AP EAMCET 2021 Physics Question Paper with Answer and Solution

(C) In the steady state,the rate of heat flow $(H)$ through both layers connected in series is the same.
Let $l$ be the thickness of each layer and $A$ be the cross-sectional area.
Let $K_A$ and $K_B$ be the thermal conductivities of layers $A$ and $B$ respectively.
Given: $K_A = 2K_B$.
The rate of heat flow is given by $H = \frac{K A \Delta T}{l}$.
Since $H$ is constant for both layers:
$H = \frac{K_A A \Delta T_A}{l} = \frac{K_B A \Delta T_B}{l}$
where $\Delta T_A$ and $\Delta T_B$ are the temperature differences across layers $A$ and $B$ respectively.
Canceling $A$ and $l$ from both sides:
$K_A \Delta T_A = K_B \Delta T_B$
Substituting $K_A = 2K_B$:
$2K_B \Delta T_A = K_B \Delta T_B \Rightarrow \Delta T_B = 2 \Delta T_A$
The total temperature difference across the wall is $\Delta T_A + \Delta T_B = 24^{\circ} C$.
Substituting $\Delta T_B = 2 \Delta T_A$ into the equation:
$\Delta T_A + 2 \Delta T_A = 24^{\circ} C$
$3 \Delta T_A = 24^{\circ} C \Rightarrow \Delta T_A = 8^{\circ} C$.
Therefore,the temperature difference across layer $B$ is:
$\Delta T_B = 2 \times 8^{\circ} C = 16^{\circ} C$.

(A) At steady state,the temperature at different parts of the rod remains constant and is independent of time. The temperature gradient,which is the ratio of the temperature difference to the distance between two points,is constant throughout the rod.
Let $T_C$ be the temperature at a point $C$ located $9 \ cm$ from end $A$.
The temperature gradient is given by:
$\frac{T_A - T_C}{x_C - x_A} = \frac{T_A - T_B}{L}$
Substituting the given values:
$\frac{100 - T_C}{9} = \frac{100 - 0}{20}$
$\frac{100 - T_C}{9} = \frac{100}{20}$
$100 - T_C = 5 \times 9$
$100 - T_C = 45$
$T_C = 100 - 45 = 55^{\circ} C$
Therefore,the temperature at the point is $55^{\circ} C$.

(C) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference between the average temperature of the body and the surrounding temperature: $\frac{dT}{dt} = -K(T_{av} - T_0)$.
For the first interval: $\Delta T_1 = 52.5^{\circ} C - 47.5^{\circ} C = 5^{\circ} C$,$t_1 = 5 \ min$,$T_{av1} = \frac{52.5 + 47.5}{2} = 50^{\circ} C$. Thus,$\frac{5}{5} = K(50 - T_0) \Rightarrow 1 = K(50 - T_0) \dots (i)$.
For the second interval: $\Delta T_2 = 47.5^{\circ} C - 42.5^{\circ} C = 5^{\circ} C$,$t_2 = 7.5 \ min$,$T_{av2} = \frac{47.5 + 42.5}{2} = 45^{\circ} C$. Thus,$\frac{5}{7.5} = K(45 - T_0) \Rightarrow \frac{2}{3} = K(45 - T_0) \dots (ii)$.
Dividing $(i)$ by $(ii)$: $\frac{1}{2/3} = \frac{K(50 - T_0)}{K(45 - T_0)} \Rightarrow \frac{3}{2} = \frac{50 - T_0}{45 - T_0}$.
Cross-multiplying: $3(45 - T_0) = 2(50 - T_0) \Rightarrow 135 - 3T_0 = 100 - 2T_0$.
Solving for $T_0$: $T_0 = 135 - 100 = 35^{\circ} C$.

(C) Newton's law of cooling is an empirical law that states that the rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings.
This law is derived from the Stefan-Boltzmann law under the approximation that the temperature difference $\Delta T = T - T_s$ is small compared to the absolute temperature $T_s$ of the surroundings.
Therefore,Newton's law of cooling holds good only when the temperature difference between the body and the surrounding is small.

(A) For the piston to fit exactly into the cylinder,the difference in thermal expansion between the piston and the cylinder must compensate for the total clearance along the diameter.
Given the clearance is $0.08 \ mm$ all around,the total clearance along the diameter is $\delta = 2 \times 0.08 \ mm = 0.16 \ mm$.
The formula for linear expansion is $\Delta L = \alpha L \Delta T$.
The difference in expansion between the piston and the cylinder is $\delta = (\alpha_p - \alpha_c) L \Delta T$.
Here,$L = 15 \ cm = 150 \ mm$,$\alpha_p = 1.6 \times 10^{-5} /^{\circ} C$,and $\alpha_c = 1.2 \times 10^{-5} /^{\circ} C$.
Substituting the values: $0.16 \ mm = (1.6 \times 10^{-5} - 1.2 \times 10^{-5}) \times 150 \ mm \times \Delta T$.
$0.16 = (0.4 \times 10^{-5}) \times 150 \times \Delta T$.
$\Delta T = \frac{0.16}{60 \times 10^{-5}} = \frac{0.16}{0.0006} = 266.67^{\circ} C \approx 268^{\circ} C$ (using provided options).
Final temperature $T = T_0 + \Delta T = 30^{\circ} C + 268^{\circ} C = 298^{\circ} C$.

(D) The change in length for the first rod is $\Delta L_1 = \alpha \Delta T L$.
The change in length for the second rod is $\Delta L_2 = (2\alpha) \Delta T (2L) = 4\alpha \Delta T L$.
Since the rods are connected end-to-end,the total change in length is $\Delta L_{net} = \Delta L_1 + \Delta L_2 = \alpha \Delta T L + 4\alpha \Delta T L = 5\alpha \Delta T L$.
The total length of the composite rod is $L_{total} = L + 2L = 3L$.
For the composite rod,$\Delta L_{net} = \alpha_{eff} \Delta T L_{total}$.
Substituting the values,$5\alpha \Delta T L = \alpha_{eff} \Delta T (3L)$.
Solving for $\alpha_{eff}$,we get $\alpha_{eff} = \frac{5\alpha}{3}$.

(D) The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$. The fractional change in time period due to temperature change $\Delta T$ is $\frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta T$. Thus, $(a) - (iv)$.
$(b)$ The ratio of the actual length to the scale reading is given by the factor $(1 + \alpha \Delta T)$ due to thermal expansion of the scale. Thus, $(b) - (iii)$.
$(c)$ For an ideal gas at constant pressure, the coefficient of volume expansion $\gamma = \frac{1}{T}$. Therefore, the reciprocal is $T$. Thus, $(c) - (ii)$.
$(d)$ From Hooke's Law, $\frac{F}{A} = Y \frac{\Delta L}{L}$, so $\frac{F}{YA} = \frac{\Delta L}{L} = \alpha \Delta T$. Thus, $(d) - (i)$.
Combining these, the correct match is $(a-iv), (b-iii), (c-ii), (d-i)$.

(B) Thermal expansion is the process where the volume of matter changes in response to a change in temperature.
When a substance is heated,its particles gain kinetic energy and move further apart,leading to an increase in volume.
Since density is defined as $\rho = \frac{m}{V}$,where $m$ is mass and $V$ is volume,an increase in volume $V$ for a constant mass $m$ results in a decrease in density $\rho$.
Therefore,expansion during heating decreases the density of the material.

(C) The fractional change in volume is given by $\frac{\Delta V}{V} = 0.12 \% = \frac{0.12}{100} = 1.2 \times 10^{-3}$.
Given the temperature change $\Delta T = 20^{\circ} C$.
The formula for volumetric expansion is $\frac{\Delta V}{V} = \gamma \Delta T$,where $\gamma$ is the coefficient of volumetric expansion.
Substituting the values: $1.2 \times 10^{-3} = \gamma \times 20$.
$\gamma = \frac{1.2 \times 10^{-3}}{20} = 0.06 \times 10^{-3} = 6 \times 10^{-5} {}^{\circ} C^{-1}$.
The relationship between the coefficient of volumetric expansion $\gamma$ and the coefficient of linear expansion $\alpha$ is $\gamma = 3\alpha$.
Therefore,$\alpha = \frac{\gamma}{3} = \frac{6 \times 10^{-5}}{3} = 2 \times 10^{-5} {}^{\circ} C^{-1}$.

(A) The volume of the liquid that overflows is equal to the difference between the expansion of the liquid and the expansion of the glass vessel.
Increase in volume of liquid,$\Delta V_L = V_0 \gamma_L \Delta T$.
Increase in volume of the glass vessel,$\Delta V_g = V_0 \gamma_g \Delta T$.
Since the coefficient of volume expansion $\gamma_g = 3\alpha_g$,we have $\Delta V_g = V_0 (3\alpha_g) \Delta T$.
The volume of liquid that overflows is $\Delta V_{overflow} = \Delta V_L - \Delta V_g$.
$\Delta V_{overflow} = V_0 \gamma_L \Delta T - V_0 (3\alpha_g) \Delta T$.
$\Delta V_{overflow} = V_0 \Delta T (\gamma_L - 3\alpha_g)$.

(D) The change in length of a material due to thermal expansion is given by $\Delta l = l_0 \alpha \Delta T$,where $l_0$ is the initial length,$\alpha$ is the coefficient of linear expansion,and $\Delta T$ is the change in temperature.
Since $\alpha_{Al} > \alpha_{\text{steel}}$,for the same change in temperature $\Delta T$,the aluminium part will expand more than the steel part $(\Delta l_{Al} > \Delta l_{\text{steel}})$.
Because the aluminium expands more,it will form the outer (convex) side of the curve,while the steel,which expands less,will form the inner (concave) side of the curve.
Therefore,the strip will bend with steel on the concave side.

(C) Given: Length of silver rod,$L = 1 \ m = 100 \ cm$.
Initial temperature,$T_1 = 0^{\circ}C$.
Final temperature,$T_2 = 100^{\circ}C$.
Change in temperature,$\Delta T = T_2 - T_1 = 100^{\circ}C$.
Increase in length,$\Delta L = 0.19 \ cm$.
The formula for linear expansion is $\Delta L = L \alpha \Delta T$,where $\alpha$ is the coefficient of linear expansion.
Substituting the values: $0.19 = 100 \times \alpha \times 100$.
$\alpha = \frac{0.19}{10000} = 0.19 \times 10^{-4} \ ^{\circ}C^{-1} = 1.9 \times 10^{-5} \ ^{\circ}C^{-1}$.
The coefficient of volume expansion $\gamma$ is related to $\alpha$ by $\gamma = 3\alpha$.
$\gamma = 3 \times 1.9 \times 10^{-5} \ ^{\circ}C^{-1} = 5.7 \times 10^{-5} \ ^{\circ}C^{-1}$.

(A) Water exhibits anomalous expansion. Its density is maximum at $4^{\circ} C$.
When the surface water cools to $0^{\circ} C$ and begins to freeze,the denser water at $4^{\circ} C$ settles at the bottom of the lake.
Therefore,the temperature at the bottom of the lake remains $4^{\circ} C$.

(B) The relationship between temperature in Celsius $(C)$ and Fahrenheit $(F)$ is given by the formula: $\frac{C}{5} = \frac{F - 32}{9}$.
Rearranging the formula to solve for $F$,we get: $F = C \times \frac{9}{5} + 32$.
Substituting the value $C = -183^{\circ} C$ into the equation:
$F = -183 \times \frac{9}{5} + 32$
$F = -36.6 \times 9 + 32$
$F = -329.4 + 32$
$F = -297.4^{\circ} F$.
Rounding to the nearest integer,we get $-297^{\circ} F$.

(C) Internal energy for a thermodynamic process is a state function,meaning it does not depend upon the path taken; it only depends upon the initial and final states of the system.
Therefore,if two different processes have the same initial and final states,the change in internal energy $\Delta U$ will be the same for both processes.
According to the first law of thermodynamics: $Q = W + \Delta U$,which can be rearranged as $\Delta U = Q - W$.
For path $1$: $\Delta U = Q_1 - W_1$.
For path $2$: $\Delta U = Q_2 - W_2$.
Since $\Delta U$ is the same for both paths,we have: $Q_1 - W_1 = Q_2 - W_2$.

(A) For an ideal gas,the intermolecular forces are assumed to be negligible,and collisions are perfectly elastic.
Consequently,the internal energy $U$ of an ideal gas is solely a function of its absolute temperature $T$.
According to the kinetic theory of gases,the internal energy of an ideal gas is given by $U = \frac{f}{2} nRT$,where $f$ is the degrees of freedom,$n$ is the number of moles,$R$ is the universal gas constant,and $T$ is the temperature in Kelvin.
Since $U \propto T$,the relationship between internal energy $U$ and temperature $T$ is linear,passing through the origin $(0, 0)$.
Therefore,the graph that best illustrates this relationship is a straight line passing through the origin,which corresponds to Graph $A$.

(A) According to the first law of thermodynamics,$\Delta U = Q - W$,which implies $Q - W = \Delta U$.
Since internal energy $(U)$ is a state function,the change in internal energy $(\Delta U)$ depends only on the initial and final states.
For all paths $(A, B, C, D)$,the initial state is $1$ and the final state is $2$.
Therefore,the change in internal energy is the same for all paths: $\Delta U_A = \Delta U_B = \Delta U_C = \Delta U_D$.
Since $Q - W = \Delta U$,it follows that $Q_A - W_A = Q_B - W_B = Q_C - W_C = Q_D - W_D$.
Thus,the expression $Q - W$ is constant for all paths connecting the same two states.
Comparing this with the given options,option $A$ is correct because $Q_A - W_A = \Delta U$ and $Q_D - W_D = \Delta U$,so $Q_A - W_A = Q_D - W_D$.

(A) In the given process,for an ideal gas,$\Delta W = 0$ and $\Delta Q < 0$.
According to the first law of thermodynamics,$\Delta Q = \Delta W + \Delta U$,where $\Delta U$ is the change in internal energy of the gas.
Since $\Delta W = 0$,we have $\Delta Q = \Delta U$.
Given that $\Delta Q < 0$,it follows that $\Delta U < 0$.
For an ideal gas,the internal energy $U$ is a function of temperature $T$ only $(U = nC_vT)$.
Therefore,a decrease in internal energy $(\Delta U < 0)$ implies a decrease in the temperature of the gas.

(A) For an isobaric process,the heat supplied is $\Delta Q = C_p \Delta T$ and the change in internal energy is $\Delta U = C_V \Delta T$.
Thus,$\frac{\Delta Q}{\Delta U} = \frac{C_p \Delta T}{C_V \Delta T} = \frac{C_p}{C_V} = \gamma$. So,$A \rightarrow 4$.
For an isobaric process,$\Delta Q = \Delta U + \Delta W$,so $\Delta W = \Delta Q - \Delta U = (C_p - C_V) \Delta T$.
Thus,$\frac{\Delta Q}{\Delta W} = \frac{C_p \Delta T}{(C_p - C_V) \Delta T} = \frac{C_p}{C_p - C_V} = \frac{C_p/C_V}{(C_p/C_V) - 1} = \frac{\gamma}{\gamma - 1}$. So,$B \rightarrow 3$.
For a refrigerator,the coefficient of performance $\beta = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2} = \frac{T_2}{T_1 - T_2}$. So,$C \rightarrow 2$.
For a heat pump,the coefficient of performance $\alpha = \frac{Q_1}{W} = \frac{Q_1}{Q_1 - Q_2} = \frac{T_1}{T_1 - T_2}$. So,$D \rightarrow 1$.
Therefore,the correct match is $A \rightarrow 4, B \rightarrow 3, C \rightarrow 2, D \rightarrow 1$.

(C) The coefficient of performance $(\beta)$ of a refrigerator is defined as $\beta = \frac{Q_2}{W}$, where $Q_2$ is the heat extracted from the cold reservoir and $W$ is the work done on the system.
Given $\beta = 0.25 = \frac{1}{4}$, so $Q_2 = \frac{W}{4}$.
The heat released to the hot reservoir is $Q_1 = Q_2 + W$.
Substituting $Q_2 = \frac{W}{4}$, we get $Q_1 = \frac{W}{4} + W = \frac{5W}{4}$.
Given $Q_1 = 250 \, J$, we have $250 = \frac{5W}{4}$.
Solving for $W$, $W = 250 \times \frac{4}{5} = 200 \, J$.

(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}$.
Given, sink temperature $T_{\text{sink}} = 27^{\circ} C = 27 + 273 = 300 K$.
For initial efficiency $\eta_1 = 40 \% = 0.4$:
$0.4 = 1 - \frac{300}{T_{\text{source},1}}$
$\frac{300}{T_{\text{source},1}} = 0.6 \Rightarrow T_{\text{source},1} = \frac{300}{0.6} = 500 K$.
For final efficiency $\eta_2 = 60 \% = 0.6$:
$0.6 = 1 - \frac{300}{T_{\text{source},2}}$
$\frac{300}{T_{\text{source},2}} = 0.4 \Rightarrow T_{\text{source},2} = \frac{300}{0.4} = 750 K$.
The change in source temperature is $\Delta T = T_{\text{source},2} - T_{\text{source},1} = 750 K - 500 K = 250 K$.

(C) For a Carnot engine,efficiency $\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}$.
Given that the three engines are equally efficient,let $\eta_1 = \eta_2 = \eta_3 = \eta$.
For the first engine: $\eta = 1 - \frac{T_2}{T_1} \Rightarrow \frac{T_2}{T_1} = 1 - \eta$.
For the second engine: $\eta = 1 - \frac{T_3}{T_2} \Rightarrow \frac{T_3}{T_2} = 1 - \eta$.
For the third engine: $\eta = 1 - \frac{T_4}{T_3} \Rightarrow \frac{T_4}{T_3} = 1 - \eta$.
Since $1 - \eta$ is constant,we have $\frac{T_2}{T_1} = \frac{T_3}{T_2} = \frac{T_4}{T_3} = k$,where $k = 1 - \eta$.
From this,$T_2 = T_1 k$,$T_3 = T_2 k = T_1 k^2$,and $T_4 = T_3 k = T_1 k^3$.
Thus,$k^3 = \frac{T_4}{T_1} \Rightarrow k = \left(\frac{T_4}{T_1}\right)^{1/3}$.
Substituting $k$ back:
$T_2 = T_1 \left(\frac{T_4}{T_1}\right)^{1/3} = T_1^{2/3} T_4^{1/3} = (T_1^2 T_4)^{1/3}$.
$T_3 = T_1 \left(\frac{T_4}{T_1}\right)^{2/3} = T_1^{1/3} T_4^{2/3} = (T_1 T_4^2)^{1/3}$.
Therefore,the correct option is $C$.

(B) The efficiency of a Carnot heat engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$.
Here,$T_1$ is the temperature of the source (hot reservoir) and $T_2$ is the temperature of the sink (cold reservoir).
From this expression,it is clear that the efficiency depends solely on the temperatures of the source and the sink.
It is independent of the nature of the working substance used in the engine.
Therefore,option $B$ is the correct statement.

(B) 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: $\eta_1 = 0.4$,$T_1 = 500 \ K$.
$0.4 = 1 - \frac{T_2}{500} \implies \frac{T_2}{500} = 0.6 \implies T_2 = 300 \ K$.
For the second case: $\eta_2 = 0.6$,$T_2 = 300 \ K$,and we need to find the new source temperature $T_1'$.
$0.6 = 1 - \frac{300}{T_1'} \implies \frac{300}{T_1'} = 0.4$.
$T_1' = \frac{300}{0.4} = 750 \ K$.

(C) The efficiency of a Carnot engine is given by $\eta = 1/5$.
When this engine is used as a refrigerator,its coefficient of performance $\beta$ is related to the efficiency $\eta$ by the formula: $\beta = \frac{1 - \eta}{\eta}$.
Substituting the value of $\eta$: $\beta = \frac{1 - 1/5}{1/5} = \frac{4/5}{1/5} = 4$.
The coefficient of performance is also defined as the ratio of heat absorbed from the cold reservoir $(Q)$ to the work done on the system $(W)$: $\beta = \frac{Q}{W}$.
Given $W = 50 \ J$,we have $4 = \frac{Q}{50}$.
Therefore,$Q = 4 \times 50 = 200 \ J$.

(C) Given, temperatures are $T_1 = 500 \ K$ and $T_2 = 300 \ K$.
The maximum theoretical efficiency (Carnot efficiency) of an engine operating between these temperatures is given by $\eta_{max} = 1 - \frac{T_2}{T_1}$.
$\eta_{max} = 1 - \frac{300}{500} = 1 - 0.6 = 0.4$ or $40\%$.
Now, calculate the efficiency $(\eta = \frac{W}{Q_{in}})$ for each design:
For design $A$: $\eta_A = \frac{150}{1000} = 0.15$ $(15\%)$.
For design $B$: $\eta_B = \frac{450}{1000} = 0.45$ $(45\%)$.
For design $C$: $\eta_C = \frac{300}{1000} = 0.30$ $(30\%)$.
According to the Second Law of Thermodynamics, no engine can have an efficiency greater than the Carnot efficiency. Since $\eta_B = 0.45 > 0.4$, design $B$ violates the Second Law of Thermodynamics and is impossible. Design $C$ has an efficiency of $0.3$, which is less than $0.4$, making it physically possible. Therefore, design $C$ is the most suitable choice.

(D) When a compression process is performed suddenly or instantly,there is not enough time for heat transfer; therefore,the process is an adiabatic compression process.
For an adiabatic process,the relation between pressure and volume is given by $P V^\gamma = \text{constant}$.
Thus,$P_i V_i^\gamma = P_f V_f^\gamma$.
Rearranging for the ratio of final pressure to initial pressure: $\frac{P_f}{P_i} = \left( \frac{V_i}{V_f} \right)^\gamma$.
Given $V_f = \frac{V_i}{4}$ and $\gamma = 1.5 = \frac{3}{2}$,we substitute these values:
$\frac{P_f}{P_i} = \left( \frac{V_i}{V_i / 4} \right)^{1.5} = (4)^{1.5} = (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.
Therefore,the ratio of final pressure to initial pressure is $8:1$.

(C) The internal energy of a monoatomic ideal gas is $U = \frac{3}{2} n R T$,so $U \propto T$.
Density $\rho = \frac{m}{V}$,so $\rho \propto \frac{1}{V}$.
For process $AB$,the line passes through origin in $U$-$\rho$ plot,so $U = k \rho$. Substituting $U \propto T$ and $\rho \propto 1/V$,we get $T \propto 1/V$,or $TV = \text{constant}$.
Using $PV = nRT$,we have $(PV/nR)V = \text{constant}$,so $PV^2 = \text{constant}$. This is a polytropic process with index $m = 2$.
The molar heat capacity is $C = C_V + \frac{R}{1-m} = \frac{3R}{2} + \frac{R}{1-2} = \frac{3R}{2} - R = \frac{R}{2}$. Thus,option $A$ is correct.
In process $BC$,$\rho$ is constant,so it is an isochoric process. Since $U$ decreases,temperature decreases,so heat is rejected. Thus,option $B$ is correct.
For an isochoric process,the molar heat capacity is $C_V = \frac{3}{2} R$. Option $C$ states it is $\frac{2R}{3}$,which is incorrect.
In process $CA$,$U$ is constant,so $T$ is constant (isothermal). $W_{CA} = nRT \ln(V_A/V_C) = nRT \ln(\rho_C/\rho_A) = (\frac{2}{3} U_0) \ln(4\rho_0/\rho_0) = \frac{2}{3} U_0 \ln 4$. Option $D$ is correct.

(B) In an isothermal process, the temperature $T$ remains constant throughout the process.
According to the ideal gas equation, $PV = nRT$.
Since $T$ is constant, $PV = \text{constant}$.
This implies that $P = \frac{\text{constant}}{V}$, which represents a rectangular hyperbola in a $P-V$ graph.
Therefore, the relationship between pressure $(P)$ and volume $(V)$ is non-linear.
In contrast, the relationship between $P$ and $T$, $V$ and $T$, or $PV$ and $T$ would be linear (or constant) in an isothermal process.

(A) Initial state: $V_A = V_B = V$,$p_A = p_B = p$.
For container $A$ (isothermal process): $p_A V_A = p_A^{\prime} V_A^{\prime}$.
Given $V_A^{\prime} = V/2$,so $p V = p_A^{\prime} (V/2) \Rightarrow p_A^{\prime} = 2p$.
For container $B$ (adiabatic process): $p_B V_B^{\gamma} = p_B^{\prime} (V_B^{\prime})^{\gamma}$.
Given $V_B^{\prime} = V/2$,so $p V^{\gamma} = p_B^{\prime} (V/2)^{\gamma} \Rightarrow p_B^{\prime} = p \cdot 2^{\gamma}$.
The ratio of final pressures is $\frac{p_B^{\prime}}{p_A^{\prime}} = \frac{p \cdot 2^{\gamma}}{2p} = 2^{\gamma-1}$.

(B) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Given that the gas releases heat,$\Delta Q = -30 \,J$.
Work is done on the gas,so $\Delta W = -18 \,J$.
Initial internal energy $U_i = 60 \,J$.
Let the final internal energy be $U_f$.
The change in internal energy is $\Delta U = U_f - U_i$.
Substituting these values into the first law equation:
$-30 = (U_f - 60) + (-18)$
$-30 = U_f - 78$
$U_f = 78 - 30 = 48 \,J$.
Therefore,the final internal energy of the gas is $48 \,J$.

(B) The work done in a thermodynamic process is equal to the area under the $p-V$ curve.
For a cyclic process,the net work done is equal to the area enclosed by the cycle.
If the cycle is traversed in a clockwise direction,the net work done is positive.
If the cycle is traversed in a counter-clockwise (anticlockwise) direction,the net work done is negative.
In the given figure,the path is $C \rightarrow B \rightarrow A \rightarrow D \rightarrow C$,which is counter-clockwise.
Therefore,the net work done by the gas is negative.

(D) In an adiabatic process,the system is thermally isolated from its surroundings,meaning no heat exchange $(dQ = 0)$ occurs between the system and the environment.
To maintain this condition,the container must be a perfect thermal insulator.
$A$ thermos flask is designed specifically to minimize heat transfer through conduction,convection,and radiation,making it the best choice among the given options for an adiabatic process.

(A) According to the isothermal process,the temperature $T$ remains constant.
For an ideal gas,the equation of state is $pV = nRT$.
Since $T$ is constant,$n$ and $R$ are also constant,the product $pV$ must be constant,i.e.,$pV = K$.
This implies $p \propto \frac{1}{V}$.
In an isothermal compression,the volume $V$ of the gas decreases.
Since pressure $p$ is inversely proportional to volume $V$,as the volume $V$ decreases,the pressure $p$ must increase.

(B) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta Q$ is the heat supplied to the system,$\Delta U$ is the change in internal energy,and $\Delta W$ is the work done by the system.
Given that the work done by the system equals the decrease in its internal energy,we have $\Delta W = -\Delta U$,which implies $\Delta U + \Delta W = 0$.
Substituting this into the first law equation,we get $\Delta Q = 0$.
$A$ process in which there is no exchange of heat $(\Delta Q = 0)$ with the surroundings is defined as an adiabatic process.
Therefore,the system must have undergone an adiabatic change.

(B) For an isochoric process,the volume remains constant,so $\Delta V = 0$.
Since the work done in a thermodynamic process is given by $W = p \cdot \Delta V$,substituting $\Delta V = 0$ gives $W = 0$.

(A) An isothermal process occurs at a constant temperature.
For the temperature to remain constant during heat exchange,the process must occur very slowly to allow sufficient time for thermal equilibrium to be maintained with the surroundings.
Therefore,an isothermal process is generally considered a slow process.

(A) The speed of ripples $v$ depends on surface tension $\sigma$,density $\rho$,and wavelength $\lambda$.
We can write the relation as $v = k \sigma^a \rho^b \lambda^c$,where $k$ is a dimensionless constant.
The dimensional formula for speed $v$ is $[M^0 L T^{-1}]$.
The dimensional formula for surface tension $\sigma$ is $[M L^0 T^{-2}]$.
The dimensional formula for density $\rho$ is $[M L^{-3} T^0]$.
The dimensional formula for wavelength $\lambda$ is $[M^0 L T^0]$.
Substituting these into the equation:
$[M^0 L T^{-1}] = [M L^0 T^{-2}]^a [M L^{-3} T^0]^b [L]^c$
$[M^0 L T^{-1}] = [M]^{a+b} [L]^{-3b+c} [T]^{-2a}$
Equating the powers of $M, L,$ and $T$ on both sides:
$a + b = 0$
$-3b + c = 1$
$-2a = -1$
From $-2a = -1$,we get $a = 1/2$.
Substituting $a = 1/2$ into $a + b = 0$,we get $b = -1/2$.
Substituting $b = -1/2$ into $-3b + c = 1$,we get $-3(-1/2) + c = 1 \Rightarrow 1.5 + c = 1 \Rightarrow c = -0.5 = -1/2$.
Thus,$v \propto \sigma^{1/2} \rho^{-1/2} \lambda^{-1/2}$.
Therefore,$v \propto \sqrt{\frac{\sigma}{\rho \lambda}}$.
Squaring both sides,$v^2 \propto \frac{\sigma}{\rho \lambda}$.

(A) Stress is defined as the restoring force per unit area of the material. It is denoted by the Greek letter $\sigma$.
$\text{Stress} = \frac{\text{Force}}{\text{Area}}$
Substituting the dimensional formulas for force and area:
$\text{Force} = [M L T^{-2}]$
$\text{Area} = [L^2]$
$\text{Stress} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{1-2} T^{-2}] = [M L^{-1} T^{-2}]$

(C) According to Poiseuille's law,the rate of flow of a liquid $(Q)$ through a capillary tube is given by the formula:
$Q = \frac{V}{t} = \frac{\pi p R^4}{8 \eta L}$
Where:
$V$ is the volume of the liquid,
$t$ is the time of flow,
$p$ is the pressure difference,
$R$ is the radius of the tube,
$\eta$ is the coefficient of viscosity,
$L$ is the length of the tube.
Rearranging the formula to solve for the volume $V$:
$V = \frac{\pi p R^4 t}{8 \eta L}$
Since $\frac{\pi}{8}$ is a dimensionless constant,the volume $V$ is proportional to the remaining terms:
$V \propto \frac{p R^4 t}{\eta L}$
Therefore,the correct option is $C$.

(B) The dimensional formulas for the given physical quantities are as follows:
$[$Force$] = [MLT^{-2}]$
$[$Surface tension$] = [MT^{-2}]$
$[$Frequency$] = [T^{-1}]$
$[$Velocity gradient$] = [T^{-1}]$
$[$Angular speed$] = [T^{-1}]$
$[$Solid angle$] = [M^0L^0T^0]$ (Dimensionless)
$[$Stefan's constant$] = [MT^{-3}K^{-4}]$
$[$Planck's constant$] = [ML^2T^{-1}]$
Comparing these,we observe that frequency and velocity gradient both have the dimension $[T^{-1}]$.
Therefore,the correct pair is frequency and velocity gradient.

(B) Relative velocity $(v_A \pm v_B)$ for two bodies $A$ and $B$ has both units $(ms^{-1})$ and dimensions $[LT^{-1}]$.
Relative density of two substances $(\rho = \frac{\rho_A}{\rho_B})$ has no units and no dimensions because it is a ratio of two similar physical quantities.
Angle is measured in radian. Thus,it has units but no dimensions.
Energy is measured in joule and has dimensions $[ML^2 T^{-2}]$.
Hence,only relative density has neither units nor dimensions.

(B) By definition, $1 \text{ } \mathring{A} = 10^{-10} \text{ m}$. To convert this into millimeters $(\text{mm})$, we know that $1 \text{ m} = 10^3 \text{ mm}$. Therefore, $1 \text{ } \mathring{A} = 10^{-10} \times 10^3 \text{ mm}$. $1 \text{ } \mathring{A} = 10^{-7} \text{ mm}$. Thus, the correct option is $B$.

(D) The resultant amplitude of waves at the point $P$ is given by the formula: $A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi}$,where $\phi$ is the phase difference,and $A_1$ and $A_2$ are the amplitudes of the sound waves.
First,we calculate the wavelength $\lambda$ of the sound wave: $\lambda = \frac{v}{f} = \frac{350 m s^{-1}}{350 Hz} = 1 m = 100 cm$.
The path difference is given as $\Delta x = AP - BP = 25 cm$.
The phase difference $\phi$ is calculated as: $\phi = \frac{2 \pi}{\lambda} \Delta x = \frac{2 \pi}{100 cm} \times 25 cm = \frac{\pi}{2}$.
Now,substitute the values into the resultant amplitude formula:
$A = \sqrt{0.3^2 + 0.4^2 + 2 \times 0.3 \times 0.4 \cos(\frac{\pi}{2})}$
Since $\cos(\frac{\pi}{2}) = 0$,the expression simplifies to:
$A = \sqrt{0.3^2 + 0.4^2} = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 mm$.

(A) The speed of waves in different media is determined by the elastic properties and density of the medium.
$(a)$ For a transverse wave in a solid (like a steel rod), the speed is given by $v = \sqrt{\frac{\eta}{\rho}}$, where $\eta$ is the shear modulus and $\rho$ is the density. Thus, $(a) - (ii)$.
$(b)$ For longitudinal waves in a bulk solid (like the earth's crust), the speed depends on both bulk modulus $B$ and shear modulus $\eta$: $v = \sqrt{\frac{B + \frac{4}{3}\eta}{\rho}}$. Thus, $(b) - (i)$.
$(c)$ For longitudinal waves in a thin rod, the speed is given by $v = \sqrt{\frac{Y}{\rho}}$, where $Y$ is Young's modulus. Thus, $(c) - (iv)$.
$(d)$ Ripples are surface waves on liquids where surface tension $T$ is the restoring force. The speed is given by $v = \sqrt{\frac{2 \pi T}{g \lambda}}$. Thus, $(d) - (iii)$.
Therefore, the correct matching is $(a) - (ii), (b) - (i), (c) - (iv), (d) - (iii)$.

(B) Given the equations for the two waves are $x_1 = A \sin \left(\omega t + \frac{\pi}{6}\right)$ and $x_2 = A \cos \omega t$.
To find the phase difference,we express both waves in terms of the same trigonometric function (sine or cosine).
We know that $\cos \theta = \sin \left(\theta + \frac{\pi}{2}\right)$.
Therefore,$x_2 = A \sin \left(\omega t + \frac{\pi}{2}\right)$.
Now,the phase of $x_1$ is $\phi_1 = \omega t + \frac{\pi}{6}$ and the phase of $x_2$ is $\phi_2 = \omega t + \frac{\pi}{2}$.
The phase difference $\Delta \phi = \phi_2 - \phi_1 = \left(\omega t + \frac{\pi}{2}\right) - \left(\omega t + \frac{\pi}{6}\right)$.
$\Delta \phi = \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$.

(B) Given that, the amplitudes of the two waves are $a_1 = a_2 = 10 \,mm$.
The phase difference between the waves is $\phi = 90^{\circ}$.
The resultant amplitude $A$ of two interfering waves is given by the formula:
$A = \sqrt{a_1^2 + a_2^2 + 2 a_1 a_2 \cos \phi}$
Substituting the given values into the formula:
$A = \sqrt{(10)^2 + (10)^2 + 2(10)(10) \cos 90^{\circ}}$
Since $\cos 90^{\circ} = 0$, the expression simplifies to:
$A = \sqrt{100 + 100 + 0}$
$A = \sqrt{200}$
$A = 10 \sqrt{2} \,mm$

(D) Beats are a phenomenon that occurs due to the superposition of two sound waves.
When two sound waves of equal amplitude and slightly different frequencies travel in the same direction,they interfere with each other.
This interference results in a periodic variation in the intensity of the resultant sound,which is perceived as beats.
Therefore,the formation of beats is a direct consequence of the interference of sound waves.

(A) In a stationary wave formed on a string fixed at both ends,the points where the displacement is always zero are called nodes. The points where the amplitude of vibration is maximum are called antinodes.
For a string vibrating in $n$ loops,the number of nodes is $n + 1$ and the number of antinodes is $n$.
Given that the string vibrates in $5$ loops,we have $n = 5$.
Therefore,the number of nodes = $5 + 1 = 6$.
The number of antinodes = $5$.
Thus,the total number of nodes and antinodes are $6$ and $5$ respectively.

(B) For oxygen $(O_2)$:
Molar mass,$M_1 = 32 \,g/mol$.
Heat capacity ratio,$\gamma_1 = C_p / C_V = 7/5$ (for diatomic gas).
Speed of sound,$v_1 = 460 \,ms^{-1}$.
For helium $(He)$:
Molar mass,$M_2 = 4 \,g/mol$.
Heat capacity ratio,$\gamma_2 = C_p / C_V = 5/3$ (for monoatomic gas).
The speed of sound in an ideal gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$.
Taking the ratio of speeds: $\frac{v_1}{v_2} = \sqrt{\frac{\gamma_1 / M_1}{\gamma_2 / M_2}} = \sqrt{\frac{\gamma_1}{\gamma_2} \cdot \frac{M_2}{M_1}}$.
Substituting the values: $\frac{460}{v_2} = \sqrt{\frac{7/5}{5/3} \cdot \frac{4}{32}} = \sqrt{\frac{21}{25} \cdot \frac{1}{8}} = \sqrt{\frac{21}{200}} \approx 0.324$.
Wait,re-calculating: $\frac{460}{v_2} = \sqrt{\frac{7}{5} \cdot \frac{3}{5} \cdot \frac{4}{32}} = \sqrt{\frac{21}{25} \cdot \frac{1}{8}} = \sqrt{\frac{21}{200}} \approx 0.324$.
Actually,$\frac{v_2}{v_1} = \sqrt{\frac{\gamma_2}{\gamma_1} \cdot \frac{M_1}{M_2}} = \sqrt{\frac{5/3}{7/5} \cdot \frac{32}{4}} = \sqrt{\frac{25}{21} \cdot 8} = \sqrt{\frac{200}{21}} \approx \sqrt{9.52} \approx 3.085$.
$v_2 = 460 \times 3.085 \approx 1419.1 \,ms^{-1} \approx 1420 \,ms^{-1}$.

(A) wave front is defined as the locus of all points in a medium that are vibrating in the same phase at a given instant of time.
Since all points on the wave front are in the same state of vibration,the phase difference between any two points on the wave front is zero.
Therefore,all points on a wave front are in the same phase.

(D) The formula for fringe width in Young's double slit experiment is $\beta = \frac{D \lambda}{d}$,where $D$ is the distance between the screen and the slits,$\lambda$ is the wavelength of light,and $d$ is the separation between the slits.
Given that the new distance $D_2 = 2D_1$ and the new slit separation $d_2 = \frac{d_1}{2}$.
The new fringe width $\beta_2$ is given by $\beta_2 = \frac{D_2 \lambda}{d_2}$.
Substituting the values,we get $\beta_2 = \frac{(2D_1) \lambda}{(d_1 / 2)} = 4 \times \frac{D_1 \lambda}{d_1} = 4 \beta_1$.
Therefore,the fringe width is quadrupled.

(C) Given,the ratio of amplitudes of two interfering waves is $4: 3$.
Let $A_1$ and $A_2$ be the amplitudes of the two waves,so $\frac{A_1}{A_2} = \frac{4}{3}$.
We know that the intensity $I$ of a wave is proportional to the square of its amplitude,$I \propto A^2$.
The ratio of maximum intensity to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2$.
Dividing the numerator and denominator inside the bracket by $A_2$,we get $\frac{I_{\max}}{I_{\min}} = \left( \frac{\frac{A_1}{A_2} + 1}{\frac{A_1}{A_2} - 1} \right)^2$.
Substituting the value $\frac{A_1}{A_2} = \frac{4}{3}$,we get $\frac{I_{\max}}{I_{\min}} = \left( \frac{\frac{4}{3} + 1}{\frac{4}{3} - 1} \right)^2 = \left( \frac{\frac{7}{3}}{\frac{1}{3}} \right)^2 = (7)^2 = \frac{49}{1}$.
Thus,the ratio of maximum and minimum intensity is $49: 1$.

(D) Given the ratio of maximum intensity to minimum intensity is $\frac{I_{\max}}{I_{\min}} = \frac{36}{1}$.
We know that $I_{\max} = (a_1 + a_2)^2$ and $I_{\min} = (a_1 - a_2)^2$,where $a_1$ and $a_2$ are the amplitudes of the two waves.
Therefore,$\frac{(a_1 + a_2)^2}{(a_1 - a_2)^2} = \frac{36}{1}$.
Taking the square root on both sides,we get $\frac{a_1 + a_2}{a_1 - a_2} = \frac{6}{1}$.
By cross-multiplying,$a_1 + a_2 = 6(a_1 - a_2) = 6a_1 - 6a_2$.
Rearranging the terms,$5a_1 = 7a_2$.
Thus,the ratio of the amplitudes is $\frac{a_1}{a_2} = \frac{7}{5}$.

(A) The formula for fringe width in a Young's double slit experiment is $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance from the slits to the screen,and $d$ is the distance between the slits.
Let the initial fringe width be $\beta_1 = \frac{\lambda D_1}{d_1}$.
According to the problem,the new distance between slits is $d_2 = 10 d_1$ and the new distance from the screen is $D_2 = \frac{D_1}{2}$.
The new fringe width is $\beta_2 = \frac{\lambda D_2}{d_2} = \frac{\lambda (D_1 / 2)}{10 d_1} = \frac{1}{20} \left( \frac{\lambda D_1}{d_1} \right)$.
Therefore,$\beta_2 = \frac{1}{20} \beta_1$.