Two wires of equal length,one of aluminium and the other of copper,have the same resistance. Which of the two wires is lighter? Hence,explain why aluminium wires are preferred for overhead power cables. $(\rho_{Al} = 2.63 \times 10^{-8} \; \Omega m, \rho_{Cu} = 1.72 \times 10^{-8} \; \Omega m, \text{Relative density of } Al = 2.7, \text{of } Cu = 8.9.)$

(A) Let the resistivity,length,area of cross-section,density,and mass of the aluminium wire be $\rho_1, l_1, A_1, d_1, m_1$ and for the copper wire be $\rho_2, l_2, A_2, d_2, m_2$ respectively.
Given: $l_1 = l_2 = l$ and $R_1 = R_2 = R$.
Since $R = \rho \frac{l}{A}$,we have $\rho_1 \frac{l}{A_1} = \rho_2 \frac{l}{A_2}$,which implies $\frac{A_1}{A_2} = \frac{\rho_1}{\rho_2} = \frac{2.63 \times 10^{-8}}{1.72 \times 10^{-8}} \approx 1.529$.
The mass of a wire is given by $m = \text{Volume} \times \text{Density} = A \cdot l \cdot d$.
Taking the ratio of masses: $\frac{m_1}{m_2} = \frac{A_1 l d_1}{A_2 l d_2} = \left( \frac{A_1}{A_2} \right) \left( \frac{d_1}{d_2} \right)$.
Substituting the values: $\frac{m_1}{m_2} = \left( \frac{2.63}{1.72} \right) \times \left( \frac{2.7}{8.9} \right) \approx 1.529 \times 0.303 \approx 0.463$.
Since $\frac{m_1}{m_2} < 1$,$m_1 < m_2$,meaning the aluminium wire is lighter.
Aluminium is preferred for overhead power cables because it is much lighter than copper for the same resistance,reducing the mechanical stress on the supporting towers.