The masses of the three wires of copper are in the ratio of `1:3:5` and their lengths are in the ratio of `5:3:1`. The ratio of their electrical resistance is

Options

A. `1:03:05`
B. `5:03:01`
C. `1 : 15 : 125`
D. `125 : 15 : 1`

Answer

D

Explanation

`R = (rhol)/(A) = (rhol^(2))/(Al) = (rhol^(2))/(V) = (rhol^(2))/(m//d) = (rhodl^(2))/(m) or R prop (l^(2))/(m)`
`R_(1) : R_(2) : R_(3) = (l_(0)^(2))/(m_(1)): (l_(1)^(2))/(m_(2)) : (l_(3)^(2))/(m_(3)) = (25)/(1) : (9)/(3) : (1)/(5) = 125 : 15 : 1`