The time period of a simple pendulum is given | vedclass

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(B) The formula for the acceleration due to gravity is $g = \frac{4 \pi^2 \ell}{T^2}$.Taking the relative error,we have $\frac{\Delta g}{g} = \frac{\D

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$3$

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(B) The formula for the acceleration due to gravity is $g = \frac{4 \pi^2 \ell}{T^2}$.Taking the relative error,we have $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}$.Given $\ell = 10 \ cm$ and $\Delta \ell = 1 \ mm = 0.1 \ cm$.For $200$ oscillations,the total time $t = 100 \ s$ with resolution $\Delta t = 1 \ s$. The time period $T = \frac{t}{200} = \frac{100}{200} = 0.5 \ s$.The error in time period is $\Delta T = \frac{\Delta t}{200} = \frac{1}{200} \ s$.Substituting these values: $\frac{\Delta g}{g} = \frac{0.1}{10} + 2 \left( \frac{1/200}{0.5} ight) = 0.01 + 2 \left( \frac{1}{100} ight) = 0.01 + 0.02 = 0.03$.Percentage error is $\frac{\Delta g}{g} 	imes 100 = 0.03 	imes 100 = 3 \%$.

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