A hollow cylindrical conductor has length of 3.14 m, while its inner and outer diameters are 4 mm and 8 mm respectively. The resistance of the conductor is $$n \times 10^{-3}$$ $$\Omega$$. If the resistivity of the material is $$2.4 \times 10^{-8}$$ $$\Omega$$ m. The value of $$n$$ is _____.

Find the resistance of a hollow cylindrical conductor. The conductor has length $$L = 3.14$$ m, inner diameter 4 mm (radius $$r_i = 2$$ mm = $$2 \times 10^{-3}$$ m), outer diameter 8 mm (radius $$r_o = 4$$ mm = $$4 \times 10^{-3}$$ m), and resistivity $$\rho = 2.4 \times 10^{-8}$$ $$\Omega$$ m.

The cross-sectional area is
$$ A = \pi(r_o^2 - r_i^2) = \pi(16 \times 10^{-6} - 4 \times 10^{-6}) = 12\pi \times 10^{-6}\;\text{m}^2 $$
and the resistance is
$$ R = \frac{\rho L}{A} = \frac{2.4 \times 10^{-8} \times 3.14}{12\pi \times 10^{-6}} = \frac{2.4 \times 10^{-8} \times 3.14}{12 \times 3.14 \times 10^{-6}} = \frac{2.4 \times 10^{-8}}{12 \times 10^{-6}} = 2 \times 10^{-3}\;\Omega $$. Therefore $$n = 2$$.

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