If T = 2 pi sqrt{frac{L}{g}},where g is a con | vedclass

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Question

(C) Given the formula $T = 2 \pi \sqrt{\frac{L}{g}}$.Taking the natural logarithm on both sides: $\ln T = \ln(2 \pi) + \frac{1}{2} \ln L - \frac{1}{2}

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(C) Given the formula $T = 2 \pi \sqrt{\frac{L}{g}}$.Taking the natural logarithm on both sides: $\ln T = \ln(2 \pi) + \frac{1}{2} \ln L - \frac{1}{2} \ln g$.Differentiating both sides,we get $\frac{dT}{T} = \frac{1}{2} \frac{dL}{L}$.The relative error in $T$ is $\frac{dT}{T}$.The percentage error in $L$ is $\frac{dL}{L} 	imes 100$.According to the problem,$\frac{dT}{T} = k 	imes (\frac{dL}{L} 	imes 100)$.Comparing this with $\frac{dT}{T} = \frac{1}{2} \frac{dL}{L}$,we have $k 	imes 100 = \frac{1}{2}$.Therefore,$k = \frac{1}{200}$.Thus,$\frac{1}{k} = 200$.

If $T = 2 \pi \sqrt{\frac{L}{g}}$,where $g$ is a constant and the relative error in $T$ is $k$ times the percentage error in $L$,then $\frac{1}{k} =$

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