Question

A clock which keep correct time at ${20}^{\circ }C$ is subjected to ${40}^{\circ }C$. If coefficient of linear expansion of the pendulum is $12\times {10}^{-6}/{}^{\circ }C$. How much will it gain or lose time?

• A. $10.3s/day$
• B. $20.6s/day$
• C. $5.3s/day$
• D. $30s/day$

Answer 1

The correct option is A $10.3s/day$

Given
$\alpha =1.2\times {10}^{-6}{/}^{\circ }C$.
$\Delta \theta ={40}^{\circ }C-{20}^{\circ }C={20}^{\circ }C$

Time period in a pendulum is given as $T=2\pi \sqrt{\frac{l}{g}}$

When temperature is increased by $\Delta \theta$ new length is $l=l_0(1+\alpha \Delta \theta )$

As $T$ is proportional to $\sqrt{l}$

So new Time period $T=T_0\sqrt{(1+\alpha \Delta \theta )}$

As $\alpha \Delta \theta <<1$ applying Binomial theorem,
$T=T_0(1+\frac{\alpha \Delta \theta }{2})$
$T-T_0=\frac{T_0\alpha \Delta \theta }{2}$

For calculating variation of time for $24hrs\Rightarrow T_0=86400s$

$\Delta T=\frac{1}{2}\alpha \Delta \theta T_0=\frac{1}{2}\times \alpha \times \Delta \theta \times 86400=\Rightarrow \Delta T=1.04s$

As there is increase in length due to rise of temperature, timeperiod of pendulum increases and clock runs slow