Answer
$1.34\times10^{-4}\Omega$
Work Step by Step
Use equation 18–3 to find the resistance, where A is the cross-sectional area of the pipe.
$$A=\pi (R_{outer}^2- R_{inner}^2)$$
$$R=\rho\frac{\mathcal{l}}{A}$$
$$R=\frac{(1.68\times10^{-8}\Omega \cdot m)(10.0m)}{\pi ((2.5\times10^{-2}m)^2- (1.5\times10^{-2}m)^2)}$$
$$R=1.34\times10^{-4}\Omega$$
