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?

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Solution

Since density is mass÷volume,
For a cylinder it becomes- mass÷(pi×radius^2×length)

Relative percent errors for-

Mass-0.003÷0.3×100=1%

Radius^2-2×0.005÷0.5×100=2%

Length-0.06÷0.6×100=1%

THE UNITS CANCEL THEMSELVES FROM THE NUMERATOR AND DENOMINATOR.

therefore,

maximum percentage error in density-1%+2%+1%=4%

so, the answer is 4%

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Similar questions

\pm

\pm

\pm

0.3 \pm 0.003 g

0.5 \pm 0.005 m m

6 \pm 0.06 c m

(0.3 \pm 0.003) g

(0.5 \pm 0.005) m m

(6 \pm 0.06) c m

l = 6 + 0.06 c m

r = 0.5 + 0.005 c m

0.3 + 0.003

(0.3 \pm 0.003)

(0.5 \pm 0.005)

(6 \pm 0.06)

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