Expansion — Set 2

Correct Answer: A. Increase

• **Increase** = The period of a simple pendulum is T = 2π√(L/g); if length L increases due to thermal expansion, T increases — the pendulum swings more slowly, causing the clock to run slow. • **T ∝ √L** — A 1% increase in length increases the period by about 0.5% — enough to cause measurable time loss in precision pendulum clocks. • 💡 Wrong-option analysis: Decrease: a longer pendulum has a longer period, not shorter; Remain unchanged: period is directly related to length, so length changes affect it; Become zero: the pendulum would still oscillate; only its period changes.

Correct Answer: B. Aluminium

• **Aluminium** = Among the given solids, aluminium has one of the highest linear expansion coefficients (α ≈ 23×10⁻⁶ K⁻¹), greater than steel (≈12), glass (≈9), and far greater than invar (≈1.2). • **α values: Al ≈ 23, Steel ≈ 12, Glass ≈ 9, Invar ≈ 1.2 (all ×10⁻⁶ K⁻¹)** — Aluminium's relatively open crystal structure leads to larger thermal expansion per degree. • 💡 Wrong-option analysis: Invar: specifically designed to have very low α (≈1.2×10⁻⁶ K⁻¹); Steel: moderate expansion (α ≈ 12×10⁻⁶ K⁻¹); Glass: also moderate (α ≈ 9×10⁻⁶ K⁻¹).

Correct Answer: B. Actual increase in volume of the liquid itself

• **Actual increase in volume of the liquid itself** = Real (or absolute) expansion of a liquid is the true increase in its volume due to heating, regardless of what its container does. • **Real = Apparent + Container expansion** — Observing liquid in a container gives apparent expansion; to get real expansion, add the container's volume expansion: γ_real = γ_apparent + γ_container. • 💡 Wrong-option analysis: Increase in mass of the liquid: mass does not change when a liquid is heated; Observed rise of liquid in a capillary only: that is apparent expansion, which is less than real; Decrease in temperature of the liquid: expansion happens on heating, not cooling.

Correct Answer: B. Thermal expansion of a liquid

• **Thermal expansion of a liquid** = A liquid thermometer works because the liquid (mercury, alcohol) expands more than the glass container when heated, causing it to rise in the narrow capillary. • **Apparent expansion** — The liquid level rises by the difference in expansion: (γ_liquid − γ_glass) × V₀ × ΔT; mercury's large apparent expansion makes it very visible. • 💡 Wrong-option analysis: Chemical reaction rate: thermometers do not rely on chemical reactions; Electrical resistance change: that describes resistance thermometers; Magnetic induction: unrelated to liquid-in-glass thermometers.

Correct Answer: A. It can vary with temperature and is treated constant over small ranges

• **It can vary with temperature and is treated constant over small ranges** = The coefficient of thermal expansion is not strictly constant — it changes with temperature; but over small temperature ranges, treating it as constant gives good results. • **Temperature dependence of α** — For precise engineering calculations, tables of α vs. T are used; for everyday calculations, a single average value is sufficient. • 💡 Wrong-option analysis: It is the same for all solids, liquids, and gases: gases have much larger expansion coefficients than solids; It is always exactly constant at all temperatures: α does vary with temperature; It is independent of the material: α is a characteristic property of each material.

Correct Answer: B. Thermal stress

• **Thermal stress** = When a heated rod tries to expand but its ends are rigidly fixed, it cannot change length; internal compressive forces develop — this is called thermal stress. • **Thermal stress = YαΔT** — where Y is Young's modulus; the stress is compressive and can be very large for stiff materials with high α. • 💡 Wrong-option analysis: No force at all: preventing expansion requires the supports to exert compressive forces on the rod; Only shear strain: thermal expansion is longitudinal; Only tensile strain: the constraint produces compressive, not tensile, stress when heating.

Correct Answer: A. Compressive

• **Compressive** = The heated rod tries to elongate; since elongation is prevented, the supports push back, creating compressive stress (the rod is being squeezed). • **σ = −YαΔT (compressive)** — The negative sign indicates compression; if the rod were cooled, it would try to shrink and develop tensile stress instead. • 💡 Wrong-option analysis: Tensile: tensile stress would arise if the rod were cooled and prevented from contracting; Purely rotational: thermal expansion is linear, not rotational; Zero: preventing free expansion requires a restraining force, so stress cannot be zero.

Correct Answer: B. The absolute temperature (in kelvin)

• **The absolute temperature (in kelvin)** = Charles' law: at constant pressure, V ∝ T (kelvin); doubling the absolute temperature doubles the volume of an ideal gas. • **V = nRT/P (ideal gas law)** — At constant n and P, V is directly proportional to T in kelvin; using Celsius gives incorrect proportionality. • 💡 Wrong-option analysis: The square of temperature in kelvin: V ∝ T, not T²; The mass of the gas only: mass determines n (moles), not directly the volume; The inverse of absolute temperature: V ∝ 1/T is the pressure law (Gay-Lussac's) not Charles' law.

Correct Answer: D. Allow thermal expansion safely

• **Allow thermal expansion safely** = Steam pipes heat up significantly; without expansion loops, thermal elongation would cause enormous stress at joints and fittings, leading to leaks or pipe failure. • **ΔL = αLΔT** — A 100-metre steel pipe heated by 200°C expands by ~0.24 m; expansion loops (U-bends) flex to absorb this without stressing the pipe. • 💡 Wrong-option analysis: Reduce steam temperature: loops do not reduce temperature; Increase pipe thickness: loops do not change wall thickness; Increase steam speed: loops are not designed to affect flow velocity.

Correct Answer: B. Invar

• **Invar** = Invar (64% iron, 36% nickel) is a special alloy designed to have an extremely low coefficient of linear expansion (α ≈ 1.2×10⁻⁶ K⁻¹), about 10 times less than steel. • **α(Invar) ≈ 1.2×10⁻⁶ K⁻¹** — Used in precision instruments, standard metre bars, clock pendulums, and satellite components where dimensional stability is critical. • 💡 Wrong-option analysis: Lead: has a high α (≈29×10⁻⁶ K⁻¹) and is soft; Aluminium: high α (≈23×10⁻⁶ K⁻¹), not suitable for precision instruments; Copper: moderate-to-high α (≈17×10⁻⁶ K⁻¹).