Two wires of equal length, one of aluminium and the other of copper have the same resistance.

Solution:

Resistivity of aluminium, ρAl = 2.63 × 10−8 Ω m

Relative density of aluminium, d1 = 2.7

Let l1 be the length of aluminium wire and m1 be its mass.

Resistance of the aluminium wire = R1

Area of cross-section of the aluminium wire = A1

Resistivity of copper, ρCu = 1.72 × 10−8 Ω m

Relative density of copper, d2 = 8.9

Let l2 be the length of copper wire and m2 be its mass.

Resistance of the copper wire = R2

Area of cross-section of the copper wire = A2

The two relations can be written as

$R_{1}=\rho_{1} \frac{l_{1}}{A_{1}}$   ...(i)

$R_{2}=\rho_{2} \frac{l_{2}}{A_{2}}$   ...(ii)

It is given that,

$R_{1}=R_{2}$

$\therefore \rho_{1} \frac{l_{1}}{A_{1}}=\rho_{2} \frac{l_{2}}{A_{2}}$

And,

$l_{1}=l_{2}$

$\therefore \frac{\rho_{1}}{A_{1}}=\frac{\rho_{2}}{A_{2}}$

$\frac{A_{1}}{A_{2}}=\frac{\rho_{1}}{\rho_{2}}$

$=\frac{2.63 \times 10^{-8}}{1.72 \times 10^{-8}}=\frac{2.63}{1.72}$

Mass of the aluminium wire,

m1 = Volume × Density

$=A_{1} l_{1} \times d_{1}=A_{1} l_{1} d_{1} \ldots$ (3)

Mass of the copper wire,

m2 = Volume × Density

$=A_{2} l_{2} \times d_{2}=A_{2} l_{2} d_{2} \ldots$ (4)

Dividing equation (3) by equation (4), we obtain

$\frac{m_{1}}{m_{2}}=\frac{A_{1} l_{1} d_{1}}{A_{2} l_{2} d_{2}}$

For $l_{1}=l_{2}$,

$\frac{m_{1}}{m_{2}}=\frac{A_{1} d_{1}}{A_{2} d_{2}}$

For $\frac{A_{1}}{A_{2}}=\frac{2.63}{1.72}$

$\frac{m_{1}}{m_{2}}=\frac{2.63}{1.72} \times \frac{2.7}{8.9}=0.46$

It can be inferred from this ratio that m1 is less than m2. Hence, aluminium is lighter than copper.

Since aluminium is lighter, it is preferred for overhead power cables over copper.