AP EAMCET 2025 Physics Question Paper with Answer and Solution

(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ is the sink temperature.
For the first case: $\eta_1 = 0.60$, $T_1 = 600 \text{ K}$.
$0.60 = 1 - \frac{T_2}{600} \implies \frac{T_2}{600} = 0.40 \implies T_2 = 240 \text{ K}$.
For the second case: $\eta_2 = 0.80$, $T_2 = 240 \text{ K}$.
$0.80 = 1 - \frac{240}{T_1'} \implies \frac{240}{T_1'} = 0.20 \implies T_1' = \frac{240}{0.20} = 1200 \text{ K}$.
Thus, the new source temperature required is $1200 \text{ K}$.

(A) For a Carnot engine,the ratio of heat exchanged is equal to the ratio of the absolute temperatures of the source and the sink: $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$.
Given:
Heat absorbed from source,$Q_1 = 600 \ J$.
Heat rejected to sink,$Q_2 = 400 \ J$.
Temperature of source,$T_1 = 127^{\circ} C = 127 + 273 = 400 \ K$.
Using the formula: $\frac{600}{400} = \frac{400}{T_2}$.
Simplifying the ratio: $1.5 = \frac{400}{T_2}$.
Solving for $T_2$: $T_2 = \frac{400}{1.5} = 266.67 \ K \approx 266.7 \ K$.

(D) The coefficient of performance $(COP)$ of a refrigerator is defined as the ratio of the heat extracted from the cold reservoir $(Q_L)$ to the work input $(W)$.
$COP = \frac{Q_L}{W} = \frac{Q_L}{Q_H - Q_L}$.
According to the second law of thermodynamics,heat cannot spontaneously flow from a colder body to a hotter body without the expenditure of external work.
This law imposes a theoretical upper limit on the efficiency of heat engines and the $COP$ of refrigerators (Carnot limit).
Therefore,the fundamental limitation to the $COP$ of a refrigerator is governed by the second law of thermodynamics.

(B) For a Carnot engine, the efficiency $\eta$ is given by $\eta = 1 - \frac{T_{low}}{T_{high}}$.
Also, the work done $W$ is given by $W = Q_{in} \cdot \eta$, where $Q_{in}$ is the heat absorbed from the source.
For engine $A$: $W_A = Q_A \left(1 - \frac{T}{600}\right)$.
For engine $B$: $W_B = Q_B \left(1 - \frac{400}{T}\right)$.
Since the engines are in series, the heat rejected by engine $A$ is the heat absorbed by engine $B$, so $Q_B = Q_A \left(\frac{T}{600}\right)$.
Given $W_A = W_B$, we have $Q_A \left(1 - \frac{T}{600}\right) = Q_A \left(\frac{T}{600}\right) \left(1 - \frac{400}{T}\right)$.
Simplifying the equation: $1 - \frac{T}{600} = \frac{T}{600} - \frac{400}{600}$.
$1 + \frac{400}{600} = \frac{2T}{600}$.
$1 + \frac{2}{3} = \frac{T}{300} \implies \frac{5}{3} = \frac{T}{300}$.
$T = \frac{5 \times 300}{3} = 500 \ K$.

(C) In a Carnot engine,the work done during isothermal expansion $(W_{exp})$ is given by $W_{exp} = nRT_H \ln(V_2/V_1)$ and the work done during isothermal compression $(W_{comp})$ is given by $W_{comp} = nRT_L \ln(V_2/V_1)$.
Given that $W_{exp} = W_{comp} + 0.25 W_{comp} = 1.25 W_{comp}$.
Substituting the expressions,we get $nRT_H \ln(V_2/V_1) = 1.25 nRT_L \ln(V_2/V_1)$.
This simplifies to $T_H = 1.25 T_L$,or $T_L/T_H = 1/1.25 = 0.8$.
The efficiency of a Carnot engine is given by $\eta = 1 - (T_L/T_H)$.
Substituting the value,$\eta = 1 - 0.8 = 0.2$.
Thus,the efficiency is $20 \%$.

(C) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink.
Given that the source temperature $T_1$ is $25 \%$ more than the sink temperature $T_2$,we have: $T_1 = T_2 + 0.25 T_2 = 1.25 T_2$.
Substituting $T_1$ into the efficiency formula: $\eta = 1 - \frac{T_2}{1.25 T_2} = 1 - \frac{1}{1.25}$.
Since $1.25 = \frac{5}{4}$,we get: $\eta = 1 - \frac{4}{5} = \frac{1}{5}$.
Converting this to percentage: $\eta = \frac{1}{5} \times 100 \% = 20 \%$.

(C) The efficiency of a Carnot heat engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$.
The coefficient of performance of a Carnot refrigerator is given by $\beta = \frac{T_2}{T_1 - T_2}$.
We need to find the ratio of the efficiency of the heat engine to the coefficient of performance of the refrigerator:
Ratio $= \frac{\eta}{\beta} = \frac{\frac{T_1 - T_2}{T_1}}{\frac{T_2}{T_1 - T_2}}$.
Simplifying the expression:
Ratio $= \frac{T_1 - T_2}{T_1} \times \frac{T_1 - T_2}{T_2} = \frac{(T_1 - T_2)^2}{T_1 T_2}$.
Thus,the correct option is $C$.

(D) The coefficient of performance $(\beta)$ of a refrigerator is given by the formula: $\beta = \frac{Q_2}{W}$, where $Q_2$ is the heat extracted from the cold reservoir and $W$ is the work done on the system.
Given: $\beta = 5$ and $Q_2 = 250 \ J$.
Substituting the values: $5 = \frac{250}{W} \implies W = \frac{250}{5} = 50 \ J$.
The heat released to the room $(Q_1)$ is the sum of the heat extracted and the work done: $Q_1 = Q_2 + W$.
$Q_1 = 250 \ J + 50 \ J = 300 \ J$.
Therefore, the heat released per cycle to the room is $300 \ J$.

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.
Let $\Delta T = T_1 - T_2$ be the temperature difference,which is constant for both engines.
Thus,$\eta = \frac{\Delta T}{T_1}$.
Given the ratio of efficiencies $\frac{\eta_A}{\eta_B} = 1.25 = \frac{5}{4}$.
Since $\eta_A = \frac{\Delta T}{T_{1A}}$ and $\eta_B = \frac{\Delta T}{T_{1B}}$,we have $\frac{\eta_A}{\eta_B} = \frac{T_{1B}}{T_{1A}} = \frac{5}{4}$.
Therefore,the ratio of the absolute temperatures of the sources $T_{1A} : T_{1B} = 4 : 5$.

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.
Given $\eta_1 = \frac{1}{3}$,so $1 - \frac{T_2}{T_1} = \frac{1}{3} \implies \frac{T_2}{T_1} = \frac{2}{3} \implies T_1 = 1.5 T_2$.
When the source temperature is decreased by $50 \ K$ and the sink temperature is increased by $25 \ K$,the new efficiency is $\eta_2 = \frac{3}{16}$.
Thus,$1 - \frac{T_2 + 25}{T_1 - 50} = \frac{3}{16} \implies \frac{T_2 + 25}{T_1 - 50} = \frac{13}{16}$.
Substituting $T_1 = 1.5 T_2$ into the equation:
$\frac{T_2 + 25}{1.5 T_2 - 50} = \frac{13}{16}$.
Cross-multiplying: $16(T_2 + 25) = 13(1.5 T_2 - 50)$.
$16 T_2 + 400 = 19.5 T_2 - 650$.
$1050 = 3.5 T_2$.
$T_2 = \frac{1050}{3.5} = 300 \ K$.

(A) In the given $P-V$ diagram:
$1$. Process $I$ is a vertical line,meaning volume $V$ is constant. This is an isochoric process. Thus,$I-c$.
$2$. Process $IV$ is a horizontal line,meaning pressure $P$ is constant. This is an isobaric process. Thus,$IV-b$.
$3$. Processes $II$ and $III$ are curves representing changes in both $P$ and $V$. Since $II$ connects the $300 \ K$ and $500 \ K$ isotherms and $III$ connects the $500 \ K$ and $700 \ K$ isotherms,they represent adiabatic processes (as they cross isotherms). Specifically,$II$ is adiabatic and $III$ is isothermal (or vice versa depending on the specific slope analysis). However,looking at the options,$I-c$ and $IV-b$ are fixed. Comparing with options,$I-c, II-a, III-d, IV-b$ fits the standard interpretation where $II$ is adiabatic and $III$ is isothermal. Therefore,the correct match is $I-c, II-a, III-d, IV-b$.

(B) sudden compression of a gas is an adiabatic process.
For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $P_1 V_1^\gamma = P_2 V_2^\gamma$,where $\gamma$ is the adiabatic index (ratio of specific heat capacities).
Given:
Initial pressure $P_1 = P$
Initial volume $V_1 = 400 \ cc$
Final volume $V_2 = 100 \ cc$
Adiabatic index $\gamma = 1.5 = 3/2$
Substituting the values into the adiabatic equation:
$P \times (400)^{3/2} = P_2 \times (100)^{3/2}$
$P_2 = P \times (400/100)^{3/2}$
$P_2 = P \times (4)^{3/2}$
$P_2 = P \times (2^2)^{3/2}$
$P_2 = P \times 2^3$
$P_2 = 8P$
Therefore,the final pressure is $8P$.

(A) For an adiabatic process,the first law of thermodynamics states that $Q = \Delta U + W$. Since the process is adiabatic,$Q = 0$,so $W = -\Delta U$.
The change in internal energy is given by $\Delta U = n C_v \Delta T$.
For a diatomic gas,the molar heat capacity at constant volume is $C_v = \frac{5}{2} R$.
Given: $n = 3$ moles,$\Delta T = -50^{\circ} C$ (since temperature decreases).
Therefore,$\Delta U = 3 \times (\frac{5}{2} R) \times (-50) = 3 \times 2.5 R \times (-50) = -375 R$.
The work done by the gas is $W = -\Delta U = -(-375 R) = 375 R$.

(D) For an ideal diatomic gas,the molar heat capacity at constant pressure is $C_p = \frac{7}{2}R$ and the molar heat capacity at constant volume is $C_v = \frac{5}{2}R$.
When heat $dQ$ is supplied at constant pressure,the total heat supplied is $dQ = n C_p dT = n (\frac{7}{2}R) dT$.
The heat utilized to increase the internal energy is $dU = n C_v dT = n (\frac{5}{2}R) dT$.
The fraction of heat utilized to increase internal energy is $f = \frac{dU}{dQ} = \frac{n (\frac{5}{2}R) dT}{n (\frac{7}{2}R) dT} = \frac{5/2}{7/2} = \frac{5}{7}$.

(C) For an adiabatic process, the relation between pressure $(P)$ and volume $(V)$ is given by $PV^{\gamma} = \text{constant}$.
Taking the logarithm on both sides, we get $\log P + \gamma \log V = \text{constant}$, which can be written as $\log P = -\gamma \log V + C$.
This is the equation of a straight line $y = mx + c$, where the slope $m = -\gamma$.
From the graph, the slope is calculated as:
$m = \frac{\log P_2 - \log P_1}{\log V_2 - \log V_1} = \frac{2.20 - 2.48}{1.4 - 1.2} = \frac{-0.28}{0.2} = -1.4$.
Since $m = -\gamma$, we have $-\gamma = -1.4$, which implies $\gamma = 1.4$.
Therefore, the ratio of the specific heat capacities is $1.4$.

(C) For an ideal gas undergoing a process at constant pressure,the work done $W$ is given by the formula $W = nR\Delta T$.
Here,the number of moles $n = 6$.
The universal gas constant $R = 8.31 \ J \ mol^{-1} \ K^{-1}$.
The change in temperature $\Delta T = 20^{\circ} C = 20 \ K$ (since the change in temperature is the same in Celsius and Kelvin scales).
Substituting these values into the formula:
$W = 6 \times 8.31 \times 20$
$W = 120 \times 8.31$
$W = 997.2 \ J$.
Therefore,the correct option is $C$.

(C) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = \text{constant}$.
Taking the natural logarithm on both sides: $\ln P + \gamma \ln V = \text{constant}$.
Differentiating both sides: $\frac{dP}{P} + \gamma \frac{dV}{V} = 0$.
Therefore,the fractional change in pressure is $\frac{dP}{P} = -\gamma \frac{dV}{V}$.
For a diatomic gas,the adiabatic exponent $\gamma = \frac{7}{5} = 1.4$.
Given the increase in volume $\frac{dV}{V} = 0.5 \% = 0.005$.
Substituting the values: $\frac{dP}{P} = -1.4 \times 0.5 \% = -0.7 \%$.
Thus,the change in pressure is $-0.7 \%$.

(A) For a monoatomic gas,the adiabatic index $\gamma = 5/3$. At $STP$,$1$ mole of gas occupies $22.4$ litres. Therefore,the number of moles $n = 5.6 / 22.4 = 0.25$ moles.
For an adiabatic process,$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given $V_1 = 5.6$ $L$,$V_2 = 0.7$ $L$,and $T_1 = 273$ $K$.
$T_2 = T_1 (V_1/V_2)^{\gamma-1} = 273 \times (5.6/0.7)^{5/3-1} = 273 \times (8)^{2/3} = 273 \times 4 = 1092$ $K$.
The work done on the gas in an adiabatic process is $W = -\Delta U = -n C_v (T_2 - T_1)$.
For a monoatomic gas,$C_v = 3R/2$.
$W = -0.25 \times (3R/2) \times (1092 - 273) = -0.375 R \times 819 = -307.125 R$.
The work done on the gas is the magnitude,which is approximately $307 R$.

(A) For an adiabatic process, the work done $W$ is given by the formula: $W = \frac{P_i V_i - P_f V_f}{\gamma - 1}$.
Since $P_i V_i^\gamma = P_f V_f^\gamma$, we have $P_f = P_i (V_i / V_f)^\gamma$.
Given $V_f = 8 V_i$ and for a monatomic gas $\gamma = 5/3$, we get $P_f = P_i (1/8)^{5/3} = P_i / 32$.
Substituting these into the work formula: $W = \frac{P_i V_i - (P_i / 32)(8 V_i)}{5/3 - 1} = \frac{P_i V_i - P_i V_i / 4}{2/3} = \frac{(3/4) P_i V_i}{2/3} = \frac{9}{8} P_i V_i$.
Given $W = 180 \text{ J}$ and $P_i = 100 \times 10^3 \text{ Pa}$, we have $180 = \frac{9}{8} \times 10^5 \times V_i$.
$V_i = \frac{180 \times 8}{9 \times 10^5} = \frac{160}{10^5} = 1.6 \times 10^{-3} \text{ m}^3$.
Converting to $\text{cm}^3$: $1.6 \times 10^{-3} \times 10^6 \text{ cm}^3 = 1600 \text{ cm}^3$.

(B) The zeroth law of thermodynamics states that if two systems are each in thermal equilibrium with a third system,then they are in thermal equilibrium with each other.
Thermal equilibrium is defined as the state in which two systems in contact with each other do not exchange any net heat.
This condition is satisfied when both systems have the same temperature.
Therefore,the physical quantity that remains the same for two bodies in thermal equilibrium is temperature.

(C) Planck's constant $(h)$ is related to the energy $(E)$ of a photon and its frequency $(
u)$ by the equation: $E = h
u$.
From this,we can write $h = \frac{E}{\nu}$.
The dimensional formula of energy $(E)$ is $[ML^2 T^{-2}]$.
The dimensional formula of frequency $(
u)$ is $[T^{-1}]$.
Substituting these into the equation for $h$:
$h = \frac{[ML^2 T^{-2}]}{[T^{-1}]} = [ML^2 T^{-2+1}] = [ML^2 T^{-1}]$.
Therefore,the correct dimensional formula for Planck's constant is $[ML^2 T^{-1}]$.

(C) To determine which pair does not have the same dimensional formula,we analyze each option:
$1$. Work and Torque: Both have the dimensional formula $[ML^2T^{-2}]$.
$2$. Angular momentum and Planck's constant: Both have the dimensional formula $[ML^2T^{-1}]$.
$3$. Stress and Linear momentum: Stress is force per unit area,$[ML^{-1}T^{-2}]$. Linear momentum is mass times velocity,$[MLT^{-1}]$. These are not the same.
$4$. Surface tension and Force constant: Both have the dimensional formula $[MT^{-2}]$.
Therefore,the pair that does not have the same dimensional formula is stress and linear momentum.

(A) According to the principle of homogeneity of dimensions,the dimensions of each term in an equation must be the same.
$1$. For the term $(t+c)$,the dimension of $c$ must be the same as the dimension of time $t$. Therefore,$[c] = [T]$.
$2$. For the term $at$,the dimension of $at$ must be equal to the dimension of velocity $v$. Since $[v] = [LT^{-1}]$ and $[t] = [T]$,we have $[a][T] = [LT^{-1}]$,which gives $[a] = [LT^{-2}]$.
$3$. For the term $\frac{b}{t+c}$,the dimension of this term must be equal to the dimension of velocity $v$. Since $[t+c] = [T]$ and $[v] = [LT^{-1}]$,we have $\frac{[b]}{[T]} = [LT^{-1}]$,which gives $[b] = [L]$.
Thus,the dimensions are $[a] = [LT^{-2}]$,$[b] = [L]$,and $[c] = [T]$.

(B) According to the rules for significant figures:
$1$. Leading zeros (zeros to the left of the first non-zero digit) are not significant. In $0.03240$,the zeros before $3$ are not significant.
$2$. Non-zero digits are always significant. Here,$3, 2, 4$ are significant.
$3$. Trailing zeros in a decimal number are significant. The zero at the end of $0.03240$ is significant.
Therefore,the significant figures are $3, 2, 4, 0$,which gives a total of $4$ significant figures.

(B) The given equation is $\text{Force} = \frac{\alpha}{\text{density} + \beta^3}$.
According to the principle of homogeneity of dimensions,quantities added together must have the same dimensions.
Therefore,the dimension of $\beta^3$ must be equal to the dimension of density.
Dimension of density = $[M L^{-3} T^0]$.
So,$[\beta^3] = [M L^{-3} T^0]$.
Taking the cube root,$[\beta] = [M^{1/3} L^{-1} T^0]$.
Now,the dimension of force is $[M L T^{-2}]$.
From the equation,$\alpha = \text{Force} \times (\text{density} + \beta^3)$.
Since $(\text{density} + \beta^3)$ has the same dimension as density,$[\alpha] = [M L T^{-2}] \times [M L^{-3} T^0] = [M^2 L^{-2} T^{-2}]$.
Thus,the dimensions are $[M^2 L^{-2} T^{-2}]$ and $[M^{1/3} L^{-1} T^0]$.

(A) The dimensions of kinetic energy $(K.E.)$ are $[M L^2 T^{-2}]$.
The dimensions of surface tension $(S)$ are $[M T^{-2}]$.
Let the required quantity be $X = \sqrt{\frac{K.E.}{S}}$.
Substituting the dimensions: $X = \sqrt{\frac{[M L^2 T^{-2}]}{[M T^{-2}]}}$.
Simplifying the expression: $X = \sqrt{[L^2]} = [L]$.
The dimension $[L]$ corresponds to length or distance.
Therefore,the correct option is $A$.

(A) Let the maximum temperature be $T_1 = 44^{\circ} C \pm 0.5^{\circ} C$ and the minimum temperature be $T_2 = 22^{\circ} C \pm 0.5^{\circ} C$.
The temperature difference is given by $\Delta T = T_1 - T_2$.
The value of the difference is $44^{\circ} C - 22^{\circ} C = 22^{\circ} C$.
According to the rule of error propagation for subtraction,when two quantities are subtracted,their absolute errors are added.
Therefore,the error in the difference is $\Delta(\Delta T) = \Delta T_1 + \Delta T_2 = 0.5^{\circ} C + 0.5^{\circ} C = 1.0^{\circ} C$.
Thus,the temperature difference is $22^{\circ} C \pm 1^{\circ} C$.

(A) Step $1$: Perform the subtraction inside the parentheses. $0.312 - 0.03 = 0.282$. According to the rules of significant figures for subtraction,the result should have the same number of decimal places as the measurement with the fewest decimal places. Here,$0.03$ has two decimal places,so the result $0.282$ is rounded to $0.28$ (two decimal places).
Step $2$: Perform the division. $\frac{0.501}{0.05} = 10.02$.
Step $3$: Multiply the results. $10.02 \times 0.28 = 2.8056$.
Step $4$: Apply the rules for multiplication. The result should have the same number of significant figures as the measurement with the fewest significant figures. $0.05$ has $1$ significant figure,and $0.28$ has $2$ significant figures. Therefore,the final result should be rounded to $1$ significant figure. Thus,the number of significant figures is $1$.

(B) The surface area $S$ of a sphere is given by $S = 4\pi r^2$,where $r$ is the radius.
The relative error in surface area is given by $\frac{\Delta S}{S} = 2 \frac{\Delta r}{r}$.
Given $\frac{\Delta S}{S} \times 100 = 1.2 \%$,we have $2 \frac{\Delta r}{r} \times 100 = 1.2 \%$,which implies $\frac{\Delta r}{r} \times 100 = 0.6 \%$.
The volume $V$ of a sphere is given by $V = \frac{4}{3}\pi r^3$.
The relative error in volume is given by $\frac{\Delta V}{V} = 3 \frac{\Delta r}{r}$.
Substituting the value of $\frac{\Delta r}{r}$,we get $\frac{\Delta V}{V} \times 100 = 3 \times 0.6 \% = 1.8 \%$.

(C) The phase difference between the two waves is given by the difference in their arguments.
Let $\phi_1 = \omega t - \frac{2 \pi x}{\lambda}$ and $\phi_2 = \omega t - \frac{2 \pi x}{\lambda} + \phi$.
The phase difference $\Delta \phi = \phi_2 - \phi_1 = \phi$.
The relationship between path difference $(\Delta x)$ and phase difference $(\Delta \phi)$ is given by the formula:
$\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.
Rearranging this formula to solve for the path difference $\Delta x$:
$\Delta x = \frac{\lambda}{2 \pi} \Delta \phi$.
Substituting $\Delta \phi = \phi$ into the equation:
$\Delta x = \frac{\lambda}{2 \pi} \phi$.

(D) The general equation of a stationary wave is given by $y = A \sin(kx) \cos(\omega t)$.
Comparing this with the given equation $y = 0.7 \sin \left(\frac{7 \pi}{4} x\right) \cos (350 \pi t)$,we get:
Wave number $k = \frac{7 \pi}{4} \ m^{-1}$
Angular frequency $\omega = 350 \pi \ rad \ s^{-1}$
The wave speed $v$ is defined as the ratio of angular frequency to wave number:
$v = \frac{\omega}{k}$
Substituting the values:
$v = \frac{350 \pi}{7 \pi / 4}$
$v = 350 \pi \times \frac{4}{7 \pi}$
$v = 50 \times 4 = 200 \ m \ s^{-1}$
Thus,the speed of the wave is $200 \ m \ s^{-1}$.

(C) The speed of a transverse wave on 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.
Linear mass density $\mu = \frac{m}{L}$.
Given: $m = 5 \times 10^{-3} \text{ kg}$, $L = 81 \text{ cm} = 0.81 \text{ m}$, $T = 50 \text{ N}$.
Calculate $\mu$: $\mu = \frac{5 \times 10^{-3}}{0.81} \text{ kg/m}$.
Now, substitute the values into the speed formula:
$v = \sqrt{\frac{50}{\frac{5 \times 10^{-3}}{0.81}}} = \sqrt{\frac{50 \times 0.81}{5 \times 10^{-3}}} = \sqrt{10 \times 0.81 \times 10^3} = \sqrt{8100} = 90 \text{ m s}^{-1}$.
Thus, the speed of the transverse wave is $90 \text{ m s}^{-1}$.

(C) The standard equation of a wave is $y = A \sin(kx + \omega t)$.
Comparing this with the given equation $y = 1.5 \sin((5 \times 10^{-3} x) + 20 t)$,we get:
Wave number $k = 5 \times 10^{-3} \ cm^{-1}$
Angular frequency $\omega = 20 \ rad/s$
The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{20}{5 \times 10^{-3}} = 4000 \ cm/s = 40 \ m/s$.
The mass per unit length $\mu$ is $\frac{m}{L} = \frac{3 \ g}{80 \ cm} = \frac{3 \times 10^{-3} \ kg}{0.8 \ m} = 3.75 \times 10^{-3} \ kg/m$.
The wave speed in a stretched string is $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension.
Therefore,$T = v^2 \mu = (40)^2 \times (3.75 \times 10^{-3}) = 1600 \times 0.00375 = 6 \ N$.

(D) The speed of sound $v = 330 \ m \ s^{-1}$.
The wavelengths are $\lambda_1 = 0.99 \ m$ and $\lambda_2 = 1.00 \ m$.
The frequencies are $f_1 = \frac{v}{\lambda_1} = \frac{330}{0.99} = \frac{33000}{99} = \frac{1000}{3} \ Hz$ and $f_2 = \frac{v}{\lambda_2} = \frac{330}{1.00} = 330 \ Hz$.
The beat frequency $f_b = |f_1 - f_2| = |\frac{1000}{3} - 330| = |\frac{1000 - 990}{3}| = \frac{10}{3} \ Hz$.
Beat frequency is defined as the number of beats per second. Thus,$f_b = \frac{\text{Number of beats}}{t}$.
Given that $10$ beats are produced in $t$ seconds,we have $\frac{10}{3} = \frac{10}{t}$.
Therefore,$t = 3 \ s$.

(D) The given equations of the two sound waves are $y_1 = 3 \sin(250 \pi t)$ and $y_2 = 2 \sin(260 \pi t)$.
Comparing these with the standard equation $y = A \sin(\omega t)$,we get the angular frequencies as $\omega_1 = 250 \pi \text{ rad/s}$ and $\omega_2 = 260 \pi \text{ rad/s}$.
The frequencies $f_1$ and $f_2$ are given by $f = \frac{\omega}{2 \pi}$.
Thus,$f_1 = \frac{250 \pi}{2 \pi} = 125 \text{ Hz}$ and $f_2 = \frac{260 \pi}{2 \pi} = 130 \text{ Hz}$.
The beat frequency $f_b$ is the difference between the two frequencies: $f_b = |f_2 - f_1| = |130 - 125| = 5 \text{ Hz}$.
The time interval between two successive maximum intensities is the time period of the beats,given by $T_b = \frac{1}{f_b}$.
Therefore,$T_b = \frac{1}{5} = 0.2 \text{ s}$.
Thus,the correct option is $D$.

(C) Let the speed of sound be $v$. The speed of the source $v_s = 0.1v$ and the speed of the observer $v_o = 0.1v$.
Since both are approaching each other,the observed frequency $f'$ is given by the Doppler effect formula:
$f' = f \left( \frac{v + v_o}{v - v_s} \right)$
Substituting the values:
$f' = f \left( \frac{v + 0.1v}{v - 0.1v} \right) = f \left( \frac{1.1v}{0.9v} \right) = f \left( \frac{11}{9} \right)$
$f' \approx 1.222f$
The change in frequency is $\Delta f = f' - f = 1.222f - f = 0.222f$.
The percentage change is $\frac{\Delta f}{f} \times 100 = 0.222 \times 100 = 22.2 \%$.

(B) For a closed organ pipe, the frequency of the $n^{th}$ harmonic is given by $f_n = n \cdot f_1$, where $n$ must be an odd integer $(n = 1, 3, 5, 7, 9, ...)$.
In a closed organ pipe of length $L$, the number of nodes $N$ formed for the $n^{th}$ harmonic is given by the formula $N = \frac{n+1}{2}$.
For the fifth harmonic $(n = 5)$:
$N_5 = \frac{5+1}{2} = \frac{6}{2} = 3$.
For the ninth harmonic $(n = 9)$:
$N_9 = \frac{9+1}{2} = \frac{10}{2} = 5$.
Therefore, the number of nodes formed are $3$ and $5$ respectively.

(D) The standard equation of a travelling wave is given by $y(x, t) = A \sin (kx - \omega t)$.
Comparing this with the given equation $y(x, t) = 0.5 \sin (70.1 x - 10 \pi t)$,we get the angular frequency $\omega = 10 \pi \text{ rad/s}$.
The relationship between angular frequency $\omega$ and frequency $f$ is given by $\omega = 2 \pi f$.
Substituting the value of $\omega$,we have $10 \pi = 2 \pi f$.
Solving for $f$,we get $f = \frac{10 \pi}{2 \pi} = 5 \text{ Hz}$.
Therefore,the frequency of the wave is $5 \text{ Hz}$.

(A) The frequency of the sound wave is $f = 500 \text{ Hz}$.
The time taken for the wave to travel from $X$ to $Y$ is $t = 2 \text{ s}$.
The number of waves $N$ produced by a source in time $t$ is given by the formula $N = f \times t$.
Substituting the given values: $N = 500 \text{ Hz} \times 2 \text{ s} = 1000$.
Thus,there are $1000$ waves between points $X$ and $Y$.

(C) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
Since $T$ and $\mu$ are constant,we have $f \propto \frac{1}{L}$,which implies $fL = \text{constant} = k$.
Therefore,$L = \frac{k}{f}$.
When the wire of length $L$ is divided into three segments of lengths $L_1, L_2, L_3$,we have $L = L_1 + L_2 + L_3$.
Substituting $L = \frac{k}{f}$,$L_1 = \frac{k}{f_1}$,$L_2 = \frac{k}{f_2}$,and $L_3 = \frac{k}{f_3}$ into the equation,we get:
$\frac{k}{f} = \frac{k}{f_1} + \frac{k}{f_2} + \frac{k}{f_3}$.
Dividing both sides by $k$,we obtain the relation: $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$.

(B) Given: Mass of the ball $m = 0.25 \ kg$,final velocity $v = 12 \ m \ s^{-1}$,initial velocity $u = 0 \ m \ s^{-1}$,and distance $s = 0.9 \ m$.
Using the third equation of motion: $v^2 = u^2 + 2as$.
Substituting the values: $(12)^2 = (0)^2 + 2 \times a \times 0.9$.
$144 = 1.8 \times a$.
$a = 144 / 1.8 = 80 \ m \ s^{-2}$.
Now,using Newton's second law of motion: $F = ma$.
$F = 0.25 \ kg \times 80 \ m \ s^{-2} = 20 \ N$.
Therefore,the net force acting on the ball is $20 \ N$.

(B) The kinetic energy $(K)$ of a particle is given by the formula $K = \frac{1}{2} m v^2$.
Since the particle moves with a constant speed $v$,the magnitude of its velocity remains unchanged throughout the motion.
Because the mass $m$ and the speed $v$ are constant,the kinetic energy $K$ remains constant at all points on the circular path.
The change in kinetic energy $(\Delta K)$ is defined as the final kinetic energy $(K_f)$ minus the initial kinetic energy $(K_i)$.
Since $K_f = K_i = \frac{1}{2} m v^2$,the change in kinetic energy is $\Delta K = K_f - K_i = 0$.

(A) The initial potential energy of the bob at the horizontal position is $PE_i = mgh$,where $h = L = 200 \ cm = 2 \ m$.
So,$PE_i = m \times 10 \times 2 = 20mg$.
When the bob reaches the mean position,its potential energy is converted into kinetic energy.
Given that $10 \%$ of the initial energy is lost,the remaining energy is $90 \%$ of the initial potential energy.
$KE_f = 0.9 \times PE_i = 0.9 \times 20mg = 18mg$.
At the mean position,$KE_f = \frac{1}{2}mv^2$.
Equating the two: $\frac{1}{2}mv^2 = 18mg$.
$v^2 = 36g = 36 \times 10 = 360$.
$v = \sqrt{360} \approx 18.97 \ m \ s^{-1}$.
Wait,re-evaluating: $PE_i = mgL$. $KE_f = 0.9 \times mgL = \frac{1}{2}mv^2$.
$v^2 = 1.8 \times g \times L = 1.8 \times 10 \times 2 = 36$.
$v = 6 \ m \ s^{-1}$.

(B) Given: Mass of body $A$ $(m_A = 20 \ kg)$,mass of body $B$ $(m_B = 5 \ kg)$,force $(F = 40 \ N)$.
Since both bodies start from rest,their kinetic energies are $K_A = K_B = K$.
Kinetic energy is given by $K = \frac{p^2}{2m}$,where $p$ is momentum.
Since $K_A = K_B$,we have $\frac{p_A^2}{2m_A} = \frac{p_B^2}{2m_B} \implies \frac{p_A}{p_B} = \sqrt{\frac{m_A}{m_B}} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2$.
From the impulse-momentum theorem,$F \cdot t = \Delta p$. Since initial momentum is zero,$p = F \cdot t$.
Thus,$\frac{p_A}{p_B} = \frac{F \cdot t_A}{F \cdot t_B} = \frac{t_A}{t_B}$.
Therefore,$\frac{t_A}{t_B} = 2$,which means $t_A: t_B = 2: 1$.

(D) Given: Mass $m = 2000 \ kg$,Power $P = 10 \ kW = 10000 \ W$,Time $t = 10 \ s$.
We know that Power $P = F \cdot v = (m \cdot a) \cdot v = m \cdot \frac{dv}{dt} \cdot v$.
So,$P \cdot dt = m \cdot v \cdot dv$.
Integrating both sides from $t = 0$ to $t = 10 \ s$ and $v = 0$ to $v = v$:
$\int_{0}^{10} P \, dt = \int_{0}^{v} m \cdot v \, dv$.
$P \cdot t = \frac{1}{2} \cdot m \cdot v^2$.
$10000 \times 10 = \frac{1}{2} \times 2000 \times v^2$.
$100000 = 1000 \times v^2$.
$v^2 = 100$.
$v = 10 \ m \ s^{-1}$.

(B) The radius of the well $r = 1 \ m$ and depth $h = 14 \ m$.
The volume of water $V = \pi r^2 h = \pi \times (1)^2 \times 14 = 14\pi \ m^3$.
The mass of water $m = \text{density} \times \text{volume} = 1000 \ kg/m^3 \times 14\pi \ m^3 = 14000\pi \ kg$.
The work done to empty the well is equal to the potential energy of the water,where the center of mass is at $h/2 = 7 \ m$.
$W = mgh_{cm} = (14000\pi) \times 10 \times 7 = 980000\pi \ J$.
Power $P = 1.4 \ kW = 1400 \ W$.
Time $t = W / P = (980000\pi) / 1400 = 700\pi \ s$.
Using $\pi \approx 3.14$,$t = 700 \times 3.14 = 2198 \ s \approx 2200 \ s$.

(B) According to the work-energy theorem,the work done by the retarding force is equal to the change in kinetic energy of the body.
$W = \Delta K$
$F \cdot d = \frac{1}{2} m v^2$
Since the retarding force $F$ is the same for both bodies,the distance $d$ travelled before coming to rest is given by:
$d = \frac{m v^2}{2F}$
For body $A$:
$d_A = \frac{m_A v_A^2}{2F} = \frac{1.5 \times (20)^2}{2F} = \frac{1.5 \times 400}{2F} = \frac{600}{2F}$
For body $B$:
$d_B = \frac{m_B v_B^2}{2F} = \frac{3 \times (15)^2}{2F} = \frac{3 \times 225}{2F} = \frac{675}{2F}$
The ratio of the distances is:
$\frac{d_A}{d_B} = \frac{600}{675} = \frac{24}{27} = \frac{8}{9}$
Thus,the ratio is $8: 9$.

(A) The potential energy of the particle is given by $U(x) = 5x(x-4) = 5x^2 - 20x \ J$.
For a particle moving in a conservative field,the force acting on it is given by $F = -\frac{dU}{dx}$.
$F = -\frac{d}{dx}(5x^2 - 20x) = -(10x - 20) = 20 - 10x$.
The speed of the particle is maximum when the net force acting on it is zero (equilibrium position).
Setting $F = 0$,we get $20 - 10x = 0$.
$10x = 20$,which implies $x = 2 \ m$.
At this position,the potential energy is minimum,and by the law of conservation of energy,the kinetic energy is maximum,meaning the speed is maximum.

(A) The total mechanical energy of the body is conserved.
Initial kinetic energy at the ground $(h=0)$ is $K_i = \frac{1}{2} m v^2 = \frac{1}{2} m (20)^2 = 200m \,J$.
Initial potential energy at the ground is $U_i = 0 \,J$.
Total energy $E = K_i + U_i = 200m \,J$.
At a height of $5 \,m$, potential energy $U_5 = mgh = m(10)(5) = 50m \,J$.
Given $U_5 = 100 \,J$, so $50m = 100$, which gives $m = 2 \,kg$.
Total energy $E = 200(2) = 400 \,J$.
At a height of $10 \,m$, potential energy $U_{10} = mgh = 2(10)(10) = 200 \,J$.
By conservation of energy, $E = K_{10} + U_{10}$.
$400 = K_{10} + 200$.
$K_{10} = 200 \,J$.

(B) Given: Mass of the train $m = 10^6 \ kg$,Speed $v = 108 \ km/h$.
First,convert the speed into $SI$ units: $v = 108 \times \frac{5}{18} \ m/s = 30 \ m/s$.
The frictional force per $100 \ kg$ is $0.5 \ N$.
Total frictional force $F = \frac{10^6 \ kg}{100 \ kg} \times 0.5 \ N = 10^4 \times 0.5 \ N = 5000 \ N$.
Since the train is moving at a constant speed,the driving force must be equal to the frictional force: $F_{drive} = F = 5000 \ N$.
Power $P = F_{drive} \times v = 5000 \ N \times 30 \ m/s = 150,000 \ W = 150 \ kW$.

(A) Let the refractive index of the glass slab be $\mu$. The angle of incidence at point $A$ is $i = 45^\circ$. Let the angle of refraction at $A$ be $r$. By Snell's Law,$1 \cdot \sin(45^\circ) = \mu \cdot \sin(r)$,so $\sin(r) = \frac{1}{\mu\sqrt{2}}$.
At point $B$,the angle of incidence is $r' = 90^\circ - r$. For total internal reflection to occur,the angle of incidence must be greater than or equal to the critical angle $C$,where $\sin(C) = \frac{1}{\mu}$.
Thus,$\sin(90^\circ - r) \ge \frac{1}{\mu}$,which implies $\cos(r) \ge \frac{1}{\mu}$.
Squaring both sides,$\cos^2(r) \ge \frac{1}{\mu^2}$,so $1 - \sin^2(r) \ge \frac{1}{\mu^2}$.
Substituting $\sin^2(r) = \frac{1}{2\mu^2}$,we get $1 - \frac{1}{2\mu^2} \ge \frac{1}{\mu^2}$,which simplifies to $1 \ge \frac{3}{2\mu^2}$.
Thus,$\mu^2 \ge \frac{3}{2}$,or $\mu \ge \sqrt{\frac{3}{2}}$. The minimum refractive index is $\sqrt{\frac{3}{2}}$.

(C) The normal least distance of distinct vision $(D)$ is $25 \ cm$. Since the boy's least distance is $35 \ cm$,he is suffering from hypermetropia (farsightedness).
To correct this,we need to use a convex lens that forms a virtual image of an object placed at $25 \ cm$ at the boy's actual near point of $35 \ cm$.
Here,object distance $u = -25 \ cm$ and image distance $v = -35 \ cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
$\frac{1}{f} = \frac{1}{-35} - \frac{1}{-25} = \frac{1}{25} - \frac{1}{35}$.
$\frac{1}{f} = \frac{7 - 5}{175} = \frac{2}{175}$.
$f = \frac{175}{2} = 87.5 \ cm$.
Since the focal length is positive,it is a convex lens.

(D) person with short-sightedness (myopia) cannot see distant objects clearly because the image forms in front of the retina. To correct this, a concave lens is used.
For a person to see objects at infinity $(u = \infty)$, the lens must form a virtual image at the person's far point $(v = -400 \ cm = -4 \ m)$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-4} - \frac{1}{\infty} = -0.25 \ m^{-1}$.
Since power $P = \frac{1}{f(\text{in meters})}$, we get $P = -0.25 \ D$.

(D) Let the focal lengths of the two lenses be $f_1$ and $f_2$.
Given that the lenses are in contact,the equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
Given $F = 4 \ cm$,so $\frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{4}$.
This simplifies to $\frac{f_1 + f_2}{f_1 f_2} = \frac{1}{4}$.
We are given $f_1 + f_2 = 18 \ cm$.
Substituting this into the equation: $\frac{18}{f_1 f_2} = \frac{1}{4}$,which gives $f_1 f_2 = 72 \ cm^2$.
We have the sum $f_1 + f_2 = 18$ and the product $f_1 f_2 = 72$.
These are the roots of the quadratic equation $x^2 - 18x + 72 = 0$.
Factoring the equation: $(x - 6)(x - 12) = 0$.
Thus,the focal lengths are $6 \ cm$ and $12 \ cm$.
Power $P$ is inversely proportional to focal length $(P = \frac{1}{f})$.
The lens with the lower power will have the larger focal length.
Therefore,the focal length of the lens with lower power is $12 \ cm$.

(D) Let the focal length of the convex lens be $f$. The magnification $m$ produced by a lens is given by $m = \frac{f}{f+u}$,where $u$ is the object distance.
For a convex lens,if the image size is the same,the magnification $m$ must be $m = 1$ (virtual image) or $m = -1$ (real image).
Case $1$: For a real image,$m = -1$. Using the formula $m = \frac{f}{f+u}$,we have $-1 = \frac{f}{f+u_1}$,which implies $f+u_1 = -f$,so $u_1 = -2f$. Given $u_1 = -20 \ cm$,we get $2f = 20 \ cm$,so $f = 10 \ cm$. However,this would mean the object is at $2f$,forming a real image of the same size.
Case $2$: If the image size is the same for two different positions,one must be a real image and one must be a virtual image. For a convex lens,the magnification $m = \frac{v}{u}$. Since the image size is the same,$|m_1| = |m_2|$.
Let the object distances be $u_1 = -20 \ cm$ and $u_2 = -10 \ cm$. The magnification is $m = \frac{f}{f+u}$.
For $u_1 = -20$,$m_1 = \frac{f}{f-20}$. For $u_2 = -10$,$m_2 = \frac{f}{f-10}$.
Since the image sizes are equal,$|m_1| = |m_2|$. Thus,$\left| \frac{f}{f-20} \right| = \left| \frac{f}{f-10} \right|$.
This implies $\frac{f}{20-f} = \frac{f}{f-10}$ (since one image is real and one is virtual,one magnification is negative).
$20-f = f-10 \implies 2f = 30 \implies f = 15 \ cm$.

(A) The lens maker's formula is given by $\frac{1}{f} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a lens in air $(\mu_a = 1)$: $\frac{1}{f_a} = (\mu_l - 1) K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$.
For a lens in liquid $(\mu_{liq})$: $\frac{1}{f_l} = (\frac{\mu_l}{\mu_{liq}} - 1) K$.
Given $\frac{f_a}{f_l} = \frac{1}{2}$,we have $\frac{f_l}{f_a} = 2$.
Dividing the two equations: $\frac{f_a}{f_l} = \frac{(\frac{\mu_l}{\mu_{liq}} - 1)}{(\mu_l - 1)} = \frac{1}{2}$.
Substituting $\mu_l = 1.5$: $\frac{(\frac{1.5}{\mu_{liq}} - 1)}{(1.5 - 1)} = \frac{1}{2}$.
$\frac{1.5}{\mu_{liq}} - 1 = 0.5 \times 0.5 = 0.25$.
$\frac{1.5}{\mu_{liq}} = 1.25$.
$\mu_{liq} = \frac{1.5}{1.25} = 1.2$.

(B) For a convex mirror,the magnification $m$ is given by $m = \frac{h_i}{h_o} = -\frac{v}{u}$.
Given that the size of the image is $1/n$ times the size of the object,we have $m = 1/n$.
Thus,$1/n = -v/u$,which implies $v = -u/n$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,where $f$ is the focal length of the convex mirror (positive).
Substituting $v = -u/n$ into the mirror formula:
$\frac{1}{-u/n} + \frac{1}{u} = \frac{1}{f}$
$-\frac{n}{u} + \frac{1}{u} = \frac{1}{f}$
$\frac{1-n}{u} = \frac{1}{f}$
$u = f(1-n)$.
Since the object distance $u$ is conventionally negative,the magnitude is $|u| = f(n-1)$.

(D) For an equilateral prism,the angle of the prism is $A = 60^{\circ}$.
When the light ray inside the prism moves parallel to the base,the ray undergoes minimum deviation,which implies that the angle of incidence $i$ is equal to the angle of emergence $e$,and the angle of refraction $r_1$ is equal to $r_2$.
Since $r_1 + r_2 = A$,we have $2r = 60^{\circ}$,which gives $r = 30^{\circ}$.
Using Snell's Law at the first surface: $\mu = \frac{\sin i}{\sin r}$.
Given $\mu = \sqrt{3}$ and $r = 30^{\circ}$,we have $\sqrt{3} = \frac{\sin i}{\sin 30^{\circ}}$.
Since $\sin 30^{\circ} = 0.5$,we get $\sin i = \sqrt{3} \times 0.5 = \frac{\sqrt{3}}{2}$.
Therefore,$i = 60^{\circ}$.

(D) For a light ray to retrace its path after reflection from the silvered second face, it must strike the second face normally (at an angle of $90^{\circ}$ to the surface).
Let $i = 60^{\circ}$ be the angle of incidence on the first face, $r_1$ be the angle of refraction at the first face, and $A = 30^{\circ}$ be the prism angle.
Since the ray strikes the second face normally, the angle of refraction at the second face is $r_2 = 0^{\circ}$.
From the geometry of the prism, we know that $A = r_1 + r_2$.
Substituting the values, $30^{\circ} = r_1 + 0^{\circ}$, so $r_1 = 30^{\circ}$.
Using Snell's Law at the first face: $\mu = \frac{\sin(i)}{\sin(r_1)}$.
Substituting the values: $\mu = \frac{\sin(60^{\circ})}{\sin(30^{\circ})}$.
$\mu = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.
Thus, the refractive index of the material of the prism is $\sqrt{3}$.

(C) The circuit consists of a $40 \ V$ $DC$ source,a series resistor $R_s = 2 \ k\Omega$,a Zener diode with breakdown voltage $V_z = 20 \ V$,and a load resistor $R_L = 5 \ k\Omega$.
First,calculate the total current $I$ flowing through the series resistor $R_s$:
$I = \frac{V_{source} - V_z}{R_s} = \frac{40 \ V - 20 \ V}{2 \ k\Omega} = \frac{20 \ V}{2 \times 10^3 \ \Omega} = 10 \times 10^{-3} \ A = 10 \ mA$.
Next,calculate the load current $I_L$ flowing through the load resistor $R_L$:
$I_L = \frac{V_z}{R_L} = \frac{20 \ V}{5 \ k\Omega} = \frac{20 \ V}{5 \times 10^3 \ \Omega} = 4 \times 10^{-3} \ A = 4 \ mA$.
The current through the Zener diode $I_z$ is given by the difference between the total current and the load current:
$I_z = I - I_L = 10 \ mA - 4 \ mA = 6 \ mA$.
Therefore,the correct option is $C$.

(A) $Zener$ $diode$ is a special type of $PN$ junction diode designed to operate in the reverse breakdown region.
When the reverse voltage across a $Zener$ $diode$ reaches the $Zener$ breakdown voltage,the voltage across it remains constant even if the current through it changes significantly.
This property makes it ideal for use as a voltage regulator in electronic circuits to maintain a stable output voltage.

(B) diode is forward biased when the potential at the $P$-terminal $(V_P)$ is higher than the potential at the $N$-terminal $(V_N)$.
In option $A$: $V_P = +2 \ V$,$V_N = +3 \ V$. Since $V_P < V_N$,it is reverse biased.
In option $B$: $V_P = +2 \ V$,$V_N = -2 \ V$. Since $V_P > V_N$,it is forward biased.
In option $C$: $V_P = -2 \ V$,$V_N = +2 \ V$. Since $V_P < V_N$,it is reverse biased.
In option $D$: $V_P = +2 \ V$,$V_N = +2 \ V$. Since $V_P = V_N$,there is no bias (or zero bias).
Therefore,the correct option is $B$.

(C) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
Given the band gap energy $E_g = 3 \ eV$.
We know that $hc \approx 1240 \ eV \cdot nm$.
Substituting the values,we get $\lambda = \frac{hc}{E_g} = \frac{1240 \ eV \cdot nm}{3 \ eV}$.
$\lambda \approx 413.33 \ nm$.
Therefore,the wavelength of light it can detect is nearly $413 \ nm$.

(C) When amplifiers are connected in series, the total voltage gain $(A_v)$ is the product of the individual voltage gains of the amplifiers.
Given: $A_1 = 8$, $A_2 = 12.5$.
Total voltage gain $A_v = A_1 \times A_2 = 8 \times 12.5 = 100$.
The input signal voltage $V_{in} = 200 \mu V = 200 \times 10^{-6} \text{ V}$.
The output voltage $V_{out}$ is given by the formula: $V_{out} = A_v \times V_{in}$.
$V_{out} = 100 \times 200 \times 10^{-6} \text{ V} = 20000 \times 10^{-6} \text{ V} = 20 \times 10^{-3} \text{ V}$.
Since $10^{-3} \text{ V} = 1 \text{ mV}$, the output voltage is $20 \text{ mV}$.

(A) For a transistor to operate in the active region,which is required for amplification,the emitter-base junction must be forward biased and the base-collector junction must be reverse biased.
In the forward-biased emitter-base junction,charge carriers are injected from the emitter into the base.
In the reverse-biased base-collector junction,these carriers are collected by the collector,allowing for current control and signal amplification.

(D) Given:
Voltage gain $(A_v)$ = $300$
Current amplification factor $(\beta)$ = $60$
Collector resistance $(R_C)$ = $5 \, k\Omega$
We know that the voltage gain $(A_v)$ in a common emitter configuration is given by:
$A_v = \beta \times \frac{R_C}{R_B}$
Rearranging the formula to find the base resistance $(R_B)$:
$R_B = \beta \times \frac{R_C}{A_v}$
Substituting the given values:
$R_B = 60 \times \frac{5 \, k\Omega}{300}$
$R_B = 60 \times \frac{5}{300} \, k\Omega$
$R_B = \frac{300}{300} \, k\Omega$
$R_B = 1 \, k\Omega$
Therefore, the base resistance is $1 \, k\Omega$.

(D) For a common emitter amplifier,the voltage gain $(A_v)$ is given by the formula: $A_v = \beta \times \frac{R_C}{R_B}$,where $\beta$ is the current amplification factor,$R_C$ is the collector resistance,and $R_B$ is the base resistance.
Given that the ratio of voltage gain $(A_v)$ to the current amplification factor $(\beta)$ is $4$,we have: $\frac{A_v}{\beta} = 4$.
Substituting the expression for $A_v$ into the given ratio: $\frac{\beta \times (R_C / R_B)}{\beta} = 4$.
This simplifies to: $\frac{R_C}{R_B} = 4$.
Therefore,the ratio of the collector resistance to the base resistance is $4 : 1$.

(C) In a standard bipolar junction transistor $(BJT)$,the three regions are designed with specific physical dimensions to optimize performance.
$1$. The $Base$ $(Y)$ is made very thin and lightly doped to allow most charge carriers to pass through to the collector.
$2$. The $Collector$ $(Z)$ is made the largest in size to handle the heat dissipation generated by the power collected from the emitter.
$3$. The $Emitter$ $(X)$ is of moderate size,larger than the base but smaller than the collector,and is heavily doped to provide a large number of charge carriers.
Therefore,the order of sizes is $Collector > Emitter > Base$,which corresponds to $Z > X > Y$.

(D) The power gain $(A_p)$ of a transistor amplifier is given by the product of the current gain $(\beta)$ and the voltage gain $(A_v)$.
$A_p = \beta \times A_v$
Voltage gain $(A_v)$ is defined as the product of the current gain $(\beta)$ and the ratio of output resistance $(R_c)$ to input resistance $(R_b)$:
$A_v = \beta \times \frac{R_c}{R_b}$
Given:
$\beta = 100$
$R_c = 3 \ k\Omega$
$R_b = 2 \ k\Omega$
Calculating $A_v$:
$A_v = 100 \times \frac{3 \ k\Omega}{2 \ k\Omega} = 100 \times 1.5 = 150$
Now,calculating power gain $(A_p)$:
$A_p = \beta \times A_v = 100 \times 150 = 15000$
Therefore,the power gain is $15000$.

(C) Given that the collector current $I_C = 0.98 I_E$.
We know that the emitter current is the sum of the base current and the collector current: $I_E = I_B + I_C$.
Substituting the value of $I_C$,we get $I_E = I_B + 0.98 I_E$.
Rearranging the terms to find $I_B$: $I_B = I_E - 0.98 I_E = 0.02 I_E$.
Now,we need to find the ratio of the base current to the collector current: $\frac{I_B}{I_C} = \frac{0.02 I_E}{0.98 I_E}$.
Simplifying the ratio: $\frac{I_B}{I_C} = \frac{2}{98} = \frac{1}{49}$.
Therefore,the ratio is $1: 49$.

(A) Let the inputs be $A=0, B=1, C=0$.
$1$. The top $NAND$ gate has inputs $A=0$ and $B=1$. Its output is $\overline{0 \cdot 1} = \overline{0} = 1$.
$2$. The second $NAND$ gate has inputs $A=0$ and the output of the first $NAND$ gate $(1)$. Its output $y_1 = \overline{0 \cdot 1} = \overline{0} = 1$.
$3$. The bottom $NOR$ gate has inputs $B=1$ and $C=0$. Its output is $\overline{1 + 0} = \overline{1} = 0$.
$4$. The second $NOR$ gate has inputs $C=0$ and the output of the first $NOR$ gate $(0)$. Its output $y_2 = \overline{0 + 0} = \overline{0} = 1$.
$5$. The final $OR$ gate has inputs $y_1=1$ and $y_2=1$. Its output $y_3 = 1 + 1 = 1$.
Thus,the values are $y_1=1, y_2=1, y_3=1$.

(C) Let the inputs be $A$ and $B$.
Each of the first two $NAND$ gates has its inputs shorted,which makes them act as $NOT$ gates.
Therefore,the output of the first $NAND$ gate is $\overline{A}$ and the output of the second $NAND$ gate is $\overline{B}$.
These outputs $\overline{A}$ and $\overline{B}$ are fed as inputs to the third $NAND$ gate.
The output $Y$ of the third $NAND$ gate is given by $Y = \overline{(\overline{A} \cdot \overline{B})}$.
Using De Morgan's theorem,$\overline{X \cdot Y} = \overline{X} + \overline{Y}$.
So,$Y = \overline{(\overline{A})} + \overline{(\overline{B})} = A + B$.
Since the output $Y = A + B$ represents an $OR$ gate,the equivalent logic gate is $OR$.

(B) The circuit consists of an $AND$ gate,an $OR$ gate,and a $NAND$ gate.
Let the inputs be $A$ and $B$.
The upper $AND$ gate receives inputs $A$ and $B$,so its output is $Y_1 = A \cdot B$.
The lower $OR$ gate receives inputs $B$ and $A$,so its output is $Y_2 = B + A$.
The final $NAND$ gate receives $Y_1$ and $Y_2$ as inputs,so the final output is $Y = \overline{Y_1 \cdot Y_2} = \overline{(A \cdot B) \cdot (A + B)}$.
Using Boolean algebra: $Y = \overline{(A \cdot B) \cdot A + (A \cdot B) \cdot B} = \overline{(A \cdot B) + (A \cdot B)} = \overline{A \cdot B}$.
This is the truth table for a $NAND$ gate:
If $A=0, B=0$,then $Y = \overline{0 \cdot 0} = 1$.
If $A=0, B=1$,then $Y = \overline{0 \cdot 1} = 1$.
If $A=1, B=0$,then $Y = \overline{1 \cdot 0} = 1$.
If $A=1, B=1$,then $Y = \overline{1 \cdot 1} = 0$.
Comparing this with the given options,option $B$ is correct.

(A) Let the inputs be $A$ and $B$.
Each input passes through a $NOT$ gate,so the inputs to the first stage $NAND$ gates become $\bar{A}$ and $\bar{B}$.
Since both inputs of the first stage $NAND$ gates are tied together,they act as $NOT$ gates. Thus,the outputs of the first stage are $\overline{\bar{A}} = A$ and $\overline{\bar{B}} = B$.
These outputs $A$ and $B$ are then fed into the final $NAND$ gate.
The output of the final $NAND$ gate is $Y = \overline{A \cdot B}$.
This is the Boolean expression for a $NAND$ gate.
Therefore,the circuit is equivalent to a $NAND$ gate.

(A) The given circuit consists of two $NOT$ gates followed by a $NOR$ gate.
Let the inputs be $A$ and $B$.
The outputs of the two $NOT$ gates are $\bar{A}$ and $\bar{B}$.
These are the inputs to the $NOR$ gate.
The output $y$ of the $NOR$ gate is given by $y = \overline{\bar{A} + \bar{B}}$.
Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
The expression $y = A \cdot B$ represents an $AND$ gate.
Therefore,the combination is equivalent to an $AND$ gate.

(C) The circuit consists of a $NOR$ gate followed by an $AND$ gate.
$1$. The output of the $NOR$ gate is $y_1 = \overline{A+B}$.
Given $A=0$ and $B=1$,we have $y_1 = \overline{0+1} = \overline{1} = 0$.
$2$. The output of the $AND$ gate is $y_2 = y_1 \cdot C$.
Given $y_1 = 0$ and $C=1$,we have $y_2 = 0 \cdot 1 = 0$.
Therefore,the values of $y_1$ and $y_2$ are $0$ and $0$ respectively.

(D) In a transistor,the three regions are the emitter,base,and collector.
$1$. The $Base$ is very thin and lightly doped to allow most charge carriers to pass through to the collector.
$2$. The $Collector$ is moderately doped and larger in size to dissipate heat.
$3$. The $Emitter$ is heavily doped to provide a large number of charge carriers.
Therefore,the order of increasing doping level is $Base < Collector < Emitter$.

(C) At absolute zero temperature $(T = 0 \ K)$,all valence electrons are tightly bound to their respective atoms in the crystal lattice.
There is no thermal energy available to excite electrons from the valence band to the conduction band.
Since there are no free charge carriers (electrons or holes) available for conduction,the intrinsic semiconductor behaves as a perfect insulator.

(B) The condition for the $n^{th}$ minima in single-slit diffraction is given by $a \sin \theta = n \lambda$,where $a$ is the slit width,$\lambda$ is the wavelength,and $n$ is the order of the minima.
For the first minima,$n = 1$,so $a \sin \theta = \lambda$.
Since the angle $\theta$ is very small,$\sin \theta \approx \tan \theta = \frac{y}{D}$,where $y$ is the distance from the central maximum and $D$ is the distance to the screen.
Thus,$y = \frac{n \lambda D}{a}$.
For the first minima $(n = 1)$,$y_1 = \frac{\lambda D}{a}$.
The distance between the first minima on either side of the central maximum is $2y_1 = \frac{2 \lambda D}{a}$.
Given: $a = 2 \ mm = 2 \times 10^{-3} \ m$,$\lambda = 500 \ nm = 500 \times 10^{-9} \ m$,$D = 1 \ m$.
Substituting the values: $2y_1 = \frac{2 \times 500 \times 10^{-9} \times 1}{2 \times 10^{-3}} = 500 \times 10^{-6} \ m = 0.5 \times 10^{-3} \ m = 0.5 \ mm$.

(C) Let the intensity of the unpolarised incident light be $I_0$.
After passing through the first polariser, the intensity becomes $I_1 = I_0 / 2$.
According to Malus' Law, the intensity of light emerging from a polariser is $I = I_{in} \cos^2 \theta$, where $\theta$ is the angle between the incident light's polarisation axis and the polariser's axis.
For the second sheet, the angle between the first and second sheet is $\theta = 30^{\circ}$. Thus, $I_2 = I_1 \cos^2(30^{\circ}) = (I_0 / 2) \times (\sqrt{3} / 2)^2 = (I_0 / 2) \times (3 / 4) = 3I_0 / 8$.
For the third sheet, the angle between the second and third sheet is also $\theta = 30^{\circ}$. Thus, $I_3 = I_2 \cos^2(30^{\circ}) = (3I_0 / 8) \times (3 / 4) = 9I_0 / 32$.
The ratio of intensities emerging from the second and third sheets is $I_2 / I_3 = (3I_0 / 8) / (9I_0 / 32) = (3 / 8) \times (32 / 9) = 4 / 3$.
Therefore, the ratio is $4: 3$.

(B) According to Rayleigh's law of scattering, the intensity of scattered light $(I)$ is inversely proportional to the fourth power of its wavelength $(\lambda)$ for particles much smaller than the wavelength of light.
Mathematically, this is expressed as $I \propto \frac{1}{\lambda^4}$.
This can be rewritten as $I \propto \lambda^{-4}$.
Comparing this with the given expression $I \propto \lambda^n$, we get $n = -4$.
Therefore, the correct option is $B$.

(D) Let $I_0$ be the intensity of the unpolarised light incident on the first polaroid.
When unpolarised light passes through the first polaroid,the intensity of the transmitted polarised light is $I_1 = \frac{I_0}{2}$.
According to Malus' Law,when this polarised light passes through the second polaroid,the intensity of the light emerging from it is $I_2 = I_1 \cos^2 \theta$,where $\theta$ is the angle between the axes of the two polaroids.
Given that $I_2 = 37.5 \% \text{ of } I_0 = 0.375 I_0$.
Substituting $I_1 = \frac{I_0}{2}$ into the equation: $0.375 I_0 = \frac{I_0}{2} \cos^2 \theta$.
$0.375 = 0.5 \cos^2 \theta$.
$\cos^2 \theta = \frac{0.375}{0.5} = 0.75 = \frac{3}{4}$.
Taking the square root on both sides: $\cos \theta = \frac{\sqrt{3}}{2}$.
Therefore,$\theta = 30^{\circ}$.

(B) The condition for the position of the $n^{\text{th}}$ bright fringe in Young's double slit experiment is given by $y_n = \frac{n \lambda D}{d}$.
For the $n^{\text{th}}$ bright fringe of red light $(\lambda_r = 7.5 \times 10^{-5} \text{ cm})$,the position is $y_r = \frac{n \lambda_r D}{d}$.
For the $(n+1)^{\text{th}}$ bright fringe of blue light $(\lambda_b = 5 \times 10^{-5} \text{ cm})$,the position is $y_b = \frac{(n+1) \lambda_b D}{d}$.
Since the fringes coincide,$y_r = y_b$,which implies $n \lambda_r = (n+1) \lambda_b$.
Substituting the given values: $n(7.5 \times 10^{-5}) = (n+1)(5 \times 10^{-5})$.
Dividing both sides by $10^{-5}$,we get $7.5n = 5(n+1)$.
$7.5n = 5n + 5$.
$2.5n = 5$.
$n = \frac{5}{2.5} = 2$.
Thus,the value of $n$ is $2$.

(D) The fringe width $\beta$ in Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the screen and the slits,and $d$ is the distance between the two slits.
Given that $\lambda' = \lambda + 0.20\lambda = 1.20\lambda$ and $d' = d - 0.25d = 0.75d$.
The new fringe width $\beta'$ is given by $\beta' = \frac{\lambda' D}{d'} = \frac{1.20\lambda D}{0.75d}$.
Substituting $\beta = \frac{\lambda D}{d}$,we get $\beta' = \frac{1.20}{0.75} \beta = 1.6 \beta$.
Given the initial fringe width $\beta = 0.3 \ mm$,the final fringe width is $\beta' = 1.6 \times 0.3 \ mm = 0.48 \ mm$.

(A) The fringe width $\beta$ in Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$.
Here,the wavelength $\lambda = 5000 \ Å = 5000 \times 10^{-10} \ m = 5 \times 10^{-7} \ m$.
The distance between the screen and the slits $D = 1 \ m$.
The distance between the slits $d = 0.2 \ cm = 0.2 \times 10^{-2} \ m = 2 \times 10^{-3} \ m$.
Substituting these values into the formula:
$\beta = \frac{5 \times 10^{-7} \times 1}{2 \times 10^{-3}}$
$\beta = 2.5 \times 10^{-4} \ m$
Converting this to millimeters:
$\beta = 2.5 \times 10^{-4} \times 10^3 \ mm = 0.25 \ mm$.
Therefore,the distance between two consecutive dark fringes is $0.25 \ mm$.

(C) The fringe width $\beta$ in Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
From the formula,we can see that the fringe width is inversely proportional to the slit separation,i.e.,$\beta \propto \frac{1}{d}$.
Let the initial slit separation be $d_1 = d$ and the initial fringe width be $\beta_1 = \beta$.
Given that the final slit separation is $d_2 = 3d$.
Therefore,the final fringe width $\beta_2$ is given by $\beta_2 = \frac{\lambda D}{3d} = \frac{\beta}{3}$.
The ratio of the initial fringe width to the final fringe width is $\frac{\beta_1}{\beta_2} = \frac{\beta}{\beta/3} = 3:1$.

(D) The formula for fringe width in Young's double slit experiment is $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance of the screen from the slits,and $d$ is the distance between the slits.
Initially,$d_1 = 2 \ mm$,$D_1 = 100 \ cm = 1000 \ mm$,and $\beta_1 = 0.36 \ mm$.
Thus,$0.36 = \frac{\lambda \times 1000}{2} \implies \lambda = \frac{0.36 \times 2}{1000} = 0.00072 \ mm$.
Now,the new distance between the slits is $d_2 = 2 \ mm - 0.5 \ mm = 1.5 \ mm$.
The new distance of the screen is $D_2 = 100 \ cm + 50 \ cm = 150 \ cm = 1500 \ mm$.
The new fringe width $\beta_2$ is given by $\beta_2 = \frac{\lambda D_2}{d_2}$.
Substituting the values: $\beta_2 = \frac{0.00072 \times 1500}{1.5} = \frac{0.00072 \times 1000}{1} = 0.72 \ mm$.

(A) Given: Power of the bulb $P = 200 \ W$,distance $r = 100 \ cm = 1 \ m$,and efficiency $\eta = 11 \% = 0.11$.
The power converted into visible radiation is $P_{vis} = \eta \times P = 0.11 \times 200 \ W = 22 \ W$.
Assuming the bulb acts as a point source,the light spreads uniformly over a sphere of radius $r$.
The intensity $I$ at a distance $r$ is given by $I = \frac{P_{vis}}{4 \pi r^2}$.
Substituting the values: $I = \frac{22}{4 \times 3.14 \times (1)^2} = \frac{22}{12.56} \approx 1.75 \ W \ m^{-2}$.

(B) The Doppler effect for light states that the fractional change in wavelength is given by $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$, where $v$ is the relative velocity of the source and $c$ is the speed of light $(3 \times 10^8 \, \text{m} \, \text{s}^{-1})$.
Given: $\lambda = 6600 \, \text{Å}$, $\Delta \lambda = 22 \, \text{Å}$.
Since the light is red-shifted, the wavelength increases, which implies the star is receding away from the earth.
Using the formula: $v = c \times \frac{\Delta \lambda}{\lambda}$.
Substituting the values: $v = (3 \times 10^8 \, \text{m} \, \text{s}^{-1}) \times \frac{22 \, \text{Å}}{6600 \, \text{Å}}$.
$v = (3 \times 10^8) \times \frac{1}{300} = 10^6 \, \text{m} \, \text{s}^{-1} = 10 \times 10^5 \, \text{m} \, \text{s}^{-1}$.
Thus, the star is receding away from the earth with a speed of $10 \times 10^5 \, \text{m} \, \text{s}^{-1}$.