A clock which keeps correct time at 20^{circ} | vedclass

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.Taking the fractional change, we have $\frac{\Delta T}{T} = \frac{1

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$10.3 ~s/	ext{day}$

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.Taking the fractional change, we have $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l} = \frac{1}{2} \alpha \Delta 	heta$.Given $\alpha = 12 	imes 10^{-6} /{ }^{\circ} C$ and $\Delta 	heta = 40^{\circ} C - 20^{\circ} C = 20^{\circ} C$.Substituting the values: $\frac{\Delta T}{T} = \frac{1}{2} 	imes 12 	imes 10^{-6} 	imes 20 = 120 	imes 10^{-6} = 1.2 	imes 10^{-4}$.The time lost or gained in a day $(T = 86400 ~s)$ is $\Delta T = T 	imes \frac{\Delta T}{T} = 86400 	imes 1.2 	imes 10^{-4} \approx 10.368 ~s$.Since the temperature increases, the length of the pendulum increases, the time period increases, and the clock loses time.

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

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