Question

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

• A. ${20}^{\circ }C$
• B. ${24}^{\circ }C$
• C. ${25}^{\circ }C$
• D. ${18}^{\circ }C$

Answer 1

The correct option is A ${20}^{\circ }C$

We know that: $\frac{\Delta T}{T}=\frac{1}{2}\alpha \Delta \theta$

Given: ${\theta }_1={15}^{\circ }C$, ${\theta }_2={30}^{\circ }C$,
$\Delta T_1=+5s$, $\Delta T_2=-10s$ $-ve$ sign just shows that watch is slow

Let ${\theta }_0$ be initial temperature
Fractional loss of time per second
$=\frac{1}{2}\alpha \Delta \theta$
$\frac{1}{2}\alpha ({\theta }_0-15)=\frac{5}{24\times 60\times 60}\dots \left(i\right)$
$\frac{1}{2}\alpha (30-{\theta }_0)=\frac{10}{24\times 60\times 60}\dots \left(ii\right)$

Divide the equation $\left(i\right)$ and $\left(ii\right)$
$\frac{{\theta }_0-15}{30-{\theta }_0}=\frac{5}{10}$
$15{\theta }_0=300$
${\theta }_0={20}^{\circ }C$
Final answer (b)