A clock with a metal pendulum beating seconds keeps correct time at `0^@C`. If it loses 12.5 s a day at `25^@C`, the coefficient of linear expansion of metal of pendulum is

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)`