Q. A clock which keeps correct time at $20℃$, is subjected to $40℃.$ If coefficient of linear expansion of the pendulum is $12 \times 10^{- 6}/℃$. How much will it gain or loose in time
A) $10.3\ seconds/day$
B) $20.6\ seconds/day$
C) $5\ seconds/day$
D) $20\ minutes/day$
Explanation
(a) Time period$\ T \propto \sqrt{l} \Rightarrow \frac{\mathrm{\Delta}T}{T} = \frac{1}{2}\frac{\mathrm{\Delta}l}{l} = \frac{1}{2}\alpha\mathrm{\Delta}\theta$
Also according to thermal expansion $l^{'} = l(1 + \alpha\mathrm{\Delta}\theta)$
$\frac{\mathrm{\Delta}l}{l} = \alpha\mathrm{\Delta}\theta$
Hence $\frac{\mathrm{\Delta}T}{T} = \frac{1}{2}\frac{\mathrm{\Delta}l}{l} = \frac{1}{2}\alpha\mathrm{\Delta}\theta$
$= \frac{1}{2} \times 12 \times 10^{- 6} \times (40 - 20) = 12 \times 10^{- 5}$
$\Rightarrow \mathrm{\Delta}T = 12 \times 10^{- 5} \times 86400\ seconds/day$
$\therefore\mathrm{\Delta}T \approx 10.3\ seconds/day$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.