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Q.

What is the accuracy of g determined by a simple pendulum of length (100 ± 0.1)cm whose period of oscillation is 2 s determined by measuring the time for 100 oscillations using a clock of 0.1 s resolution?

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a

0.2 %

b

0.5%

c

0.1%

d

2%

answer is A.

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Detailed Solution

T=tn and ΔT=Δtn∴ ΔTT×100=Δtt×100=0.12×100×100=0.05%Δll×100=0.1cm100cm×100=0.1%Now, T=2πlgor    g=4π2lT2∝lT2So % error in g isΔgg×100=Δll+2ΔTT×100soΔgg×100=(0.1+2×0.05)=0.2 %
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