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Q.

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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a

1.2%

b

1.6%

c

1.3%

d

1.4%

answer is D.

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Detailed Solution

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

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

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

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