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A penduluin clock of an iron rod is conn...

A penduluin clock of an iron rod is connected to a small, heavy bob. If it shows correct time at `20^(@)C`, how fast or slow will it go in one day at `40^(@)C` ? (Coefficient of linear expansion for steel is `12xx10^(-6)//""^(@)C`.)

Text Solution

Verified by Experts

We know `(DeltaT)/T=1/2alphaDeltatrArr(T_(2)-T_(1))/T_(1)=1/2alpha(t_(2)-t_(1))`
Which is the fractional loss of time. As the tempe rature period increases and the clock loses time or goes slow. The time lost in one day (or in 24 hours)
`(1/2alpha(t_(2)-t_(1)))=(86400s)1/2alpha(t_(2)-t_(1))=43200alpha(t_(2)-t_(1))s`
In this problem, `alpha=12xx10^(-6)//""^(@)C,t_(1)=20^(@)C,t_(2)=40^(@)C`
The time lost in one day = `43200xx12xx10^(-6)(40-20)`
= 10.368 s.
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