A pendulum clock shows the correct time at 20 | vedclass

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(D) The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.When the temperature increases,the length $L$ of the pendulum increases as

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$8$

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(D) The time period of a pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.When the temperature increases,the length $L$ of the pendulum increases as $L' = L(1 + \alpha \Delta 	heta)$.The change in time $\Delta t$ for a total time $t$ is given by the formula $\Delta t = \frac{1}{2} \alpha \Delta 	heta 	imes t$.Here,$\alpha = 10^{-5} {^{\circ}C}^{-1}$,$\Delta 	heta = (40 - 20) = 20^{\circ}C$,and $t = 86400 \, s$ (number of seconds in a day).Substituting the values:$\Delta t = \frac{1}{2} 	imes 10^{-5} 	imes 20 	imes 86400$$\Delta t = 10^{-5} 	imes 10 	imes 86400$$\Delta t = 10^{-4} 	imes 86400 = 8.64 \, s$.

$A$ pendulum clock shows the correct time at $20^{\circ}C$. During summer days,when the temperature rises to $40^{\circ}C$,what will be the time change in the clock in one day (in $.64$)? (Given: $\alpha = 10^{-5} {^{\circ}C}^{-1}$)

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