Answer:
1.24
Explanation:
The resistivity of copper[tex]\rho_1=2.65\times 10^{-8}\ \Omega-m[/tex]
The resistivity of Aluminum,[tex]\rho_2=1.72\times 10^{-8}\ \Omega-m[/tex]
The wires have same resistance per unit length.
The resistance of a wire is given by :
[tex]R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}[/tex]
According to given condition,
[tex]\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24[/tex]
So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.
