A clock with a metallic pendulum is 5 seconds fast each day at a temperature of  ${15}^{\circ }\text{c}$ and 10 seconds slow each day at a temperature of  ${30}^{\circ }\text{c}$, find the coefficient of linear expansion of the metal?

Graduation temperature= ${\text{T}}_{\text{o}}$

change in time due to coefficient of linear expansion of material of pendulum is

$$\begin{gathered} ∆time = \frac{1}{2}\alpha (temperature difference)x86400 \\ here \alpha is coefficient of linear expansion of metal of pendulum \end{gathered}$$
$\text{5=}\frac{\text{1}}{\text{2}}\text{.α.}\left({\text{T}}_{\text{o}}\text{-15}\right)\text{×86400}$----(1)
$\text{10=}\frac{\text{1}}{\text{2}}\text{.α.}\left({\text{30-T}}_{\text{o}}\right)\text{×86400}$----(2)

$$\begin{gathered} on dividing eqn \left(1\right) with eqn\left(2\right) \\ \frac{1}{2}=\frac{T_0-15}{30-T_0} \\ \Rightarrow T_0={20}^{0}C \\ substitute T_0 vallue in eqn\left(1\right) \\ 5=\frac{1}{2}\alpha (20-15)86400\Rightarrow \alpha =2.31 x {10}^{-5} {}^{0}C^{-1}=coefficient of linear expansion of metal \end{gathered}$$