A silver wire has a 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

We need to find the maximum percentage error in the measurement of density of the silver wire.

The density is given by:

$$\rho = \frac{m}{V} = \frac{m}{\pi r^2 l}$$

The percentage error in density is:

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

Substituting the given values:

$$\frac{\Delta m}{m} \times 100 = \frac{0.006}{0.6} \times 100 = 1\%$$

$$2 \times \frac{\Delta r}{r} \times 100 = 2 \times \frac{0.005}{0.5} \times 100 = 2\%$$

$$\frac{\Delta l}{l} \times 100 = \frac{0.04}{4} \times 100 = 1\%$$

Therefore, the maximum percentage error in density:

$$\frac{\Delta \rho}{\rho} \times 100 = 1\% + 2\% + 1\% = 4\%$$

The correct answer is Option C.