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 $\left(\right.$ in $\left.^{\circ} C \right)$ does it give the correct time-

Answer 1

Time period of simple pendulum $(T)=2 \pi$ $\sqrt{\frac{\ell}{ g }}$ on increasing temp.( $\left.\theta\right)$, 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)
$\frac{5}{10} =\frac{\theta-15}{\theta-30}$
$\theta-30 =2 \theta-30$
$3 \theta =60$
$\theta =20^{\circ} C$