A wire of length $l = 6 + 0.06 c m$ and radius $r = 0.5 + 0.005 c m$ has mass m = $0.3 + 0.003$ g. Maximum percentage error in density is

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Solution

The correct option is A $4$
$d = \frac{M}{V}$
$d = \frac{M}{π r^{2} h}$
$%$ error in mass $= \frac{0.003}{0.3} \times 100 = 1 %$
$%$ error in length $= \frac{0.006}{0.6} \times 100 = 1 %$
$%$ error in radius $= \frac{0.005}{0.5} \times 100 = 1 %$
$%$ error in mass $= 1 % + 2 \times 1 % + 1 % = 4 %$

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Similar questions

A wire has a mass (0.3 +/- 0.003) , RADIUS(0.5 +/- 0.005) MM and length (6+/-0.06) cm. the maximum percentage error in measurements of density is?

l = 6 \pm 0.06

r = 0.5 \pm 0.005

m = 0.3 \pm 0.003

\pm

\pm

\pm

0.3 \pm 0.003 g

0.5 \pm 0.005 m m

6 \pm 0.06 c m

l = 5 \pm 0.05 c m

r = 0.5 \pm 0.005 c m

m = 0.4 + 0.004 g

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