Masses of 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 are

Options

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

Answer

C

Explanation

(c) : Their volume are in the ratio 1 : 3 : 5
`pir_(1)^(2)l_(1) : pir_(2)^(2)l_(2) : pir_(3)^(2)l_(3)` = 1 : 3 : 5
`pir_(1)^(2)5l : pir_(2)^(2)3l : pir_(3)^(2)l = 1 : 3 : 5 rArr 5r_(1)^(2) : 3r_(2)^(2) : lr_(3)^(2) = 1:3:5`
`(5r_(1)^(2))/1 = (3r_(2)^(2))/3 = (r_(3)^(2))/5 = k rArr r_(1)^(2) = k/5, r_(2)^(2) = k, r_(3)^(2) = 5k`
`R_(1):R_(2):R_(3) = l_(1)/(pir_(1)^(2)):l_(2)/(pir_(2)^(2)):l_(3)/(pir_(3)^(2)) = (5l)/(pir_(1)^(2)):(3l)/(pir_(2)^(2)):(l)/(pir_(3)^(2))`
`rArr (5l)/(k//5):(3l)/(k):(l)/(5k) rArr 125:15:1`