A silver wire has mass (0.6 pm 0.006) ; g,rad | vedclass

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(A) Density $\rho = \frac{m}{V} = \frac{m}{\pi r^2 l}$.The relative error in density is given by $\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\fr

Vedclass · Accepted Answer

$4$

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(A) Density $ho = \frac{m}{V} = \frac{m}{\pi r^2 l}$.The relative error in density is given by $\frac{\Delta ho}{ho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$.Given values:$m = 0.6 \; g, \Delta m = 0.006 \; g$$r = 0.5 \; mm, \Delta r = 0.005 \; mm$$l = 4 \; cm, \Delta l = 0.04 \; cm$Calculating percentage error:$\frac{\Delta ho}{ho} 	imes 100 = \left( \frac{0.006}{0.6} + 2 	imes \frac{0.005}{0.5} + \frac{0.04}{4} ight) 	imes 100$$= (0.01 + 2 	imes 0.01 + 0.01) 	imes 100$$= (0.01 + 0.02 + 0.01) 	imes 100 = 0.04 	imes 100 = 4 \%$.

$A$ silver wire has mass $(0.6 \pm 0.006) \; g$,radius $(0.5 \pm 0.005) \; mm$ and length $(4 \pm 0.04) \; cm$. The maximum percentage error in the measurement of its density will be $......\,\%$

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