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The period of oscillation of a simple pendulum is $T = 2 π \sqrt{L / g} .$ Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1 s resolution. What is percentage error the determination of g ? (Nearly)

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Detailed Solution

$\frac{Δ g}{g} = \frac{Δ L}{L} + 2 \frac{Δ T}{T}$
$= \frac{0.1}{20.0} + (\frac{1}{90}) [\frac{Δ T}{T} = \frac{Δ t}{t} = \frac{1}{90}]$
$= 0.027 = 0.03$

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