Q136 / 160 · TS EAMCET 2007 · Physics

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

  • Correct answer
    A
    $10.3 \mathrm{~s} /$ day
  • B
    20.6 s/day
  • C
    5 s/day
  • D
    $20 \mathrm{~min} /$ day

Solution

$T=2 \pi \sqrt{\frac{I}{g}}$ $\begin{aligned} \frac{\Delta T}{T} & =\frac{1}{2} \frac{\Delta l}{l}=\frac{1}{2} \alpha \Delta \theta \\ & =\frac{1}{2} \times 62 \times 10^{-6}(40-20)=12 \times 10^{-5} \\ \Delta T & =T \times 12 \times 10^{-5} \\ & =24 \times 60 \times 60 \times 12 \times 10^{-5} \\ & =10.3 \mathrm{~s} / \text { day }\end{aligned}$

Hence, the correct answer is option: 'A'.

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