AP EAMCET 2024 Physics Question Paper with Answer and Solution

(B) Given: $P_1 = 10^5 \text{ Nm}^{-2}$,$V_1 = 16 \text{ L} = 16 \times 10^{-3} \text{ m}^3$,$V_2 = 2 \text{ L} = 2 \times 10^{-3} \text{ m}^3$,$C_v = \frac{3R}{2}$.
For an ideal gas,$C_p = C_v + R = \frac{3R}{2} + R = \frac{5R}{2}$.
The adiabatic index $\gamma = \frac{C_p}{C_v} = \frac{5/2 R}{3/2 R} = \frac{5}{3}$.
For an adiabatic process,$P_1 V_1^\gamma = P_2 V_2^\gamma$.
$P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma = 10^5 \left( \frac{16}{2} \right)^{5/3} = 10^5 \times (8)^{5/3} = 10^5 \times (2^3)^{5/3} = 10^5 \times 2^5 = 32 \times 10^5 \text{ Nm}^{-2}$.
Work done by the gas $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$.
$W = \frac{(10^5 \times 16 \times 10^{-3}) - (32 \times 10^5 \times 2 \times 10^{-3})}{5/3 - 1} = \frac{1600 - 6400}{2/3} = \frac{-4800}{2/3} = -4800 \times \frac{3}{2} = -7200 \text{ J} = -7.2 \text{ kJ}$.
Since the work done by the gas is $-7.2 \text{ kJ}$,the work done on the gas is $7.2 \text{ kJ}$.

(D) The work done in an isothermal process $(T = \text{constant})$ is given by $W = nRT \ln \left( \frac{P_i}{P_f} \right)$.
For gas $A$, the pressure decreases by $50 \%$, so $P_f = P_i - 0.5 P_i = 0.5 P_i$. Thus, $\frac{P_i}{P_f} = \frac{1}{0.5} = 2$.
For gas $B$, the pressure decreases by $75 \%$, so $P_f = P_i - 0.75 P_i = 0.25 P_i$. Thus, $\frac{P_i}{P_f} = \frac{1}{0.25} = 4$.
Given that the work done by both gases is equal $(W_1 = W_2)$ and the number of moles $(n)$ is the same:
$nRT_1 \ln(2) = nRT_2 \ln(4)$
$T_1 \ln(2) = T_2 \ln(2^2)$
$T_1 \ln(2) = 2 T_2 \ln(2)$
Dividing both sides by $\ln(2)$:
$T_1 = 2 T_2 \Rightarrow \frac{T_1}{T_2} = \frac{2}{1}$.
Therefore, the ratio $T_1: T_2$ is $2: 1$.

(C) For an adiabatic process,the work done is given by $W = -\Delta U = -\frac{f}{2} nR \Delta T = \frac{f}{2} nR (T_i - T_f)$.
For the monatomic gas: $n_1 = 2$,$f_1 = 3$,$T_{i1} = 80^{\circ} C$,$T_{f1} = 50^{\circ} C$.
$W = \frac{3}{2} \times 2 \times R \times (80 - 50) = 3R \times 30 = 90R$.
For the diatomic gas: $n_2 = 3$,$f_2 = 5$,$T_{i2} = 50^{\circ} C$,$T_{f2} = 20^{\circ} C$.
$W' = \frac{5}{2} \times 3 \times R \times (50 - 20) = \frac{15}{2} R \times 30 = 15R \times 15 = 225R$.
Now,find the ratio: $\frac{W'}{W} = \frac{225R}{90R} = \frac{225}{90} = 2.5$.
Therefore,$W' = 2.5 W$.

(A) $1$. Thermal conductivity $(k)$: From the formula $\frac{Q}{t} = \frac{kA(\theta_1 - \theta_2)}{l}$,we have $[k] = \frac{[Q][l]}{[t][A][\Delta\theta]} = \frac{[ML^2T^{-2}][L]}{[T][L^2][K]} = [MLT^{-3}K^{-1}]$. Thus,$(A) - (i)$.
$2$. Boltzmann constant $(k_B)$: From $PV = Nk_BT$,we have $[k_B] = \frac{[PV]}{[N][T]} = \frac{[ML^2T^{-2}]}{[K]} = [ML^2T^{-2}K^{-1}]$. Thus,$(B) - (iii)$.
$3$. Latent heat $(L)$: From $Q = mL$,we have $[L] = \frac{[Q]}{[m]} = \frac{[ML^2T^{-2}]}{[M]} = [M^0L^2T^{-2}]$. Thus,$(C) - (iv)$.
$4$. Specific heat $(s)$: From $Q = ms\Delta\theta$,we have $[s] = \frac{[Q]}{[m][\Delta\theta]} = \frac{[ML^2T^{-2}]}{[M][K]} = [M^0L^2T^{-2}K^{-1}]$. Thus,$(D) - (ii)$.
Therefore,the correct match is $(A) - (i), (B) - (iii), (C) - (iv), (D) - (ii)$.

(A) The dimensions of the given physical quantities are:
$E = [M L^2 T^{-2}]$
$m = [M]$
$L = [M L^2 T^{-1}]$
$G = [M^{-1} L^3 T^{-2}]$
Substituting these into the expression $\frac{E L^2}{m^5 G^2}$:
$\left[\frac{E L^2}{m^5 G^2}\right] = \frac{[M L^2 T^{-2}] [M L^2 T^{-1}]^2}{[M]^5 [M^{-1} L^3 T^{-2}]^2}$
$= \frac{[M L^2 T^{-2}] [M^2 L^4 T^{-2}]}{[M^5] [M^{-2} L^6 T^{-4}]}$
$= \frac{[M^3 L^6 T^{-4}]}{[M^3 L^6 T^{-4}]}$
$= [M^0 L^0 T^0]$
Since the dimensions are $[M^0 L^0 T^0]$,the quantity is dimensionless,which corresponds to the dimensions of an $\text{Angle}$.

(A) Given the equation: $F = a \sqrt{x} + b t^2$.
According to the principle of homogeneity of dimensions,the dimensions of each term on both sides of the equation must be the same.
Therefore,$[F] = [a \sqrt{x}]$ and $[F] = [b t^2]$.
From $[F] = [a \sqrt{x}]$,we get $[a] = \frac{[F]}{[\sqrt{x}]} = \frac{[MLT^{-2}]}{[L^{1/2}]} = [ML^{1/2}T^{-2}]$.
From $[F] = [b t^2]$,we get $[b] = \frac{[F]}{[t^2]} = \frac{[MLT^{-2}]}{[T^2]} = [MLT^{-4}]$.
Now,we find the dimensional formula for $\frac{a}{b}$:
$\left[\frac{a}{b}\right] = \frac{[ML^{1/2}T^{-2}]}{[MLT^{-4}]} = [M^{1-1} L^{1/2-1} T^{-2-(-4)}] = [M^0 L^{-1/2} T^2]$.
Thus,the correct option is $A$.

(C) Let the time period $T$ be proportional to $R^a M^b G^c$,so $T = K R^a M^b G^c$.
Writing the dimensional formulas for each quantity:
$[T] = [T]^1$
$[R] = [L]^1$
$[M] = [M]^1$
$[G] = [M^{-1} L^3 T^{-2}]$
Substituting these into the equation: $[T]^1 = [L]^a [M]^b [M^{-1} L^3 T^{-2}]^c = [M]^{b-c} [L]^{a+3c} [T]^{-2c}$.
Comparing the powers of $M$,$L$,and $T$ on both sides:
For $T$: $-2c = 1 \Rightarrow c = -1/2$.
For $M$: $b - c = 0 \Rightarrow b = c = -1/2$.
For $L$: $a + 3c = 0 \Rightarrow a = -3c = -3(-1/2) = 3/2$.
Substituting these values back into the expression: $T = K R^{3/2} M^{-1/2} G^{-1/2} = K \sqrt{\frac{R^3}{GM}}$.

(B) According to the principle of homogeneity of dimensions,only physical quantities of the same dimension can be added or subtracted.
In the given equation $(P+\frac{a}{V^2})(V-b)=RT$,the term $\frac{a}{V^2}$ is added to pressure $P$.
Therefore,the dimensions of $\frac{a}{V^2}$ must be equal to the dimensions of pressure $P$.
$[P] = [\frac{a}{V^2}] \implies [a] = [P][V^2]$.
The dimension of pressure $P$ is $[ML^{-1} T^{-2}]$ and the dimension of volume $V$ is $[L^3]$.
Substituting these,we get $[a] = [ML^{-1} T^{-2}] \times [L^3]^2 = [ML^{-1} T^{-2}] \times [L^6] = [ML^5 T^{-2}]$.

(D) The length of the side of the cube is $l = 1.2 \times 10^{-2} \ m$.
The volume of a cube is given by $V = l^3$.
$V = (1.2 \times 10^{-2} \ m)^3 = 1.728 \times 10^{-6} \ m^3$.
According to the rules of significant figures,the result of a multiplication or division should have the same number of significant figures as the measurement with the fewest significant figures.
The given length $1.2 \times 10^{-2} \ m$ has $2$ significant figures.
Therefore,the volume should be rounded to $2$ significant figures.
Rounding $1.728$ to $2$ significant figures gives $1.7$.
Thus,the volume is $1.7 \times 10^{-6} \ m^3$.

(A) For a wave on a string, the speed $v$ is given by $v = \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density.
Since $\mu$ is constant, $v \propto \sqrt{T}$.
Given $v_1 = 150 \,ms^{-1}$ and $T_1 = 120 \,N$.
We want to increase the speed by $20 \%$, so $v_2 = v_1 + 0.20 v_1 = 1.2 v_1$.
Using the proportionality $v \propto \sqrt{T}$, we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $1.2 = \sqrt{\frac{T_2}{120}}$.
Squaring both sides: $1.44 = \frac{T_2}{120}$.
$T_2 = 1.44 \times 120 = 172.8 \,N$.
The percentage increase in tension is given by $\frac{T_2 - T_1}{T_1} \times 100$.
$\% \Delta T = \frac{172.8 - 120}{120} \times 100 = \frac{52.8}{120} \times 100 = 44 \%$.

(C) The beat frequency is given by $|n_A - n_B| = 4 \ Hz$.
When the tension in string $A$ is increased,its frequency $n_A$ increases because $n \propto \sqrt{T}$.
Since the number of beats increases from $4$ to $7$ upon increasing $n_A$,it implies that $n_A$ must be greater than $n_B$ (i.e.,$n_A - n_B = 4$).
If $n_A$ were less than $n_B$,increasing $n_A$ would decrease the beat frequency towards zero before increasing it again,but the problem implies a direct increase.
Therefore,$n_A = n_B + 4$.
Given $n_B = 480 \ Hz$,we have $n_A = 480 + 4 = 484 \ Hz$.

(A) Given: Frequency of horn $f = 1000 \,Hz$,Speed of sound $v = 340 \,ms^{-1}$.
When the car is approaching the observer,the observed frequency is $f_1 = \left(\frac{v}{v-v_s}\right) f$.
When the car is moving away from the observer,the observed frequency is $f_2 = \left(\frac{v}{v+v_s}\right) f$.
The ratio of frequencies is given as $\frac{f_1}{f_2} = \frac{11}{9}$.
Substituting the expressions: $\frac{f_1}{f_2} = \frac{v+v_s}{v-v_s} = \frac{11}{9}$.
$\frac{340+v_s}{340-v_s} = \frac{11}{9}$.
$9(340+v_s) = 11(340-v_s)$.
$3060 + 9v_s = 3740 - 11v_s$.
$20v_s = 680$.
$v_s = 34 \,ms^{-1}$.

(B) For an open organ pipe,the frequency of the $n^{th}$ harmonic is given by $f_n = \frac{n v}{2L}$,where $n = 1, 2, 3, ...$ is the harmonic mode.
Given:
Length $L = 30 \ cm = 0.3 \ m$
Frequency $f = 1.65 \ kHz = 1650 \ Hz$
Velocity of sound $v = 330 \ m/s$
Substituting the values into the formula:
$1650 = \frac{n \times 330}{2 \times 0.3}$
$1650 = \frac{n \times 330}{0.6}$
$1650 = n \times 550$
$n = \frac{1650}{550} = 3$
Therefore,the pipe resonates in the $3^{rd}$ harmonic mode.

(D) For figure $(a)$,$L = \frac{\lambda_1}{4} \Rightarrow \lambda_1 = 4L$.
For figure $(b)$,$L = \frac{\lambda_2}{4} + \frac{\lambda_2}{2} + \frac{\lambda_2}{4} = \lambda_2 \Rightarrow \lambda_2 = L$.
For figure $(c)$,$L = \frac{\lambda_3}{4} + \frac{\lambda_3}{4} + \frac{\lambda_3}{2} = \lambda_3 \Rightarrow \lambda_3 = L$ (Note: Based on the diagram,it represents a pipe open at both ends with one node in the middle,so $L = \lambda_3/2 \Rightarrow \lambda_3 = 2L$).
For figure $(d)$,$L = \frac{\lambda_4}{2} + \frac{\lambda_4}{4} = \frac{3\lambda_4}{4} \Rightarrow \lambda_4 = \frac{4L}{3}$.
Since frequency $f \propto \frac{1}{\lambda}$,we have $f_1 : f_2 : f_3 : f_4 = \frac{1}{\lambda_1} : \frac{1}{\lambda_2} : \frac{1}{\lambda_3} : \frac{1}{\lambda_4}$.
Substituting the values: $f_1 : f_2 : f_3 : f_4 = \frac{1}{4L} : \frac{1}{L} : \frac{1}{2L} : \frac{3}{4L}$.
Multiplying by $4L$,we get $1 : 4 : 2 : 3$.

(C) For an open organ pipe of length $L$,the fundamental frequency is given by:
$f_0 = \frac{v}{2L} = 100 \ Hz$
When the bottom end is closed and $1/3$ of the pipe is filled with water,the effective length of the air column becomes $L' = L - \frac{L}{3} = \frac{2L}{3}$.
For a closed organ pipe of length $L'$,the fundamental frequency is:
$f' = \frac{v}{4L'}$
Substituting $L' = \frac{2L}{3}$:
$f' = \frac{v}{4(2L/3)} = \frac{3v}{8L}$
We can rewrite this as:
$f' = \frac{3}{4} \left( \frac{v}{2L} \right)$
Since $\frac{v}{2L} = 100 \ Hz$,we have:
$f' = \frac{3}{4} \times 100 = 75 \ Hz$

(A) For an open organ pipe,the frequency of the $n^{th}$ harmonic is given by $f_n = \frac{n v}{2 l_o}$.
For the third harmonic $(n=3)$,$f_{open} = \frac{3 v}{2 l_o}$.
Given $l_o = 72 \ cm$,so $f_{open} = \frac{3 v}{2 \times 72} = \frac{v}{48}$.
For a closed organ pipe,the frequency of the $n^{th}$ harmonic (where $n$ is odd) is given by $f_n = \frac{n v}{4 l_c}$.
For the fifth harmonic $(n=5)$,$f_{closed} = \frac{5 v}{4 l_c}$.
According to the problem,$f_{closed} = f_{open}$.
$\frac{5 v}{4 l_c} = \frac{3 v}{2 l_o}$.
Substituting $l_o = 72 \ cm$:
$\frac{5}{4 l_c} = \frac{3}{2 \times 72} = \frac{3}{144} = \frac{1}{48}$.
$4 l_c = 5 \times 48 = 240$.
$l_c = \frac{240}{4} = 60 \ cm$.

(B) The speed of a wave $v$ is given by $v = f \lambda$. Since the medium is not changed,the speed $v$ remains constant.
Therefore,$f_1 \lambda_1 = f_2 \lambda_2$,which implies $f \propto \frac{1}{\lambda}$.
Given that the frequency is increased by $25 \%$,the new frequency $f_2 = f_1 + 0.25 f_1 = 1.25 f_1 = \frac{5}{4} f_1$.
Using the inverse relationship,$\frac{\lambda_2}{\lambda_1} = \frac{f_1}{f_2} = \frac{f_1}{1.25 f_1} = \frac{1}{1.25} = \frac{4}{5} = 0.8$.
This means $\lambda_2 = 0.8 \lambda_1$.
The percentage change in wavelength is $\frac{\lambda_2 - \lambda_1}{\lambda_1} \times 100 = \frac{0.8 \lambda_1 - \lambda_1}{\lambda_1} \times 100 = -0.2 \times 100 = -20 \%$.
The negative sign indicates a $20 \%$ decrease in wavelength.

(C) At the top of the hemisphere,the forces acting on the disc are the gravitational force $mg$ acting downwards and the normal reaction $N$ acting upwards.
The net centripetal force required for circular motion is provided by the difference between the gravitational force and the normal reaction:
$mg - N = \frac{mv^2}{R}$
For the disc to leave the hemisphere surface immediately,the normal reaction $N$ must become zero at the top point.
Setting $N = 0$ in the equation:
$mg - 0 = \frac{mv^2}{R}$
$mg = \frac{mv^2}{R}$
$v^2 = gR$
$v = \sqrt{gR}$
Thus,the smallest horizontal velocity required is $\sqrt{gR}$.

(D) Let $l$ be the total length of the inclined plane and $\theta$ be the angle of inclination.
According to the work-energy theorem,the total work done on the body is equal to the change in its kinetic energy.
Since the body starts from rest and comes to rest at the bottom,the change in kinetic energy $\Delta K = 0$.
The forces acting on the body are gravity,friction,and the normal force.
$1$. Work done by gravity $(W_g)$: $W_g = mgh = mg(l \sin \theta)$.
$2$. Work done by friction $(W_f)$: Friction acts only on the lower part of length $l(1 - 1/n)$. Thus,$W_f = -f_k \cdot d = -(\mu_k mg \cos \theta) \cdot l(1 - 1/n)$.
$3$. Work done by the normal force $(W_N)$: Since the normal force is always perpendicular to the displacement,$W_N = 0$.
Applying the work-energy theorem: $W_g + W_f + W_N = 0$.
$mg l \sin \theta - \mu_k mg \cos \theta \cdot l \left(\frac{n-1}{n}\right) = 0$.
Dividing by $mg l \cos \theta$ (assuming $\cos \theta \neq 0$):
$\tan \theta = \mu_k \left(\frac{n-1}{n}\right)$.
Therefore,$\theta = \tan^{-1}\left[\left(\frac{n-1}{n}\right) \mu_k\right]$.

(C) The useful work done by the machine (potential energy gained by the block) is given by the efficiency multiplied by the energy input:
$U = \eta \times E_{\text{in}} = \frac{2}{3} \times 12 \text{ J} = 8 \text{ J}$.
Since $U = mgh$,we have $8 = 2 \times 10 \times h$,which gives $h = \frac{8}{20} = 0.4 \text{ m}$.
When the block falls freely,by the law of conservation of energy,the loss in potential energy equals the gain in kinetic energy:
$mgh = \frac{1}{2}mv^2$.
$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.4} = \sqrt{8} = 2\sqrt{2} \text{ m/s}$.

(D) Given: Mass $m = 50 \ kg$,Range $R = 8 \ m$,Angle $\theta = 45^{\circ}$,Acceleration due to gravity $g = 9.8 \ m/s^2$.
The formula for the horizontal range of a projectile is $R = \frac{u^2 \sin(2\theta)}{g}$.
Substituting the given values:
$8 = \frac{u^2 \sin(2 \times 45^{\circ})}{9.8} = \frac{u^2 \sin(90^{\circ})}{9.8} = \frac{u^2}{9.8}$.
Therefore,$u^2 = 8 \times 9.8 = 78.4 \ m^2/s^2$.
The initial kinetic energy $(K.E.)$ is given by $\frac{1}{2} m u^2$.
$(K.E.) = \frac{1}{2} \times 50 \times 78.4 = 25 \times 78.4 = 1960 \ J$.

(A) The formula for average power is $P_{av} = \frac{W}{t} = \frac{mgh}{t}$.
Given:
Mass $m = 90 \ kg$
Height $h = 600 \ m$
Time $t = 90 \ minutes = 90 \times 60 \ s = 5400 \ s$
Acceleration due to gravity $g = 10 \ ms^{-2}$.
Substituting the values:
$P_{av} = \frac{90 \times 10 \times 600}{90 \times 60} = \frac{540000}{5400} = 100 \ W$.

(A) The work done by a variable force is given by the integral of the force with respect to displacement: $W = \int_{x_1}^{x_2} F(x) dx$.
Given $F(x) = (6x^2 - 4x + 3) \text{ N}$,$x_1 = 2 \text{ m}$,and $x_2 = 5 \text{ m}$.
$W = \int_{2}^{5} (6x^2 - 4x + 3) dx$
$W = [\frac{6x^3}{3} - \frac{4x^2}{2} + 3x]_{2}^{5}$
$W = [2x^3 - 2x^2 + 3x]_{2}^{5}$
Substituting the limits:
$W = [2(5)^3 - 2(5)^2 + 3(5)] - [2(2)^3 - 2(2)^2 + 3(2)]$
$W = [2(125) - 2(25) + 15] - [2(8) - 2(4) + 6]$
$W = [250 - 50 + 15] - [16 - 8 + 6]$
$W = 215 - 14 = 201 \text{ J}$.

(B) Given force $\vec{F} = (4 \hat{i} + 2 \hat{j} + \hat{k}) \text{ N}$.
Initial position $\vec{r_1} = (2 \hat{i} + 2 \hat{j} + \hat{k}) \text{ m}$.
Final position $\vec{r_2} = (4 \hat{i} + 3 \hat{j} + 2 \hat{k}) \text{ m}$.
Displacement $\vec{S} = \vec{r_2} - \vec{r_1} = (4-2) \hat{i} + (3-2) \hat{j} + (2-1) \hat{k} = (2 \hat{i} + \hat{j} + \hat{k}) \text{ m}$.
Work done $W = \vec{F} \cdot \vec{S}$.
$W = (4 \hat{i} + 2 \hat{j} + \hat{k}) \cdot (2 \hat{i} + \hat{j} + \hat{k})$.
$W = (4 \times 2) + (2 \times 1) + (1 \times 1) = 8 + 2 + 1 = 11 \text{ J}$.

(D) In an $n-p-n$ transistor,the relationship between emitter current $(I_E)$,collector current $(I_C)$,and base current $(I_B)$ is given by $I_E = I_B + I_C$.
Given that $95\%$ of the emitted electrons reach the collector,we have $I_C = 0.95 \times I_E$.
Given $I_C = 10 \text{ mA}$,we can find $I_E$ as:
$I_E = \frac{I_C}{0.95} = \frac{10}{0.95} \approx 10.53 \text{ mA}$.
Now,calculating the base current:
$I_B = I_E - I_C = 10.53 \text{ mA} - 10 \text{ mA} = 0.53 \text{ mA}$.

(C) Given: Current gain $\beta = 80$, Collector resistance $R_C = 5 \text{ k}\Omega$, Base resistance $R_B = 1 \text{ k}\Omega$, Input voltage $V_I = 2 \text{ mV} = 2 \times 10^{-3} \text{ V}$.
The voltage gain $A_V$ for a common emitter transistor is given by the formula:
$A_V = \frac{V_O}{V_I} = \beta \times \frac{R_C}{R_B}$
Substituting the given values:
$V_O = V_I \times \beta \times \frac{R_C}{R_B}$
$V_O = (2 \times 10^{-3} \text{ V}) \times 80 \times \frac{5 \text{ k}\Omega}{1 \text{ k}\Omega}$
$V_O = 2 \times 10^{-3} \times 80 \times 5$
$V_O = 800 \times 10^{-3} \text{ V}$
$V_O = 0.8 \text{ V}$.

(D) The voltage gain $A_V$ of a transistor amplifier in common emitter configuration is defined as the ratio of output voltage $V_O$ to input voltage $V_I$.
$A_V = \frac{V_O}{V_I} = \frac{V_C}{V_B}$
Since $V_C = I_C R_C$ and $V_B = I_B R_B$,we can write:
$A_V = \frac{I_C R_C}{I_B R_B}$
Rearranging the terms,we get:
$A_V = \left(\frac{I_C}{I_B}\right) \left(\frac{R_C}{R_B}\right)$
Given that the current amplification factor $\beta = \frac{I_C}{I_B}$,substituting this into the equation gives:
$A_V = \beta \cdot \frac{R_C}{R_B}$

(A) The circuit consists of a $NAND$ gate with inputs $B$ and $C$,and a $NOR$ gate with inputs $\overline{A}$ and $y_1$.
First,calculate $y_1$:
$y_1 = \overline{B \cdot C} = \overline{1 \cdot 1} = \overline{1} = 0$.
Next,calculate $y_2$:
The inputs to the $NOR$ gate are $\overline{A}$ and $y_1$.
$\overline{A} = \overline{1} = 0$.
$y_2 = \overline{\overline{A} + y_1} = \overline{0 + 0} = \overline{0} = 1$.
Thus,the values are $y_1 = 0$ and $y_2 = 1$.

(A) From the figure,the output $y_1$ is the output of a $NAND$ gate with inputs $(A \cdot B)$ and $B$. Thus,$y_1 = \overline{(A \cdot B) \cdot B}$.
Given $A=0$ and $B=1$,$A \cdot B = 0 \cdot 1 = 0$.
So,$y_1 = \overline{0 \cdot 1} = \overline{0} = 1$.
The output $y_2$ is the output of a $NOR$ gate with inputs $(B+C)$ and $B$. Thus,$y_2 = \overline{(B+C) + B}$.
Given $B=1$ and $C=1$,$B+C = 1+1 = 1$.
So,$y_2 = \overline{1 + 1} = \overline{1} = 0$.
Therefore,the values of $y_1$ and $y_2$ are $1$ and $0$ respectively.

(C) The circuit consists of a $NOR$ gate,an $AND$ gate,and a $NAND$ gate.
Let the inputs be $A$ and $B$.
The output of the $NOR$ gate is $\overline{A+B}$.
The output of the $AND$ gate is $A \cdot B$.
These two outputs are fed into a $NAND$ gate.
Therefore,the final output $Y$ is given by:
$Y = \overline{(\overline{A+B}) \cdot (A \cdot B)}$
Using De Morgan's Law,$\overline{X \cdot Z} = \overline{X} + \overline{Z}$:
$Y = \overline{(\overline{A+B})} + \overline{(A \cdot B)}$
$Y = (A+B) + (\overline{A} + \overline{B})$
$Y = (A + \overline{A}) + (B + \overline{B})$
Since $A + \overline{A} = 1$ and $B + \overline{B} = 1$:
$Y = 1 + 1 = 1$
Thus,for any combination of inputs $(A, B)$,the output $Y$ is always $1$.

(C) The given circuit consists of two $NOR$ gates acting as $NOT$ gates,followed by a $NOR$ gate.
$1$. The first two gates are $NOR$ gates with both inputs tied together. $A$ $NOR$ gate with inputs $A$ and $A$ gives output $\overline{A+A} = \overline{A}$.
$2$. Similarly,the second $NOR$ gate with inputs $B$ and $B$ gives output $\overline{B+B} = \overline{B}$.
$3$. These two outputs are fed into the final $NOR$ gate.
$4$. The output $Y$ of the final $NOR$ gate is $Y = \overline{\overline{A} + \overline{B}}$.
$5$. By De Morgan's Law,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$.
$6$. Therefore,the output $Y = A \cdot B$,which is the Boolean expression for an $AND$ gate.

(C) In a full-wave rectifier, the output consists of two pulses for every cycle of the input $AC$ signal.
Therefore, the fundamental frequency of the output ripple is twice the input frequency.
Given, input frequency $f_{in} = 50 \text{ Hz}$.
Fundamental frequency of output ripple $f_0 = 2 \times f_{in} = 2 \times 50 \text{ Hz} = 100 \text{ Hz}$.

(C) In a semiconductor at thermal equilibrium,the product of the concentration of electrons $(n_e)$ and the concentration of holes $(n_h)$ is equal to the square of the intrinsic carrier concentration $(n_i^2)$.
This is expressed by the law of mass action: $n_e n_h = n_i^2$.
Therefore,the intrinsic carrier concentration is given by $n_i = \sqrt{n_e n_h}$.

(D) For an intrinsic semiconductor,the law of mass action states that $n_e n_h = n_i^2$.
Given,$n_e = n_h = n_i = 1.5 \times 10^{16} \ m^{-3}$.
Therefore,$n_i^2 = (1.5 \times 10^{16})^2 = 2.25 \times 10^{32} \ m^{-6}$.
When the hole concentration is increased to $n_h' = 3 \times 10^{22} \ m^{-3}$,the new electron concentration $n_e'$ is given by:
$n_e' = \frac{n_i^2}{n_h'} = \frac{2.25 \times 10^{32}}{3 \times 10^{22}}$.
$n_e' = 0.75 \times 10^{10} \ m^{-3} = 7.5 \times 10^9 \ m^{-3}$.

(C) Visible light corresponds to the wavelength range of approximately $380 \, nm$ to $750 \, nm$.
Using the relation $E = \frac{hc}{\lambda}$, where $h = 6.63 \times 10^{-34} \, J \cdot s$, $c = 3 \times 10^8 \, m/s$, and $1 \, eV = 1.6 \times 10^{-19} \, J$.
For the longest wavelength of visible light $(\lambda = 750 \, nm)$, the energy is $E = \frac{1240 \, eV \cdot nm}{750 \, nm} \approx 1.65 \, eV$.
In practice, to emit visible light efficiently, the semiconductor material used for the fabrication of visible LEDs must have a band gap of at least $1.8 \, eV$.

(D) The permeability of free space $\mu_0$ has the $SI$ unit of Henry per meter $(H \cdot m^{-1})$.
From the formula for self-inductance $L = \frac{\mu_0 N^2 A}{l}$,we get the unit of $\mu_0$ as $\frac{H \cdot m}{m^2} = H \cdot m^{-1}$. Thus,option $A$ is a unit.
Since $L = \frac{\phi}{I}$,the unit of $L$ is $Wb \cdot A^{-1}$. Substituting this into the expression for $\mu_0$,we get $\frac{Wb \cdot A^{-1} \cdot m}{m^2} = Wb \cdot A^{-1} \cdot m^{-1}$. Thus,option $B$ is a unit.
Since $L = \frac{e}{dI/dt}$,the unit of $L$ is $\frac{V}{A \cdot s^{-1}} = V \cdot s \cdot A^{-1} = \Omega \cdot s$. Substituting this into the expression for $\mu_0$,we get $\frac{\Omega \cdot s \cdot m}{m^2} = \Omega \cdot s \cdot m^{-1}$. Thus,option $C$ is a unit.
Option $D$ $(Volt \cdot second \cdot meter^{-1})$ is equivalent to $V \cdot s \cdot m^{-1}$. Comparing this with the derived units,it lacks the $A^{-1}$ factor required for permeability. Therefore,it is not a unit of permeability.

(B) Diffraction is the phenomenon of bending of light around the corners of an obstacle or aperture of the size comparable to the wavelength of light.
For significant diffraction to occur,the size of the aperture or slit width '$a$' must be comparable to or smaller than the wavelength '$\lambda$' of the incident light.
Mathematically,the condition is expressed as $a \leq \lambda$.
Dividing both sides by '$\lambda$',we get $\frac{a}{\lambda} \leq 1$.

(C) For single-slit diffraction,the condition for the $n^{th}$ minima is given by $x \sin \theta = n \lambda$.
For the first minima,$n = 1$.
Given: $\lambda = 6500 \text{ Å} = 6500 \times 10^{-10} \text{ m} = 6.5 \times 10^{-7} \text{ m}$ and $\theta = 30^{\circ}$.
Substituting these values into the formula:
$x \sin 30^{\circ} = 1 \times 6.5 \times 10^{-7} \text{ m}$.
Since $\sin 30^{\circ} = 0.5$,we have:
$x \times 0.5 = 6.5 \times 10^{-7} \text{ m}$.
$x = \frac{6.5 \times 10^{-7}}{0.5} \text{ m} = 13 \times 10^{-7} \text{ m} = 1.3 \times 10^{-6} \text{ m}$.
Therefore,$x = 1.3 \mu \text{m}$.

(D) The intensity of unpolarized light incident on the polariser is $I$.
After passing through the polariser,the intensity of the polarized light becomes $I_1 = \frac{I}{2}$.
According to Malus' Law,the intensity of light emerging from the analyser is $I_2 = I_1 \cos^2 \theta$.
Given $\theta = 45^{\circ}$,we have $I_2 = \frac{I}{2} \cos^2 45^{\circ}$.
Substituting the value of $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we get $I_2 = \frac{I}{2} \times (\frac{1}{\sqrt{2}})^2 = \frac{I}{2} \times \frac{1}{2} = \frac{I}{4}$.

(B) The formula for fringe width in Young's double-slit experiment is $\beta = \frac{\lambda D}{d}$.
Given initial fringe width $\beta_1 = \frac{\lambda D}{d} = 3 \text{ mm}$.
According to the problem, the new distance $D' = D + 0.5D = 1.5D$ and the new slit separation $d' = d - 0.1d = 0.9d$.
The new fringe width $\beta_2$ is given by $\beta_2 = \frac{\lambda D'}{d'} = \frac{\lambda (1.5D)}{0.9d}$.
Simplifying this, $\beta_2 = \frac{1.5}{0.9} \times \frac{\lambda D}{d} = \frac{15}{9} \times \beta_1 = \frac{5}{3} \times 3 \text{ mm}$.
Therefore, $\beta_2 = 5 \text{ mm}$.

(D) In Young's double slit experiment $(YDSE)$,the path difference is given as $\Delta x = \frac{\lambda}{6}$.
Assuming the intensities of the two slits are equal,let $I_1 = I_2 = I_s$.
The phase difference $\Delta \phi$ is given by $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x = \frac{2 \pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}$.
The resultant intensity $I$ at any point is given by $I = I_1 + I_2 + 2 \sqrt{I_1 I_2} \cos(\Delta \phi)$.
Substituting the values: $I = I_s + I_s + 2 \sqrt{I_s I_s} \cos(\frac{\pi}{3}) = 2 I_s + 2 I_s (\frac{1}{2}) = 2 I_s + I_s = 3 I_s$.
The maximum intensity $I_0$ occurs when $\cos(\Delta \phi) = 1$,so $I_0 = I_1 + I_2 + 2 \sqrt{I_1 I_2} = (\sqrt{I_s} + \sqrt{I_s})^2 = (2 \sqrt{I_s})^2 = 4 I_s$.
Therefore,the ratio $\frac{I}{I_0} = \frac{3 I_s}{4 I_s} = \frac{3}{4}$.

(C) In Young's double slit experiment $(YDSE)$,the fringe width $\beta$ is given by the formula: $\beta = \frac{\lambda D}{d}$.
Given values are:
Slit separation,$d = 2 \ mm = 2 \times 10^{-3} \ m$.
Distance to screen,$D = 2 \ m$.
Wavelength of light,$\lambda = 400 \ nm = 400 \times 10^{-9} \ m$.
Substituting these values into the formula:
$\beta = \frac{(400 \times 10^{-9} \ m) \times (2 \ m)}{2 \times 10^{-3} \ m}$.
$\beta = \frac{800 \times 10^{-9}}{2 \times 10^{-3}} \ m$.
$\beta = 400 \times 10^{-6} \ m = 0.4 \times 10^{-3} \ m$.

(D) Given intensities are $I_1 = I$ and $I_2 = 2I$.
Path difference $\Delta x = 12.5 \% \text{ of } \lambda = \frac{12.5}{100} \lambda = \frac{\lambda}{8}$.
Phase difference $\phi = \frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{8} = \frac{\pi}{4}$.
The resultant intensity $I_R$ is given by the formula $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
Substituting the values: $I_R = I + 2I + 2\sqrt{I \cdot 2I} \cos(\frac{\pi}{4})$.
$I_R = 3I + 2\sqrt{2I^2} \cdot \frac{1}{\sqrt{2}}$.
$I_R = 3I + 2I \cdot \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 3I + 2I = 5I$.

(A) The electric potential $V$ at any point at a distance $r$ from a point charge $q$ is given by $V = \frac{kq}{r}$.
Since points $A, B$ and $C$ lie on a semicircle with the charge $q$ at its centre,all these points are at the same distance $r$ (the radius of the semicircle) from the charge $q$.
Therefore,the electric potential at points $A, B$ and $C$ is the same,i.e.,$V_A = V_B = V_C = V$.
The work done in moving a charge $q_0$ from point $P$ to a point $X$ is given by $W = q_0(V_X - V_P)$.
Thus,the work done to move the charge to points $A, B$ and $C$ are:
$W_A = q_0(V_A - V_P) = q_0(V - V_P)$
$W_B = q_0(V_B - V_P) = q_0(V - V_P)$
$W_C = q_0(V_C - V_P) = q_0(V - V_P)$
Since $V_A = V_B = V_C$,it follows that $W_A = W_B = W_C$.
Since point $P$ is at a different distance from the charge $q$ compared to the points on the semicircle,$V_P \neq V$,therefore $W_A = W_B = W_C \neq 0$.

(B) The rate of mass consumption is given by $\frac{\Delta m}{\Delta t} = 1 \times 10^{-3} \text{ g s}^{-1} = 10^{-6} \text{ kg s}^{-1}$.
According to Einstein's mass-energy equivalence principle,the power $P$ generated is given by $P = \frac{\Delta E}{\Delta t} = \left(\frac{\Delta m}{\Delta t}\right) c^2$.
Substituting the values,where $c = 3 \times 10^8 \text{ m s}^{-1}$:
$P = 10^{-6} \text{ kg s}^{-1} \times (3 \times 10^8 \text{ m s}^{-1})^2$.
$P = 10^{-6} \times 9 \times 10^{16} \text{ W}$.
$P = 9 \times 10^{10} \text{ W}$.
Since $1 \text{ kW} = 10^3 \text{ W}$,we convert the power to kW:
$P = \frac{9 \times 10^{10}}{10^3} \text{ kW} = 9 \times 10^7 \text{ kW}$.