In a simple pendulum experiment, the maximum percentage error in the measurement of length is $2 %$ and that in the observation of the time-period is $3 %$ . Then, the maximum percentage error in determination of the acceleration due to gravity $g$ is

5 %

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6 %

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7 %

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8 %

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10 %

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Solution

The correct option is E $8 %$
Time period, $T = 2 π \sqrt{\frac{l}{g}}$
or $g = \frac{4 π^{2} l}{T^{2}}$
The maximum percentage error
$\frac{Δ g}{g} = \frac{Δ l}{l} + 2 ∣ ∣ ∣ \frac{Δ T}{T} ∣ ∣ ∣$
where $\frac{Δ l}{l} = 2 %$ and $\frac{Δ T}{T} = 3 %$
$\frac{Δ g}{g} = 2 % + 2 \times 3 %$
$= 2 % + 6 % = 8 %$

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In a simple pendulum experiment, the maximum percentage error in the measurement of length is 2% and that in the observation of the time-period is 3%. Then, the maximum percentage error in determination of the acceleration due to gravity g is

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