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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.

(ρ_{Al} = 2.63 × 10^{−8} Ω m, ρ_{Cu} = 1.72 × 10^{−8} Ω m, Relative density of Al = 2.7, of Cu = 8.9.)

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#### 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 = ρ_1"l"_1/"A"_1` ..........(1)

`"R"_2 = ρ_2"l"_2/"A"_2` ............(2)

It is given that

R_{1 }= R_{2}

∴ `ρ_1"l"_1/"A"_1 = ρ_2"l"_2/"A"_2`

And

I_{1 }= I_{2}

∴ `ρ_1/"A"_1 = ρ_2/"A"_2`

`"A"_1/"A"_2 = ρ_1/ρ_2`

= `(2.63 xx 10^-8)/(1.72 xx 10^-8)`

= `2.63/1.72`

Mass of the aluminium wire,

m_{1} = Volume × Density

= A_{1}l_{1} × d_{1 }= A_{1} l_{1}d_{1} ….......(3)

Mass of the copper wire,

m_{2} = Volume × Density

= A_{2}l_{2} × d_{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 I_{1 }= I_{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 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.

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