An experiment is performed to obtain the value of acceleration due to gravity \(g\) by using a simple pendulum of length \(L\). In this experiment time for \(100\, oscillations\) is measured by using a watch of \(1\, second\) least count and the value is \(90.0\, seconds\). The length \(L\) is measured by using a meter scale of least count \(1\, mm\) and the value is \(20.0\, cm\). The error in the determination of \(g\) would be ........... \(\%\)

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Answer & Solution

Correct Answer

(B) \(2.7\)

Detailed explanation

Here, \(T=2 \pi \sqrt{\frac{L}{g}}\) or \(T^{2}=4 \pi^{2}(L / g)\)

So, \(g=\frac{4 \pi^{2} L}{T^{2}}\)

Thus, \(\frac{\Delta g}{g}=\frac{\Delta L}{L}+2 \frac{\Delta T}{T}\)

% error in \(g=\frac{\Delta g}{g} \times 100\)

\(=\left(\frac{\Delta L}{L}+2 \frac{\Delta T}{T}\right) \times 100\)

\(=\left(\frac{(1 / 10)}{20}+2 \times \frac{1}{90}\right) \times 100=2.72 \%\)

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