A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep correct time at `20^(0) C`, how fast or slow will it go in 24 hours at `40^(0) C?` Coefficient of linear expansion of iron `= 1.2 xx 10^(-5)C^(-1)`.

Explanation

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.