A wire has a mass $0.3 \pm 0.003 g$ and length $6 \pm 0.06 c m$ . The maximum percentage error in the measurements of its linear mass density is:-

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Solution

The correct option is D 4
Mass $= 0.3 \pm 0.003 g$
Length $= 6 \pm 0.06 c m$
$ρ = \frac{m}{V}$
$d e n s i t y = \frac{m a s s}{v o l u m e}$
Fr a cylinder it be ___ mass $\div (P π \times r a d i u s^{2} \times l e n g t h)$
Relative percent errors for mass $= 0.003 \div \times 100 = 1 %$
$R a d i u s^{2} = 2 \times 0.005 \div 0.5 \times 100 = 2 %$
Length $= 0.06 \div 0.6 \times 100 = 1 %$
Maximum percentage error in density $= 1 % + 2 % + 1 % = 4 %$

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\pm

\pm

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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

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 + 0.06 c m

r = 0.5 + 0.005 c m

0.3 + 0.003

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