1. Write the formula for density

For a wire (cylindrical shape),

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

So,

$\rho \propto \frac{m}{r^2 l}$

2. Use maximum fractional error rule

If

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

then maximum relative error is

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

(Note: $\pi$ is a constant, so it has no error.)

3. Calculate percentage error in each quantity

• For mass:

$\frac{\Delta m}{m} = \frac{0.003}{0.60} = 0.005 = 0.5\%$

• For radius:

$\frac{\Delta r}{r} = \frac{0.01}{0.50} = 0.02 = 2\%$

Since radius is squared, contribution becomes:

$2 \times 2\% = 4\%$

• For length:

$\frac{\Delta l}{l} = \frac{0.05}{10.00} = 0.005 = 0.5\%$

4. Add the maximum percentage errors

$\text{Maximum percentage error in density} = 0.5\% + 4\% + 0.5\% = 5\%$

5. Match with options

$5\%$ corresponds to Option C.

6. Compare with stored correct answer

Stored correct answer = C

My derived answer = C

So they agree.