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

Options

A. `1:3:5`
B. `5:3:1`
C. `1:15:25`
D. `125:15:1`

Answer

D

Explanation

Let `A_(1),A_(2),A_(3)` be the area of cross-section of three wires of copper of masses `m_(1),m_(2),m_(3)` and length `l_(1),l_(2),l_(3)` respectively.
Given, `m_(1)=m,m_(2)=3m,m_(3)=5m,l_(1)=5l,l_(2)=3l,l_(3)=l`
Mass = Volume `xx` density
`m=A_(1)xx5lxxd` (`because` Volume = Area `xx` length) ...(i)
`3m=A_(2)xx3lxxd" "...(ii)`
`5m=A_(3)xxlxxd" "...(iii)`
From (i) and (ii), we get `A_(2)=5A_(1)`
From (i) and (iii), we get `A_(3)=25A_(1)`
`therefore R_(1)=(rhol_(1))/(A_(1))=(rho5l)/(A_(1))`
`R_(2)=(rhol_(2))/(A_(2))=(rhoxx3l)/(5A_(1))=(3)/(25)R_(1)`
`R_(3)=(rhol_(3))/(A_(3))=(rhoxxl)/(25A_(1))=(R_(1))/(125)`
`therefore R_(1):R_(2):R_(3)=R_(1):(3)/(25)R_(1):(R_(1))/(125)=125:15:1`