The massses of the three wires of copper are in the ratio 1 : 3 : 5. And their lengths are in th 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

`m = d A_1 5 l rArr A_1 = k//5`
`3 m = d A_2 3l rArr A_2 = k`
`5 m = d A_3 l rArr A_3 = 5 k`
`R_1 : R_2 : R_3 = (5l)/(A_1) : ( 3l)/(A_2) : (l)/(A_3)`
=`(5l)/(k//5) : (3 l)/(k) : (l)/(5 k)`
=`125 : 15 : 1`.