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EEJ MAIN Physics QUESTION #7099
Question 1
A pendulum clock loses $12\text{ s}$ per day at $40^{\circ}\text{C}$ and gains $4\text{ s}$ per day at $20^{\circ}\text{C}$. Determine the correct time temperature and the linear expansion coefficient ($\alpha$) [cite: 446].
  • $60^{\circ}\text{C}$, $\alpha=1.85 \times 10^{-4}/^{\circ}\text{C}$
  • $30^{\circ}\text{C}$, $\alpha=1.85 \times 10^{-3}/^{\circ}\text{C}$
  • $55^{\circ}\text{C}$, $\alpha=1.85 \times 10^{-2}/^{\circ}\text{C}$
  • $25^{\circ}\text{C}$, $\alpha=1.85 \times 10^{-5}/^{\circ}\text{C}$✔️
Correct Answer Explanation

The error in time per day is $\Delta t = \frac{1}{2} \alpha \Delta \theta \times 86400$ [cite: 446].

  • Case 1: $12 = \frac{1}{2} \alpha (40 - \theta) \times 86400$
  • Case 2: $-4 = \frac{1}{2} \alpha (20 - \theta) \times 86400$
  • Dividing gives $\frac{12}{-4} = \frac{40-\theta}{20-\theta} \implies \theta = 25^{\circ}\text{C}$
  • Substituting back gives $\alpha = 1.85 \times 10^{-5}/^{\circ}\text{C}$[cite: 446].