Question

A pendulum clock keeps correct time at $0^{\circ }$. Its mean coefficient of linear expansions is $\alpha {/}^{\circ }$, then the loss in seconds per day by the clock if the temperature rises by $t^{\circ }$ is

• A. $\frac{\frac{1}{2}\alpha t\times 864000}{1-\frac{\alpha t}{2}}$
• B. $\frac{1}{2}\alpha t\times 86400$
• C. $\frac{\frac{1}{2}\alpha t\times 86400}{{(1-\frac{\alpha t}{2})}^2}$
• D. $\frac{\frac{1}{2}\alpha t\times 846400}{1+\frac{\alpha t}{2}}$

Answer 1

The correct option is B

$\frac{1}{2}\alpha t\times 86400$

Loss in time per second $\frac{△T}{T}=\frac{1}{2}\alpha △\theta =\frac{1}{2}\alpha (t-0)$

$\Rightarrow$loss in time per day

$△T=\left(\frac{1}{2}\alpha t\right)t=\frac{1}{2}\alpha t\times (24\times 60\times 60)=\frac{1}{2}\alpha t\times 86400$