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

We have dicussed that time lost or gained per second by a pendulum clock is given by
`deta t=1/2 alpha DeltaT`
Here temperature is higher then graduation temperature thus clock will loose time and if it is lower then graduation temperature will gain time.
Thus tume lost or gained per day is
`delta t=1/2 alpha Deltaxx86400` [As 1 day = 86400 s]
If graduation temperature of clock of `T_(0)` then we have
As `15^(@)C` clock is gaining time, thus
`5=1/2. alpha. (T_(0)-15)xx86400`............1.27
At `30^(@)` clock is loosing time thus
`10=1/2 alpha(30-T_(0))xx86400` ...1.28
Dividing equqtion `(1.28)` by( 1.27) we get
`2(T_(0)-15)=(30-T_(0))`
or `T_(0)=20^(@)C`
Thus from equation (1.27)
`5=1/2xxalphaxx[20-15]xx86400alpha`
`=2.31xx10^(-5).^(@)C^(-1)`