In the measurement of the acceleration due to gravity using a simple pendulum, the percentage error in the measured value of the length of the pendulum is $2 %$ and that in the time period of oscillation is $1 %$ . The percentage error in the measured value of the acceleration due to gravity is

$4 %$

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$5 %$

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$6 %$

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$3 %$

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Solution

The correct option is B
$4 %$

$T = 2 π \sqrt{\frac{l}{g}} \Rightarrow g = \frac{4 π^{2} l}{T^{2}}$

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

$\Rightarrow (% \frac{Δ g}{g}) = (% \frac{Δ l}{l}) + 2 (% \frac{Δ T}{T})$

$\Rightarrow \frac{Δ g}{g} (in %) = 2 % + 2 \times 1 % = 4 %$

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