The time period T of a simple pendulum of len | vedclass

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(A) Given,$T=2 \pi \sqrt{\frac{l}{g}}$.Taking the natural logarithm on both sides,we get $\ln T = \ln(2 \pi) + \frac{1}{2} \ln l - \frac{1}{2} \ln g$.

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$0.5$

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(A) Given,$T=2 \pi \sqrt{\frac{l}{g}}$.Taking the natural logarithm on both sides,we get $\ln T = \ln(2 \pi) + \frac{1}{2} \ln l - \frac{1}{2} \ln g$.Differentiating both sides with respect to $l$,we get $\frac{1}{T} \frac{dT}{dl} = \frac{1}{2l}$.Thus,the relative change is $\frac{dT}{T} = \frac{1}{2} \frac{dl}{l}$.Given that the length is increased by $1 \%$,we have $\frac{dl}{l} 	imes 100 = 1 \%$.Therefore,the percentage change in the time period is $\frac{dT}{T} 	imes 100 = \frac{1}{2} 	imes \left( \frac{dl}{l} 	imes 100 ight) = \frac{1}{2} 	imes 1 \% = 0.5 \%$.Hence,the approximate change in the time period is $0.5 \%$.

The time period $T$ of a simple pendulum of length $l$ is given by $T=2 \pi \sqrt{\frac{l}{g}}$,where $g$ denotes the acceleration due to gravity. If the length of the pendulum is increased by $1 \%$,then the approximate change in its time period is (in $\%$)

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