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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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$6$

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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 $\ell = 20 \ cm$ and $\Delta \ell = 2 \ mm = 0.2 \ cm$.Given $T_{total} = 40 \ s$ for $50$ oscillations,so $T = \frac{40}{50} = 0.8 \ s$. The resolution $\Delta T_{total} = 1 \ s$,so $\Delta T = \frac{1}{50} = 0.02 \ s$.Substituting the values: $\frac{\Delta g}{g} = \frac{0.2}{20} + 2 \left( \frac{0.02}{0.8} ight)$.$\frac{\Delta g}{g} = 0.01 + 2 \left( 0.025 ight) = 0.01 + 0.05 = 0.06$.Percentage error $N = 0.06 	imes 100 = 6 \%$.

The measured value of the length of a simple pendulum is $20 \ cm$ with $2 \ mm$ accuracy. The time for $50$ oscillations was measured to be $40 \ s$ with $1 \ s$ resolution. From these measurements,the accuracy in the measurement of acceleration due to gravity is $N \%$. The value of $N$ is:

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