A wire of length $\ell = 6 \pm 0.06\,\text{cm}$ and radius $r = 0.5 \pm 0.005\,\text{cm}$ and mass $m = 0.3 \pm 0.003\,\text{gm}$. Maximum percentage error in density is

Step 1: Identify the given values.

The measurements and their absolute errors are:

$$\ell = 6 \pm 0.06\ \text{cm}$$ $$r = 0.5 \pm 0.005\ \text{cm}$$ $$m = 0.3 \pm 0.003\ \text{gm}$$

Step 2: Write the formula for density and its relative error.

The density $\rho$ of a cylindrical wire is given by:

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

Taking the natural logarithm on both sides:

$$\ln\rho = \ln m - 2\ln r - \ln \ell - \ln\pi$$

Differentiating to find the maximum fractional error:

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

Step 3: Calculate the individual percentage errors.

For mass $m$:

$$\frac{\Delta m}{m} \times 100\% = \frac{0.003}{0.3} \times 100\% = 1\%$$

For radius $r$:

$$2\left(\frac{\Delta r}{r}\right) \times 100\% = 2 \times \frac{0.005}{0.5} \times 100\% = 2\%$$

For length $\ell$:

$$\frac{\Delta \ell}{\ell} \times 100\% = \frac{0.06}{6} \times 100\% = 1\%$$

Step 4: Calculate the maximum percentage error in density.

Sum the individual percentage errors:

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

Answer: The maximum percentage error in density is $4\%$ (Option 1).