The length of a pendulum is measured as 1.01 | vedclass

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

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$10$

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(C) 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: $l = 1.01 \ m$, $\Delta l = 0.01 \ m$, $T_{total} = 63 \ s$, $\Delta T_{total} = 3 \ s$.The time period $T = \frac{T_{total}}{30} = \frac{63}{30} = 2.1 \ s$.The error in time period $\Delta T = \frac{\Delta T_{total}}{30} = \frac{3}{30} = 0.1 \ s$.Substituting these values into the error formula:$\frac{\Delta g}{g} = \frac{0.01}{1.01} + 2 	imes \frac{0.1}{2.1} \approx 0.0099 + 0.0952 \approx 0.1051$.Percentage error = $0.1051 	imes 100 \% \approx 10.5 \%$.Rounding to the nearest provided option, the percentage error is $10 \%$.

The length of a pendulum is measured as $1.01 \ m$ and the time for $30$ oscillations is measured as $1 \ minute \ 3 \ s$. The error in length is $0.01 \ m$ and the error in time is $3 \ s$. The percentage error in the measurement of acceleration due to gravity is: (in $\%$)

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