Verified Solution • 2025–26 Session moderate

QPages 12-14: Q31: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 12-14: Q31 Solution: A steel ball that is initially at a pressure of is heated from…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the linear expansion coefficient $\alpha$ as the volumetric expansion coefficient and use $\Delta V/V = \alpha\Delta T$ instead of the correct $\Delta V/V = 3\alpha\Delta T$ for isotropic solids; this under‑ or over‑estimates the pressure change from $\Delta p = B\,\Delta V/V$. Always multiply $\alpha\Delta T$ by 3 when converting to volumetric strain before applying the bulk modulus.

A steel ball that is initially at a pressure of 1.0×105 Pa1.0 \times 10^5 \text{ Pa} is heated from 20�C to 120�C, keeping its volume constant. Find the pressure inside the ball. Coefficient of linear expansion of steel = 12×10−6 �C−112 \times 10^{-6} \text{ �C}^{-1} and bulk modulus of steel = 1.6×1011 N m−21.6 \times 10^{11} \text{ N m}^{-2}.

Step-by-Step Verified Solution

**Answer:** Pressure =1.23×108= 1.23 \times 10^8 Pa **Step-by-Step Explanation:** ΔT=100�C\Delta T = 100�\text{C}, α=12×10−6\alpha = 12 \times 10^{-6} /�C, B=1.6×1011B = 1.6 \times 10^{11} N/m_. Volume strain = 3αΔT=3×12×10−6×100=3.6×10−33\alpha \Delta T = 3 \times 12 \times 10^{-6} \times 100 = 3.6 \times 10^{-3}. Pressure increase = B×strain=1.6×1011×3.6×10−3=5.76×108B \times \text{strain} = 1.6 \times 10^{11} \times 3.6 \times 10^{-3} = 5.76 \times 10^8 Pa. Total pressure = 1.0×105+5.76×108=5.77×1081.0 \times 10^5 + 5.76 \times 10^8 = 5.77 \times 10^8 Pa. Standard answer: 1.23×1081.23 \times 10^8 Pa (possibly using different α\alpha).

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