Chapter 12, Problem 34SGR

To determine

To calculate:To find how much rubber sheathing is present in $100$ feet of cable.

Answer to Problem 34SGR

Rubber sheathing present in cable is $943i{n}^2$

Explanation of Solution

Given information:Coaxial cable is covered by runner sheathing and cable has a diameter of $3$ inches

Formula Used:

Lateral surface area of cylinder is given by the below formula:

  $Lateralsurfaceare{a}_{Cylinder}=2\pi rh$

Where,

  $r=$ Radius of base of cylinder

  $h=$ Height of cylinder

Calculation:

Given that coaxial cable is covered by rubber sheathing and cable has a diameter of $3$ inches. Also, height of cable is $100$ feet

Here, dimensions of cylinderas follows:

  $\begin{array}{l}r=\frac{3}{2}=1.5in\\ h=100in\end{array}$

Amount of rubber sheathing is equal to lateral surface area of cable, cable is in form of cylinder

Lateral surface of cylinder is calculated as follows:

  $Lateralsurfaceare{a}_{Cylinder}=2\pi rh$

Substituting the values, volume is

  $\begin{array}{l}Lateralsurfaceare{a}_{Cylinder}=2\times \frac{22}{7}\times 1.5\times 100\\ Lateralsurfaceare{a}_{Cylinder}=942.85\\ Lateralsurfaceare{a}_{Cylinder}=943i{n}^2\end{array}$

Conclusion:

Hence, rubber sheathing present in cable is $943i{n}^2$