Verified Solution • 2025–26 Session moderate

QPages 12-14: Q27: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 12-14: Q27 Solution: A steel rod is clamped at its two ends and rests on a fixed…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often compute the thermal strain as a positive value and forget that a clamped rod cannot expand, so the actual longitudinal strain is compressive (negative). Remember to apply the sign convention: if expansion is prevented, the strain equals $-\alpha\Delta T$.

A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is unstrained at 20�C. Find the longitudinal strain developed in the rod if the temperature rises to 50�C. Coefficient of linear expansion of steel = 1.2×10−5 �C−11.2 \times 10^{-5} \text{ �C}^{-1}.

Step-by-Step Verified Solution

**Answer:** Strain =−3.6×10−4= -3.6 \times 10^{-4} **Step-by-Step Explanation:** ΔT=30�C\Delta T = 30�\text{C}, α=1.2×10−5\alpha = 1.2 \times 10^{-5} /�C. Thermal strain if free = αΔT=1.2×10−5×30=3.6×10−4\alpha \Delta T = 1.2 \times 10^{-5} \times 30 = 3.6 \times 10^{-4}. Since the rod is clamped, it cannot expand, so compressive strain = −3.6×10−4-3.6 \times 10^{-4}.

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