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Q.

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

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a

9

b

5

c

8

d

7

answer is A.

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Detailed Solution

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