A wire of length l=6±0.06 cm and radius of cross-section r=0.5±0.005 cm and mass m=0.3±0.003 gm. Maximum percentage error in density is -----

ρ=mlπr2Δρρ=Δmm+2Δrr+Δll=0.0030.3+2×0.0050.5+0.066=4100=4%

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A wire of length $l = 6 \pm 0.06 c m$ and radius of cross-section $r = 0.5 \pm 0.005 c m$ and mass $m = 0.3 \pm 0.003 g m .$ Maximum percentage error in density is -----

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answer is 4.

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Detailed Solution

$ρ = \frac{m}{l π r^{2}}$

$\frac{Δ ρ}{ρ} = \frac{Δ m}{m} + \frac{2 Δ r}{r} + \frac{Δ l}{l} = \frac{0.003}{0.3} + \frac{2 \times 0.005}{0.5} + \frac{0.06}{6} = \frac{4}{100} = 4 %$

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