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QPages 12-14: Q11: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 12-14: Q11 Solution: A metre scale made of steel is calibrated at 20�C to give…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often take ΔT as a positive magnitude even when the object cools, applying $L=L_0(1+\alpha\Delta T)$ with $\Delta T=+10\,^\circ\!C$ and obtaining an expanded length. Remember that $\Delta T=T_{\text{actual}}-T_{\text{calibrated}}$; for a drop from 20 °C to 10 °C, $\Delta T=-10\,^\circ\!C$, giving a shorter actual distance between marks.

A metre scale made of steel is calibrated at 20�C to give correct reading. Find the distance between the 50 cm mark and the 51 cm mark if the scale is used at 10�C. Coefficient of linear expansion of steel is 1.1×10−5 �C−11.1 \times 10^{-5} \text{ �C}^{-1}.

Step-by-Step Verified Solution

**Answer:** d=1.00011 cmd = 1.00011 \text{ cm} **Step-by-Step Explanation:** At 20�C, 1 cm on scale = 1 cm actual. At 10�C, the steel scale contracts, so each scale-cm corresponds to a smaller actual distance. Actual distance = scale reading _ (1+αΔT)(1 + \alpha \Delta T) where ΔT=20−10=10�C\Delta T = 20 - 10 = 10�\text{C}:
d=1×(1+1.1×10−5×10)=1+1.1×10−4=1.00011 cmd = 1 \times (1 + 1.1 \times 10^{-5} \times 10) = 1 + 1.1 \times 10^{-4} = 1.00011 \text{ cm}

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