A wire has a mass 0.4±0.004g and length 8±0.08(cm) The maximum percentage error in the measurement of its density is 4%. The radius of the wire is r±Δr, find Δr.

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0.02r

0.01r

0.03r

0.1r

answer is B.

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Detailed Solution

ρ=mπr2LΔρρ×100=Δmm+2Δrr+ΔLL×100 =0.01+2Δrr+0.01×100=4 (given) or 100Δrr=1 ⇒Δr=0.01r

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