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 `1mm`. Find the ratio of resistance `R_(A)` to `R_(B)`.

Explanation

Let l= length of conductors A and B and `rho`= resistivity of the material of conductors A and B. The resistance of conductor A is given by:
`R_(A) = (rho l)/((pi d_(A)^(2))/(4)) = (4l rho)/(pi d_(A)^(2)) = (4 rho l)/(pi (1 xx 10^(-3))^(2))`
The resistance of conductor B is given by:
`R_(B) = (rho l)/(pi ((d_(2)^(2))/(4) - (d_(1)^(2))/(4))) = (4 rho l)/(pi (d_(2)^(2) - d_(1)^(2))) = (4 rho l)/(pi (4 xx 10^(-6)-1 xx 10^(-6)))`
`= (4 rho l)/(pi xx 3 xx 10^(-6))`
`:. (R_(A))/(R_(B)) = (4 rho l)/(pi xx 10^(-6)) xx (pi xx 3 xx 10^(-6))/(4 rho l) = (3)/(1)`
`rArr R_(A) : R_(B) = 3:1`