A simple pendulum keeps correct time at 0^{ci | vedclass

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(A) The time lost by a pendulum clock due to a change in temperature is given by the formula: $\Delta t = \frac{1}{2} \alpha \Delta 	heta T$,where $\

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$\frac{1}{86400} \, /^{\circ}C$

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(A) The time lost by a pendulum clock due to a change in temperature is given by the formula: $\Delta t = \frac{1}{2} \alpha \Delta 	heta T$,where $\Delta t$ is the time lost,$\alpha$ is the coefficient of linear expansion,$\Delta 	heta$ is the change in temperature,and $T$ is the total time in a day $(86400 \, s)$.Given: $\Delta t = 12.5 \, s$,$\Delta 	heta = 25^{\circ}C - 0^{\circ}C = 25^{\circ}C$,and $T = 86400 \, s$.Substituting the values: $12.5 = \frac{1}{2} 	imes \alpha 	imes 25 	imes 86400$.$12.5 = \alpha 	imes 1080000$.$\alpha = \frac{12.5}{1080000} = \frac{12.5}{12.5 	imes 86400} = \frac{1}{86400} \, /^{\circ}C$.

$A$ simple pendulum keeps correct time at $0^{\circ}C$. At $25^{\circ}C$,it loses $12.5 \, s$ in a day. What is the coefficient of linear expansion of the metal of the pendulum?

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