Welcome to Lecture 109 (Master Index Entry: L103) of our comprehensive JEE Physics Master Series! This milestone lecture delivers a grand, unified Comprehensive Revision of Units 7, 8, and 9: Properties of Matter, Thermodynamics, and Kinetic Theory of Gases. Spanning elasticity, hydrostatics, fluid dynamics, surface tension, viscosity, thermal expansion, calorimetry, heat transfer, the laws of thermodynamics, indicator cycles, molecular speeds, degrees of freedom, equipartition of energy, and mean free path, this master module bridges mechanical and thermal physics. We synthesize key governing relations, highlight tricky JEE edge cases, solve 4 multi-concept graded problems connecting cross-unit phenomena, and present a 10-question full-spectrum practice worksheet with complete solutions.
Detailed Concept Card: Properties of Matter & Thermal Physics Master Blueprint
1. Unit 7: Properties of Solids and Liquids (Elasticity & Fluids)
- Hooke’s Law & Moduli of Elasticity: $$\text{Stress} = Y \times \text{Strain} \implies \frac{F}{A} = Y \frac{\Delta L}{L}$$ – Bulk Modulus: $B = -V \frac{\Delta P}{\Delta V} = \frac{1}{\kappa}$ (where $\kappa$ is compressibility). – Shear Modulus: $G = \frac{F/A}{\theta}$. – Poisson’s Ratio: $\sigma = -\frac{\Delta d / d}{\Delta L / L}$ (theoretical range: $-1 \le \sigma \le 0.5$; real materials: $0 < \sigma < 0.5$). - Elastic Strain Energy Density: $u = \frac{1}{2} \times \text{stress} \times \text{strain} = \frac{1}{2} Y (\text{strain})^2 = \frac{\sigma^2}{2Y}$.
- Fluid Statics & Hydrodynamics: – Pressure variation in accelerated fluids: $P_2 – P_1 = \rho g_{\text{eff}} h$; free surface tilt: $\tan\theta = a_x / g$. – Archimedes’ Principle: $F_B = V_{\text{sub}} \rho_{\text{fluid}} g$. – Equation of Continuity: $A_1 v_1 = A_2 v_2 = \text{constant}$ (incompressible flow). – Bernoulli’s Equation: $P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant}$. – Torricelli’s Law: Efflux velocity $v = \sqrt{2gh}$; range $R = 2\sqrt{h(H – h)}$; maximum range $R_{\text{max}} = H$ when $h = H/2$.
- Viscosity & Surface Tension: – Stokes’ Viscous Drag: $F_v = 6 \pi \eta r v$. – Terminal Velocity: $v_t = \frac{2}{9} \frac{r^2 (\rho – \sigma) g}{\eta}$. – Poiseuille’s Volume Flow Rate: $Q = \frac{\pi \Delta P r^4}{8 \eta L}$. – Surface Energy & Tension: $\Delta U = S \Delta A$. – Excess Pressure: Inside spherical droplet $\Delta P = \frac{2S}{R}$; inside soap bubble in air $\Delta P = \frac{4S}{R}$. – Capillary Rise: $h = \frac{2S \cos\theta}{\rho g r}$ (Jurin’s Law).
- Thermal Expansion & Heat Transfer: – Expansions: $\Delta L = L_0 \alpha \Delta T, \Delta A = A_0 (2\alpha) \Delta T, \Delta V = V_0 (3\alpha) \Delta T$. – Thermal Stress in Clamped Rod: $\sigma = Y \alpha \Delta T \implies F = Y A \alpha \Delta T$. – Heat Conduction: $\frac{dQ}{dt} = \frac{K A \Delta T}{L}$. – Stefan-Boltzmann Law: $P_{\text{net}} = e \sigma A (T^4 – T_0^4)$. – Newton’s Law of Cooling (Small $\Delta T$): $-\frac{dT}{dt} = k(T – T_0)$. – Wien’s Displacement Law: $\lambda_{\text{max}} T = b \approx 2.898 \times 10^{-3}\text{ m}\cdot\text{K}$.
2. Unit 8: Thermodynamics
- First Law of Thermodynamics: $$\mathbf{\Delta Q = \Delta U + W}$$ – Internal Energy is a state function: $\mathbf{\Delta U = n C_v \Delta T}$ (universal for all ideal gas paths). – Work done by gas: $W = \int P dV$ (area under $P-V$ indicator curve).
- Master Path Analysis Table:
- Isochoric ($\Delta V = 0$): $W = 0 \implies Q = \Delta U = n C_v \Delta T, C = C_v$.
- Isobaric ($\Delta P = 0$): $W = P \Delta V = n R \Delta T \implies Q = n C_p \Delta T, \Delta U = n C_v \Delta T, C = C_p$.
- Isothermal ($\Delta T = 0$): $\Delta U = 0 \implies Q = W = n R T \ln\left(\frac{V_f}{V_i}\right) = n R T \ln\left(\frac{P_i}{P_f}\right), C \to \infty$.
- Adiabatic ($Q = 0$): $P V^\gamma = \text{const}, T V^{\gamma-1} = \text{const}, P^{1-\gamma} T^\gamma = \text{const}$. $$W = -\Delta U = \frac{P_i V_i – P_f V_f}{\gamma – 1} = \frac{n R (T_i – T_f)}{\gamma – 1}, \quad C = 0$$
- Polytropic ($P V^x = \text{const}$): $$C = C_v + \frac{R}{1 – x}, \quad W = \frac{n R \Delta T}{1 – x}$$
- Second Law & Heat Engines: – Thermal Efficiency: $\eta = \frac{W}{Q_{\text{in}}} = 1 – \frac{|Q_{\text{out}}|}{Q_{\text{in}}}$. – Carnot Engine Maximum Efficiency: $\eta_{\text{Carnot}} = 1 – \frac{T_C}{T_H}$. – Refrigerator Coefficient of Performance: $\beta = \frac{Q_C}{W} = \frac{T_C}{T_H – T_C} \implies \eta \beta = 1 – \eta$.
3. Unit 9: Kinetic Theory of Gases (KTG)
- Pressure & Speeds: $$P = \frac{1}{3} \rho v_{\text{rms}}^2 \implies v_{\text{rms}} = \sqrt{\frac{3 R T}{M}} = \sqrt{\frac{3 k_B T}{m}}$$ – Average Speed: $\bar{v} = \sqrt{\frac{8 R T}{\pi M}} \approx 0.921 v_{\text{rms}}$. – Most Probable Speed: $v_{\text{mp}} = \sqrt{\frac{2 R T}{M}} \approx 0.816 v_{\text{rms}}$. – Speed Ordering: $\mathbf{v_{\text{rms}} > \bar{v} > v_{\text{mp}}}$ ($\sqrt{3} : \sqrt{8/\pi} : \sqrt{2}$).
- Equipartition of Energy & Degrees of Freedom ($f$): $$\mathbf{\langle \epsilon \rangle = \frac{1}{2} k_B T \quad \text{per quadratic degree of freedom}}$$ – Monoatomic ($f = 3$): $C_v = \frac{3}{2}R, C_p = \frac{5}{2}R, \gamma = \frac{5}{3} \approx 1.67$. – Diatomic rigid ($f = 5$): $C_v = \frac{5}{2}R, C_p = \frac{7}{2}R, \gamma = \frac{7}{5} = 1.40$. – Diatomic with vibration ($f = 7$): $C_v = \frac{7}{2}R, C_p = \frac{9}{2}R, \gamma = \frac{9}{7} \approx 1.29$. – Non-linear polyatomic ($f = 6$): $C_v = 3R, C_p = 4R, \gamma = \frac{4}{3} \approx 1.33$. – Mayer’s Relation: $C_p – C_v = R$, specific heats $c_p – c_v = R / M$.
- Gas Mixtures: $$C_{v,\text{mix}} = \frac{\sum n_i C_{vi}}{\sum n_i}, \quad C_{p,\text{mix}} = C_{v,\text{mix}} + R, \quad \frac{\sum n_i}{\gamma_{\text{mix}} – 1} = \sum \frac{n_i}{\gamma_i – 1}$$
- Mean Free Path & Collisions: $$\mathbf{\lambda = \frac{1}{\sqrt{2} n \pi d^2} = \frac{k_B T}{\sqrt{2} \pi d^2 P} \propto \frac{V}{N}}$$ – Isochoric: $\lambda = \text{constant}$ (independent of $T, P$ in a sealed vessel!). – Collision Frequency: $Z_1 = \bar{v} / \lambda \propto \sqrt{T}$ at constant volume.
Four Graded Multi-Concept Solved Examples
Example 1 (JEE Main Level: Fluid Statics Coupled with Thermal Expansion)
Problem Statement: A solid sphere floats in a liquid at temperature $T_1 = 0^\circ\text{C}$ with exactly a fraction $f_1 = 0.75$ ($75\%$) of its volume submerged. The coefficient of volume expansion of the sphere material is $\gamma_s = 1.2 \times 10^{-5}\text{ K}^{-1}$, and the coefficient of volume expansion of the liquid is $\gamma_l = 8.0 \times 10^{-4}\text{ K}^{-1}$. Determine the temperature $T_2$ in $^\circ\text{C}$ at which the sphere floats completely submerged ($f_2 = 1.0$) in neutral equilibrium.
Solution:
Step 1: Flotation Condition at Initial Temperature.
At $T_1 = 0^\circ\text{C}$, the buoyant force balances the weight of the solid sphere:
$$F_{B, 1} = m_s g \implies (f_1 V_{s, 1}) \rho_{l, 1} g = (\rho_{s, 1} V_{s, 1}) g \implies \frac{\rho_{s, 1}}{\rho_{l, 1}} = f_1 = 0.75$$
Step 2: Flotation Condition at Final Temperature.
At temperature $T_2 = T_1 + \Delta T$, the densities of the solid and liquid decrease according to volumetric thermal expansion:
$$\rho_{s, 2} = \frac{\rho_{s, 1}}{1 + \gamma_s \Delta T}, \qquad \rho_{l, 2} = \frac{\rho_{l, 1}}{1 + \gamma_l \Delta T}$$
For the sphere to be completely submerged ($f_2 = 1.0$):
$$\rho_{s, 2} = \rho_{l, 2} \implies \frac{\rho_{s, 1}}{1 + \gamma_s \Delta T} = \frac{\rho_{l, 1}}{1 + \gamma_l \Delta T}$$
$$\frac{\rho_{s, 1}}{\rho_{l, 1}} = \frac{1 + \gamma_s \Delta T}{1 + \gamma_l \Delta T} \implies 0.75 = \frac{1 + \gamma_s \Delta T}{1 + \gamma_l \Delta T}$$
Step 3: Solve for Temperature Rise ($\Delta T$).
$$0.75 (1 + \gamma_l \Delta T) = 1 + \gamma_s \Delta T$$
$$0.75 + 0.75 \gamma_l \Delta T = 1 + \gamma_s \Delta T \implies (0.75 \gamma_l – \gamma_s) \Delta T = 0.25$$
$$\Delta T = \frac{0.25}{0.75(8.0 \times 10^{-4}) – 1.2 \times 10^{-5}} = \frac{0.25}{6.0 \times 10^{-4} – 0.12 \times 10^{-4}} = \frac{0.25}{5.88 \times 10^{-4}} \approx \mathbf{425.2\text{ K}}$$
Since $T_1 = 0^\circ\text{C}$, the final temperature is:
$$T_2 = 0 + 425.2^\circ\text{C} = \mathbf{425.2^\circ\text{C}}$$
Example 2 (Mathematical Cross-Unit Linkage: Elastic Strain Energy & Thermal Stress)
Problem Statement: A uniform steel wire of length $L = 1.0\text{ m}$ and cross-sectional area $A = 2.0\text{ mm}^2 = 2.0 \times 10^{-6}\text{ m}^2$ has Young’s modulus $Y = 2.0 \times 10^{11}\text{ N/m}^2$, linear thermal expansion coefficient $\alpha = 1.2 \times 10^{-5}\text{ K}^{-1}$, and mass density $\rho = 7800\text{ kg/m}^3$. The wire is clamped firmly between two rigid immovable walls at $T_1 = 50^\circ\text{C}$ with zero initial tension. The temperature is cooled to $T_2 = 10^\circ\text{C}$ ($\Delta T = 40\text{ K}$).
1. Calculate the tension $F$ developed in the wire.
2. Calculate the total elastic strain energy $U_{\text{elastic}}$ stored in the wire.
3. If the wire suddenly snaps at its midpoint, find the instantaneous speed $v$ with which the snapped ends retract.
Solution:
Step 1: Tension Developed via Thermal Stress.
Thermal strain prevented by rigid clamps:
$$\epsilon = \alpha \Delta T = (1.2 \times 10^{-5}\text{ K}^{-1})(40\text{ K}) = 4.8 \times 10^{-4}$$
Tensile stress developed:
$$\sigma = Y \epsilon = (2.0 \times 10^{11}\text{ N/m}^2)(4.8 \times 10^{-4}) = \mathbf{9.6 \times 10^7\text{ N/m}^2 = 96\text{ MPa}}$$
Tension force:
$$F = \sigma A = (9.6 \times 10^7\text{ N/m}^2)(2.0 \times 10^{-6}\text{ m}^2) = \mathbf{192\text{ N}}$$
Step 2: Total Elastic Strain Energy.
Strain energy density:
$$u = \frac{1}{2} \sigma \epsilon = \frac{1}{2} (9.6 \times 10^7\text{ N/m}^2)(4.8 \times 10^{-4}) = 2.304 \times 10^4\text{ J/m}^3$$
Volume of the wire: $V = A L = (2.0 \times 10^{-6}\text{ m}^2)(1.0\text{ m}) = 2.0 \times 10^{-6}\text{ m}^3$.
$$U_{\text{elastic}} = u V = (2.304 \times 10^4\text{ J/m}^3)(2.0 \times 10^{-6}\text{ m}^3) = \mathbf{0.04608\text{ J} \approx 46.1\text{ mJ}}$$
Step 3: Retraction Speed Upon Snapping.
By conservation of mechanical energy, the stored elastic strain energy converts entirely into the kinetic energy of the retracting wire halves:
$$\frac{1}{2} m v^2 = U_{\text{elastic}} \implies \frac{1}{2} (\rho A L) v^2 = \frac{1}{2} \left(\frac{\sigma^2}{Y}\right) (A L) \implies v = \frac{\sigma}{\sqrt{\rho Y}}$$
$$\rho Y = (7800\text{ kg/m}^3)(2.0 \times 10^{11}\text{ N/m}^2) = 1.56 \times 10^{15}\text{ kg}^2/(\text{m}^4\cdot\text{s}^2)$$
$$\sqrt{\rho Y} \approx 3.9497 \times 10^7\text{ kg/(m}^2\cdot\text{s)}$$
$$v = \frac{9.6 \times 10^7\text{ N/m}^2}{3.9497 \times 10^7} \approx \mathbf{2.43\text{ m/s}}$$
Example 3 (JEE Advanced Level: Multi-Stage Indicator Cycle Analysis)
Problem Statement: One mole of a monoatomic ideal gas ($C_v = \frac{3}{2}R, \gamma = 5/3$) undergoes a reversible cyclic process $A \to B \to C \to A$ on a $P-V$ plane:
– $A \to B$: Isobaric expansion at pressure $P_A = 2.0 \times 10^5\text{ Pa}$ from $V_A = 0.010\text{ m}^3$ to $V_B = 0.030\text{ m}^3$.
– $B \to C$: Reversible adiabatic expansion from $(P_B, V_B)$ to $V_C = 0.060\text{ m}^3$.
– $C \to A$: A straight-line compression path returning directly from $(P_C, V_C)$ to $(P_A, V_A)$.
1. Find pressure $P_C$ at state $C$.
2. Calculate the work done in each leg ($W_{AB}, W_{BC}, W_{CA}$) and the total net work $W_{\text{net}}$.
3. Find the total heat absorbed $Q_{\text{in}}$ and the cycle thermal efficiency $\eta$.
Solution:
Step 1: Adiabatic State $C$.
For adiabatic expansion $B \to C$:
$$P_C V_C^\gamma = P_B V_B^\gamma \implies P_C = P_B \left(\frac{V_B}{V_C}\right)^{5/3} = (2.0 \times 10^5) \left(\frac{0.030}{0.060}\right)^{5/3} = \frac{2.0 \times 10^5}{2^{5/3}}$$
Since $2^{5/3} = 2 \times 2^{2/3} \approx 2 \times 1.5874 = 3.1748$:
$$P_C = \frac{200000}{3.1748} \approx \mathbf{62996\text{ Pa} \approx 6.30 \times 10^4\text{ Pa}}$$
Step 2: Work Calculations for Each Leg.
– Leg $A \to B$ (Isobaric):
$$W_{AB} = P_A (V_B – V_A) = (2.0 \times 10^5\text{ Pa})(0.030 – 0.010\text{ m}^3) = \mathbf{4000\text{ J}}$$
– Leg $B \to C$ (Adiabatic):
$$W_{BC} = \frac{P_B V_B – P_C V_C}{\gamma – 1} = \frac{(2.0 \times 10^5 \times 0.030) – (62996 \times 0.060)}{5/3 – 1} = \frac{6000 – 3779.8}{2/3} = \mathbf{3330.3\text{ J}}$$
– Leg $C \to A$ (Linear Path):
$$W_{CA} = \frac{P_C + P_A}{2} (V_A – V_C) = \frac{62996 + 200000}{2} (0.010 – 0.060) = (131498)(-0.050) = \mathbf{-6574.9\text{ J}}$$
– Net Work Done:
$$W_{\text{net}} = W_{AB} + W_{BC} + W_{CA} = 4000 + 3330.3 – 6574.9 = \mathbf{755.4\text{ J}}$$
Step 3: Heat Input and Cycle Efficiency.
During isobaric leg $A \to B$:
$$\Delta U_{AB} = \frac{3}{2} (P_B V_B – P_A V_A) = \frac{3}{2} (6000 – 2000) = 6000\text{ J}$$
$$Q_{\text{in}} = Q_{AB} = \Delta U_{AB} + W_{AB} = 6000 + 4000 = \mathbf{10000\text{ J}}$$
Cycle efficiency:
$$\eta = \frac{W_{\text{net}}}{Q_{\text{in}}} = \frac{755.4\text{ J}}{10000\text{ J}} \approx \mathbf{7.55\%}$$
Example 4 (JEE Advanced Level: Surface Tension, Evaporation & Molecular Collision Rate)
Problem Statement: A spherical water droplet of initial radius $R_0 = 1.0\text{ mm} = 1.0 \times 10^{-3}\text{ m}$ is suspended in dry ambient air at temperature $T = 300\text{ K}$ and pressure $P = 1.0 \times 10^5\text{ Pa}$. The surface tension of water is $S = 0.072\text{ N/m}$, density $\rho = 1000\text{ kg/m}^3$, and the effective molar mass of air is $M_{\text{air}} = 29.0\text{ g/mol}$.
1. Calculate the excess pressure $\Delta P$ inside the droplet.
2. If the droplet evaporates slowly such that its mass decreases at a rate proportional to its surface area ($dm/dt = -\beta \cdot 4\pi R^2$), show that the radius decreases linearly with time: $R(t) = R_0 – \frac{\beta}{\rho} t$.
3. Using kinetic theory, estimate the total collision frequency ($\Phi$) of surrounding air molecules striking the outer surface of the droplet.
Solution:
Step 1: Excess Pressure Inside Droplet.
A liquid droplet in air has one liquid-gas interface:
$$\Delta P = \frac{2S}{R_0} = \frac{2(0.072\text{ N/m})}{1.0 \times 10^{-3}\text{ m}} = \mathbf{144\text{ Pa}}$$
Step 2: Radius Evolution During Evaporation.
The mass of the droplet is $m = \frac{4}{3} \pi R^3 \rho$. Differentiating:
$$\frac{dm}{dt} = 4\pi R^2 \rho \frac{dR}{dt}$$
Equating to $-\beta (4\pi R^2)$:
$$4\pi R^2 \rho \frac{dR}{dt} = -4\pi R^2 \beta \implies \frac{dR}{dt} = -\frac{\beta}{\rho} = \text{constant}$$
Integrating with initial condition $R(0) = R_0$:
$$\mathbf{R(t) = R_0 – \frac{\beta}{\rho} t}$$
Step 3: Collision Rate with the Droplet Surface.
From kinetic theory, the molecular collision flux (number of molecules striking unit area per unit time) is:
$$J = \frac{1}{4} n \bar{v}$$
Molecular number density of ambient air:
$$n = \frac{P}{k_B T} = \frac{1.0 \times 10^5\text{ Pa}}{(1.38 \times 10^{-23}\text{ J/K})(300\text{ K})} \approx 2.415 \times 10^{25}\text{ molecules/m}^3$$
Average speed of air molecules:
$$\bar{v} = \sqrt{\frac{8 R T}{\pi M_{\text{air}}}} = \sqrt{\frac{8 \times 8.314 \times 300}{\pi \times 0.0290\text{ kg/mol}}} \approx 468.0\text{ m/s}$$
Collision flux:
$$J = \frac{1}{4} (2.415 \times 10^{25}\text{ m}^{-3})(468.0\text{ m/s}) \approx 2.826 \times 10^{27}\text{ collisions/(m}^2\cdot\text{s)}$$
Surface area of the droplet:
$$A = 4\pi R_0^2 = 4\pi (1.0 \times 10^{-3}\text{ m})^2 \approx 1.257 \times 10^{-5}\text{ m}^2$$
Total rate of collisions:
$$\Phi = J \cdot A = (2.826 \times 10^{27}\text{ m}^{-2}\cdot\text{s}^{-1})(1.257 \times 10^{-5}\text{ m}^2) \approx \mathbf{3.55 \times 10^{22}\text{ collisions/second}}$$
Practice Worksheet: 10 Cross-Unit JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. A metal wire of Young’s modulus $Y$ is stretched producing a longitudinal strain $\epsilon$. The elastic potential energy stored per unit volume of the wire is:
(A) $Y \epsilon$
(B) $\frac{1}{2} Y \epsilon^2$
(C) $\frac{1}{2} Y^2 \epsilon$
(D) $2 Y \epsilon^2$
Q2. In an isobaric expansion of a monoatomic ideal gas ($\gamma = 5/3$), the percentage of the heat energy supplied that is converted into internal energy is:
(A) $40\%$
(B) $50\%$
(C) $60\%$
(D) $75\%$
Q3. The ratio of the root mean square speed ($v_{\text{rms}}$) to the average speed ($\bar{v}$) of an ideal gas at temperature $T$ is:
(A) $\sqrt{\frac{8}{3\pi}}$
(B) $\frac{\sqrt{3}}{\sqrt{2}}$
(C) $\sqrt{\frac{3\pi}{8}}$
(D) $\sqrt{\frac{\pi}{8}}$
Q4. The excess pressure inside a spherical soap bubble of radius $R$ and surface tension $S$ blown in air is:
(A) $\frac{S}{R}$
(B) $\frac{2S}{R}$
(C) $\frac{4S}{R}$
(D) $\frac{8S}{R}$
Q5. An ideal diatomic gas ($\gamma = 1.4 = 7/5$) undergoes a reversible adiabatic expansion in which its volume increases by a factor of $32$ ($V_f = 32 V_i$). The ratio of its final absolute temperature to initial absolute temperature ($T_f / T_i$) is:
(A) $\frac{1}{4}$
(B) $\frac{1}{8}$
(C) $\frac{1}{16}$
(D) $\frac{1}{32}$
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. Which of the following statements regarding fluid dynamics, heat transfer, and kinetic theory is/are correct?
(A) In steady horizontal streamline flow of an ideal incompressible fluid, the pressure is lowest where the fluid speed is highest.
(B) The terminal falling speed of a small sphere in a viscous fluid is proportional to the square of its radius ($v_t \propto r^2$).
(C) According to the Dulong-Petit law, the classical molar heat capacity of 3D crystalline solids at high temperatures approaches $3R$.
(D) In a rigid sealed vessel of fixed volume, heating the gas causes its mean free path to remain strictly constant while collision frequency increases as $\sqrt{T}$.
Q7. One mole of a monoatomic ideal gas ($C_v = \frac{3}{2}R$) undergoes an expansion process along the path $P = \alpha V$, where $\alpha$ is a positive constant. Which of the following statements is/are correct?
(A) The process represents a polytropic path $P V^n = \text{constant}$ with polytropic exponent $n = -1$.
(B) The molar heat capacity of the gas during this process is $C = 2R$.
(C) The fraction of heat supplied that converts into internal energy is $\frac{\Delta U}{Q} = \frac{3}{4}$.
(D) The fraction of heat supplied that converts into boundary work is $\frac{W}{Q} = \frac{1}{4}$.
Q8. Two identical reversible Carnot engines operate in series between heat reservoirs at temperatures $T_H$ and $T_C$. Engine 1 absorbs heat $Q_H$ from $T_H$ and rejects heat $Q_M$ to an intermediate reservoir at $T_M$. Engine 2 absorbs $Q_M$ and rejects heat $Q_C$ to $T_C$. Which of the following statements is/are correct?
(A) If both engines perform equal amounts of work ($W_1 = W_2$), the intermediate temperature is the arithmetic mean: $T_M = \frac{T_H + T_C}{2}$.
(B) If both engines operate with equal thermal efficiencies ($\eta_1 = \eta_2$), the intermediate temperature is the geometric mean: $T_M = \sqrt{T_H T_C}$.
(C) The overall efficiency of the two-engine combined system is $\eta = 1 – \frac{T_C}{T_H}$.
(D) The efficiency of a Carnot engine depends exclusively on the reservoir temperatures, independent of whether the working substance is monoatomic or diatomic.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. A uniform metal rod of cross-sectional area $A = 1.0\text{ cm}^2 = 1.0 \times 10^{-4}\text{ m}^2$ and Young’s modulus $Y = 1.0 \times 10^{11}\text{ N/m}^2$ is clamped tightly between two rigid, immovable supports. The coefficient of linear thermal expansion of the metal is $\alpha = 2.0 \times 10^{-5}\text{ K}^{-1}$. When the temperature of the rod is increased by $\Delta T = 50\text{ K}$, the compressive thermal force developed in the rod in kilo-Newtons ($\text{kN}$) is $F$. Determine the integer value of $F$.
Q10. A reversible heat engine operates on a Carnot cycle between temperatures $T_H = 600\text{ K}$ and $T_C = 300\text{ K}$. When this machine is reversed to operate as a refrigerator between the same two temperature reservoirs, it consumes an external work input of $W_{\text{in}} = 200\text{ J}$ per cycle. Determine the heat energy extracted from the cold reservoir per cycle ($Q_{\text{extracted}}$) in Joules.
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (B) $\frac{1}{2} Y \epsilon^2$ | Elastic Potential Energy Density |
| Q2 | (C) $60\%$ | Isobaric Heat Partitioning ($\Delta U/Q = 1/\gamma$) |
| Q3 | (C) $\sqrt{3\pi/8}$ | Ratio of RMS Speed to Average Speed |
| Q4 | (C) $4S/R$ | Excess Pressure in Soap Bubble |
| Q5 | (A) $\frac{1}{4}$ | Adiabatic Temperature-Volume Scaling ($T V^{\gamma-1} = \text{const}$) |
| Q6 | (A), (B), (C), (D) | Cross-Unit Principles (Bernoulli, Stokes, Dulong-Petit, KTG) |
| Q7 | (A), (B), (C), (D) | Linear $P \propto V$ Polytropic Heat Capacity Analysis |
| Q8 | (A), (B), (C), (D) | Carnot Engine Series Operating Principles |
| Q9 | 10 | Thermal Stress Force ($F = Y A \alpha \Delta T = 10\text{ kN}$) |
| Q10 | 200 | Carnot Refrigerator COP & Heat Extraction ($\beta = 1 \implies Q_C = 200\text{ J}$) |
Complete Step-by-Step Solutions
Solution 1:
Elastic strain energy per unit volume is $u = \frac{1}{2} \times \text{stress} \times \text{strain}$.
Since $\text{stress} = Y \times \text{strain} = Y \epsilon$:
$$u = \frac{1}{2} (Y \epsilon) \epsilon = \mathbf{\frac{1}{2} Y \epsilon^2}$$
Correct Option: (B)
Solution 2:
In any isobaric process of an ideal gas:
$$\Delta U = n C_v \Delta T, \qquad Q = n C_p \Delta T$$
$$\frac{\Delta U}{Q} = \frac{C_v}{C_p} = \frac{1}{\gamma}$$
For a monoatomic gas, $\gamma = 5/3$:
$$\frac{\Delta U}{Q} = \frac{1}{5/3} = \frac{3}{5} = \mathbf{0.60 = 60\%}$$
Correct Option: (C)
Solution 3:
$$v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}, \qquad \bar{v} = \sqrt{\frac{8 R T}{\pi M}}$$
Dividing the two expressions:
$$\frac{v_{\text{rms}}}{\bar{v}} = \frac{\sqrt{3}}{\sqrt{8/\pi}} = \mathbf{\sqrt{\frac{3\pi}{8}} \approx 1.085}$$
Correct Option: (C)
Solution 4:
A soap bubble has two spherical liquid-air surfaces (an inner surface and an outer surface). Each surface contributes $2S/R$ to the excess pressure:
$$\Delta P = \frac{2S}{R} + \frac{2S}{R} = \mathbf{\frac{4S}{R}}$$
Correct Option: (C)
Solution 5:
For a reversible adiabatic expansion:
$$T_i V_i^{\gamma – 1} = T_f V_f^{\gamma – 1} \implies \frac{T_f}{T_i} = \left(\frac{V_i}{V_f}\right)^{\gamma – 1}$$
Given $\gamma = 1.4 \implies \gamma – 1 = 0.4 = 2/5$, and $V_f / V_i = 32 = 2^5$:
$$\frac{T_f}{T_i} = \left(\frac{1}{32}\right)^{2/5} = \left(\frac{1}{2^5}\right)^{2/5} = \left(\frac{1}{2}\right)^2 = \mathbf{\frac{1}{4}}$$
Correct Option: (A)
Solution 6:
(A) Correct: Bernoulli’s principle $P + \frac{1}{2}\rho v^2 = \text{const}$ directly links higher velocity with lower static pressure.
(B) Correct: Balancing gravity against buoyancy and Stokes’ drag yields $v_t = \frac{2}{9}\frac{r^2 (\rho – \sigma) g}{\eta} \propto r^2$.
(C) Correct: In 3D crystals, each atom has 3 kinetic and 3 potential quadratic terms, yielding $C = 6 \times \frac{1}{2}R = 3R$.
(D) Correct: In a rigid sealed vessel ($V = \text{const}$), number density $n$ is constant, so $\lambda = \frac{1}{\sqrt{2} n \pi d^2}$ is invariant, while $Z_1 = \bar{v}/\lambda \propto \sqrt{T}$.
Correct Options: (A), (B), (C), (D)
Solution 7:
(A) Correct: $P \propto V \implies P V^{-1} = \text{const} \implies n = -1$.
(B) Correct: $C = C_v + \frac{R}{1 – n} = \frac{3}{2}R + \frac{R}{1 – (-1)} = \frac{3}{2}R + \frac{1}{2}R = 2R$.
(C) Correct: $\Delta U = n C_v \Delta T = \frac{3}{2} n R \Delta T$; $Q = n C \Delta T = 2 n R \Delta T \implies \frac{\Delta U}{Q} = \frac{3/2}{2} = \frac{3}{4}$.
(D) Correct: By conservation of energy, $\frac{W}{Q} = 1 – \frac{\Delta U}{Q} = 1 – \frac{3}{4} = \frac{1}{4}$.
Correct Options: (A), (B), (C), (D)
Solution 8:
(A) Correct: $W_1 = Q_H (1 – T_M/T_H)$ and $W_2 = Q_M (1 – T_C/T_M)$. Since $Q_M = Q_H (T_M/T_H)$, setting $W_1 = W_2$ gives $T_H – T_M = T_M – T_C \implies T_M = \frac{T_H + T_C}{2}$.
(B) Correct: $\eta_1 = 1 – \frac{T_M}{T_H}$ and $\eta_2 = 1 – \frac{T_C}{T_M}$. Equating gives $\frac{T_M}{T_H} = \frac{T_C}{T_M} \implies T_M^2 = T_H T_C \implies T_M = \sqrt{T_H T_C}$.
(C) Correct: Overall heat absorbed is $Q_H$ and rejected is $Q_C$, so $\eta = 1 – \frac{Q_C}{Q_H} = 1 – \frac{T_C}{T_H}$.
(D) Correct: Carnot efficiency $\eta = 1 – T_C/T_H$ depends solely on reservoir temperatures.
Correct Options: (A), (B), (C), (D)
Solution 9:
Thermal stress in the clamped rod:
$$\sigma = Y \alpha \Delta T = (1.0 \times 10^{11}\text{ N/m}^2) \times (2.0 \times 10^{-5}\text{ K}^{-1}) \times (50\text{ K}) = 1.0 \times 10^8\text{ N/m}^2$$
The thermal force exerted on the rigid walls is:
$$F = \sigma A = (1.0 \times 10^8\text{ N/m}^2) \times (1.0 \times 10^{-4}\text{ m}^2) = 10000\text{ N} = \mathbf{10\text{ kN}}$$
Answer: 10
Solution 10:
The coefficient of performance ($\beta$) of a Carnot refrigerator operating between $T_C = 300\text{ K}$ and $T_H = 600\text{ K}$ is:
$$\beta = \frac{T_C}{T_H – T_C} = \frac{300\text{ K}}{600\text{ K} – 300\text{ K}} = \frac{300}{300} = 1.0$$
By definition of COP:
$$\beta = \frac{Q_{\text{extracted}}}{W_{\text{in}}} \implies Q_{\text{extracted}} = \beta \cdot W_{\text{in}} = 1.0 \times 200\text{ J} = \mathbf{200\text{ J}}$$
Answer: 200