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Question

The time period of oscillation of simple pendulum is given by T=2πlg ; The length of the pendulum is measured as l = 10±0.1  cm and time period as T = 0.5±0.02 s.  The percentage error in measuring ‘g’ is

Moderate
Solution

T=2πlg          g=4π2lT2
Δgg=Δll+2ΔTT=0.110+2×0.020.5=0.01+0.08=0.09
gg%=9%

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