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Question

Physics Question on Units and measurement

A silver wire has a mass (0.6+0.006)(0.6 \overset{+}{-} 0.006) g, radius (0.5+0.005)(0.5 \overset{+}{-} 0.005) mm and length (4+0.04)(4 \overset{+}{-} 0.04) cm. The maximum percentage error in the measurement of its density will be

A

4%

B

3%

C

6%

D

7%

Answer

4%

Explanation

Solution

ρ=mv=mπr2×l\rho = \frac{m}{v} = \frac{m}{\pi r^2 \times l}

∴ Percentage error in

ρ=(0.0060.6+2×0.0050.5+0.044)×100ρ = \bigg(\frac{0.006}{0.6}+2\times \frac{0.005}{0.5}+\frac{0.04}{4}\bigg)\times 100

= 4%4 \%