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