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Question
The period of oscillation of a simple pendulum is $T = 2\pi \sqrt {{L \over g}} $. Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be :
1.30%
1.33%
1.13%
1.03%

Solution

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