Q. 164.2( 16 Votes )

Two wires of equa

Answer :

Resistivity of aluminium, p1 = 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, p2 = 1.72 × 10−8 Ω metre


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


We can write,


R1 = (p1 × l1)/A1


R2 = (p2 × l2)/A2


Since R1 = R2


(p1 × l1)/A1 = (p2 × l2)/A2


And l1 = l2


Therefore p1/A1 = p2/A2


A1/A2 = (2.63 × 10-8)/(1.72 × 10–8)


A1/A2 = 2.63/1.72


Mass of the aluminium wire,


m1 = Volume × Density


= A1 × l1 × d1 = A1 l1 d1 .....1


Mass of the copper wire, m2 = Volume × Density


= A2 × l2 × d2 = A2 l2 d2 .......2


Dividing equation (1) by equation (2), we obtain


m1/m2 = (A1 l1 d1)/(A2 l2 d2)


Since l1 = l2


m1/m2 = A1 d1/A2 d2


Since A1/A2 = 2.63/1.72 calculated above


m1/m2 = (2.63 × 2.7)/(1.72 × 8.9)


m1/m2 = 0.46


Since m1 is smaller than m2 Hence, aluminium is lighter than copper.


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


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