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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.Squaring both sides,we get $T^2 = 4\pi^2 \frac{L}{g}$.Rearranging f

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$(\alpha + 2\beta) 	imes 100$

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.Squaring both sides,we get $T^2 = 4\pi^2 \frac{L}{g}$.Rearranging for $g$,we get $g = 4\pi^2 \frac{L}{T^2}$.The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.Given that the relative error in length $\frac{\Delta L}{L} = \alpha$ and the relative error in time period $\frac{\Delta T}{T} = \beta$.Substituting these values,we get $\frac{\Delta g}{g} = \alpha + 2\beta$.To find the percentage error,we multiply by $100$: $	ext{Percentage error} = (\alpha + 2\beta) 	imes 100$.

The acceleration due to gravity is measured on the surface of the Earth by using a simple pendulum. If $\alpha$ and $\beta$ are relative errors in the measurement of length and time period respectively,then the percentage error in the measurement of acceleration due to gravity is ................

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