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

Answer

D

Explanation

Mass of wire, M = Volume `xx` Density
`=Al xx d`
or `A= (M)/(ld)`
Resistance of wire, `R= (rho l)/(A) = (rho l)/((M//ld)) or R= (rho l^(2)d)/(m)`
Since all three wires are made up of same material, therefore `rho` and d remain the same for all the three wires
`:. R prop (l^(2))/(M)`
Hence, `R_(1) : R_(2) : R_(3) =(l_(1)^(2))/(M_(1)) : (l_(2)^(2))/(M_(2)): (l_(3)^(3))/(M_(3))= (5^(2))/(1) : (3^(2))/(3): (1^(2))/(5)`
`=25:3: (1)/(5)= 125: 15: 1`