In an experiment for the measurement of g usi | vedclass

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

Vedclass · Accepted Answer

$0.9$

Vedclass · Answer

(C) The time period of a simple pendulum is given by the formula: $T = 2\pi \sqrt{\frac{L}{g}}$.Squaring both sides,we get: $T^2 = 4\pi^2 \frac{L}{g}$.Rearranging for $g$: $g = 4\pi^2 \frac{L}{T^2}$.The relative error in $g$ is given by: $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.Given that the percentage error in length $\frac{\Delta L}{L} 	imes 100 = 0.5 \%$ and the percentage error in time period $\frac{\Delta T}{T} 	imes 100 = 0.2 \%$.Substituting these values into the error equation:Percentage error in $g = 0.5 \% + 2(0.2 \%) = 0.5 \% + 0.4 \% = 0.9 \%$.

In an experiment for the measurement of $g$ using a simple pendulum,the time period was measured with an accuracy of $0.2 \%$ while the length was measured with an accuracy of $0.5 \%$. The percentage accuracy in the value of $g$ thus obtained is: (in $\%$)

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