The period of oscillation of a simple pendulum is \(T=2\pi \sqrt {\frac{l}{g}} \). 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 \(90\ s\) using a wrist watch of \(1\; s\) resolution. The accuracy in the determination of \(g\) is ........ \(\%\)

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(A) \(3\)

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Detailed explanation

AS, \(g = 4\,{\pi ^2}\frac{l}{{{T^2}}}\)
So, \(\frac{{\Delta g}}{g} \times 100 = \frac{{\Delta l}}{L} \times 100 + 2\frac{{\Delta T}}{T} \times 100\)
\( = \frac{{0.1}}{{20}} \times 100 + 2 \times \frac{1}{{90}} \times 100 = 2.72 \simeq 3\% \)

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