91Ó°ÊÓ

Wires$1$and $2$are made of the same metal. Wire $2$has twice the length and twice the diameter of wire with ${\mathrm{ÒÏ}}_2/{\mathrm{ÒÏ}}_1$ of the resistivity.

The resistivity of a material is an intensive property. Hence, it does not rely on the dimensions of the material.

Therefore $\frac{{\mathrm{ÒÏ}}_2}{{\mathrm{ÒÏ}}_1}$ is always given by:
$\frac{{\mathrm{ÒÏ}}_2}{{\mathrm{ÒÏ}}_1}=1$.

The ratio of the resistivity is $\frac{{\mathrm{ÒÏ}}_2}{{\mathrm{ÒÏ}}_1}=1$.

Wires $1$and $2$are made of the same metal. Wire $2$has twice the length and twice the diameter of wire $1$with $R_2/R_1$of the resistances of the two wires.

To solve for $\frac{R_2}{R_1}$, and the equation for the resistance of a material which is given by $R=\frac{\mathrm{ÒÏ}\mathrm{â„“}}{A}$. Hence, $\frac{R_2}{R_1}$ as,
$\frac{R_2}{R_1}=\frac{\frac{{\mathrm{ÒÏ}}_2{\mathrm{â„“}}_2}{A_2}}{\frac{{\mathrm{ÒÏ}}_1{\mathrm{â„“}}_1}{A_1}}$

Since, $A=\mathrm{Ï€}{r}^2$, plugging-that into the equation,
$\frac{R_2}{R_1}=\frac{\frac{{\mathrm{â„“}}_2}{\left(\mathrm{Ï€}{r}_2^2\right)}}{\frac{{\mathrm{â„“}}_1}{\left(\mathrm{Ï€}{r}_1^2\right)}}$

$=\frac{\frac{{\mathrm{â„“}}_2}{r_2^2}}{\frac{{\mathrm{â„“}}_1}{r_1^2}}$

The wire $2$has twice the length and diameter of wire l, hence the relationships between ${\mathrm{â„“}}_1$with ${\mathrm{â„“}}_2$and $r_1$with $r_2$is $2{\mathrm{â„“}}_1=\mathrm{â„“}2$and $2r_1=r_2$. Substituting that into the equation, then:
$\frac{R_2}{R_1}=\frac{\left(2{\mathrm{â„“}}_1\right)r_1^2}{{\mathrm{â„“}}_1{\left(2r_1\right)}^2}$
$=\frac{2}{4}$
$=0.5$

The ratio of $R_2/R_1$of the resistances of the two wires is $\frac{R_2}{R_1}=0.5$.