A wire has a mass 0.3 pm 0.003 text{ g},radiu | vedclass

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(C) Density $\rho$ is given by the formula $\rho = \frac{M}{V} = \frac{M}{\pi r^2 L}$.The relative error in density is given by $\frac{\Delta \rho}{\r

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(C) Density $ho$ is given by the formula $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:$M = 0.3 	ext{ g}, \Delta M = 0.003 	ext{ g} \implies \frac{\Delta M}{M} = \frac{0.003}{0.3} = 0.01$.$r = 0.5 	ext{ mm}, \Delta r = 0.005 	ext{ mm} \implies \frac{\Delta r}{r} = \frac{0.005}{0.5} = 0.01$.$L = 6 	ext{ cm}, \Delta L = 0.06 	ext{ cm} \implies \frac{\Delta L}{L} = \frac{0.06}{6} = 0.01$.Substituting these values into the error formula:$\frac{\Delta ho}{ho} = 0.01 + 2(0.01) + 0.01 = 0.01 + 0.02 + 0.01 = 0.04$.The percentage error is $\frac{\Delta ho}{ho} 	imes 100 \% = 0.04 	imes 100 \% = 4 \%$.

$A$ wire has a mass $0.3 \pm 0.003 \text{ g}$,radius $0.5 \pm 0.005 \text{ mm}$ and length $6 \pm 0.06 \text{ cm}$. The maximum percentage error in the measurement of its density is (in $\%$)

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