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

The time difference occurred in `24` hours `(86400` seconds) is given by
`Deltat = (1)/(2) alpha Delta theta t`
`=(1)/(2) xx 1.2 xx 10^(-6) xx 20 xx 86400 = 1.04 sec`.
This is loss of time as `theta` is greater than `theta_(0)`. As the temperature increases, the time period also increases. Thus, the clock goes slow.