MCQ
A pendulum clock keeps correct time at $0^{\circ} \mathrm{C}$. Its mean coefficient of linear expansions is $\alpha /{ }^{\circ} \mathrm{C}$, then the loss in seconds per day by the clock if the temperature rises by $t^{\circ} \mathrm{C}$ is
  • A
    $\frac{\frac{1}{2} \alpha t \times 864000}{1-\frac{\alpha t}{2}}$
  • $\frac{1}{2} \alpha t \times 86400$
  • C
    $\frac{\frac{1}{2} \alpha t \times 86400}{\left(1-\frac{\alpha t}{2}\right)^2}$
  • D
    $\frac{\frac{1}{2} \alpha t \times 86400}{1+\frac{\alpha t}{2}}$

Answer

Correct option: B.
$\frac{1}{2} \alpha t \times 86400$
(b) Loss in time per second$\frac{\Delta T}{T}=\frac{1}{2} \alpha \Delta \theta=\frac{1}{2} \alpha(t-0)$$\Rightarrow$ loss in time per day$\Delta t=\left(\frac{1}{2} \alpha t\right) t=\frac{1}{2} \alpha t \times(24 \times 60 \times 60)=\frac{1}{2} \alpha t \times 86400$

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