Question

A pendulum clock loses $12s$ a day if the temperature is ${40}^o$C and gains $4s$ a day if the temperature is ${20}^o$C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion $\left(\alpha \right)$ of the metal of the pendulum shaft are respectively

• A. ${25}^o$C$;\alpha =1.85\times {10}^{-5}{/}^o$C
• B. ${60}^o$C $;\alpha =1.85\times {10}^{-4}{/}^o$C
• C. ${30}^o$C$;\alpha =1.85\times {10}^{-3}{/}^o$C
• D. ${55}^o$C$;\alpha =1.85\times {10}^{-2}{/}^o$C

Answer 1

The correct option is A ${25}^o$C$;\alpha =1.85\times {10}^{-5}{/}^o$C

The time period of a pendulum is given by,
$t=2\pi \sqrt{\frac{l}{g}}$
$\Delta t=\frac{2\pi }{\sqrt{g}}\frac{\Delta l}{2\sqrt{l}}$
$\Delta l=\alpha \Delta \theta$
$\Delta t=\frac{\pi }{\sqrt{gl}}\alpha \Delta T$
$\therefore \frac{\Delta t_1}{\Delta t_2}=\frac{\Delta T_1}{\Delta T_2}$

$\Rightarrow \frac{12}{4}=\frac{40-T}{T-20}$
$\Rightarrow 3T-60=40-T$
$\Rightarrow T={25}^{\circ }C$

From $\Delta t=\frac{\pi }{\sqrt{gl}}\alpha \Delta T$,
$\alpha =1.85\times {10}^{-5}/°C$