A clock which keeps correct time at {20^o}C, | vedclass

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Question

(a) Time period $T \propto \sqrt l $

==>$\frac{{\Delta T}}{T} = \frac{1}{2}\frac{{\Delta l}}{l} = \frac{1}{2}\alpha \Delta 	heta $
Also accord

Vedclass · Accepted Answer

$10.3$ seconds / day

Vedclass · Answer

(a) Time period $T \propto \sqrt l $

==>$\frac{{\Delta T}}{T} = \frac{1}{2}\frac{{\Delta l}}{l} = \frac{1}{2}\alpha \Delta 	heta $
Also according to thermal expansion $l' = (1 + \alpha \Delta 	heta )$
$\frac{{\Delta l}}{l} = \alpha + 	heta $. Hence $\frac{{\Delta T}}{T} = \frac{1}{2}\frac{{\Delta l}}{l} = \frac{1}{2}\alpha \Delta 	heta $
$ = \frac{1}{2} 	imes 12 	imes {10^{ - 6}} 	imes (40 - 20) = 12 	imes {10^{ - 5}}$
$ \Rightarrow \Delta T = 12 	imes {10^{ - 5}} 	imes 86400\, seconds / day$

$\Delta T \approx 10.3\, seconds/day$

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

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