Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1 mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistances `R_A` to `R_B`

Explanation

Here ` rho_1 = rho_2 =rho ` and ` l_1 = l_2 = l`
Conductor A is a solid wire of diameter D=1 mm, its cross-section area `A_1 = (pi D^2)/(4)`
Conductor B is a hollow tube of outer diameter `D_1` = 2 mm and inner diameter `D_2` = 1 mm, hence its cross-section area `A_2 = pi/4 (D_1^2 - D_2^2)`
As `R = (rho l)/(A) `, hence `(R_A)/(R_B) = A_2/A_1 = (pi/3 (D_1^2 - D_2^2) )/(pi/4D^2) = (D_1^2 - D_2^2)/(D^2) = ( (2mm)^2 - (1mm)^2)/((1mm)^2) = 3/1`
` rArr R_A = 3R_B`