The period of oscillation of a simple pendulu | vedclass

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(C) Given the formula for the time period: $T = 2\pi \sqrt{\frac{l}{g}}$.Squaring both sides,we get: $T^2 = 4\pi^2 \frac{l}{g}$,which implies $g = 4\p

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$0.2$

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(C) Given the formula for the time period: $T = 2\pi \sqrt{\frac{l}{g}}$.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 $l = 100 \, cm = 1000 \, mm$ and $\Delta l = 1 \, mm$,the percentage error in $l$ is: $\frac{\Delta l}{l} 	imes 100 = \frac{1}{1000} 	imes 100 = 0.1 \%$.For $100$ oscillations,the total time $t = 100T = 200 \, s$. The least count of the stopwatch is $\Delta t = 0.1 \, s$.The error in the period $T$ is $\Delta T = \frac{\Delta t}{100} = \frac{0.1}{100} = 0.001 \, s$.The percentage error in $T$ is: $\frac{\Delta T}{T} 	imes 100 = \frac{0.001}{2} 	imes 100 = 0.05 \%$.Therefore,the percentage error in $g$ is: $\frac{\Delta g}{g} 	imes 100 = (0.1 \%) + 2 	imes (0.05 \%) = 0.1 \% + 0.1 \% = 0.2 \%$.

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