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. `15:1:125`
B. `1:125:15`
C. `125:1:15`
D. `125:15:1`

Answer

D

Explanation

Resistance in terms of length and mass is given by,
`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))`
`R_(1):R_(2):R_(3)=(25l_(1)^(2))/(m_(1)):(9l_(1)^(2))/(3m_(1)):(l_(1)^(2))/(5m_(1))`
`=25:3:1/5`
`=125:15:1`.