Welcome to Lecture 89 of our comprehensive JEE Physics Master Series. In this lecture, we begin our exploration of thermal physics: Heat, Temperature, and Thermal Expansion. We deconstruct the microscopic lattice origin of expansion from asymmetric potential energy curves, derive the fundamental relations between the linear ($\alpha$), superficial ($\beta$), and cubical ($\gamma$) expansion coefficients, analyze thermal stress in constrained structures, evaluate the apparent expansion of liquids in deformable vessels, and formulate the mechanics of bimetallic strips and pendulum clocks.
1. Detailed Concept Card
1.1 Microscopic Origin of Thermal Expansion
Thermal expansion is a direct macroscopic manifestation of the asymmetry of intermolecular potential energy curves ($U(r)$ versus interatomic separation $r$):
- At absolute zero ($T = 0\text{ K}$), atoms occupy the potential energy minimum at equilibrium separation $r_0$.
- As temperature increases, atoms absorb thermal kinetic energy and oscillate about their equilibrium positions.
- If the potential well were purely harmonic (parabolic: $U(r) = \frac{1}{2}k(r – r_0)^2$), atomic oscillations would be strictly symmetric. The time-averaged mean separation $\langle r \rangle$ would remain identically equal to $r_0$ at all temperatures, resulting in zero thermal expansion!
- In real solids, the potential well is anharmonic (asymmetric): it rises steeply at $r < r_0$ due to Pauli electron repulsion, while flattening gradually at $r > r_0$ due to Van der Waals attraction.
- Consequently, as thermal oscillation amplitude increases with temperature, the mean interatomic separation shifts outward: $$\langle r \rangle_T > \langle r \rangle_0 \implies \Delta r > 0$$
- This microscopic increase in mean separation across billions of lattice planes manifests macroscopically as Thermal Expansion.
1.2 Coefficients of Thermal Expansion
A. Linear Expansion ($\alpha$)
For a long, thin rod of initial length $L_0$ at temperature $T_0$, the increase in length $\Delta L$ when heated by $\Delta T$ is:
$$\Delta L = L_0 \, \alpha \, \Delta T \implies L(T) = L_0 (1 + \alpha \Delta T)$$
where $\alpha$ is the Coefficient of Linear Expansion:
$$\alpha = \frac{1}{L_0} \frac{dL}{dT} \quad [\text{Unit: } \text{K}^{-1} \text{ or } ^\circ\text{C}^{-1}, \quad \text{Dimensions: } [\text{M}^0 \text{L}^0 \text{T}^0 \Theta^{-1}]]$$
B. Superficial / Areal Expansion ($\beta$)
For a thin laminar sheet of initial area $A_0$:
$$\Delta A = A_0 \, \beta \, \Delta T \implies A(T) = A_0 (1 + \beta \Delta T)$$
For an isotropic solid with initial dimensions $L_1$ and $L_2$ ($A_0 = L_1 L_2$):
$$A(T) = L_1(1 + \alpha \Delta T) \cdot L_2(1 + \alpha \Delta T) = A_0 (1 + 2\alpha \Delta T + \alpha^2 \Delta T^2)$$
Because $\alpha \approx 10^{-5}\text{ K}^{-1}$, the second-order term $\alpha^2 \Delta T^2 \ll 1$ is negligible:
$$\mathbf{\beta = 2\alpha}$$
C. Cubical / Volume Expansion ($\gamma$)
For a three-dimensional body of initial volume $V_0$:
$$\Delta V = V_0 \, \gamma \, \Delta T \implies V(T) = V_0 (1 + \gamma \Delta T)$$
For an isotropic body ($V_0 = L_1 L_2 L_3$):
$$V(T) = V_0 (1 + \alpha \Delta T)^3 \approx V_0 (1 + 3\alpha \Delta T) \implies \mathbf{\gamma = 3\alpha}$$
The Golden Isotropic Ratio:
$$\mathbf{\alpha : \beta : \gamma = 1 : 2 : 3 \implies \alpha = \frac{\beta}{2} = \frac{\gamma}{3}}$$
Anisotropic Crystals: If the linear coefficients along the principal orthogonal crystal axes are $\alpha_x, \alpha_y, \alpha_z$:
$$\mathbf{\gamma = \alpha_x + \alpha_y + \alpha_z}$$
D. Expansion of Cavities (The Photographic Enlargement Principle)
When a hollow object (such as a metal washer with a central hole, or a hollow spherical shell) is heated, the cavity expands identically to a solid body of the same material filling that cavity:
- The perimeter, diameter, and volume of any internal hole increase upon heating.
- Thermal expansion acts like a geometric photographic magnification centered at any point in the solid.
1.3 Thermal Stress in Constrained Structures
If a rod of length $L_0$, cross-sectional area $A$, and Young’s modulus $Y$ is free to expand, heating by $\Delta T$ produces free expansion $\Delta L_{\text{free}} = L_0 \alpha \Delta T$ with zero internal stress.
However, if the rod is clamped rigidly between two immovable unyielding supports:
- The rigid supports exert an inward compressive force $F$ that prevents the expansion, compressing the rod back by $\Delta L = L_0 \alpha \Delta T$.
- The compressive Thermal Strain is: $$\epsilon_{\text{thermal}} = \frac{\Delta L}{L_0} = \alpha \Delta T$$
- The resulting Thermal Stress is: $$\mathbf{\sigma_{\text{thermal}} = Y \cdot \epsilon_{\text{thermal}} = Y \, \alpha \, \Delta T}$$
- The compressive reaction force exerted by the rod against each wall is: $$\mathbf{F = \sigma_{\text{thermal}} \cdot A = Y \, A \, \alpha \, \Delta T}$$
Critical Insight: The thermal stress $\sigma = Y\alpha \Delta T$ depends only on material properties ($Y, \alpha$) and temperature rise ($\Delta T$). It is completely independent of the length $L_0$ and cross-sectional area $A$ of the rod!
1.4 Real and Apparent Expansion of Liquids
Liquids do not have a fixed shape, so only cubical (volumetric) expansion is meaningful. When measuring liquid expansion, the solid container holding the liquid also expands:
- Real Coefficient of Volume Expansion ($\gamma_r$): The true fractional change in volume of the liquid per unit temperature rise: $$\Delta V_{\text{real}} = V_0 \, \gamma_r \, \Delta T$$
- Apparent Coefficient of Volume Expansion ($\gamma_a$): The observed overflow or meniscus rise relative to the graduated scale of the vessel: $$\Delta V_{\text{apparent}} = V_0 \, \gamma_a \, \Delta T$$
- Volume Expansion of the Vessel ($\gamma_g = 3\alpha_g$): $$\Delta V_{\text{vessel}} = V_0 (3\alpha_g) \Delta T$$
From volume conservation ($\Delta V_{\text{real}} = \Delta V_{\text{apparent}} + \Delta V_{\text{vessel}}$):
$$\mathbf{\gamma_r = \gamma_a + 3\alpha_g \implies \gamma_a = \gamma_r – 3\alpha_g}$$
Regimes of Observed Liquid Meniscus in a Heated Vessel:
- If $\gamma_r > 3\alpha_g \implies \gamma_a > 0$: The liquid expands faster than the vessel, so the liquid level rises (or overflows if initially full).
- If $\gamma_r = 3\alpha_g \implies \gamma_a = 0$: The liquid and vessel expand identically; the liquid level remains completely stationary.
- If $\gamma_r < 3\alpha_g \implies \gamma_a < 0$: The vessel expands faster than the liquid; the liquid level drops upon heating!
Variation of Density with Temperature:
Because mass $m = \rho V$ is conserved:
$$\rho(T) = \frac{m}{V(T)} = \frac{\rho_0 V_0}{V_0 (1 + \gamma \Delta T)} = \frac{\rho_0}{1 + \gamma \Delta T} \approx \mathbf{\rho_0 (1 – \gamma \Delta T)}$$
1.5 Anomalous Expansion of Water
Almost all liquids expand continuously as temperature increases. Water exhibits a striking and biologically vital anomaly between $0^\circ\text{C}$ and $4^\circ\text{C}$:
- When water is heated from $0^\circ\text{C}$ to $4^\circ\text{C}$, it contracts in volume ($\gamma < 0$).
- At $4^\circ\text{C}$, water attains its minimum volume and maximum density ($\rho_{\text{max}} = 1000\text{ kg/m}^3 = 1.000\text{ g/cm}^3$).
- Above $4^\circ\text{C}$, water expands normally with a positive expansion coefficient ($\gamma > 0$).
- Lattice Mechanism: Ice has an open cage-like hexagonal hydrogen-bonded crystal structure. When ice melts at $0^\circ\text{C}$, fragmented hydrogen-bonded cages collapse, allowing water molecules to pack more closely until $4^\circ\text{C}$, where normal thermal agitation overtakes packing collapse.
- Ecological Significance: In winter, surface lake water cools to $4^\circ\text{C}$, becomes densest, and sinks to the bottom. Further cooling of surface water below $4^\circ\text{C}$ decreases its density, keeping near-freezing water at the top. Ice freezes from the surface downward, forming an insulating lid that keeps the deep bottom water at $4^\circ\text{C}$, sustaining aquatic life.
1.6 Bimetallic Strips & Pendulum Clock Error Dynamics
A. Bimetallic Strip Curvature
A bimetallic strip consists of two dissimilar thin metal strips of equal initial length $L_0$ and thickness $t_1 = t_2 = t$ firmly bonded (riveted or welded) along their lengths. Let $\alpha_2 > \alpha_1$ (e.g., brass and steel).
- When heated by $\Delta T$, strip 2 expands more than strip 1. Because they are bonded, the composite strip bends into a circular arc with the more expansive metal on the outer (convex) side.
- The radius of curvature $R$ of the neutral interface between the strips is: $$\mathbf{R \approx \frac{d}{(\alpha_2 – \alpha_1) \Delta T} = \frac{2t}{(\alpha_2 – \alpha_1) \Delta T}}$$ where $d = 2t$ is the total thickness of the composite strip.
- When cooled below bonding temperature ($\Delta T < 0$), the strip bends in the opposite direction (strip 2 on the inner concave side).
B. Fractional Error in Pendulum Clocks
The time period of a simple pendulum of effective length $L$ is $T_0 = 2\pi \sqrt{\frac{L}{g}}$. Differentiating logarithmically:
$$\frac{\Delta T_{\text{period}}}{T_0} = \frac{1}{2} \frac{\Delta L}{L_0} = \mathbf{\frac{1}{2} \alpha \Delta \theta}$$
where $\Delta \theta$ is the change in ambient temperature. In a total elapsed time $t$ (e.g., one day $= 86400\text{ s}$), the clock loses or gains a total time $\Delta t$:
$$\mathbf{\Delta t = \frac{1}{2} \alpha \Delta \theta \cdot t = \frac{1}{2} \alpha \Delta \theta \times 86400\text{ seconds/day}}$$
- In summer ($\Delta \theta > 0$): Length increases, time period increases, clock ticks slower $\implies$ clock loses time.
- In winter ($\Delta \theta < 0$): Length decreases, time period decreases, clock ticks faster $\implies$ clock gains time.
2. Four Graded Solved Examples
Example 1: Thermal Stress in Clamped Steel Rail (JEE Main Level)
Problem Statement: A structural steel rail of length $L_0 = 1.0\text{ m}$ and cross-sectional area $A = 2.0\text{ cm}^2$ is fitted rigidly between two immovable concrete abutments at $T_1 = 20^\circ\text{C}$ with zero initial stress. The temperature rises to $T_2 = 70^\circ\text{C}$. Taking the coefficient of linear expansion of steel as $\alpha = 1.2 \times 10^{-5}\text{ K}^{-1}$ and Young’s modulus as $Y = 2.0 \times 10^{11}\text{ N/m}^2$, calculate:
- (a) The thermal strain developed in the rail.
- (b) The compressive thermal stress generated within the rail.
- (c) The total compressive reaction force exerted by the rail against each abutment.
Step-by-Step Solution:
Step 1: Calculate temperature change $\Delta T$:
$$\Delta T = T_2 – T_1 = 70^\circ\text{C} – 20^\circ\text{C} = 50\text{ K}$$
Step 2: Calculate thermal strain:
$$\epsilon = \frac{\Delta L_{\text{free}}}{L_0} = \alpha \Delta T = (1.2 \times 10^{-5}\text{ K}^{-1}) \times 50\text{ K} = 6.0 \times 10^{-4}$$
Step 3: Calculate compressive thermal stress $\sigma$:
$$\sigma = Y \cdot \epsilon = Y \, \alpha \, \Delta T = (2.0 \times 10^{11}\text{ N/m}^2) \times (6.0 \times 10^{-4}) = 1.20 \times 10^8\text{ N/m}^2 = 120\text{ MPa}$$
Step 4: Calculate reaction force $F$:
Convert area to SI units: $A = 2.0\text{ cm}^2 = 2.0 \times 10^{-4}\text{ m}^2$.
$$F = \sigma \cdot A = (1.20 \times 10^8\text{ N/m}^2) \times (2.0 \times 10^{-4}\text{ m}^2) = 24000\text{ N} = 24.0\text{ kN}$$
Key Takeaway: A modest $50\text{ K}$ temperature rise in a rigidly constrained rail creates a massive compressive load ($24\text{ kN} \approx 2.4\text{ metric tons force}$), illustrating why railway tracks must incorporate expansion joints or pre-stressing.
Example 2: Apparent Liquid Expansion & Mercury Overflow (Volumetric Dynamics)
Problem Statement: A Pyrex glass flask has an internal volume of $V_0 = 1000\text{ cm}^3$ at $0^\circ\text{C}$ and is completely filled with mercury to the brim. The system is heated to $100^\circ\text{C}$. The coefficient of real cubical expansion of mercury is $\gamma_{\text{Hg}} = 1.80 \times 10^{-4}\text{ K}^{-1}$, and the coefficient of linear expansion of Pyrex glass is $\alpha_g = 9.0 \times 10^{-6}\text{ K}^{-1}$. Determine:
- (a) The coefficient of apparent expansion of mercury in this glass flask.
- (b) The volume of mercury that overflows from the flask.
- (c) The mass of overflowing mercury (density of mercury at $0^\circ\text{C}$ is $\rho_0 = 13.6\text{ g/cm}^3$).
Step-by-Step Solution:
Step 1: Calculate volume expansion coefficient of the glass flask:
$$\gamma_g = 3\alpha_g = 3 \times (9.0 \times 10^{-6}\text{ K}^{-1}) = 2.70 \times 10^{-5}\text{ K}^{-1} = 0.27 \times 10^{-4}\text{ K}^{-1}$$
Step 2: Calculate coefficient of apparent expansion $\gamma_a$:
$$\gamma_a = \gamma_{\text{Hg}} – \gamma_g = 1.80 \times 10^{-4} – 0.27 \times 10^{-4} = 1.53 \times 10^{-4}\text{ K}^{-1}$$
Step 3: Calculate the volume of overflowing mercury $\Delta V_{\text{overflow}}$:
$$\Delta V_{\text{overflow}} = V_0 \, \gamma_a \, \Delta T = (1000\text{ cm}^3) \times (1.53 \times 10^{-4}\text{ K}^{-1}) \times (100\text{ K}) = 15.30\text{ cm}^3$$
Step 4: Calculate mass of overflowing mercury:
At $100^\circ\text{C}$, the density of mercury is:
$$\rho_{100} = \frac{\rho_0}{1 + \gamma_{\text{Hg}}\Delta T} = \frac{13.6}{1 + 1.80 \times 10^{-4} \times 100} = \frac{13.6}{1.018} \approx 13.36\text{ g/cm}^3$$
$$m_{\text{overflow}} = \Delta V_{\text{overflow}} \times \rho_{100} = 15.30\text{ cm}^3 \times 13.36\text{ g/cm}^3 \approx 204.4\text{ g}$$
Key Takeaway: The volume of liquid that overflows is governed exclusively by the apparent expansion coefficient $\gamma_a = \gamma_{\text{liquid}} – 3\alpha_{\text{vessel}}$. To convert this volume to mass, one must evaluate density at the final temperature ($100^\circ\text{C}$).
Example 3: Pendulum Clock Error & Thermal Compensation (JEE Advanced Level)
Problem Statement: A pendulum clock constructed with a brass suspension rod ($\alpha_b = 1.90 \times 10^{-5}\text{ K}^{-1}$) keeps precise astronomical time at $T_0 = 20.0^\circ\text{C}$.
- (a) During a heatwave, the average daily temperature rises to $40.0^\circ\text{C}$. How many seconds does the clock gain or lose each day ($24\text{ hours}$)?
- (b) To make the clock thermally invariant, a composite pendulum rod is constructed using a steel rod of length $L_s$ ($\alpha_s = 1.20 \times 10^{-5}\text{ K}^{-1}$) and a zinc tube of length $L_z$ ($\alpha_z = 2.40 \times 10^{-5}\text{ K}^{-1}$) in an opposing arrangement. Find the ratio $L_s / L_z$ required so that the effective length remains strictly independent of temperature.
Step-by-Step Solution:
Step 1: Calculate daily time error:
$$\Delta \theta = 40.0^\circ\text{C} – 20.0^\circ\text{C} = 20.0\text{ K}$$
Because $\Delta \theta > 0$, the pendulum expands, its period increases, and the clock loses time (runs slow):
$$\Delta t = \frac{1}{2} \alpha_b \Delta \theta \times 86400\text{ s} = \frac{1}{2} \times (1.90 \times 10^{-5}) \times (20.0) \times 86400$$
$$\Delta t = (1.90 \times 10^{-4}) \times 86400 = 16.416\text{ seconds/day}$$
The clock loses approximately $16.42\text{ s}$ per day.
Step 2: Condition for thermal compensation in composite pendulum:
In a compensated gridiron or concentric tube pendulum, the steel rod expands downward while the zinc tube expands upward. The net effective length is $L_{\text{eff}} = L_s – L_z$.
For $L_{\text{eff}}$ to be independent of temperature:
$$\frac{dL_{\text{eff}}}{dT} = 0 \implies \frac{dL_s}{dT} – \frac{dL_z}{dT} = 0$$
$$L_s \alpha_s \Delta T = L_z \alpha_z \Delta T \implies L_s \alpha_s = L_z \alpha_z$$
$$\frac{L_s}{L_z} = \frac{\alpha_z}{\alpha_s} = \frac{2.40 \times 10^{-5}}{1.20 \times 10^{-5}} = 2.0$$
Key Takeaway: Perfect thermal compensation requires the length ratio of the opposing elements to be inversely proportional to their thermal expansion coefficients ($L_s / L_z = \alpha_z / \alpha_s = 2.0$).
Example 4: Curvature Derivation of a Bimetallic Thermostat Strip
Problem Statement: A bimetallic strip is formed by welding together two strips of brass ($\alpha_2 = 1.90 \times 10^{-5}\text{ K}^{-1}$) and invar steel ($\alpha_1 = 1.20 \times 10^{-6}\text{ K}^{-1}$), each of uniform thickness $t = 0.50\text{ mm}$ and initial length $L_0 = 10.0\text{ cm}$. The strip is heated by $\Delta T = 100\text{ K}$.
- (a) Derive the radius of curvature $R$ of the bimetallic arc from fundamental strain geometry.
- (b) Calculate the numerical value of $R$.
- (c) Calculate the deflection $\delta$ of the free tip if one end is rigidly clamped.
Step-by-Step Solution:
Step 1: Geometric derivation of radius of curvature $R$:
Let the bent strip subtend an angle $\phi$ at its center of curvature. The centerline of strip 1 is at radius $R – t/2$, and the centerline of strip 2 is at radius $R + t/2$:
$$L_1 = \left(R – \frac{t}{2}\right) \phi = L_0 (1 + \alpha_1 \Delta T)$$
$$L_2 = \left(R + \frac{t}{2}\right) \phi = L_0 (1 + \alpha_2 \Delta T)$$
Subtracting the two equations:
$$L_2 – L_1 = t \phi = L_0 (\alpha_2 – \alpha_1) \Delta T$$
Dividing by $L_0 \approx R \phi$:
$$\frac{t}{R} = (\alpha_2 – \alpha_1) \Delta T \implies \mathbf{R = \frac{t}{(\alpha_2 – \alpha_1) \Delta T}}$$
If each strip has thickness $t$ (total composite thickness $d = 2t$):
$$R = \frac{2t}{2(\alpha_2 – \alpha_1) \Delta T} = \frac{d}{2(\alpha_2 – \alpha_1)\Delta T}$$
Using the standard thin strip relation $R = \frac{2t}{(\alpha_2 – \alpha_1)\Delta T}$:
Step 2: Numerical calculation of $R$:
$$2t = 2 \times (0.50 \times 10^{-3}\text{ m}) = 1.0 \times 10^{-3}\text{ m}$$
$$\alpha_2 – \alpha_1 = 1.90 \times 10^{-5} – 0.12 \times 10^{-5} = 1.78 \times 10^{-5}\text{ K}^{-1}$$
$$R = \frac{1.0 \times 10^{-3}}{(1.78 \times 10^{-5}) \times 100} = \frac{1.0 \times 10^{-3}}{1.78 \times 10^{-3}} = \frac{1}{1.78} \approx 0.562\text{ m} = 56.2\text{ cm}$$
Step 3: Calculate tip deflection $\delta$:
For small deflections ($L_0 \ll R$), the circular arc is approximated by a parabola:
$$\delta \approx \frac{L_0^2}{2R} = \frac{(0.10\text{ m})^2}{2 \times 0.562\text{ m}} = \frac{0.010}{1.124} \approx 8.90 \times 10^{-3}\text{ m} = 8.90\text{ mm}$$
Key Takeaway: Bimetallic strips amplify microscopic differential thermal expansion into large, measurable tip deflections ($8.9\text{ mm}$), forming the operational basis for electrical circuit breakers and thermostats.
3. Practice Worksheet (10 Questions)
Questions 1 to 5: JEE Main Level (Single Correct Option)
Q1. For an isotropic homogeneous solid, the coefficients of linear expansion ($\alpha$), superficial expansion ($\beta$), and cubical expansion ($\gamma$) are related by:
- (A) $\alpha : \beta : \gamma = 1 : 2 : 3$
- (B) $\alpha : \beta : \gamma = 3 : 2 : 1$
- (C) $\alpha : \beta : \gamma = 1 : 4 : 9$
- (D) $\alpha : \beta : \gamma = 1 : \sqrt{2} : \sqrt{3}$
Q2. A uniform circular metal plate has a concentric circular hole cut in its center. When the plate is heated uniformly:
- (A) The diameter of the plate increases, but the diameter of the hole decreases.
- (B) The diameter of the plate increases, and the diameter of the hole increases.
- (C) The diameter of the plate increases, while the diameter of the hole remains unchanged.
- (D) Both the diameter of the plate and the diameter of the hole decrease.
Q3. A pendulum clock keeps correct time at $20^\circ\text{C}$. If the clock is placed in an environment maintained at $35^\circ\text{C}$:
- (A) It gains time because its length decreases.
- (B) It loses time because its length increases.
- (C) It gains time because its frequency increases.
- (D) Its timekeeping remains unaltered because gravity is constant.
Q4. Water is heated continuously from $0^\circ\text{C}$ to $10^\circ\text{C}$. Its volume:
- (A) Increases monotonically throughout the heating range.
- (B) Decreases monotonically throughout the heating range.
- (C) First decreases up to $4^\circ\text{C}$ and then increases.
- (D) First increases up to $4^\circ\text{C}$ and then decreases.
Q5. An anisotropic crystal has linear coefficients of expansion along three mutually orthogonal axes given by $\alpha_x = 2.0 \times 10^{-5}\text{ K}^{-1}$, $\alpha_y = 3.0 \times 10^{-5}\text{ K}^{-1}$, and $\alpha_z = 5.0 \times 10^{-5}\text{ K}^{-1}$. The coefficient of volume expansion $\gamma$ of this crystal is:
- (A) $3.33 \times 10^{-5}\text{ K}^{-1}$
- (B) $5.00 \times 10^{-5}\text{ K}^{-1}$
- (C) $1.00 \times 10^{-4}\text{ K}^{-1}$
- (D) $3.00 \times 10^{-4}\text{ K}^{-1}$
Questions 6 to 8: JEE Advanced Level (One or More than One Correct)
Q6. A metal rod of length $L$, cross-sectional area $A$, Young’s modulus $Y$, and coefficient of linear expansion $\alpha$ is clamped rigidly at both ends between two unyielding immovable walls. The temperature of the rod is increased by $\Delta T$. Which of the following statements are correct?
- (A) The thermal strain developed in the rod is $\epsilon = \alpha \Delta T$.
- (B) The compressive thermal stress developed in the rod is $\sigma = Y \alpha \Delta T$.
- (C) The thermal stress depends inversely on the original length $L$ of the rod.
- (D) The compressive force exerted on each wall is $F = Y A \alpha \Delta T$.
Q7. A glass vessel of volumetric expansion coefficient $\gamma_g = 3\alpha_g$ contains a liquid of real volumetric expansion coefficient $\gamma_r$. Which of the following statements regarding the behavior of the liquid level upon heating are correct?
- (A) The coefficient of apparent expansion of the liquid is $\gamma_a = \gamma_r – 3\alpha_g$.
- (B) If $\gamma_r > 3\alpha_g$, the liquid level in the vessel rises.
- (C) If $\gamma_r = 3\alpha_g$, the liquid level in the vessel remains unchanged.
- (D) If $\gamma_r < 3\alpha_g$, the liquid level in the vessel falls.
Q8. A bimetallic strip made of two metals $A$ and $B$ (coefficients $\alpha_A > \alpha_B$) is straight at room temperature. Which of the following statements are correct?
- (A) When heated above room temperature, it bends into an arc with metal $A$ on the convex side.
- (B) When cooled below room temperature, it bends into an arc with metal $A$ on the concave side.
- (C) The radius of curvature of the bent strip is inversely proportional to the temperature change $\Delta T$.
- (D) If the thickness of each strip is doubled, the radius of curvature becomes halved.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. Two metal rods of initial lengths $L_1$ and $L_2 = 30.0\text{ cm}$ have linear expansion coefficients $\alpha_1 = 1.8 \times 10^{-5}\text{ K}^{-1}$ and $\alpha_2 = 1.2 \times 10^{-5}\text{ K}^{-1}$ respectively. If the difference in their lengths $(L_2 – L_1)$ remains constant at all temperatures, determine the length $L_1$ (in $\text{cm}$).
Q10. A steel wire of cross-sectional area $A = 1.0\text{ mm}^2$ is stretched tightly without tension between two rigid fixed posts at $40.0^\circ\text{C}$. The temperature subsequently drops to $20.0^\circ\text{C}$. Given that $Y = 2.0 \times 10^{11}\text{ N/m}^2$ and $\alpha = 1.2 \times 10^{-5}\text{ K}^{-1}$, calculate the tension force (in $\text{N}$) developed in the wire.
4. Detailed Solutions & Answer Key
Quick Answer Key
| Question | Correct Answer | Topic / Category |
|---|---|---|
| Q1 | (A) | Expansion Coefficients Ratio |
| Q2 | (B) | Expansion of Internal Cavity |
| Q3 | (B) | Pendulum Timekeeping in Heat |
| Q4 | (C) | Anomalous Expansion of Water |
| Q5 | (C) | Anisotropic Volume Expansion |
| Q6 | (A), (B), (D) | Thermal Stress Independence of Length |
| Q7 | (A), (B), (C), (D) | Apparent Expansion Regimes in Liquids |
| Q8 | (A), (B), (C) | Bimetallic Strip Curvature Principles |
| Q9 | 20 | Thermal Compensation Length Invariance |
| Q10 | 48 | Thermal Tension Under Cooling |
Step-by-Step Solutions
Solution 1:
For an isotropic solid:
$$\beta = 2\alpha, \quad \gamma = 3\alpha \implies \alpha : \beta : \gamma = \alpha : 2\alpha : 3\alpha = 1 : 2 : 3$$
Correct Option: (A).
Solution 2:
Thermal expansion causes all linear dimensions of a solid to expand by the factor $(1 + \alpha \Delta T)$. The distance between any two material points increases. Thus, the inner hole expands exactly as if it were a solid disk of the same metal. Both outer and inner diameters increase.
Correct Option: (B).
Solution 3:
As temperature rises from $20^\circ\text{C}$ to $35^\circ\text{C}$, the pendulum length increases: $L = L_0(1 + \alpha \Delta T)$. The time period $T = 2\pi\sqrt{L/g}$ increases, meaning each swing takes longer. Consequently, the pendulum executes fewer oscillations per day and loses time.
Correct Option: (B).
Solution 4:
Due to the anomalous expansion of water, heating water from $0^\circ\text{C}$ to $4^\circ\text{C}$ breaks open hydrogen-bonded clusters, causing volume to decrease to a minimum at $4^\circ\text{C}$. Beyond $4^\circ\text{C}$, normal thermal agitation dominates and volume increases.
Correct Option: (C).
Solution 5:
For an anisotropic solid with orthogonal expansion coefficients:
$$\gamma = \alpha_x + \alpha_y + \alpha_z = (2.0 + 3.0 + 5.0) \times 10^{-5}\text{ K}^{-1} = 10.0 \times 10^{-5}\text{ K}^{-1} = 1.00 \times 10^{-4}\text{ K}^{-1}$$
Correct Option: (C).
Solution 6:
- (A) is Correct: $\epsilon = \Delta L / L_0 = \alpha \Delta T$.
- (B) is Correct: $\sigma = Y \cdot \epsilon = Y \alpha \Delta T$.
- (C) is Incorrect: The stress formula $\sigma = Y \alpha \Delta T$ is completely independent of $L$.
- (D) is Correct: Force $F = \sigma \cdot A = Y A \alpha \Delta T$.
Correct Options: (A), (B), (D).
Solution 7:
- (A) is Correct: $\gamma_a = \gamma_r – 3\alpha_g$ by conservation of volume.
- (B) is Correct: If $\gamma_r > 3\alpha_g$, apparent expansion is positive and the level rises.
- (C) is Correct: If $\gamma_r = 3\alpha_g$, $\gamma_a = 0$ and the liquid level is invariant.
- (D) is Correct: If $\gamma_r < 3\alpha_g$, $\gamma_a < 0$ and the liquid level falls.
Correct Options: (A), (B), (C), (D).
Solution 8:
- (A) is Correct: On heating, metal $A$ expands more, forming the longer outer (convex) perimeter.
- (B) is Correct: On cooling, metal $A$ contracts more, forming the shorter inner (concave) perimeter.
- (C) is Correct: $R = \frac{d}{(\alpha_A – \alpha_B)\Delta T} \propto \frac{1}{\Delta T}$.
- (D) is Incorrect: $R \propto d$; doubling the thickness doubles the radius of curvature (makes it stiffer and less curved).
Correct Options: (A), (B), (C).
Solution 9:
For $(L_2 – L_1)$ to be invariant with temperature:
$$\Delta L_2 = \Delta L_1 \implies L_2 \alpha_2 \Delta T = L_1 \alpha_1 \Delta T$$
$$L_1 = L_2 \left(\frac{\alpha_2}{\alpha_1}\right) = (30.0\text{ cm}) \times \left(\frac{1.2 \times 10^{-5}}{1.8 \times 10^{-5}}\right) = 30.0 \times \frac{2}{3} = 20.0\text{ cm}$$
Final Answer: 20
Solution 10:
When cooled, the wire tries to contract by $\Delta L = L_0 \alpha \Delta T$. Because the posts are rigid, a tensile stress develops:
$$\sigma = Y \alpha \Delta T$$
Given: $A = 1.0\text{ mm}^2 = 1.0 \times 10^{-6}\text{ m}^2$, $\Delta T = 40.0 – 20.0 = 20.0\text{ K}$.
$$F = Y A \alpha \Delta T = (2.0 \times 10^{11}) \times (1.0 \times 10^{-6}) \times (1.2 \times 10^{-5}) \times 20.0$$
$$F = (2.0 \times 10^5) \times (2.4 \times 10^{-4}) = 48.0\text{ N}$$
Final Answer: 48
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