Question

A clock which keeps correct time at $20^{\circ} C$, is subjected to $40^{\circ} C .$ If coefficient of linear expansion of the pendulum is $12 \times 10^{-6} /{ }^{\circ} C$. How much will it gain or lose time ?

• A. 10.3 s/day
• B. 20.6 s/day
• C. 5 s/day
• D. 20 min/day

Answer 1

Answer: (A)

10.3 s/day

Answer 2

$T=2 \pi \sqrt{\frac{I}{g}}$ $\frac{\Delta T}{T}= \frac{1}{2} \frac{\Delta l}{l}$ $=\frac{1}{2} \alpha \Delta \theta$ $=\frac{1}{2} \times 62 \times 10^{-6}(40-20)$ $=12 \times 10^{-5}$ $\Delta T=T \times 12 \times 10^{-5}$ $=24 \times 60 \times 60 \times 12 \times 10^{-5}$ $=10.3\, s / day$