In an experiment of a simple pendulum,the tim | vedclass

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(A) The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.Squaring both sides and rearranging for $g$,we get $g = \fra

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$0.5$

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(A) The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.Squaring both sides and rearranging for $g$,we get $g = \frac{4\pi^2 l}{T^2}$.The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T}$.Given: $l = 100 \, cm$,$\Delta l = 0.1 \, cm$,$T_{total} = 50 \, s$,$\Delta T_{total} = 0.1 \, s$.Note that $T$ is the time period of one vibration,$T = \frac{T_{total}}{25} = \frac{50}{25} = 2 \, s$.The error in $T$ is $\Delta T = \frac{\Delta T_{total}}{25} = \frac{0.1}{25} = 0.004 \, s$.Substituting these values: $\frac{\Delta g}{g} = \frac{0.1}{100} + 2 \left( \frac{0.004}{2} ight) = 0.001 + 0.004 = 0.005$.Percentage error = $0.005 	imes 100 = 0.5 \%$.

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