A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper wire to that of the tungsten wire.

directly proportional to its length i.e[tex]R\propto l[/tex]
inversely proportional to its cross section area i.e[tex]R\propto \frac{1}{A}[/tex]

Therefore

[tex]R=\rho\frac{l}{A}[/tex]

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

[tex]A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2[/tex]

[tex]R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }[/tex]

[tex]\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }[/tex]......(1)

Again for tungsten:

[tex]R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }[/tex]

[tex]\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }[/tex]........(2)

Given that [tex]R_1=R_2[/tex] and [tex]l_1=l_2[/tex]

Dividing the equation (1) and (2)

[tex]\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}[/tex]

[tex]\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}[/tex] [since [tex]R_1=R_2[/tex] and [tex]l_1=l_2[/tex]]

[tex]\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}[/tex]

[tex]\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}[/tex]

[tex]\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}[/tex]

Therefore the ratio of diameter of the copper to that of the tungsten is

[tex]\sqrt{3} :\sqrt{10}[/tex]