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 $\frac{12\times {10}^{-6}}{{}^{\circ }C}$. How much time will it gain or lose per day

• A. 10.3 seconds / day
• B. 20.6 seconds / day
• C. 5 seconds / day
• D. 20 minutes / day

Answer 1

The correct option is A

10.3 seconds / day

Time period $T\propto \sqrt{l}\Rightarrow \frac{\Delta T}{T}=\frac{1}{2}\frac{\Delta l}{l}=\frac{1}{2}\alpha \Delta \theta$

$=\frac{1}{2}\times 12\times {10}^{-6}\times (40-20)=12\times {10}^{-5}$

$\Rightarrow \Delta T=12\times {10}^{-5}\times 86400$ seconds /day

$\therefore \Delta T\approx 10.3$ seconds /day