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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

a
9
b
5
c
8
d
7

detailed solution

Correct option is A

T=2πlg                 g=4π2lT2Δgg=Δll+2ΔTT=0.110+2×0.020.5=0.01+0.08=0.09∆gg%=9%

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