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(B) The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$.Squaring both sides,we get $T^2 = 4\pi^2 \frac{L}{g}$,which

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$2.7$

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(B) The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$.Squaring both sides,we get $T^2 = 4\pi^2 \frac{L}{g}$,which implies $g = 4\pi^2 \frac{L}{T^2}$.The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.Given: $L = 20.0$ cm,$\Delta L = 1$ mm = $0.1$ cm.$T_{total} = 90.0$ s,$\Delta T_{total} = 1$ s. Since $T = \frac{T_{total}}{100}$,$\Delta T = \frac{\Delta T_{total}}{100} = \frac{1}{100} = 0.01$ s.Substituting the values:$\frac{\Delta g}{g} = \frac{0.1}{20.0} + 2 	imes \frac{1}{90} = 0.005 + 0.0222 = 0.0272$.Percentage error = $0.0272 	imes 100 = 2.72 \% \approx 2.7 \%$.

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