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. (ρAl = 2.63 × 10–8 Ω m, ρCu = 1.72 × 10–8 Ω m, Relative density of Al = 2.7, of Cu = 8.9.)

Answer –

According to the question statement, some given properties of aliminium are –

Length = l₁

Resistance = R

Resistivity ,ρ_AI =ρ₁= 2.63×10⁻⁸ Ωm

Relative density , d₁ = 2.7

Area of cross-section = A₁

Similarly, some given properties of copper are –

Length = l₂

Resistance = R₂

Resistivity, ρ_cu =ρ₂=1.72 × 10^–8 Ω m

Relative density, d₂ = 8.9

Area of cross-section = A₂

We can write the resistances of both in the following manner –

${{R}_{1}}={{\rho }_{1}}\frac{{{l}_{1}}}{{{A}_{1}}}\to (1)$

\[{{R}_{2}}={{\rho }_{2}}\frac{{{l}_{2}}}{{{A}_{2}}}\to (2)\]

Also, it has been given that R₁ = R₂, so from 1 and 2 we get –

\[{{\rho }_{1}}\frac{{{l}_{1}}}{{{A}_{1}}}={{\rho }_{2}}\frac{{{l}_{2}}}{{{A}_{2}}}\]

Given, l₁ = l₂ we have –

\[\]\[\frac{{{\rho }_{1}}}{{{A}_{1}}}=\frac{{{\rho }_{2}}}{{{A}_{2}}}\]

\[\therefore \frac{{{A}_{1}}}{{{A}_{2}}}=\frac{{{\rho }_{1}}}{{{\rho }_{2}}}\]

=  (2.63×10⁻⁸) / (1.72 × 10^–8)
​= 1.52

Mass of aluminiumcan be expressed as,

m₁ = Volume x density

= A₁ l₁ x d₁

Similarly, Mass of copper is given by

 m₂ = Volume x density

= A₂ l₂ x d₂

Therefore by dividing both masses, we get the relation –

m₁/m₂ =(A₁l₁ x d₁) / ( A₂l₂ x d₂)
and since l₁ = l₂ the relation becomes –

m₁/m₂ =(A₁ d₁) / ( A₂d₂)

m₁/m₂ = (1.52) x (2.7/8.9) = (1.52) x (0.303)

m₁/m₂ = 0.46

Therefore, the calculated mass ratio of aluminium to copper is 0.46. Since aluminium is lighter, it is preferred for long suspensions of cables.