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

Time lost or gained per second by a pendulum clock is given by
`deltat=(1)/(2)alphaDeltaT`
Here temperature is higher than graduation temperature thus clock will lose time and if it is lower than graduation temperature, clock will gain time.
Thus time lost or gained per day is
`deltat=(1)/(2)alphaDeltaTxx86400`
If graduation temperature of clock is `T_0` then we have
At `15^@C`, clock is gaining 5 s. Thus
`5=(1)/(2)alpha(T_0-15)xx86400` .(i)
At `30^@C` clock is losing 10 s. Thus
`10=(1)/(2)alpha(30-T_0)xx86400` ..(ii)
Dividing Eq. (i) by Eq. (ii), we get
`2(T_0-15)=(30-T_0)`
or `T_0=20^@C`
From Eq. (i)
`5=(1)/(2)xxalphaxx[20-15]xx86400`
`alpha=2.31xx(10^-5^@)/(C)`