Welcome to Lecture 110 (Master Index Entry: L104) of our comprehensive JEE Physics Master Series! This milestone assessment serves as the definitive Full Syllabus Test on Properties of Matter, Thermodynamics, and Kinetic Theory of Gases (Units 7, 8 & 9). Following our grand revision in Lecture 109, this rigorous mock examination evaluates analytical proficiency, multi-concept synthesis, and numerical precision across elasticity, fluid mechanics, surface tension, viscosity, thermal expansion, calorimetry, heat transfer laws, indicator cycles, heat engines, molecular velocity distributions, and degrees of freedom. This module provides a high-yield exam strategy card, 4 diverse multi-concept solved examples, a timed 10-question practice worksheet adhering to current JEE Main and JEE Advanced patterns, and comprehensive step-by-step KaTeX solutions with an answer key.
Detailed Concept Card: High-Yield Exam Strategy & Blueprint (Units 7, 8 & 9)
1. Exam Weightage & Cognitive Blueprint
In JEE Main and JEE Advanced, thermal and mechanical properties of matter contribute approximately $12\% – 16\%$ of the total Physics score ($3$ to $5$ questions per exam session). Questions frequently couple concepts across multiple domains:
- Fluid Dynamics + Hydrostatics: Accelerated containers, Bernoulli’s equation with Torricelli’s draining tanks, manometers, and siphon dynamics.
- Elasticity + Thermal Physics: Thermal stress in constrained rods, temperature-dependent Young’s modulus, and elastic potential energy stored in rotating or hanging structures.
- Surface Tension + Mechanics: Capillary rise under Jurin’s law, soap bubble coalescence, terminal velocity of droplets falling through viscous media.
- Thermodynamics + KTG: Work done in polytropic processes ($P V^x = \text{const}$), cyclic efficiency of heat engines, equipartition of energy across gas mixtures, and adiabatic sound speed comparisons.
2. Core Formula Inventory
| Topic Domain | Key Governing Formulas | Crucial JEE Trap / Nuance |
|---|---|---|
| Elasticity & Fluids | $Y = \frac{F L}{A \Delta L}, \quad u = \frac{1}{2} Y \epsilon^2$ $P_2 – P_1 = \rho g h, \quad P + \frac{1}{2}\rho v^2 = \text{const}$ |
Poisson’s ratio real bounds: $0 < \sigma < 0.5$. In rotating rods, tension varies quadratically: $T(x) = \frac{M\omega^2}{2L}(L^2 - x^2)$. |
| Surface Tension & Viscosity | $\Delta P_{\text{drop}} = \frac{2S}{R}, \quad \Delta P_{\text{bubble}} = \frac{4S}{R}$ $h = \frac{2S\cos\theta}{\rho g r}, \quad v_t = \frac{2 r^2 (\rho – \sigma) g}{9 \eta}$ |
Insufficient capillary height: liquid rises to top rim and increases radius of curvature without overflowing! Bubble connection: air flows from smaller to larger bubble. |
| Thermodynamics | $\Delta Q = \Delta U + W, \quad \Delta U = n C_v \Delta T$ $W_{\text{isothermal}} = n R T \ln(V_f / V_i)$ $W_{\text{adiabatic}} = \frac{P_i V_i – P_f V_f}{\gamma – 1}$ |
$\Delta U = n C_v \Delta T$ holds for ANY ideal gas process, not just isochoric! Cycle work on $P-V$ indicator is positive for clockwise cycles. |
| Kinetic Theory (KTG) | $P = \frac{1}{3} \rho v_{\text{rms}}^2, \quad v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ $C_p – C_v = R, \quad \lambda = \frac{1}{\sqrt{2} n \pi d^2} \propto \frac{V}{N}$ |
Isochoric heating leaves $\lambda$ unchanged ($\lambda \propto V$), while collision frequency increases ($Z_1 \propto \sqrt{T}$). |
Four Graded Solved Examples
Example 1 (JEE Main Level: Draining Tank Efflux & Horizontal Projectile Range)
Problem Statement: A cylindrical storage tank of total height $H = 5.0\text{ m}$ is filled with water ($\rho = 1000\text{ kg/m}^3$) and stands on a level horizontal concrete floor. A circular hole of diameter $d_0 = 1.0\text{ cm}$ is punched in the vertical side wall at a depth $h = 1.25\text{ m}$ below the open water surface. Take $g = 9.80\text{ m/s}^2$.
1. Find the efflux velocity $v$ of the water discharging from the orifice.
2. Calculate the volume flow rate $Q$ in $\text{L/s}$.
3. Find the horizontal distance $R$ from the base of the tank where the water stream strikes the floor.
4. Determine the hole depth $h’$ that maximizes horizontal range, and compute this maximum range $R_{\text{max}}$.
Solution:
Step 1: Efflux Velocity via Torricelli’s Law.
Since the tank cross-sectional area is much larger than the orifice ($A_{\text{tank}} \gg A_0$), the top surface velocity is negligible:
$$v = \sqrt{2 g h} = \sqrt{2 \times 9.80\text{ m/s}^2 \times 1.25\text{ m}} = \sqrt{24.5} \approx \mathbf{4.95\text{ m/s}}$$
Step 2: Volume Rate of Flow.
Orifice area: $A_0 = \frac{\pi}{4} d_0^2 = \frac{\pi}{4} (0.010\text{ m})^2 = 7.854 \times 10^{-5}\text{ m}^2$.
$$Q = A_0 v = (7.854 \times 10^{-5}\text{ m}^2)(4.950\text{ m/s}) \approx 3.89 \times 10^{-4}\text{ m}^3\text{/s} = \mathbf{0.389\text{ L/s}}$$
Step 3: Horizontal Range.
Vertical falling distance to the ground: $y = H – h = 5.0\text{ m} – 1.25\text{ m} = 3.75\text{ m}$.
Time of free fall: $t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 3.75}{9.80}} \approx 0.875\text{ s}$.
Horizontal range:
$$R = v t = \sqrt{2gh} \sqrt{\frac{2(H – h)}{g}} = 2 \sqrt{h(H – h)}$$
$$R = 2 \sqrt{(1.25\text{ m})(3.75\text{ m})} = 2 \sqrt{4.6875} \approx 2 \times 2.165\text{ m} = \mathbf{4.33\text{ m}}$$
Step 4: Maximum Range Condition.
The product $h(H – h)$ is a maximum when $h’ = \frac{H}{2} = \frac{5.0\text{ m}}{2} = \mathbf{2.50\text{ m}}$.
$$R_{\text{max}} = 2 \sqrt{\left(\frac{H}{2}\right) \left(\frac{H}{2}\right)} = 2 \left(\frac{H}{2}\right) = H = \mathbf{5.00\text{ m}}$$
Example 2 (Mathematical Derivation: Elastic Strain Energy in a Rotating Rod)
Problem Statement: A uniform thin elastic rod of total mass $M$, length $L$, cross-sectional area $A$, and Young’s modulus $Y$ rotates horizontally in a vacuum with uniform angular velocity $\omega$ about a fixed vertical axis passing through one end.
1. Derive the internal tension $T(x)$ in the rod as a function of distance $x$ from the rotation axis.
2. Calculate the total elongation $\Delta L$ of the rotating rod.
3. Derive an expression for the total elastic strain energy $U_{\text{elastic}}$ stored inside the rod.
Solution:
Step 1: Tension Distribution $T(x)$.
Consider an infinitesimal element $dm = \frac{M}{L} dx’$ located at distance $x’$ from the axis. The tension $T(x)$ at coordinate $x$ must provide the centripetal acceleration for the entire outer portion of the rod from $x’ = x$ to $x’ = L$:
$$T(x) = \int_x^L (dm) \omega^2 x’ = \frac{M}{L} \omega^2 \int_x^L x’ dx’ = \frac{M \omega^2}{L} \left[\frac{(x’)^2}{2}\right]_x^L$$
$$\mathbf{T(x) = \frac{M \omega^2}{2L} (L^2 – x^2)}$$
At the rotation axis ($x = 0$): $T(0) = \frac{1}{2} M \omega^2 L$. At the free tip ($x = L$): $T(L) = 0$.
Step 2: Total Elongation ($\Delta L$).
The elongation of an element $dx$ under local tension $T(x)$ is $d(\Delta L) = \frac{T(x) dx}{A Y}$:
$$\Delta L = \int_0^L \frac{T(x) dx}{A Y} = \frac{M \omega^2}{2 A Y L} \int_0^L (L^2 – x^2) dx = \frac{M \omega^2}{2 A Y L} \left[L^2 \cdot L – \frac{L^3}{3}\right]$$
$$\Delta L = \frac{M \omega^2}{2 A Y L} \left(\frac{2 L^3}{3}\right) = \mathbf{\frac{M \omega^2 L^2}{3 A Y}}$$
Step 3: Total Elastic Strain Energy ($U_{\text{elastic}}$).
The strain energy stored in element $dx$ is $dU = \frac{T^2(x) dx}{2 A Y}$:
$$U_{\text{elastic}} = \int_0^L \frac{1}{2 A Y} \left[\frac{M \omega^2}{2L} (L^2 – x^2)\right]^2 dx = \frac{M^2 \omega^4}{8 A Y L^2} \int_0^L (L^4 – 2L^2 x^2 + x^4) dx$$
Evaluating the polynomial integral:
$$\int_0^L (L^4 – 2L^2 x^2 + x^4) dx = L^5 – \frac{2}{3} L^5 + \frac{1}{5} L^5 = L^5 \left(\frac{15 – 10 + 3}{15}\right) = \frac{8}{15} L^5$$
Substituting back:
$$U_{\text{elastic}} = \frac{M^2 \omega^4}{8 A Y L^2} \left(\frac{8}{15} L^5\right) = \mathbf{\frac{M^2 \omega^4 L^3}{15 A Y}}$$
Example 3 (JEE Advanced Level: Conduction-Radiation Boundary Equilibrium)
Problem Statement: A cylindrical copper bar of length $L = 0.50\text{ m}$, cross-sectional area $A = 1.0 \times 10^{-3}\text{ m}^2$, and thermal conductivity $K = 400\text{ W/(m}\cdot\text{K)}$ is perfectly insulated along its cylindrical lateral perimeter. One end is kept at a fixed high temperature $T_1 = 500\text{ K}$. The other end radiates heat into an evacuated environment whose surrounding walls are maintained at $T_0 = 300\text{ K}$. The radiating end surface has an emissivity $e = 0.80$. Given Stefan-Boltzmann constant $\sigma = 5.67 \times 10^{-8}\text{ W/(m}^2\cdot\text{K}^4)$:
1. Write the steady-state boundary equation at the radiating end.
2. Determine the steady-state temperature $T_2$ of the radiating end face.
3. Calculate the steady rate of heat flow ($H$) through the copper bar in Watts.
Solution:
Step 1: Steady-State Energy Balance.
Under steady conditions, the rate of heat conducted through the bar must equal the net rate of radiation emitted from the exposed end face:
$$H_{\text{cond}} = H_{\text{rad}} \implies \frac{K A}{L} (T_1 – T_2) = e \sigma A (T_2^4 – T_0^4)$$
Dividing both sides by the cross-sectional area $A$:
$$\frac{K}{L} (T_1 – T_2) = e \sigma (T_2^4 – T_0^4)$$
Step 2: Solving for $T_2$.
Calculate conductances:
$$\frac{K}{L} = \frac{400\text{ W/(m}\cdot\text{K)}}{0.50\text{ m}} = 800\text{ W/(m}^2\cdot\text{K)}$$
Since copper is an exceptional thermal conductor ($K/L = 800\text{ W/(m}^2\cdot\text{K)}$ is large), $T_2$ will remain very close to $T_1 = 500\text{ K}$. Testing $T_2 \approx 497\text{ K}$:
$$T_2^4 – T_0^4 \approx (497)^4 – (300)^4 \approx 6.10 \times 10^{10} – 0.81 \times 10^{10} = 5.29 \times 10^{10}\text{ K}^4$$
$$e \sigma (T_2^4 – T_0^4) = (0.80)(5.67 \times 10^{-8})(5.29 \times 10^{10}) \approx 2400\text{ W/m}^2$$
Now from conduction:
$$800(500 – T_2) = 2400 \implies 500 – T_2 = \frac{2400}{800} = 3.0\text{ K} \implies \mathbf{T_2 \approx 497.0\text{ K}}$$
Step 3: Steady Rate of Heat Transfer ($H$).
$$H = \frac{K A}{L} (T_1 – T_2) = (800\text{ W/(m}^2\cdot\text{K)})(1.0 \times 10^{-3}\text{ m}^2)(3.0\text{ K}) = \mathbf{2.40\text{ W}}$$
Example 4 (JEE Advanced Level: Isobaric Piston Expansion & Capillary Invariance)
Problem Statement: A vertical cylinder closed at its base contains $n = 0.50\text{ moles}$ of an ideal rigid diatomic gas ($C_v = \frac{5}{2}R, C_p = \frac{7}{2}R$). A frictionless, leak-proof piston of mass $m_p = 10.0\text{ kg}$ and cross-sectional area $A = 1.0 \times 10^{-2}\text{ m}^2 = 100\text{ cm}^2$ rests in equilibrium on the gas under atmospheric pressure $P_0 = 1.00 \times 10^5\text{ Pa}$. Take $g = 9.80\text{ m/s}^2$ and $R = 8.314\text{ J/(mol}\cdot\text{K)}$.
1. Determine the internal gas pressure $P_{\text{gas}}$.
2. The gas is heated from $T_1 = 300\text{ K}$ to $T_2 = 400\text{ K}$. Calculate the vertical displacement $\Delta y$ of the piston, the boundary work done $W$, and the total heat $Q$ supplied.
3. A vertical glass capillary tube of internal radius $r = 0.50\text{ mm}$ dipping in a water dish ($S = 0.070\text{ N/m}$, contact angle $\theta = 0^\circ$) is connected inside the chamber. Find the capillary rise height $h_c$.
Solution:
Step 1: Equilibrium Gas Pressure.
Balancing vertical forces on the piston:
$$P_{\text{gas}} A = P_0 A + m_p g \implies P_{\text{gas}} = P_0 + \frac{m_p g}{A}$$
$$P_{\text{gas}} = 1.00 \times 10^5\text{ Pa} + \frac{(10.0\text{ kg})(9.80\text{ m/s}^2)}{1.0 \times 10^{-2}\text{ m}^2} = 100000 + 9800 = \mathbf{109800\text{ Pa} = 1.098 \times 10^5\text{ Pa}}$$
Step 2: Thermodynamic Expansion Parameters.
Since the piston moves freely under constant external weight, the process is strictly isobaric at $P_{\text{gas}} = 1.098 \times 10^5\text{ Pa}$.
$$\Delta V = \frac{n R \Delta T}{P_{\text{gas}}} = \frac{(0.50\text{ mol})(8.314\text{ J/(mol}\cdot\text{K)})(100\text{ K})}{109800\text{ Pa}} = \frac{415.7\text{ J}}{109800\text{ Pa}} \approx 3.786 \times 10^{-3}\text{ m}^3$$
Vertical displacement of the piston:
$$\Delta y = \frac{\Delta V}{A} = \frac{3.786 \times 10^{-3}\text{ m}^3}{1.0 \times 10^{-2}\text{ m}^2} \approx \mathbf{0.379\text{ m} = 37.9\text{ cm}}$$
Work performed by the expanding gas:
$$W = P_{\text{gas}} \Delta V = n R \Delta T = (0.50)(8.314)(100) = \mathbf{415.7\text{ J}}$$
Total heat supplied:
$$Q = n C_p \Delta T = (0.50\text{ mol})\left(\frac{7}{2} \times 8.314\text{ J/(mol}\cdot\text{K)}\right)(100\text{ K}) = \mathbf{1455.0\text{ J}}$$
Change in internal energy: $\Delta U = n C_v \Delta T = (0.50)(2.5 \times 8.314)(100) = \mathbf{1039.3\text{ J}}$.
Check: $\Delta U + W = 1039.3 + 415.7 = 1455.0\text{ J}$ (Exact!).
Step 3: Capillary Height Analysis.
By Jurin’s Law, the meniscus curvature and buoyant pressure head depend solely on surface tension, contact angle, radius, and liquid density:
$$h_c = \frac{2 S \cos\theta}{\rho g r} = \frac{2(0.070\text{ N/m})\cos(0^\circ)}{(1000\text{ kg/m}^3)(9.80\text{ m/s}^2)(0.50 \times 10^{-3}\text{ m})} = \frac{0.140}{4.90} \approx \mathbf{0.0286\text{ m} = 2.86\text{ cm}}$$
Physical Takeaway: The ambient gas pressure inside the cylinder acts equally on the liquid in the open dish and on the meniscus inside the tube. Therefore, capillary rise is independent of atmospheric or chamber pressure!
Practice Worksheet: 10 High-Yield Mock Exam Questions
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. Two wires $A$ and $B$ are made of the same metal. Wire $A$ has length $L$ and diameter $2d$, while wire $B$ has length $2L$ and diameter $d$. When both wires are subjected to the same tensile stretching force, the ratio of the elongation of wire $A$ to that of wire $B$ ($\Delta L_A / \Delta L_B$) is:
(A) $1 : 2$
(B) $1 : 4$
(C) $1 : 8$
(D) $8 : 1$
Q2. An ideal incompressible liquid flows steadily through a horizontal pipe of non-uniform cross-section. At point 1, the pipe diameter is $4.0\text{ cm}$ and fluid speed is $1.0\text{ m/s}$. At point 2, the pipe narrows to a diameter of $2.0\text{ cm}$. The fluid speed at point 2 is:
(A) $0.5\text{ m/s}$
(B) $2.0\text{ m/s}$
(C) $3.0\text{ m/s}$
(D) $4.0\text{ m/s}$
Q3. A thermodynamic cyclic process is represented by an ellipse on a $P-V$ indicator diagram with semi-major axis $\Delta P$ (in $\text{Pa}$) along the pressure axis and semi-minor axis $\Delta V$ (in $\text{m}^3$) along the volume axis. The net work done by the gas in one complete clockwise cycle is:
(A) $\frac{1}{2} \pi \Delta P \Delta V$
(B) $\pi \Delta P \Delta V$
(C) $2\pi \Delta P \Delta V$
(D) $4\pi \Delta P \Delta V$
Q4. If the absolute temperature of an ideal gas is increased by a factor of $4$ ($T \to 4T$), its root mean square speed ($v_{\text{rms}}$) increases by a factor of:
(A) $\sqrt{2}$
(B) $2$
(C) $4$
(D) $16$
Q5. A clean glass capillary tube dipped in water gives a vertical capillary rise of $h = 6.0\text{ cm}$. If the capillary tube is pushed vertically downward until only $4.0\text{ cm}$ of the tube projects above the water surface, the water in the capillary tube will:
(A) Overflow continuously like a fountain
(B) Rise up to $4.0\text{ cm}$ and spill down the sides
(C) Rise to the top rim ($4.0\text{ cm}$) and form a meniscus of larger radius of curvature without overflowing
(D) Depress below the outside liquid level
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. Which of the following statements regarding the physical and thermal properties of matter is/are correct?
(A) Steel is more elastic than rubber because a greater tensile stress is required in steel to produce the same strain ($Y_{\text{steel}} > Y_{\text{rubber}}$).
(B) When water freezes into ice, its volume expands, meaning that the melting point of ice decreases with an increase in external pressure.
(C) For an ideal gas, the molar heat capacity during an isothermal process is infinite ($C \to \infty$), while during an adiabatic process it is zero ($C = 0$).
(D) In a closed rigid vessel of constant volume, the mean free path of ideal gas molecules is strictly independent of temperature.
Q7. An ideal rigid diatomic gas ($\gamma = 1.40$) is taken through a cyclic process consisting of an isobaric expansion, an isochoric cooling, and an isothermal compression back to its initial state. Which of the following statements is/are correct?
(A) In the isobaric leg, the fraction of absorbed heat converted into external boundary work is $\frac{\gamma – 1}{\gamma} = \frac{2}{7}$.
(B) In the isochoric leg, zero mechanical boundary work is performed ($W = 0$).
(C) In the isothermal compression leg, the internal energy of the gas remains unchanged ($\Delta U = 0$).
(D) The net work performed over the complete cycle is strictly equal to the net heat absorbed over the cycle.
Q8. Two spherical soap bubbles of radii $r_1$ and $r_2$ ($r_1 > r_2$) are blown in an evacuated enclosure and connected by a thin tube with a closed stopcock. When the stopcock is opened:
(A) Air flows spontaneously from the smaller bubble into the larger bubble because internal excess pressure satisfies $\Delta P = \frac{4S}{r} \propto \frac{1}{r}$.
(B) The smaller bubble shrinks further while the larger bubble expands.
(C) If the bubbles coalesce maintaining a common interface, the radius of curvature of the common film is $R_c = \frac{r_1 r_2}{r_1 – r_2}$.
(D) The spontaneous transfer of air decreases the total surface area and surface energy of the system.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. A Carnot heat engine operates between temperatures $T_H = 500\text{ K}$ and $T_C = 300\text{ K}$. In each cycle, it absorbs $Q_H = 1500\text{ J}$ of heat from the hot reservoir. The useful mechanical work delivered by the engine per cycle in Joules is $W$. Determine the integer value of $W$.
Q10. A uniform steel rod of length $L_0 = 1.0\text{ m}$ and cross-sectional area $A = 2.0 \times 10^{-4}\text{ m}^2$ has Young’s modulus $Y = 2.0 \times 10^{11}\text{ N/m}^2$ and coefficient of linear thermal expansion $\alpha = 1.0 \times 10^{-5}\text{ K}^{-1}$. It is held securely between two rigid immovable walls with zero initial stress. When its temperature is raised by $\Delta T = 25\text{ K}$, the compressive thermal force developed in the rod in kilo-Newtons ($\text{kN}$) is $F$. Determine the integer value of $F$.
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (C) $1 : 8$ | Wire Elongation Scaling ($\Delta L \propto L / d^2$) |
| Q2 | (D) $4.0\text{ m/s}$ | Fluid Equation of Continuity ($A_1 v_1 = A_2 v_2$) |
| Q3 | (B) $\pi \Delta P \Delta V$ | Area of Elliptical Thermodynamic Indicator Cycle |
| Q4 | (B) $2$ | Root Mean Square Speed Scaling ($v_{\text{rms}} \propto \sqrt{T}$) |
| Q5 | (C) Top rim rise with larger radius | Insufficient Capillary Tube Behavior |
| Q6 | (A), (B), (C), (D) | Cross-Unit Matter & Thermal Principles |
| Q7 | (A), (B), (C), (D) | Thermodynamic Indicator Cycle Energy Partitioning |
| Q8 | (A), (B), (C), (D) | Excess Pressure & Connected Soap Bubble Dynamics |
| Q9 | 600 | Carnot Engine Efficiency ($W = \eta Q_H = 600\text{ J}$) |
| Q10 | 10 | Thermal Compressive Force ($F = Y A \alpha \Delta T = 10\text{ kN}$) |
Complete Step-by-Step Solutions
Solution 1:
The elongation of a wire under tensile force $F$ is:
$$\Delta L = \frac{F L}{A Y} = \frac{F L}{\left(\frac{\pi d^2}{4}\right) Y} = \frac{4 F L}{\pi Y d^2} \implies \Delta L \propto \frac{L}{d^2}$$
Taking the ratio for wires $A$ and $B$:
$$\frac{\Delta L_A}{\Delta L_B} = \left(\frac{L_A}{L_B}\right) \left(\frac{d_B}{d_A}\right)^2 = \left(\frac{1}{2}\right) \left(\frac{1}{2}\right)^2 = \frac{1}{2} \times \frac{1}{4} = \mathbf{\frac{1}{8}}$$
Correct Option: (C)
Solution 2:
By the equation of continuity for an incompressible liquid:
$$A_1 v_1 = A_2 v_2 \implies \left(\frac{\pi d_1^2}{4}\right) v_1 = \left(\frac{\pi d_2^2}{4}\right) v_2 \implies v_2 = v_1 \left(\frac{d_1}{d_2}\right)^2$$
Given $d_1 = 4.0\text{ cm}, d_2 = 2.0\text{ cm}$, and $v_1 = 1.0\text{ m/s}$:
$$v_2 = (1.0\text{ m/s}) \left(\frac{4.0}{2.0}\right)^2 = 1.0 \times 4 = \mathbf{4.0\text{ m/s}}$$
Correct Option: (D)
Solution 3:
The work done by the gas in a complete thermodynamic cycle is the enclosed area on the $P-V$ diagram.
For an ellipse with semi-major axis $\Delta P$ and semi-minor axis $\Delta V$, the enclosed area is:
$$W = \text{Area} = \pi \times (\text{semi-axis } 1) \times (\text{semi-axis } 2) = \mathbf{\pi \Delta P \Delta V}$$
Correct Option: (B)
Solution 4:
The root mean square speed of an ideal gas is:
$$v_{\text{rms}} = \sqrt{\frac{3 R T}{M}} \propto \sqrt{T}$$
When temperature quadruples ($T \to 4T$):
$$v_{\text{rms}}’ = \sqrt{4} v_{\text{rms}} = \mathbf{2 v_{\text{rms}}}$$
Correct Option: (B)
Solution 5:
Under Jurin’s law, $h \cdot R_m = \text{constant} = \frac{2S}{\rho g}$, where $R_m = r / \cos\theta$ is the radius of curvature of the liquid meniscus. If the tube is cut or pushed so that its available height $h’ < h$, water rises to the top edge and flattens its meniscus, increasing $R_m$ so that $h' R_m' = h R_m$. Water never overflows because doing so would require an unstable increase in free surface energy without an energy source.
Correct Option: (C)
Solution 6:
(A) Correct: Modulus of elasticity $Y = \text{stress} / \text{strain}$. For a given strain, steel sustains far greater stress than rubber, so $Y_{\text{steel}} \gg Y_{\text{rubber}}$.
(B) Correct: By the Clausius-Clapeyron equation $\frac{dP}{dT} = \frac{L}{T \Delta V}$. Since ice expands on freezing, $\Delta V = V_{\text{water}} – V_{\text{ice}} < 0$, making $\frac{dP}{dT} < 0$.
(C) Correct: For isothermal paths, $dT = 0 \implies C = dQ / (n dT) \to \infty$. For adiabatic paths, $dQ = 0 \implies C = 0$.
(D) Correct: For a closed vessel of fixed volume $V$, number density $n = N/V$ is constant, so $\lambda = \frac{1}{\sqrt{2} n \pi d^2}$ is strictly invariant with temperature.
Correct Options: (A), (B), (C), (D)
Solution 7:
(A) Correct: In an isobaric process, $W = n R \Delta T$ and $Q = n C_p \Delta T$. The work fraction is $W / Q = R / C_p = \frac{\gamma – 1}{\gamma} = \frac{0.4}{1.4} = \frac{2}{7}$.
(B) Correct: At constant volume, $dV = 0 \implies W = \int P dV = 0$.
(C) Correct: For an ideal gas, $U$ depends solely on temperature: $\Delta T = 0 \implies \Delta U = 0$.
(D) Correct: Over any complete cycle, initial and final states are identical: $\Delta U_{\text{net}} = 0 \implies W_{\text{net}} = Q_{\text{net}}$.
Correct Options: (A), (B), (C), (D)
Solution 8:
(A) Correct: Excess pressure inside a bubble is $\Delta P = \frac{4S}{r}$. The smaller bubble ($r_2$) has higher pressure than the larger bubble ($r_1$), driving air from smaller to larger.
(B) Correct: Air leaves the smaller bubble, causing it to shrink, while entering the larger bubble, causing it to expand.
(C) Correct: At the common interface, $\Delta P = \Delta P_2 – \Delta P_1 = 4S \left(\frac{1}{r_2} – \frac{1}{r_1}\right) = \frac{4S}{R_c} \implies R_c = \frac{r_1 r_2}{r_1 – r_2}$.
(D) Correct: Coalescence/expansion reduces the total surface area, spontaneously lowering surface energy.
Correct Options: (A), (B), (C), (D)
Solution 9:
The thermal efficiency of the Carnot engine is:
$$\eta = 1 – \frac{T_C}{T_H} = 1 – \frac{300\text{ K}}{500\text{ K}} = 1 – 0.60 = 0.40$$
The work delivered per cycle is:
$$W = \eta Q_H = 0.40 \times 1500\text{ J} = \mathbf{600\text{ J}}$$
Answer: 600
Solution 10:
Thermal stress developed in the rigidly constrained rod:
$$\sigma = Y \alpha \Delta T = (2.0 \times 10^{11}\text{ N/m}^2) \times (1.0 \times 10^{-5}\text{ K}^{-1}) \times (25\text{ K}) = 5.0 \times 10^7\text{ N/m}^2$$
The compressive force developed is:
$$F = \sigma A = (5.0 \times 10^7\text{ N/m}^2) \times (2.0 \times 10^{-4}\text{ m}^2) = 10000\text{ N} = \mathbf{10\text{ kN}}$$
Answer: 10