Verified Solution • 2025–26 Session moderate

QPages 12-14: Q19: HC Verma: Concepts of Physics - Volume 2 Heat and Temperature Pages 12-14: Q19 Solution: A pendulum clock shows correct time at 20�C at a place where .…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often write $L = L_0(1+\alpha T)$ treating $T$ as the absolute temperature, forgetting that the linear expansion formula uses the temperature change $\Delta T$ from the reference length. Use $L_T = L_{20}[1+\alpha (T-20)]$ (or $L = L_{ref}(1+\alpha\Delta T)$) to avoid over‑ or under‑estimating the required temperature.

A pendulum clock shows correct time at 20�C at a place where g=9.800 m s−2g = 9.800 \text{ m s}^{-2}. The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where g=9.788 m s−2g = 9.788 \text{ m s}^{-2}. At what temperature will the clock show correct time? Coefficient of linear expansion of steel = 12×10−6 �C−112 \times 10^{-6} \text{ �C}^{-1}.

Step-by-Step Verified Solution

**Answer:** T=10�CT = 10�\text{C} **Step-by-Step Explanation:** At 20�C, g=9.800g = 9.800 m/s_, clock keeps correct time. At g′=9.788g' = 9.788 m/s_, clock must again keep correct time. T=2πL/gT = 2\pi\sqrt{L/g}. For the clock to keep correct time, L/gL/g must be the same:
L20g=LTg′\frac{L_{20}}{g} = \frac{L_T}{g'}
L0(1+12×10−6×20)9.800=L0(1+12×10−6×T)9.788\frac{L_0(1 + 12 \times 10^{-6} \times 20)}{9.800} = \frac{L_0(1 + 12 \times 10^{-6} \times T)}{9.788}
Solving: T≈10�CT \approx 10�\text{C}.

All Questions in this Chapter

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