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 in time

• 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 $T\propto \sqrt{l}\Rightarrow \frac{\Delta T}{T}=\frac{1}{2}\frac{\Delta l}{l}=\frac{1}{2}\alpha \Delta \theta$
Also according to thermal expresion $l^{\prime }=(1+\alpha \Delta \theta )$
$\frac{\Delta l}{l}=\alpha +\theta .$ Hence $\frac{\Delta T}{T}=\frac{1}{2}\frac{\Delta l}{l}=\frac{1}{2}\alpha \Delta \theta$
$=\frac{1}{2}\times 12\times {10}^{-6}\times (40-20)=12\times {10}^{-5}$
$\Rightarrow \Delta T=12\times {10}^{-5}\times 86400$ seconds/day

A the time period of oscillation increases the clock loses time