Answer –
According to the question statement, some given properties of aliminium are –
Length = l1
Resistance = R
Resistivity ,ρAI =ρ1= 2.63×10−8 Ωm
Relative density , d1 = 2.7
Area of cross-section = A1
Similarly, some given properties of copper are –
Length = l2
Resistance = R2
Resistivity, ρcu =ρ2=1.72 × 10–8 Ω m
Relative density, d2 = 8.9
Area of cross-section = A2
We can write the resistances of both in the following manner –
${{R}_{1}}={{\rho }_{1}}\frac{{{l}_{1}}}{{{A}_{1}}}\to (1)$
\[{{R}_{2}}={{\rho }_{2}}\frac{{{l}_{2}}}{{{A}_{2}}}\to (2)\]
Also, it has been given that R1 = R2, so from 1 and 2 we get –
\[{{\rho }_{1}}\frac{{{l}_{1}}}{{{A}_{1}}}={{\rho }_{2}}\frac{{{l}_{2}}}{{{A}_{2}}}\]
Given, l1 = l2 we have –
\[\]\[\frac{{{\rho }_{1}}}{{{A}_{1}}}=\frac{{{\rho }_{2}}}{{{A}_{2}}}\]
\[\therefore \frac{{{A}_{1}}}{{{A}_{2}}}=\frac{{{\rho }_{1}}}{{{\rho }_{2}}}\]
= (2.63×10−8) / (1.72 × 10–8)
= 1.52
Mass of aluminiumcan be expressed as,
m1 = Volume x density
= A1 l1 x d1
Similarly, Mass of copper is given by
m2 = Volume x density
= A2 l2 x d2
Therefore by dividing both masses, we get the relation –
m1/m2 =(A1l1 x d1) / ( A2l2 x d2)
and since l1 = l2 the relation becomes –
m1/m2 =(A1 d1) / ( A2d2)
m1/m2 = (1.52) x (2.7/8.9) = (1.52) x (0.303)
m1/m2 = 0.46
Therefore, the calculated mass ratio of aluminium to copper is 0.46. Since aluminium is lighter, it is preferred for long suspensions of cables.