Chapter 62, Problem 1A

In order to make a conical duct a circular sheet of metal is to be cut in the shape of a sector of a circle. It the radius of the sector is 32.5 in. and the arc length of the sector is 264 in., what will be the volume of the cone? Round the answer to 1 decimal place.

To determine

The volume of the cone.

Answer to Problem 1A

Weight of the pipe, $V=11325.07i{n}^3$

Explanation of Solution

Given information:

  $\begin{array}{l}Radius\text{ of the sector }\left(r\right)=32.5in\\ Arc\text{ length of the sector }\left(l\right)=26.4in\end{array}$

Calculations:

Volume of the sector is defined as, $V=A\times l\mathrm{............}\left(1\right)$

  $\begin{array}{l}\begin{array}{l}\begin{array}{l}\text{Where , }\\ \text{A = Area of the sector}\end{array}\hfill \\ \text{l = length of the sector}\hfill \\ \text{And length of the sector is defined as –}\hfill \end{array}\\ l=r\left(\theta \times \frac{\pi }{{180}^o}\right)\\ Where,\\ \theta \to \text{central angle of the sector}\\ \text{Substituting values,}\\ 26.4=32.5\left(\theta \times \frac{\pi }{{180}^o}\right)\\ \theta =\frac{26.5}{32.5}\\ {46.54}^o\end{array}$

Hence, the area of the sector: -

  $\begin{array}{l}A=\pi r^2\times \left(\theta \times \frac{\theta }{{360}^o}\right)\\ A=\frac{\pi {\left(32.5\right)}^2\left(46.54\right)}{{360}^o}\\ A=428.98i{n}^2\end{array}$

  $\begin{array}{l}FromEq.\left(1\right),\\ V=A\times l=\left(428.98i{n}^2\right)\times \left(26.4in\right)\\ V=11325.07i{n}^3\end{array}$

Conclusion:

The volume of cone is $V=11325.07i{n}^3.$