The period of oscillation of a simple pendulu | vedclass

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(D) The formula for the acceleration due to gravity is $g = 4 \pi^2 L / T^2$.Taking the relative error,we have $\frac{\Delta g}{g} = \frac{\Delta L}{L

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$3$

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(D) The formula for the acceleration due to gravity is $g = 4 \pi^2 L / T^2$.Taking the relative error,we have $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.Given $L = 20.0 \; cm$ and $\Delta L = 1 \; mm = 0.1 \; cm$. Thus,$\frac{\Delta L}{L} = \frac{0.1}{20.0} = 0.005$.For the time period,$T = \frac{t}{n}$,where $t = 90 \; s$ and $n = 100$. The resolution of the watch is $\Delta t = 1 \; s$.Since $T = t/n$,the relative error is $\frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{1}{90}$.Substituting these into the error formula: $\frac{\Delta g}{g} = 0.005 + 2 \left( \frac{1}{90} ight) = 0.005 + 0.0222 = 0.0272$.The percentage error is $\frac{\Delta g}{g} 	imes 100 = 0.0272 	imes 100 \approx 3 \%$.

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