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`.)

Explanation

We known `(Delta T)/(T) = (1)/(2) alpha Delta t`
`rArr (T_(2) - T_(1))/(T_(1)) = (1)/(2) alpha (t_(2) - t_(1))`
Which is the fractional loss of time . As the temperature period increases and the clock loses time or goes slow.
The time lost in one day (or in 24 hours)
`((1)/(2) alpha (t_(2) - t_(1)) ) = (86400 s) (1)/(2) alpha (t_(2) - t_(1))`
= 43200 `alpha (t_(2) - t_(1)) s `
In this problem, `alpha = 12 xx 10^(-6)//^(0)C`,
`t_(1) = 20^(0)C , t_(2) = 40^(0)C `
The time lost in one day = 43200 `xx 12 xx 10^(-6) (40 - 20)` = 10.368 s.