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

Loss or gain per day `=dT=(1)/(2)alphadtxx86400`
Since `T=86400s` for each day
At `15^@C`,`5=(1)/(2)alpha(t-15)xx86400`
At `30^@C`,`10=(1)/(2)alpha(30-t)xx86400`
`(30-t)/(t-15)=2implies3t=60^@Cimpliest=20^@C`
`alpha=(10)/((t-15)xx86400)=(10)/(5xx86400)=0.000023`
`2.3xx(10^-5^@)/(C)=2`