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Q.

The period of oscillation of a simple  pendulum is T=2πLg. Measured value of L is 20 cm known to  1mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist-watch of 1 s resolution. The accuracy in the determination of g is

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a

1%

b

2%

c

3%

d

4%

answer is C.

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Detailed Solution

Given, T=2πLgor g=4π2LT2Hence, T=tn  and  ΔT=Δtn∴           ΔTT=ΔttThe errors in both L and t are the least count errors. Therefore, (Δgg)=(ΔLL)+2(ΔTT)=0.120.0+2(190)=0.027The percentage error in g is(Δgg)×100=(ΔLL)×100+2×(ΔTT)×100=3%
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