Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A pendulum clock (fitted with a small heavy bob that is connected with a metal rod) is 5 seconds fast each day at a temperature of $15^{\circ} C$ and 10 seconds slow at a temperature of $30^{\circ} C$. The temperature at which it is designed to give correct time, is

Thermal Properties of Matter

Solution:

Fractional loss of time per second $=\frac{1}{2} \alpha \Delta T$
Therefore $\frac{1}{2} \alpha\left(T_{0}-15\right) \times(24\, hrs )=5$
and $\frac{1}{2} \alpha\left(30-T_{0}\right) \times(24\, hrs )=10$
on solving $T_{0}=20^{\circ} C$