A pendulum clock keeps correct time at 0 ^oC. | vedclass

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(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.Taking the derivative,the fractional change in time period is $\fra

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$\frac{1}{2} \alpha t 	imes 86400$

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(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.Taking the derivative,the fractional change in time period is $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L}$.Since $\Delta L = L \alpha \Delta 	heta$,where $\Delta 	heta = t$,we have $\frac{\Delta L}{L} = \alpha t$.Therefore,the fractional change in time is $\frac{\Delta T}{T} = \frac{1}{2} \alpha t$.This represents the time lost per second.Since there are $86400$ seconds in a day $(24 	imes 60 	imes 60 = 86400)$,the total time lost per day is given by $\Delta T_{day} = \frac{1}{2} \alpha t 	imes 86400$ seconds.

$A$ pendulum clock keeps correct time at $0\ ^oC$. The thermal coefficient of linear expansion of the material of the pendulum is $\alpha$. If the temperature rises to $t\ ^oC$,then the clock loses per day by (in seconds):

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