The maximum percentage error in the measurement of density of a wire is

[Given, mass of wire = (0.60 ± 0.003)g; radius of wire = (0.50 ± 0.01)cm; length of wire = (10.00 ± 0.05)cm]

Given Data:

Mass of wire:

• $$m = (0.60 \pm 0.003) \text{ g}$$

Percentage error in mass:

• $$\frac{\Delta m}{m} \times 100 = \frac{0.003}{0.60} \times 100 = 0.5\%$$

Radius of wire:

• $$r = (0.50 \pm 0.01) \text{ cm}$$

Percentage error in radius:

• $$\frac{\Delta r}{r} \times 100 = \frac{0.01}{0.50} \times 100 = 2\%$$

Length of wire:

• $$l = (10.00 \pm 0.05) \text{ cm}$$

Percentage error in length:

• $$\frac{\Delta l}{l} \times 100 = \frac{0.05}{10.00} \times 100 = 0.5\%$$

Step 1: Formula for Density

Density $\rho$ is given by:

$$\rho = \frac{\text{Mass}}{\text{Volume}}$$

For a cylindrical wire, volume:

$$V = \pi r^2 l$$

Thus, density:

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

Step 2: Maximum Percentage Error in Density

Using the error propagation formula:

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

Substituting values:

$$\frac{\Delta \rho}{\rho} \times 100 = 0.5\% + 2(2\%) + 0.5\%$$

 $$= 0.5\% + 4\% + 0.5\%$$ $$= 5\%$$