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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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1.7%

4.4%

2.7%

2.27%

answer is C.

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Detailed Solution

$\begin{array}{l} g = 4 π^{2} \frac{ℓ}{T^{2}} \Rightarrow \frac{Δg}{g} = \frac{Δℓ}{ℓ} + \frac{2 ΔT}{T} \\ = \frac{0.1}{20} \times 100 + 2 [\frac{\frac{1}{\frac{100}{90}}}{100}] \times 100 \\ = 0.5 + 2.22 = 2.72 = 2.7 % \end{array}$

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