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. `25 : 1 : 125`
B. `1 : 125 : 25`
C. `125 : 1 : 25`
D. `125 : 25 : 1`

Explanation

Given , Ratio of masses = `1 : 3 : 5`
As density will remain same
Ratio of volumes = `1 : 3 : 5 " " …. (i)`
and Ratio of length `5 : 3 : 1 " " …. (ii)`
For ratio of resistance
`R_(1) : R_(2) : R_(3) = (rho l_(1))/(A_(1)) : (rho l_(2))/(A_(2) ) : (rho l_(3))/(A_(3))`
Since , resistivity remain same
` = (l_(1))/(A_(1)) : (l_(2))/(A_(2)) : (l_(3))/(A_(3)) = (l_(1)^(2))/(A_(1) l_(1)) : (l_(2)^(2))/(A_(2) l_(2)) : (l_(3)^(2))/(A_(3) l_(3))`
From Eqs . (i) and (ii)
`R_(1) : R_(2) : R_(3) = ([5]^(2))/(1) : ([3]^(2))/(3) : ([1]^(2))/(5) = 25 : 3 : (1)/(5) = 125 : 15 : 1`