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(B) Let the quantity be $X = \frac{L}{T^2}$.The relative error in $X$ is given by $\frac{\Delta X}{X} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.Giv

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$7$

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(B) Let the quantity be $X = \frac{L}{T^2}$.The relative error in $X$ is given by $\frac{\Delta X}{X} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.Given that the percentage error in $L$ is $\frac{\Delta L}{L} 	imes 100 = 3 \%$ and the percentage error in $T$ is $\frac{\Delta T}{T} 	imes 100 = 2 \%$.The maximum percentage error in $X$ is $\frac{\Delta X}{X} 	imes 100 = \left( \frac{\Delta L}{L} 	imes 100 ight) + 2 \left( \frac{\Delta T}{T} 	imes 100 ight)$.Substituting the given values: $	ext{Percentage Error} = 3 \% + 2 	imes (2 \%) = 3 \% + 4 \% = 7 \%$.

In an experiment of a simple pendulum, the errors in the measurement of length of the pendulum $(L)$ and time period $(T)$ are $3 \%$ and $2 \%$, respectively. The maximum percentage error in the value of $\frac{L}{T^2}$ is (in $\%$)

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