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A clock which keeps correct time at ${20^o}C$, is subjected to ${40^o}C$. If coefficient of linear expansion of the pendulum is $12 \times {10^{ - 6}}/^\circ C$. How much will it gain or loose in time
$10.3$ seconds / day
$20.6$ seconds / day
$5$ seconds / day
$20$ minutes / day
Solution
(a) Time period $T \propto \sqrt l $
==>$\frac{{\Delta T}}{T} = \frac{1}{2}\frac{{\Delta l}}{l} = \frac{1}{2}\alpha \Delta \theta $
Also according to thermal expansion $l' = (1 + \alpha \Delta \theta )$
$\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$
$\Delta T \approx 10.3\, seconds/day$
