If the length of a cylinder is measured to be 4.28cm with an error of 0.01cm, then the percentage error in the measurement will be
A
0.4%
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B
0.5%
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C
0.2%
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D
0.1%
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Solution
The correct option is C0.2% Given, L=(4.28±0.01)cm On comparing with L=Lmean±ΔLmean, we get, Lmean=4.28cm and ΔLmean=0.01cm We know, percentage error, δa=ΔLmeanLmean×100% ⇒δa=0.014.28×100%=0.2%