Density of the wire is given by $\rho = \frac{\text{mass}}{\text{volume}} = \frac{m}{\pi r^2 l}$. The maximum relative error in density is given by: $\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$ Percentage error in density = $\left( \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l} \right) \times 100\%$ Given: $m = 0.4\text{ g}$, $\Delta m = 0.002\text{ g}$ $r = 0.3\text{ mm}$, $\Delta r = 0.001\text{ mm}$ $l = 5\text{ cm}$, $\Delta l = 0.02\text{ cm}$ Substituting the values: Percentage error = $\left( \frac{0.002}{0.4} + 2 \times \frac{0.001}{0.3} + \frac{0.02}{5} \right) \times 100\%$ $= \left( \frac{2}{400} + \frac{2}{300} + \frac{2}{500} \right) \times 100\%$ $= \left( \frac{1}{200} + \frac{1}{150} + \frac{1}{250} \right) \times 100\%$ $= \left( \frac{1}{2} + \frac{2}{3} + \frac{2}{5} \right)\%$ $= (0.5 + 0.666... + 0.4)\%$ $= 1.566...\% \approx 1.6\%$