Time period of 10 oscillations of a simple pendulum is found to be 20 sec by a stopwatch of resolution 0.25 sec. Length of the string is measured to be 30 cm by a scale of least count 0.1 cm. Percentage error in measurement of acceleration due to gravity g by simple pendulum is? $T = 2 π \sqrt{\frac{l}{g}}$

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Solution

$T = 2 π \sqrt{\frac{l}{g}} \Rightarrow T^{2} = 4 π^{2} \frac{l}{g}$
$g = 4 π^{2} \frac{l}{T^{2}} e r r o r = \frac{Δ g}{g} \times 100 % = (\frac{Δ l}{l} \times 100 + \frac{2 Δ T}{T} \times 100) %$
$= (\frac{0.1}{30} \times 100 + 2 \times \frac{0.25}{20} \times 100) %$
$= (0.33 + 2.5) % = 2.83 %$

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