Question

A pendulum clock is $5\sec$ fast at temperature of $15^{\circ }C$ and $10\sec$ slow at a temperature of $30^{\circ }C$. At what temperature $($ in ${}^{\circ }C)$ does it give the correct time-

Answer 1

Answer: 20.00

Time period of simple pendulum $(T)=2\pi$ $\sqrt{\frac{\ell }{g}}$ on increasing temp.( $\theta )$, length increases hence time period increases so clock becomes slow and vice-versa. Change in time period
$\Delta t=\frac{1}{2}\alpha \Delta \theta ,T$
Let clock gives the correct time at temperature $\theta$, then
$5=\frac{1}{2}\alpha (\theta -15).T$ ...(1)
$10=\frac{1}{2}\alpha (30-\theta ).T$ ...(2)
Using equation (1) and (2)
$\frac{5}{10}=\frac{\theta -15}{\theta -30}$
$\theta -30=2\theta -30$
$3\theta =60$
$\theta =20^{\circ }C$