A current of 5 A passes through a copper conductor (resistivity = $$1.7 \times 10^{-8}$$ Ω m) of radius of cross-section 5 mm. Find the mobility of the charges if their drift velocity is $$1.1 \times 10^{-3}$$ ms$$^{-1}$$.

We have a current $$I = 5\ \text{A}$$ passing through a copper wire whose resistivity is given as $$\rho = 1.7 \times 10^{-8}\ \Omega\,\text{m}$$. The radius of the circular cross-section is $$r = 5\ \text{mm}$$, which in SI units is

$$r = 5\ \text{mm} = 5 \times 10^{-3}\ \text{m}.$$

The drift velocity of the charge carriers is provided as $$v_d = 1.1 \times 10^{-3}\ \text{m s}^{-1}$$. We wish to find the mobility $$\mu$$ of these carriers.

First, we recall the relation between mobility, drift velocity and electric field:

$$\mu = \frac{v_d}{E}.$$

Therefore, we must determine the electric field $$E$$ inside the conductor. Ohm’s microscopic form gives us

$$E = \rho J,$$

where $$J$$ is the current density. Current density itself is defined by

$$J = \frac{I}{A},$$

with $$A$$ being the cross-sectional area of the wire. Because the cross-section is a circle, its area is

$$A = \pi r^2.$$

Substituting the numerical value of the radius, we get

$$A = \pi \left(5 \times 10^{-3}\ \text{m}\right)^2 = \pi \times 25 \times 10^{-6}\ \text{m}^2 = 25\pi \times 10^{-6}\ \text{m}^2.$$ Numerically,

$$A = 25 \times 3.1416 \times 10^{-6}\ \text{m}^2 \approx 78.54 \times 10^{-6}\ \text{m}^2 = 7.85 \times 10^{-5}\ \text{m}^2.$$

Now we find the current density:

$$J = \frac{I}{A} = \frac{5\ \text{A}}{7.85 \times 10^{-5}\ \text{m}^2} \approx 6.37 \times 10^{4}\ \text{A m}^{-2}.$$

Next, we calculate the electric field using $$E = \rho J$$:

$$E = (1.7 \times 10^{-8}\ \Omega\,\text{m}) \left(6.37 \times 10^{4}\ \text{A m}^{-2}\right) = 1.7 \times 6.37 \times 10^{-8 + 4}\ \text{V m}^{-1} = 10.829 \times 10^{-4}\ \text{V m}^{-1} \approx 1.08 \times 10^{-3}\ \text{V m}^{-1}.$$

Finally, we obtain the mobility:

$$\mu = \frac{v_d}{E} = \frac{1.1 \times 10^{-3}\ \text{m s}^{-1}} {1.08 \times 10^{-3}\ \text{V m}^{-1}} \approx 1.02\ \text{m}^2 \text{V}^{-1} \text{s}^{-1}.$$

Rounding to one significant figure, $$\mu \approx 1.0\ \text{m}^2 \text{V}^{-1} \text{s}^{-1}$$, which matches the third option in the list.

Hence, the correct answer is Option C.