A cylindrical wire of mass (0.4 pm 0.01) , g | vedclass

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(C) The density $\rho$ of a cylindrical wire is given by the formula $\rho = \frac{m}{V} = \frac{m}{\pi r^2 l}$.Taking the relative error,we have $\fr

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$4$

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(C) The density $ho$ of a cylindrical wire is given by the formula $ho = \frac{m}{V} = \frac{m}{\pi r^2 l}$.Taking the relative error,we have $\frac{\Delta ho}{ho} = \frac{\Delta m}{m} + 2 \frac{\Delta r}{r} + \frac{\Delta l}{l}$.Given values are $m = 0.4 \, g, \Delta m = 0.01 \, g$,$l = 8 \, cm, \Delta l = 0.04 \, cm$,and $r = 6 \, mm, \Delta r = 0.03 \, mm$.Substituting these values into the error formula:$\frac{\Delta ho}{ho} = \frac{0.01}{0.4} + 2 \left( \frac{0.03}{6} ight) + \frac{0.04}{8}$.Calculating each term:$\frac{\Delta ho}{ho} = 0.025 + 0.01 + 0.005 = 0.04$.To find the percentage error,multiply by $100 \%$: $	ext{Percentage error} = 0.04 	imes 100 \% = 4 \%$.

$A$ cylindrical wire of mass $(0.4 \pm 0.01) \, g$ has length $(8 \pm 0.04) \, cm$ and radius $(6 \pm 0.03) \, mm$. The maximum error in its density will be $...... \, \%$.

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