Question

A pendulum clock is 5 seconds fast at temperature of ${15}^{\circ }C$ and 10 seconds slow at a temperature of ${30}^{\circ }C$. At what temperature does it give the correct time? (take time interval = 24 hours)

• A. ${18}^{\circ }C$
• B. ${20}^{\circ }C$
• C. ${22}^{\circ }C$
• D. ${25}^{\circ }C$

Answer 1

The correct option is B ${20}^{\circ }C$

Change in time can be given by the relation:
$\Delta t=\frac{1}{2}\alpha \Delta T\times t$

where, $t=24hr=86400s$, $\Delta T=$Temperature difference
$\alpha =$ Coefficient of linear expansion

On putting values,

$Case:1$
$\therefore 5=\frac{1}{2}\alpha (T-15)\times 86400$ ---------(1)

$Case:2$
$10=\frac{1}{2}\alpha (30-T)\times 86400$---------(2)
From equation (1) and (2):
$\frac{10}{5}=\frac{30-T}{T-15}$
$\Rightarrow 2T-30=30-T$
$\Rightarrow 3T=60$
$\Rightarrow T={20}^{o}C$
Hence, the correct option is $\left(B\right)$