The masses of three wires of copper are in the ratio of \(1: 3: 5\) and their lengths are in the ratio of \(5: 3: 1\). The ratio of their electrical resistance is: (A) \(1: 1: 1\) (B) \(1: 3: 5\) (C) \(5: 3: 1\) (D) \(125: 15: 1\)

For a wire, the electrical resistance (R) is given by the formula: \(R = \rho \frac{L}{A}\) Where: - R = electrical resistance - \(\rho\) = resistivity (a constant for a given material, in this case, copper) - L = length of the wire - A = cross-sectional area of the wire However, the problem gives us the mass ratios of the wires, so we need to find a way to relate the mass to the cross-sectional area.

The mass (m) of the wire can be written as the volume (V) multiplied by the density (\(\delta\)) of the material: \(m = \delta V\) The volume (V) of the wire can be written as the product of its length (L) and cross-sectional area (A): \(V = A \cdot L\) Now we can rewrite the mass formula in terms of A and L: \(m = \delta (A \cdot L)\) We can then solve for the cross-sectional area (A) in terms of the mass (m) and length (L): \(A = \frac{m}{\delta \cdot L}\)

Now we'll substitute the expression for the cross-sectional area (A) from step 2 into the resistance formula from step 1: \(R = \rho \frac{L}{\frac{m}{\delta \cdot L}}\) We can now simplify the formula: \(R = \rho \frac{L^2 \cdot \delta}{m}\)

According to the problem, the mass ratios of the three wires are 1:3:5, and the length ratios are 5:3:1. We will assign variables to the masses (m1, m2, m3) and lengths (L1, L2, L3) of the wires: \(m1 : m2 : m3 = 1 : 3 : 5\) \(L1 : L2 : L3 = 5 : 3 : 1\) Now we can write the resistances (R1, R2, R3) of the three wires using the simplified formula from step 3: \(R1 = \rho \frac{L1^2 \cdot \delta}{m1}\) \(R2 = \rho \frac{L2^2 \cdot \delta}{m2}\) \(R3 = \rho \frac{L3^2 \cdot \delta}{m3}\)

To find the ratio of the resistances, we will form a proportion between the three resistances: \(\frac{R1}{R2} = \frac{\rho \frac{L1^2 \cdot \delta}{m1}}{\rho \frac{L2^2 \cdot \delta}{m2}}\) and \(\frac{R2}{R3} = \frac{\rho \frac{L2^2 \cdot \delta}{m2}}{\rho \frac{L3^2 \cdot \delta}{m3}}\) Simplifying and substituting the given mass and length ratios: \(\frac{R1}{R2} = \frac{(5)^2}{(3)^2} \cdot \frac{3}{1} = \frac{25}{9} \cdot 3 = \frac{75}{9}\) \(\frac{R2}{R3} = \frac{(3)^2}{(1)^2} \cdot \frac{5}{3} = \frac{9}{1} \cdot \frac{5}{3} = \frac{45}{3}\) Now we can find the ratio of all three resistances: \(R1 : R2 : R3 = \frac{75}{9} : \frac{45}{3} : 1 = 125 : 15 : 1\) Hence, the ratio of their electrical resistances is: (D) \(125: 15: 1\)