Question

A clock which keeps correct time at ${20}^{\circ }C$, is subjected to ${40}^{\circ }C$. If coefficient of linear expansion of the pendulum is $12\times {10}^{-6}{/}^{\circ }C$, then how much will it gain or loss in time?

• A. $5s/day$
• B. $10.3s/day$
• C. $20.6s/day$
• D. $20min/day$

Answer 1

Answer: (B)

$10.3s/day$

As, $T_2=T_1\sqrt{I_2/I_1}$
$=T_1\times \sqrt{\frac{I_1\left[\{1+12\times {10}^{-6}\times (40-20\left)\right\}\right]}{I_1}}$
$=T_1\sqrt{1+240\times {10}^{-6}}$
$\therefore \frac{T_2-T_1}{T_1}=(1+240\times {10}^{-6}{)}^{1/2}-1$
$120\times {10}^{-6}$
or $T_2-T_1=120\times {10}^{-6}\times (24\times 60\times 60)s$
$=10.36s/day$
It will be a loss as $T_2>T_1$.