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

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$4$

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(C) The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{\ell}{g}}$.Squaring both sides and rearranging for $g$,we get $g = \frac{4\pi^2 \ell}{T^2}$.The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}$.Given: $\ell = 25.0 \; cm$,so $\Delta \ell = 0.1 \; cm$. The time for $40$ oscillations is $t = 50 \; s$,so the time period $T = \frac{50}{40} = 1.25 \; s$. The resolution of the stopwatch is $\Delta t = 1 \; s$,so $\Delta T = \frac{\Delta t}{40} = \frac{1}{40} \; s$.Substituting these values: $\frac{\Delta g}{g} = \frac{0.1}{25.0} + 2 	imes \frac{1/40}{50/40} = \frac{0.1}{25} + 2 	imes \frac{1}{50} = 0.004 + 0.04 = 0.044$.Therefore,the percentage error is $0.044 	imes 100 = 4.4 \%$.

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