{\sqrt{2}} \approx MB \times 0.707$.
The ratio of the torques is $\frac{\tau_2}{\tau_1} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2} \approx 1.414$.
Therefore,$\tau_2 = 1.414 \tau_1$.
The percentage increase in torque is given by $\frac{\tau_2 - \tau_1}{\tau_1} \times 100 = (1.414 - 1) \times 100 = 41.4 \%$.
Thus,the torque increases by $41.4 \%$.

(D) In the decay of a nucleus with mass number $A = 236$,the $\alpha$-particle $(m_{\alpha} = 4)$ and the daughter nucleus $(m_d = 232)$ are produced.
Given the kinetic energy of the $\alpha$-particle is $(KE)_{\alpha} = E$.
Since the total momentum is conserved,the magnitudes of the momenta of the $\alpha$-particle and the daughter nucleus are equal: $P_{\alpha} = P_d = P$.
The kinetic energy is given by $KE = \frac{P^2}{2m}$,which implies $KE \propto \frac{1}{m}$.
Therefore,the ratio of kinetic energies is $\frac{(KE)_d}{(KE)_{\alpha}} = \frac{m_{\alpha}}{m_d} = \frac{4}{232} = \frac{1}{58}$.
Thus,$(KE)_d = \frac{E}{58}$.
The total energy released $(Q)$ is the sum of the kinetic energies of the products: $Q = (KE)_{\alpha} + (KE)_d = E + \frac{E}{58} = \frac{59 E}{58}$.

(D) Energy released per fission $E = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
Power $P = 128 \text{ W} = 128 \text{ J/s}$.
Let $n$ be the number of fissions per second.
Then,$P = n \times E$.
$n = \frac{P}{E} = \frac{128}{3.2 \times 10^{-11}}$.
$n = \frac{1280}{3.2} \times 10^{10} = 400 \times 10^{10} = 4 \times 10^{12} \text{ fissions/s}$.

(C) The neutron multiplication factor $K$ (also denoted as $k$) is defined as the ratio of the number of neutrons produced in a given generation to the number of neutrons produced in the preceding generation.
When $K = 1$,the rate of neutron production is equal to the rate of neutron loss,meaning the chain reaction is self-sustaining at a constant power level.
This state is known as the critical state of a nuclear reactor.
If $K < 1$,the reactor is subcritical,and the chain reaction dies out.
If $K > 1$,the reactor is supercritical,and the power level increases exponentially.

(A) The average mass of an element is calculated by taking the weighted average of the masses of its isotopes based on their relative abundances.
Given the ratio of abundances is $2:3:5$,the total parts are $2 + 3 + 5 = 10$.
The fractional abundances are:
For isotope $A$: $2/10 = 0.2$
For isotope $B$: $3/10 = 0.3$
For isotope $C$: $5/10 = 0.5$
The average mass is given by the formula: $\text{Average Mass} = (\text{fraction}_A \times m_1) + (\text{fraction}_B \times m_2) + (\text{fraction}_C \times m_3)$.
Substituting the values: $\text{Average Mass} = 0.2 m_1 + 0.3 m_2 + 0.5 m_3$.

(D) Given half-life $T_{1/2} = 12 \text{ minutes}$.
At $28 \%$ decay, the remaining amount is $100 - 28 = 72 \%$.
At $82 \%$ decay, the remaining amount is $100 - 82 = 18 \%$.
We know that the amount of radioactive substance reduces to half in one half-life period.
Starting from $72 \%$, after one half-life $(12 \text{ minutes})$, the amount becomes $72 / 2 = 36 \%$.
After another half-life $(12 \text{ minutes})$, the amount becomes $36 / 2 = 18 \%$.
Therefore, the total time gap is $12 + 12 = 24 \text{ minutes}$.

(A) The mass of a radioactive substance remaining after time $t$ is given by $M = M_0 (1/2)^{t/T_{1/2}}$,where $M_0$ is the initial mass and $T_{1/2}$ is the half-life.
For material $A$: $M_A = M_{0A} (1/2)^{t/T}$.
For material $B$: $M_B = M_{0B} (1/2)^{t/2T}$.
Given the ratio of initial masses $M_{0A} / M_{0B} = 8/1$ and the ratio of remaining masses $M_A / M_B = 4/1$.
Therefore,$\frac{M_A}{M_B} = \frac{M_{0A}}{M_{0B}} \cdot \frac{(1/2)^{t/T}}{(1/2)^{t/2T}} = 4/1$.
Substituting the values: $8 \cdot (1/2)^{t/T - t/2T} = 4$.
$8 \cdot (1/2)^{t/2T} = 4$.
$(1/2)^{t/2T} = 4/8 = 1/2$.
$(1/2)^{t/2T} = (1/2)^1$.
Comparing the exponents,$t/2T = 1$,which gives $t = 2T$.

(A) Let the number of $\alpha$-particles emitted be $n_{\alpha}$ and the number of $\beta^{-}$-particles emitted be $n_{\beta}$.
For the mass number: $238 = 214 + 4n_{\alpha} \implies 4n_{\alpha} = 24 \implies n_{\alpha} = 6$.
For the atomic number: $92 = 82 + 2n_{\alpha} - 1n_{\beta}$.
Substituting $n_{\alpha} = 6$: $92 = 82 + 2(6) - n_{\beta} \implies 92 = 82 + 12 - n_{\beta} \implies 92 = 94 - n_{\beta} \implies n_{\beta} = 2$.
Thus,the number of $\alpha$-particles is $6$ and the number of $\beta^{-}$-particles is $2$.

(B) The $\beta$-decay of a nucleus is represented as:
${ }_{Z}^{A} X \longrightarrow{ }_{Z+1}^{A} Y+{ }_{-1}^{0} e + \bar{\nu}$
In this process,the mass number $A$ remains constant,while the atomic number $Z$ increases by $1$.
Since the mass number $A$ is the same for both the parent nucleus $X$ and the daughter nucleus $Y$,they are isobars.
Therefore,the correct option is $B$.

(A) The time period of oscillation of a current-carrying coil in an external magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic dipole moment,and $B$ is the magnetic field.
Since $M = NIA$ (where $N$ is the number of turns,$I_{curr}$ is the current,and $A$ is the area),we have $M \propto I_{curr}$.
Therefore,$T \propto \frac{1}{\sqrt{M}} \propto \frac{1}{\sqrt{I_{curr}}}$.
Given $I_{curr1} = 2.5 \ A$ and $I_{curr2} = 10 \ A$,we have $\frac{T_2}{T_1} = \sqrt{\frac{I_{curr1}}{I_{curr2}}} = \sqrt{\frac{2.5}{10}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,$T_2 = \frac{T_1}{2} = \frac{T}{2}$.

(A) For a glass slab,the equivalent focal length $F$ is $\infty$.
For two thin lenses of focal lengths $f_1$ and $f_2$ separated by a distance $d$,the equivalent focal length $F$ is given by the formula:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$
Since the combination acts as a glass slab,we set $F = \infty$,which implies $\frac{1}{F} = 0$.
Substituting this into the formula:
$0 = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$
$0 = \frac{f_2 + f_1}{f_1 f_2} - \frac{d}{f_1 f_2}$
$\frac{d}{f_1 f_2} = \frac{f_1 + f_2}{f_1 f_2}$
$d = f_1 + f_2$
Thus,the distance of separation between the lenses must be $f_1 + f_2$.

(B) Given,power of the lens $P = 2 \ D$.
The focal length of the lens is $f = \frac{1}{P} = \frac{100}{2} = 50 \ cm$.
For a person with a near point defect (hypermetropia),the lens forms an image of an object placed at the standard near point $(u = -25 \ cm)$ at the person's actual near point $(v)$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{50} = \frac{1}{v} - \frac{1}{-25}$.
$\frac{1}{50} = \frac{1}{v} + \frac{1}{25}$.
$\frac{1}{v} = \frac{1}{50} - \frac{1}{25} = \frac{1 - 2}{50} = -\frac{1}{50}$.
Therefore,$v = -50 \ cm$.
The magnitude of the near point is $50 \ cm$.

(C) For a thin convex lens,$P = +4 \ D$,$\mu_{\ell} = \frac{3}{2}$,$\mu_m = \frac{5}{3}$.
Since $P = \frac{1}{f_a}$,we have $f_a = \frac{1}{4} \ m = 25 \ cm$.
In air,the lens maker's formula is $\frac{1}{f_a} = (\mu_{\ell} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ ... $(i)$.
In a liquid medium,the focal length $f_m$ is given by $\frac{1}{f_m} = \left(\frac{\mu_{\ell}}{\mu_m} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ ... (ii).
Dividing equation (ii) by $(i)$,we get $\frac{f_a}{f_m} = \frac{(\mu_{\ell} - 1)}{(\frac{\mu_{\ell}}{\mu_m} - 1)}$.
Substituting the values: $\frac{f_a}{f_m} = \frac{(\frac{3}{2} - 1)}{(\frac{3/2}{5/3} - 1)} = \frac{1/2}{(9/10 - 1)} = \frac{1/2}{-1/10} = -5$.
Therefore,$f_m = -5 \times f_a = -5 \times 25 \ cm = -125 \ cm$.
The negative sign indicates that the lens behaves like a concave lens with a focal length of $125 \ cm$.

(C) For a plano-convex lens,the radius of curvature $R$ is related to the aperture radius $r$ and the thickness $t$ by the geometry: $R^2 = r^2 + (R-t)^2$. Since $t$ is very small,$R^2 = r^2 + R^2 - 2Rt + t^2 \approx r^2 + R^2 - 2Rt$,which simplifies to $R = \frac{r^2}{2t}$.
Using the Lens Maker's Formula: $\frac{1}{f} = (n-1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. For a plano-convex lens,$R_1 = R$ and $R_2 = \infty$,so $\frac{1}{f} = (n-1) \frac{1}{R}$,which gives $R = f(n-1)$.
Equating the two expressions for $R$: $f(n-1) = \frac{r^2}{2t}$.
Given $f = 73.5 \ cm$,$n = \frac{5}{3}$,and diameter $d = 8.4 \ cm$,so $r = \frac{d}{2} = 4.2 \ cm$.
Substituting the values: $73.5 \times \left(\frac{5}{3} - 1\right) = \frac{(4.2)^2}{2t}$.
$73.5 \times \frac{2}{3} = \frac{17.64}{2t} \Rightarrow 49 = \frac{8.82}{t}$.
$t = \frac{8.82}{49} \ cm = 0.18 \ cm = 1.8 \ mm$.

(C) For a convex lens,the displacement method states that for a fixed distance between the object and the screen,there are two positions of the lens where a sharp image is formed on the screen.
Let $h_0$ be the height of the object,$h_1$ be the height of the first image,and $h_2$ be the height of the second image.
The relationship between these heights is given by the formula $h_0 = \sqrt{h_1 \times h_2}$.
Given: $h_0 = 12 \ cm$ and $h_1 = 18 \ cm$.
Substituting the values into the formula:
$12 = \sqrt{18 \times h_2}$
Squaring both sides:
$144 = 18 \times h_2$
$h_2 = \frac{144}{18} = 8 \ cm$.
Therefore,the height of the second image is $8 \ cm$.

(B) Given:
Diameter of the objective,$d = 3.6 \ m$
Wavelength of light,$\lambda = 540 \ nm = 540 \times 10^{-9} \ m$
The limit of resolution $(d\theta)$ of a telescope is given by the formula:
$d\theta = \frac{1.22 \lambda}{d}$
Substituting the values:
$d\theta = \frac{1.22 \times 540 \times 10^{-9} \ m}{3.6 \ m}$
$d\theta = \frac{658.8 \times 10^{-9}}{3.6} \ rad$
$d\theta = 183 \times 10^{-9} \ rad$
$d\theta = 1.83 \times 10^{-7} \ rad$

(D) For a prism,the relationship between the angles is given by $A + \delta = i + e$,where $A$ is the angle of the prism,$\delta$ is the angle of deviation,$i$ is the angle of incidence,and $e$ is the angle of emergence.
Given that the prism is equilateral,the angle of the prism $A = 60^{\circ}$.
The angle of incidence $i$ and the angle of emergence $e$ are both equal to $70 \%$ of the angle of the prism.
Therefore,$i = e = 0.70 \times 60^{\circ} = 42^{\circ}$.
Substituting these values into the formula $A + \delta = i + e$:
$60^{\circ} + \delta = 42^{\circ} + 42^{\circ}$
$60^{\circ} + \delta = 84^{\circ}$
$\delta = 84^{\circ} - 60^{\circ} = 24^{\circ}$.
Thus,the angle of deviation is $24^{\circ}$.

(A) For a thin prism,the angle of minimum deviation $\delta$ is given by the formula $\delta = A(\mu - 1)$,where $A$ is the prism angle and $\mu$ is the refractive index of the prism material relative to the surrounding medium.
$1$. When the prism is in air:
The refractive index of the prism with respect to air is $\mu_1 = \frac{3}{2}$.
Therefore,the angle of minimum deviation is $\delta_1 = A(\frac{3}{2} - 1) = A(\frac{1}{2}) = \frac{A}{2}$.
$2$. When the prism is in water:
The refractive index of water is $\mu_w = \frac{4}{3}$. The refractive index of the prism with respect to water is $\mu_2 = \frac{\mu_g}{\mu_w} = \frac{3/2}{4/3} = \frac{9}{8}$.
Therefore,the angle of minimum deviation is $\delta_2 = A(\frac{9}{8} - 1) = A(\frac{1}{8}) = \frac{A}{8}$.
$3$. The ratio of the angles of minimum deviation is:
$\frac{\delta_1}{\delta_2} = \frac{A/2}{A/8} = \frac{8}{2} = \frac{4}{1}$.
Thus,the ratio is $4: 1$.

(B) Given: Refractive index $\mu = 1.6$,Angle of minimum deviation $\delta_m = 4.2^{\circ}$.
For a small-angled prism,the relationship between the angle of deviation $\delta$,refractive index $\mu$,and prism angle $A$ is given by the formula: $\delta = (\mu - 1)A$.
Substituting the given values into the formula:
$4.2^{\circ} = (1.6 - 1)A$
$4.2^{\circ} = (0.6)A$
$A = \frac{4.2^{\circ}}{0.6} = 7^{\circ}$.
Therefore,the angle of the prism is $7^{\circ}$.

(A) Step $1$: Current through Load Resistor $(I_L)$
Since the Zener diode regulates the voltage at $30 \text{ V}$,the voltage across the load resistor $(R_L = 6 \text{ k}\Omega)$ will also be $30 \text{ V}$.
The current through the load resistor,$I_L$,is:
$I_L = \frac{V_Z}{R_L} = \frac{30 \text{ V}}{6 \text{ k}\Omega} = 5 \text{ mA}$
Step $2$: Total Current in the Circuit $(I_{\text{total}})$
The total current supplied by the source,$I_{\text{total}}$,is given by:
$I_{\text{total}} = \frac{V_{\text{in}} - V_Z}{R_{\text{total}}}$
where $R_{\text{total}} = 5 \text{ k}\Omega + 2 \text{ k}\Omega = 7 \text{ k}\Omega$.
For maximum $I_{\text{total}}$,we use the maximum input voltage $V_{\text{in}} = 100 \text{ V}$:
$I_{\text{total}} = \frac{100 \text{ V} - 30 \text{ V}}{7 \text{ k}\Omega} = \frac{70 \text{ V}}{7 \text{ k}\Omega} = 10 \text{ mA}$
Step $3$: Maximum Zener Diode Current $(I_{Z \max})$
The Zener diode current $I_Z$ is the difference between the total current and the load current:
$I_{Z \max} = I_{\text{total}} - I_L = 10 \text{ mA} - 5 \text{ mA} = 5 \text{ mA}$
Final Answer:
The maximum current through the Zener diode is $5 \text{ mA}$.

(A) transistor operates as an amplifier in its active region.
In the active region,the emitter-base junction is forward-biased and the collector-base junction is reverse-biased.
This configuration allows the transistor to provide current and voltage gain,making it suitable for amplification purposes.
In contrast,the cut-off region acts as an open switch ($OFF$ state),and the saturation region acts as a closed switch ($ON$ state).

(A) For a transistor in common emitter configuration, the voltage gain $A_V$ is given by the product of current gain $\beta$ and resistance gain $R_C/R_B$.
$A_V = \beta \times \frac{R_C}{R_B}$
Given: $A_V = 160$, $R_B = 1 \text{ k}\Omega$, $R_C = 4 \text{ k}\Omega$, and $\Delta I_B = 100 \mu A$.
We know that $\beta = \frac{\Delta I_C}{\Delta I_B}$.
Substituting this into the voltage gain formula:
$A_V = \left(\frac{\Delta I_C}{\Delta I_B}\right) \times \frac{R_C}{R_B}$
$160 = \left(\frac{\Delta I_C}{100 \times 10^{-6} \text{ A}}\right) \times \left(\frac{4 \text{ k}\Omega}{1 \text{ k}\Omega}\right)$
$160 = \left(\frac{\Delta I_C}{100 \times 10^{-6}}\right) \times 4$
$40 = \frac{\Delta I_C}{100 \times 10^{-6}}$
$\Delta I_C = 40 \times 100 \times 10^{-6} \text{ A} = 4000 \times 10^{-6} \text{ A} = 4 \times 10^{-3} \text{ A} = 4 \text{ mA}$.
Thus, the change in output current is $4 \text{ mA}$.

(B) The circuit consists of a $NOR$ gate followed by an $AND$ gate and an $OR$ gate.
Let the output of the $NOR$ gate be $C = \overline{A + B}$.
Given $A = 1$ and $B = 1$,the output of the $NOR$ gate is $C = \overline{1 + 1} = \overline{1} = 0$.
The output $y_1$ is the output of an $AND$ gate with inputs $A$ and $C$. Thus,$y_1 = A \cdot C = 1 \cdot 0 = 0$.
The output $y_2$ is the output of an $OR$ gate with inputs $B$ and $C$. Thus,$y_2 = B + C = 1 + 0 = 1$.
Therefore,the values are $y_1 = 0$ and $y_2 = 1$.

(B) The circuit consists of a $NOR$ gate and a $NAND$ gate.
$1$. The inputs to the $NOR$ gate are $A=1$ and $B=0$. The output of the $NOR$ gate is $y_2 = \overline{A+B} = \overline{1+0} = \overline{1} = 0$.
$2$. The inputs to the $NAND$ gate are $A=1$ and the output of the $NOR$ gate $y_2=0$. The output of the $NAND$ gate is $y_1 = \overline{A \cdot y_2} = \overline{1 \cdot 0} = \overline{0} = 1$.
Therefore,the values are $y_1=1$ and $y_2=0$.

(D) rectifier converts $AC$ into pulsating $DC$. To remove the ripples and obtain a smooth or steady $DC$ output,a capacitor is connected in parallel across the output terminals. The capacitor acts as a filter by charging during the peak of the pulse and discharging during the valley,thereby smoothing out the voltage fluctuations.

(D) rectifier converts $AC$ into a pulsating $DC$. To obtain a steady $DC$ output,a filter circuit is used. $A$ capacitor connected in parallel to the output terminals acts as a filter because it charges during the peak of the pulsating $DC$ and discharges when the voltage drops,thereby smoothing the output and providing a steady $DC$ voltage.

(C) For an intrinsic semiconductor, the law of mass action states that $n_i^2 = n_e n_h$, where $n_i$ is the intrinsic carrier concentration, $n_e$ is the electron concentration, and $n_h$ is the hole concentration.
Given $n_i = 6 \times 10^{15} \,m^{-3}$ and the doped electron concentration $n_e = 4 \times 10^{22} \,m^{-3}$.
Substituting these values into the formula:
$(6 \times 10^{15})^2 = (4 \times 10^{22}) \times n_h$
$36 \times 10^{30} = 4 \times 10^{22} \times n_h$
$n_h = \frac{36 \times 10^{30}}{4 \times 10^{22}}$
$n_h = 9 \times 10^8 \,m^{-3}$.

(C) An $n$-type semiconductor is formed by doping an intrinsic semiconductor with pentavalent atoms. Although the number of free electrons exceeds the number of holes,the total number of negative charges (electrons) is exactly balanced by the total number of positive charges (protons in the nuclei and the ionized donor atoms). Therefore,the net charge of an $n$-type semiconductor is zero,making it electrically neutral.

(C) Brewster's law states that the Brewster angle $i_p$ is related to the refractive index $\mu$ of the medium by the formula: $i_p = \tan ^{-1}(\mu)$.
Given that the refractive index of glass is $\mu = 1.5$.
Substituting the value of $\mu$ into the formula,we get $i_p = \tan ^{-1}(1.5)$.
Since $1.5 = \frac{3}{2}$,the Brewster angle is $i_p = \tan ^{-1}\left(\frac{3}{2}\right)$.

(D) Given: Power of the bulb $P = 60 \ W$,Efficiency $\eta = 16 \% = 0.16$,Distance $r = 2 \ m$.
First,calculate the intensity of the electromagnetic radiation emitted by the bulb at distance $r$:
The power radiated as electromagnetic waves is $P_{rad} = \eta \times P = 0.16 \times 60 = 9.6 \ W$.
The intensity $I$ at distance $r$ is given by $I = \frac{P_{rad}}{4 \pi r^2} = \frac{9.6}{4 \pi (2)^2} = \frac{9.6}{16 \pi} = \frac{0.6}{\pi} \ W/m^2$.
The intensity of an electromagnetic wave is also related to the peak electric field $E_0$ by $I = \frac{1}{2} c \epsilon_0 E_0^2$.
Equating the two expressions for intensity: $\frac{0.6}{\pi} = \frac{1}{2} c \epsilon_0 E_0^2$.
We know that $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$ and $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9$,so $\epsilon_0 = \frac{1}{36 \pi \times 10^9}$.
Using $c = 3 \times 10^8 \ m/s$,we have $E_0^2 = \frac{2 \times I}{c \epsilon_0} = \frac{2 \times 0.6 / \pi}{3 \times 10^8 \times (1 / (36 \pi \times 10^9))}$.
$E_0^2 = \frac{1.2}{\pi} \times \frac{36 \pi \times 10^9}{3 \times 10^8} = 1.2 \times 12 \times 10 = 144$.
Therefore,$E_0 = \sqrt{144} = 12 \ Vm^{-1}$.

(A) The intensity of light in a double slit experiment is given by $I = I_0 \cos^2(\phi/2)$,where $\phi$ is the phase difference.
Alternatively,using the formula $I = I_{max} \cos^2(\frac{\pi \Delta x}{\lambda})$,where $\Delta x$ is the path difference.
For the first point,$\Delta x_1 = 2000 \ Å$ and $\lambda = 6000 \ Å$.
Phase difference $\phi_1 = \frac{2\pi}{\lambda} \Delta x_1 = \frac{2\pi}{6000} \times 2000 = \frac{2\pi}{3} = 120^{\circ}$.
Intensity $I_1 = I_{max} \cos^2(\frac{120^{\circ}}{2}) = I_{max} \cos^2(60^{\circ}) = I_{max} (1/2)^2 = I_{max}/4$.
For the second point,$\Delta x_2 = 1000 \ Å$.
Phase difference $\phi_2 = \frac{2\pi}{\lambda} \Delta x_2 = \frac{2\pi}{6000} \times 1000 = \frac{\pi}{3} = 60^{\circ}$.
Intensity $I_2 = I_{max} \cos^2(\frac{60^{\circ}}{2}) = I_{max} \cos^2(30^{\circ}) = I_{max} (\sqrt{3}/2)^2 = 3I_{max}/4$.
Therefore,the ratio $\frac{I_1}{I_2} = \frac{I_{max}/4}{3I_{max}/4} = \frac{1}{3}$.
Thus,$I_1: I_2 = 1: 3$.

(A) The intensity in an interference pattern is given by $I_{res} = I_{max} \cos^2(\frac{\phi}{2})$,where $\phi$ is the phase difference.
For a path difference $\Delta x = \lambda$,the phase difference is $\phi_1 = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$.
The intensity is $I = I_{max} \cos^2(\frac{2\pi}{2}) = I_{max} \cos^2(\pi) = I_{max}(1)^2 = I_{max}$.
For a path difference $\Delta x = \frac{\lambda}{3}$,the phase difference is $\phi_2 = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3}$.
The intensity at this point is $I' = I_{max} \cos^2(\frac{\phi_2}{2}) = I_{max} \cos^2(\frac{2\pi/3}{2}) = I_{max} \cos^2(\frac{\pi}{3})$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we have $I' = I_{max} (\frac{1}{2})^2 = \frac{I_{max}}{4}$.
Since $I_{max} = I$,the intensity is $\frac{I}{4}$.

(A) The acceleration of a charged particle in a uniform electric field $E$ is given by $a = \frac{qE}{m}$.
Since the particles start from rest $(u = 0)$,the velocity after time $t$ is $v = at = \left(\frac{qE}{m}\right)t$.
According to the work-energy theorem,the work done $W$ by the electric field is equal to the change in kinetic energy $\Delta K$:
$W = \Delta K = \frac{1}{2}mv^2$.
Substituting the expression for $v$:
$W = \frac{1}{2}m\left(\frac{qEt}{m}\right)^2 = \frac{q^2 E^2 t^2}{2m}$.
Since $E$ and $t$ are the same for both particles,$W \propto \frac{q^2}{m}$.
For a proton,$q_p = e$ and $m_p = m$. For an $\alpha$-particle,$q_\alpha = 2e$ and $m_\alpha = 4m$.
Therefore,the ratio is:
$\frac{W_p}{W_\alpha} = \left(\frac{q_p}{q_\alpha}\right)^2 \left(\frac{m_\alpha}{m_p}\right) = \left(\frac{e}{2e}\right)^2 \left(\frac{4m}{m}\right) = \left(\frac{1}{4}\right) \times 4 = 1: 1$.