Solution:

Given, masses of three wires of copper are in the ratio 1:3:5 and their lengths are in the ratio 5:3:1.

Let the masses of the three wires be m1, m2, and m3, and their lengths be l1, l2, and l3, respectively.

Given, m1:m2:m3 = 1:3:5 and l1:l2:l3 = 5:3:1

Let the density of copper be ρ.

Then, the volumes of the wires are given by V1 = (m1/ρ), V2 = (m2/ρ), and V3 = (m3/ρ).

Let the resistivities of the wires be ρ1, ρ2, and ρ3, respectively.

Then, the electrical resistances of the wires are given by R1 = (ρ1*l1)/(π*(d1/2)^2), R2 = (ρ2*l2)/(π*(d2/2)^2), and R3 = (ρ3*l3)/(π*(d3/2)^2).

Here, d1, d2, and d3 are the diameters of the wires.

As the wires are made of the same material (copper), their resistivities are the same. Therefore, we can write ρ1 = ρ2 = ρ3 = ρ.

Now, we have to find the ratio of their electrical resistances.

We know that R1:R2:R3 = (ρ1*l1)/(π*(d1/2)^2) : (ρ2*l2)/(π*(d2/2)^2) : (ρ3*l3)/(π*(d3/2)^2)

Substituting ρ1 = ρ2 = ρ3 = ρ, l1:l2:l3 = 5:3:1, and simplifying, we get:

R1:R2:R3 = (d2/d1)^2 : 1 : (d1/d3)^2

Now, we have to find the ratio of the diameters of the wires.

We know that the masses of the wires are proportional to their volumes. Therefore,

m1/m2 = V1/V2 = (d1/d2)^2 * (l1/l2) = (d1/d2)^2 * (5/3)
m2/m3 = V2/V3 = (d2/d3)^2 * (l2/l3) = (d2/d3)^2 * (3/1)

From these equations, we can find the values of (d1/d2)^2 and (d2/d3)^2.

(d1/d2)^2 = (m1/m2) * (3/5) = (1/3) * (3/5) = 1/5
(d2/d3)^2 = (m2/m3) * (1/3) = (3/5) * (1/3) = 1/5

Therefore, (d1/d2)^2 = (d2/d3)^2 = 1/5.

Substituting these values in the equation R1:R2:R3 = (d2/d1)^2 : 1 : (d1/d3)^2