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

Options

A. `1:3:5`
B. `5:3:1`
C. `1:15:125`
D. `125:15:1`

Answer

D

Explanation

`R = (rho l)/A = (rho l^(2))/(Al) = (rho l^(2) xx "density")/("Volume x density") rArr R prop l^(2)/m`
`R_(1) : R_(2) :R_(3) = l_(1)^(2)/m_(1) : l_(2)^(2)/m_(2) : l_(3)^(2)/m_(3) = 25/1 : 9/3 : 1/5`
`rArr R_(1) : R_(2) : R_(3) = 25 : 3 :1/5 = 125 : 15 : 1`