The period of oscillation of a simple pendulum is T=2πLg. Measured value of L is 20.0 cm known to 1 mm accuracy. And time for 100 oscillations of the pendulum is found to be 90 sec using a wrist watch of 1 sec resolution. The accuracy in the determination of g is
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a
5%
b
2%
c
3%
d
1%
answer is C.
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Detailed Solution
If T=2πLg then g=4π2LT2∆gg=∆LL+2∆TT=0.120+2190∆gg=0.9+4180=4.9180∆gg×100=490180=2.7≈3