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 prefered for overhead power cables ? Given `rho_(Al)=2.63xx10^(-8) Omega m`, `rho_(Cu)=0.72xx10^(-8) Omega m` , relative density of `Al=2.7` and that of `Cu =8.9`.

Explanation

Resistivity of aluminium, `rho_(Al) = 2.63 xx 10^(-8) Omega` 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 copper, `rho_(Cu)=1.72 xx 10^(-8) Omega` m
Relative density of copper, `rho_(Cu) =1.72 xx 10^(-8) Omega` 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 relation can be written as
`R_(1) = rho_(1)(l_(1))/(A_(1))`....(1)
`R_(2) = rho_(2) (l_(2))/(A_(2))`
It is given that
`R_(1)=R_(2)` ltbr. `rho_(1)(l_(2))/(A_(1))=rho_(2)(l_(2))/(A_(2))`
And, `l_(1)=l_(2)`
`:. (rho_(1))/(A_(1))=(rho_(2))/(A_(2))`
`=(2.63 xx 10^(-8))/(1.72 xx 10^(-8))=(2.63)/(1.72)`
Mass of the aluminium wire,
`m_(1)` = Volume `xx` Density
`= A_(1)l_(1)xxd_(1)=A_(1)l_(1)d_(1)`...(3)
Mass of the copper wire,
`m_(2)`= Volume `xx` Density
`= A_(2)l_(2)xxd_(2)=A_(2)l_(2)d_(2)`....(4)
Dividing equation (3) by equation (4), we obtain
`(m_(1))/(m_(2))=(A_(1)l_(1)d_(1 ))/(A_(2)l_(2)d_(2))`
For `l_(1)=l_(2)`
`(m_(1))/(m_(2))=(A_(1)d_(1))/(A_(2)d_(2))`
For `(A_(1))/(A_(2))=(2.63)/(1.72)`
`(m_(1))/(m_(2))=(2.63)/(1.72) xx(2.7)/(8.9) = 0.46`
It can be inferred 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.