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QPages 12-14: Q29: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 12-14: Q29 Solution: A steel rod is rigidly clamped at its two ends. The rod is…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the given $2.00\,\text{mm}^2$ directly into $F=\sigma A$ without converting to $\text{m}^2$, producing a force off by a factor of $10^6$. Always change the area to SI units ($2.00\times10^{-6}\,\text{m}^2$) before multiplying by the thermal stress.

A steel rod is rigidly clamped at its two ends. The rod is under zero tension at 20�C. If the temperature rises to 100�C, what force will the rod exert on one of the clamps? Area of cross-section of the rod is 2.00 mm2^2. Coefficient of linear expansion of steel is 12.0×10−6 �C−112.0 \times 10^{-6} \text{ �C}^{-1} and Young's modulus of steel is 2.00×1011 N m−22.00 \times 10^{11} \text{ N m}^{-2}.

Step-by-Step Verified Solution

**Answer:** Force =384 N= 384 \text{ N} **Step-by-Step Explanation:** A=2.00×10−6A = 2.00 \times 10^{-6} m_, ΔT=80�C\Delta T = 80�\text{C}, α=12.0×10−6\alpha = 12.0 \times 10^{-6} /�C, Y=2.00×1011Y = 2.00 \times 10^{11} N/m_. Thermal stress = YαΔT=2.00×1011×12.0×10−6×80=1.92×108Y \alpha \Delta T = 2.00 \times 10^{11} \times 12.0 \times 10^{-6} \times 80 = 1.92 \times 10^8 N/m_. Force = Stress _ Area = 1.92×108×2.00×10−6=3841.92 \times 10^8 \times 2.00 \times 10^{-6} = 384 N.

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