Find the resistance of a hollow cylindrical conductor of length 1.0m and inner and outer radii 1.0mm and 2.0mm respectively. The resistivity of the material is `2.0xx10^(-8)(Omega)m`.

Explanation

The area of cross section of the conductor through which the charges will flow is
`A=(pi)(2.0mm)^(2)-(pi)(1.0mm)^(2)`.
`=3.0xx(pi)mm^(2).`
the resistance of the wire is therefore,
`R=(rho)(l)/(A)=((2.0xx10^(-8)(Omega)m))/(3.0xx(pi)xx10^(-6)m^(2))=2.1xx10^(-3)(Omega).`