Chapter 8.2, Problem 4P

To determine

To calculate: Conductance of a wire with length $\text{75 m}$ and diameter $2.5\text{ mm}$ .

Answer to Problem 4P

Conductance of a wire with length $\text{75 m}$ and diameter $2.5\text{ mm}$ is $0.125\text{ mho}$ .

Explanation of Solution

Given information:

The following variation:

The conductance of a wire varies directly as square of the wire’s diameter and inversely as its length.

When $\text{length}=50\text{ m}$ and $\text{diameter = 2 mm}$ , $\text{conductance = 0}\text{.12 mho}$ .

Calculation:

Let C be the conductance of the wire, l be the length of the wire and d be diameter of the wire. Then, C varies directly as $d^2$ and inversely as l ,

  $C=k\frac{d^2}{l}$

where k is a constant.

Substituting $C=0.12\text{ mho}$ , $l=50\text{ m}=50000\text{ mm}$ and $d=2\text{ mm}$ ,

  $\begin{array}{l}C=k\frac{d^2}{l}[\text{Equation]}\\ 0.12=\frac{k{\left(2\right)}^2}{50000}[\text{Substitute]}\\ 0.12=\frac{4k}{50000}[\text{Find square]}\\ 4k=6000[\text{Multiply both sides by 50000]}\\ k=1500[\text{Divide both sides by 4]}\end{array}$

Therefore, an equation of the given variation is $C=1500\frac{d^2}{l}$

To find C when $l=75\text{ m}=75000\text{ mm}$ and $d=2.5\text{ mm}$ , substitute in $C=1500\frac{d^2}{l}$ .

  $\begin{array}{l}C=1500\frac{d^2}{l}[\text{Equation]}\\ C=\frac{1500{\left(2.5\right)}^2}{75000}[\text{Substitute]}\\ C=\frac{1500\left(6.25\right)}{75000}[\text{Find square]}\\ C=\frac{9375}{75000}[\text{Multilply]}\\ C=0.125[\text{Divide]}\end{array}$