Question

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

Answer 1

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.