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(B) 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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$5$

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(B) 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 $g = 4 \pi^2 \frac{\ell}{T^2}$.The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}$.Given: $\ell = 10 \ cm = 100 \ mm$ and $\Delta \ell = 1 \ mm$.The total time for $100$ oscillations is $t = 100 	imes 0.5 \ s = 50 \ s$. The resolution of the watch is $\Delta t = 1 \ s$.Thus,the time period $T = 0.5 \ s$ and the error in time period $\Delta T = \frac{\Delta t}{100} = \frac{1 \ s}{100} = 0.01 \ s$.Substituting the values: $\frac{\Delta g}{g} = \frac{1 \ mm}{100 \ mm} + 2 	imes \frac{0.01 \ s}{0.5 \ s} = 0.01 + 2 	imes 0.02 = 0.01 + 0.04 = 0.05$.Therefore,the percentage error is $0.05 	imes 100 = 5 \%$.Thus,$x = 5$.

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