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

The density of a cylindrical wire is: $$\rho = \frac{m}{\pi r^2 L}$$

The maximum percentage error in density is:

$$ \frac{\Delta\rho}{\rho} \times 100 = \frac{\Delta m}{m} \times 100 + 2\frac{\Delta r}{r} \times 100 + \frac{\Delta L}{L} \times 100 $$

Substituting the given values:

$$ = \frac{0.01}{0.4} \times 100 + 2 \times \frac{0.03}{6} \times 100 + \frac{0.04}{8} \times 100 $$

$$ = 2.5\% + 1.0\% + 0.5\% = 4\% $$

The maximum error in density is 4%.