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QPages 12-14: Q28: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 12-14: Q28 Solution: A steel wire of cross-sectional area 0.5 mm is held between two …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently forget that the wire’s contraction is fully restrained, so the thermal strain becomes a stress given by $sigma=Y\alpha\Delta T$; they instead compute force from an unrestrained $Delta L$ or omit the area conversion, yielding an incorrect tension. Always treat the thermal strain as a prevented strain, calculate stress with $Y\alpha\Delta T$, then multiply by the correctly converted cross‑sectional area.

A steel wire of cross-sectional area 0.5 mm2^2 is held between two fixed supports. If the wire is just taut at 20�C, determine the tension when the temperature falls to 0�C. Coefficient of linear expansion of steel is 1.2×10−5 �C−11.2 \times 10^{-5} \text{ �C}^{-1} and its Young's modulus is 2.0×1011 N m−22.0 \times 10^{11} \text{ N m}^{-2}.

Step-by-Step Verified Solution

**Answer:** Tension =24 N= 24 \text{ N} **Step-by-Step Explanation:** A=0.5×10−6A = 0.5 \times 10^{-6} m_, ΔT=20�C\Delta T = 20�\text{C}, α=1.2×10−5\alpha = 1.2 \times 10^{-5} /�C, Y=2.0×1011Y = 2.0 \times 10^{11} N/m_. Thermal stress = YαΔT=2.0×1011×1.2×10−5×20=4.8×107Y \alpha \Delta T = 2.0 \times 10^{11} \times 1.2 \times 10^{-5} \times 20 = 4.8 \times 10^7 N/m_. Tension = Stress _ Area = 4.8×107×0.5×10−6=244.8 \times 10^7 \times 0.5 \times 10^{-6} = 24 N.

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