ln a simple pendulum experiment, the maximum percentage error in the measurement of length is 2% and that in the observation of the time-period is 3%. Then what is the maximum percentage error in determination of the acceleration due to gravity g ?

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answer is 8.

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Detailed Solution

As we know, time period of a simple pendulum

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

The maximum percentage error in g

$\begin{array}{l} \frac{Δg}{g} \times 100 = \frac{ΔL}{L} \times 100 + 2 (\frac{ΔT}{T} \times 100) \\ = 2 % + 2 (3 %) = 8 % \end{array}$

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