Question

A seconds metallic pendulum keeps correct time at $0^{\circ }\text{C}$. If it loses $12.5s$ a day at ${25}^{\circ }\text{C}$, the coefficient of linear expansion of metal pendulum is

• A. $\frac{1}{86400}{/}^{\circ }\text{C}$
• B. $\frac{1}{43200}{/}^{\circ }\text{C}$
• C. $\frac{1}{14400}{/}^{\circ }\text{C}$
• D. $\frac{1}{28800}{/}^{\circ }\text{C}$

Answer 1

The correct option is A $\frac{1}{86400}{/}^{\circ }\text{C}$

Time period of simple pendulum $T=2\pi \sqrt{\frac{L}{g}}$

As $g$ is constant, differentiating above equation we get,

$\frac{dT}{T}=\frac{dL}{2L}$

For finite change in $T,L$
$\frac{\Delta T}{T}=\frac{\Delta L}{2L}$

Due to thermal expansion $\Delta L=L\alpha \Delta \theta$

$\Rightarrow \frac{\Delta T}{T}=\frac{1}{2}\alpha \Delta \theta$

Number of seconds lost in a day
$\Delta T=\frac{1}{2}\alpha \Delta \theta \times 86400$

The coefficient of linear expansion of metal pendulum
$\alpha =\frac{\Delta T}{\Delta \theta \times 86400}=\frac{12.5\times 2}{25\times 86400}$

$\alpha =\frac{1}{86400}{/}^{\circ }\text{C}$