A clock with a metallic pendulum gains 5 s each day at a temperature of `15^@C` and loses 10 s each day at a temperature of `30^@C`. Find the coefficient of thermal expansion of the pendulum metal.

Explanation

The time lost or gained per day is
`Deltat = (1)/(2) alphaDeltaT xx 86400 [as 1 day = 86400 s`
If graduction temperature of clock is `T_(0)` then gain in time at `15^(0)C` is
`5 = (1)/(2) (alpha)(T_(0)-15) xx 86400)` ………..(i)
At `30^(@)C` clock is loosing time thus
`10 = (1)/(2) alpha (30-T_(0)) 86400 ......(ii)`
Dividing equation `(ii)` by `(i)`, we get
`2(T_(0)-15) = (30-T_(0))` or `T_(0) = 20^(@)C`
Thus form equation (i)
`5 = (1)/(2) alpha [20 - 15] 86400`
`alpha = 2.31 xx 10^(-5)//@C`