Question

26. Wires $1$ and $2$ are made of the same metal. Wire$2$ has twice the length and twice the diameter of wire $1$. What are the ratios (a) ${\rho }_2/{\rho }_1$ of the resistivities and (b) $R_2/R_1$ of the resistances of the two wires?

Short answer

(a) The ratio of ${\rho }_2/{\rho }_1$of the resistivity is $\frac{{\rho }_2}{{\rho }_1}=1$.

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

Step-by-step solution

Given explanation Part (a)

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

Explanation Part (a)

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

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

Final answer Part (a)

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

Given information Part (b)

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.

Explanation Part (b)

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

Since, $A=\pi r^2$, plugging-that into the equation,
$\frac{R_2}{R_1}=\frac{\frac{{\ell }_2}{\left(\pi r_2^2\right)}}{\frac{{\ell }_1}{\left(\pi r_1^2\right)}}$

$=\frac{\frac{{\ell }_2}{r_2^2}}{\frac{{\ell }_1}{r_1^2}}$

$=\frac{{\ell }_2{r}_1^2}{{\ell }_1{r}_2^2}$

Explanation Part (b)

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

Final answer Part (b)

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