Question

A clock which keep 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 loss time?

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

Answer 1

The correct option is A

10.3 s/day

$T=2\pi \sqrt{\frac{l}{g}}$

$\frac{\Delta T}{T}=\frac{1}{2}\frac{\Delta l}{l}=\frac{1}{2}\alpha \left(\Delta T\right)$

$=\frac{1}{2}\times 12\times {10}^{-6}\times (40-20)$

$\Delta T=24\times 60\times 60\times 12\times {10}^{-5}=10.3s/day$