Question

A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep correct time at ${20}^{\circ }C$, then how fast or slow will it go in $24$ hours at ${40}^{\circ }C$ ? Co-efficient of linear expansion of iron = $1.2\times {10}^{-6}{/}^{\circ }C$.

• A. $2.08s$, slow
• B. $2.08s$, fast
• C. $1.04s$, slow
• D. $1.04s$, fast

Answer 1

The correct option is C $1.04s$, slow

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, time period of pendulum increases and clock runs slow.