**Question:**

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. $\left(\rho_{A l}=2.63 \times 10^{-8} \Omega \mathrm{m}, \rho_{C u}=1.72 \times 10^{-8} \Omega \mathrm{m}\right.$, Relative density of Al $=$ $2.7$, of $\mathrm{Cu}=8.9$.)

**Solution:**

Resistivity of aluminium, *ρ*Al = 2.63 × 10−8 Ω m

Relative density of aluminium, *d*1 = 2.7

Let *l*1 be the length of aluminium wire and *m*1 be its mass.

Resistance of the aluminium wire = *R*1

Area of cross-section of the aluminium wire =* A*1

Resistivity of copper, *ρ*Cu = 1.72 × 10−8 Ω m

Relative density of copper, *d*2 = 8.9

Let *l*2 be the length of copper wire and *m*2 be its mass.

Resistance of the copper wire =* R*2

Area of cross-section of the copper wire =* A*2

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,

*m*1 = Volume × Density

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

Mass of the copper wire,

*m*2 = 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 *m*1 is less than *m*2. Hence, aluminium is lighter than copper.

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