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(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}} = 2\pi l^{1/2} g^{-1/2}$.Using the formula for propagation of errors

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$\pm 3 \%$

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(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}} = 2\pi l^{1/2} g^{-1/2}$.Using the formula for propagation of errors,the maximum relative error in $T$ is given by:$\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l} + \frac{1}{2} \frac{\Delta g}{g}$.To find the maximum percentage error,multiply by $100$:$\left( \frac{\Delta T}{T} 	imes 100 ight) \% = \frac{1}{2} \left( \frac{\Delta l}{l} 	imes 100 ight) \% + \frac{1}{2} \left( \frac{\Delta g}{g} 	imes 100 ight) \%$.Given $\left( \frac{\Delta l}{l} 	imes 100 ight) \% = 2\%$ and $\left( \frac{\Delta g}{g} 	imes 100 ight) \% = 4\%$.Substituting these values:$	ext{Max } \% 	ext{ error in } T = \frac{1}{2} (2\%) + \frac{1}{2} (4\%) = 1\% + 2\% = 3\%$.Thus,the maximum percentage error in the time period is $\pm 3\%$.

In a simple pendulum experiment,the maximum percentage error in the measurement of length is $2\%$ and that in acceleration due to gravity $g$ is $4\%$. Then the maximum percentage error in determination of the time-period is

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