There are two wires A and B of same mass and of the same material. The diameter of wire A is one-third the diameter of wire B. If the resistance of wire A is `30 Omega`, find the resistance of wire B.

Explanation

Here, `R_(A) =30 Omega`. Let `l^(A), D^(A)` be the length and diameter of wire A and `l_(B),D^(B)` be the length and diameter of wire B. Let d be the density of the material of wires A and B.
Mass of the wire = volume `xx` density
= area of cross-section `xx` lenght `xx` density
As mass of two wires is same to
`m=pi D_(A)^(2)/4 xx l_(A) xx d = pi D_(B)^(2)/4 xx l_(B) xx d`
or `D_(A)^(2)l_(A)=D_(B)^(2)l_(B)`
or `l_(B)/l_(A) =D_(A)^(2)/D_(B)^(2)=(D_(B//3))^(2)/D_(B)^(2)=1/9`
`R_(B)/R_(A) = (4 rho l_(B)//(pi D_(B)^(2)))/(4 rho l_(A)//(pi D_(A)^(2))) = l_(B)/l_(A) xx D_(A)^(2)/D_(B)^(2) = 1/9 xx 1/9 =1/81`
`R_(B)=R_(A)/81 =30/81=0.37 Omega`