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EEJ MAIN Physics QUESTION #7081
Question 1
A pendulum clock loses $12\text{ s/day}$ at $40^{\circ}\text{C}$ and gains $4\text{ s/day}$ at $20^{\circ}\text{C}$. Find the temperature for correct time and the coefficient of linear expansion $\alpha$.
  • $60^{\circ}\text{C}$, $\alpha=1.85 \times 10^{-4}/^{\circ}\text{C}$
  • $30^{\circ}\text{C}$, $\alpha=1.85 \times 10^{-3}/^{\circ}\text{C}$
  • $55^{\circ}\text{C}$, $\alpha=1.85 \times 10^{-2}/^{\circ}\text{C}$
  • $25^{\circ}\text{C}$, $\alpha=1.85 \times 10^{-5}/^{\circ}\text{C}$✔️
Correct Answer Explanation

The change in time is given by $\Delta t = \frac{1}{2} \alpha \Delta \theta \times t$.

  • $12 = \frac{1}{2} \alpha (40 - \theta) \times 86400$
  • $-4 = \frac{1}{2} \alpha (20 - \theta) \times 86400$
  • Dividing the equations gives $\theta = 25^{\circ}\text{C}$.
  • Substituting $\theta$ back gives $\alpha = 1.85 \times 10^{-5}/^{\circ}\text{C}$.