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A clock with iron pendulum keeps correct time at $15^{\circ} \mathrm{C}$. If the room temperature is $20^{\circ} \mathrm{C}$, the error in second per day will be nearly (coefficient of linear expansion of iron is $\alpha=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}$ )
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$2.6 \mathrm{~s}$
Time period of the pendulum is, $T=2 \pi \sqrt{\frac{l}{g}}$
$\therefore \Delta T=\frac{2 \pi}{\sqrt{g}} \times \frac{\Delta l}{2 \sqrt{l}}$
We know, $\frac{\Delta l}{l}=\alpha \Delta \theta$ the change in length due to thermal expansion.
$\therefore \frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta l}{l}=\frac{1}{2} \alpha \Delta \theta$
Given, $\quad \alpha=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}, \quad g=9.8 \mathrm{~m} / \mathrm{s}, \quad T=86,400 \mathrm{~s} \quad$ and $\Delta T=\alpha \Delta \theta T \approx 2.6 \mathrm{sec}$
$\therefore \Delta T=\frac{2 \pi}{\sqrt{g}} \times \frac{\Delta l}{2 \sqrt{l}}$
We know, $\frac{\Delta l}{l}=\alpha \Delta \theta$ the change in length due to thermal expansion.
$\therefore \frac{\Delta T}{T}=\frac{1}{2} \frac{\Delta l}{l}=\frac{1}{2} \alpha \Delta \theta$
Given, $\quad \alpha=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}, \quad g=9.8 \mathrm{~m} / \mathrm{s}, \quad T=86,400 \mathrm{~s} \quad$ and $\Delta T=\alpha \Delta \theta T \approx 2.6 \mathrm{sec}$
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