A metal wire has mass 0.4±0.002g, radius 0.3±0.001mm, and length 5±0.02 cm. The maximum possible percentage error in the measurement of density will nearly be

We know, d=MV=Mπr2lProportional error, Δdd=ΔMM+2Δrr+Δll =0.0020.4+2×0.0010.3+0.025 =0.024+0.023+0.025 =0.3+0.4+0.2460=0.9460 % error=0.9460×100=9460 ≈1.6%

A metal wire has mass 0.4±0.002g, radius 0.3±0.001mm, and length 5±0.02 cm. The maximum possible percentage error in the measurement of density will nearly be

We know, d=MV=Mπr2lProportional error, Δdd=ΔMM+2Δrr+Δll =0.0020.4+2×0.0010.3+0.025 =0.024+0.023+0.025 =0.3+0.4+0.2460=0.9460 % error=0.9460×100=9460 ≈1.6%

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A metal wire has mass $(0.4 \pm 0.002) g,$ radius $(0.3 \pm 0.001) m m,$ and length $(5 \pm 0.02) c m .$ The maximum possible percentage error in the measurement of density will nearly be

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1.2%

1.6%

1.3%

1.4%

answer is D.

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Detailed Solution

We know, $d = \frac{M}{V} = \frac{M}{π r^{2} l}$

Proportional error, $\frac{Δd}{d} = \frac{ΔM}{M} + \frac{2 Δr}{r} + \frac{Δl}{l}$

$\begin{aligned} = \frac{0.002}{0.4} + \frac{2 \times 0.001}{0.3} + \frac{0.02}{5} \\ = \frac{0.02}{4} + \frac{0.02}{3} + \frac{0.02}{5} \\ = \frac{0.3 + 0.4 + 0.24}{60} = \frac{0.94}{60} \\ % error = \frac{0.94}{60} \times 100 = \frac{94}{60} \\ \approx 1.6 % \end{aligned}$