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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+2ΔTT=0.120.0+2190=0.027The percentage error in g isΔgg×100=ΔLL×100+2×ΔTT×100=3%

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