Question

A clock with an iron pendulum keeps correct time at ${20}^{\circ }C$. How much will it lose or gain in 1 day if temperature changes to ${40}^{\circ }C$? [Given cubical expansion of iron $=36\times {10}^{-6}{}^{\circ }{C}^{-1}]$

• A. 10.368 sec gain
• B. 10.368 sec loss
• C. 5.184 sec gain
• D. 5.184 sec loss

Answer 1

The correct option is B 10.368 sec loss

$\frac{\Delta t}{t}=\frac{1}{2}\alpha \Delta T$ (t = time)
$\gamma =3\alpha$
$\alpha =\frac{\gamma }{3}$
Temperature $\uparrow \Rightarrow$ length $\uparrow \Rightarrow$ = time period $\uparrow$
$\therefore$ loss
$\Delta t=\frac{1}{2}\times \frac{\gamma }{3}\times \Delta T\times t$
$\Delta t=\frac{1}{2}\times \frac{36}{3}\times {10}^{-6}\times (40-20)\times (24\times 60\times 60)$
$\Delta t=\frac{1}{2}\times \frac{36}{3}\times {10}^{-6}\times 20\times 86400$
$=12\times 10\times 86400\times {10}^{-6}$
$=10368000\times {10}^{-6}$
$\Delta t=10.368\text{sec}\text{loss}$.