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In an experiment to determine the value of acceleration due to gravity g using a simple pendulum, the measured value of length of the pendulum is 31.4cm known to 1 mm accuracy and the time period for 100 oscillations of pendulum is 112.0s known to 0.01s accuracy. Find the accuracy in determining the value of g.

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2.3 %

0.5 %

0.23 %

0.79 %

answer is A.

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Detailed Solution

$g \propto \frac{l}{T^{2}} \Rightarrow \frac{Δ g}{g} = \frac{Δ l}{l} + 2 \frac{Δ T}{T}$

$= \frac{0.1}{31.4} + 2 (\frac{0.01}{1.12}) = 0.023 = 2.3 %$

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