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In 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 the maximum percentage error in determination of the acceleration due to gravity g is
Explanation
As we know, time period of a simple pendulum $T = 2\pi\sqrt{L/g} \Rightarrow g = 4\pi^{2}L/T^{2}$ The maximum percentage error in g $\Delta g/g \times 100 = \Delta L/L \times 100 + 2(\Delta T/T \times 100) = 2\% + 2(3\%) = 8\%$