Chapter 61, Problem 12A

To determine

To compare:

The weight of the larger and smaller piece of a right circular cone.

Answer to Problem 12A

  ${\text{V}}_{\text{2}}\text{=}{\text{4V}}_{\text{1}}$

Explanation of Solution

Given:

Height $\text{h = 6}\text{in}\text{.}$

Base diameter of smaller piece ${\text{D}}_{\text{1}}\text{ = 2}\text{in}\text{.}$

Base diameter of larger piece ${\text{D}}_2\text{ = 4}\text{in}\text{.}$

Concept used:

The volume of a right circular cone is

  $\text{V =}\frac{\text{1}}{\text{3}}{\text{A}}_{\text{B}}\text{h}$

Where, h = height, and A_B = Base area.

Calculation:

The base area is $A_b=\pi r^2$

Now, the base area of smaller cone is

  $\begin{array}{l}{A}_{b1}=\frac{22}{7}\times {\left(\frac{D_1}{2}\right)}^2\left[\therefore r=\frac{D}{2}\right]\\ A_{b1}=\frac{22}{7}\times {\left(\frac{2}{2}\right)}^2\\ A_{b1}=\frac{22}{7}in{.}^2\end{array}$

Now, the base area of larger cone is

  $\begin{array}{l}{A}_{b2}=\frac{22}{7}\times {\left(\frac{D_2}{2}\right)}^2\left[\therefore r=\frac{D}{2}\right]\\ A_{b2}=\frac{22}{7}\times {\left(\frac{4}{2}\right)}^2\\ A_{b2}=\frac{22}{7}\times {\left(2\right)}^2\\ A_{b2}=\frac{88}{7}in{.}^2\end{array}$

Thus, volume of smaller cone is

  $\begin{array}{l}{\text{V}}_1\text{ =}\frac{\text{1}}{\text{3}}\text{×}\frac{22}{7}\text{×6}\text{or,}\\ {\text{V}}_1\text{ =}\frac{22}{7}\text{×2}\\ {\text{V}}_1\text{ = }\frac{44}{7}\text{in}{\text{.}}^{\text{3}}\\ \end{array}$

And, the volume of larger cone is

  $\begin{array}{l}{\text{V}}_2\text{ =}\frac{\text{1}}{\text{3}}\text{×}\frac{88}{7}\text{×6}\text{or,}\\ {\text{V}}_2\text{ =}\frac{88}{7}\text{×2}\\ {\text{V}}_2\text{ = }\frac{44}{7}\text{×4}\text{in}{\text{.}}^{\text{3}}\\ {\text{V}}_2\text{ =}{\text{4V}}_1\\ \end{array}$

Conclusion:

Hence, volume of cone of larger diameter is 4 times heavier than the cone of smaller diameter.