Question

A pendulum clock gains 6 s per day when the temperature is $296\ \mathrm{K}$ and gains 9 s per day when the temperature is $291\ \mathrm{K}$. The temperature at which the pendulum clock loses 12 s per day is:

• A. $476\ \mathrm{K}$
• B. $236\ \mathrm{K}$
• C. $53\ \mathrm{K}$
• D. $326\ \mathrm{K}$

Hint:

When given two readings and asked to extrapolate, a linear model is often implied — solve for intercept and slope.
Be mindful of sign convention: gains are positive, losses negative.
Physical interpretation: thermal expansion changes pendulum length altering period and timekeeping.
Check units (K) consistently when solving linear relations.

Answer 1

The Correct Option is D

• Let the daily time error (gain positive, loss negative) be a linear function of temperature: $E(T)=a+bT$.
• Given $E(296)=6$ s/day and $E(291)=9$ s/day. Solve for $a,b$.
• Slope $b = \dfrac{9-6}{291-296} = \dfrac{3}{-5} = -0.6\ \text{s day}^{-1}\mathrm{K}^{-1}$.
• Use $E(296)=6$: $a -0.6\times296 = 6 \Rightarrow a = 6 + 177.6 = 183.6$.
• We want $T$ such that $E(T) = -12$ (loses 12 s/day): \[ -12 = 183.6 - 0.6T \Rightarrow -0.6T = -195.6 \Rightarrow T = \frac{195.6}{0.6} = 326\ \mathrm{K}. \] • Hence the clock loses 12 s/day at $326\ \mathrm{K$}.