Chapter 63, Problem 13A

A vessel is in the shape of a right circular cone. This vessel contains liquid to a depth of 12.8 centimeters, as shown. How many liters of liquid must be added in order to fill the vessel? One liter contains 1000 cubic centimeters.

To determine

The amount of liquid needs to be added to fill the vessel completely.

Answer to Problem 13A

  $\text{0}\text{.80}\text{L}$

Explanation of Solution

Given:

  

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 the cone is

  $\begin{array}{l}{A}_b=3.14\times {\left(\frac{D}{2}\right)}^2\left[\therefore r=\frac{D}{2}\right]\\ A_b=3.14\times {\left(\frac{15}{2}\right)}^2\\ A_b=3.14\times {\left(7.5\right)}^2\\ A_b=176.625c{m}^2\end{array}$

The total volume of the vessel is

  $\begin{array}{l}\text{V =}\frac{\text{1}}{\text{3}}\times 176.625\times 19.2\\ \text{V =}\frac{3391.2}{\text{3}}\\ \text{V=}\text{1130}\text{.4}{\text{cm}}^3\end{array}$

Now, the base area of cone up to it contains liquid is

  $\begin{array}{l}{A}_{b1}=3.14\times {\left(\frac{D_1}{2}\right)}^2\left[\therefore r=\frac{D}{2}\right]\\ A_{b1}=3.14\times {\left(\frac{10}{2}\right)}^2\\ A_{b1}=3.14\times {\left(5\right)}^2\\ A_{b1}=78.5c{m}^2\end{array}$

The volume of the vessel is

  $\begin{array}{l}{\text{V}}_1\text{ =}\frac{\text{1}}{\text{3}}\times 78.5\times 12.8\\ {\text{V}}_1\text{ =}\frac{1004.8}{\text{3}}\\ {\text{V}}_1\text{=}335{\text{cm}}^3\end{array}$

Thus, the remaining amount of liquid that is needed to fill the vessel is

  $V_r$ = Total volume − volume of vessel up to it contains liquid

  $\begin{array}{l}{\text{V}}_{\text{r}}\text{ =}{\text{V-V}}_{\text{1}}\text{or,}\\ {\text{V}}_{\text{r}}\text{ =}\text{1130}\text{.4}\text{-}\text{335}\\ {\text{V}}_{\text{r}}\text{ = 795}\text{.4}{\text{cm}}^{\text{3}}\\ \end{array}$

Thus, the amount of liquid is to be added in liters is

  $\begin{array}{l}{\text{V}}_r\text{ =}\frac{\text{795}\text{.4}}{1000}\text{or,}\\ {\text{V}}_r\text{ =}\text{0}\text{.79}\approx \text{0}\text{.80}\\ {\text{V}}_{\text{r}}\text{ =}\text{0}\text{.80}\text{L}\end{array}$

Conclusion:

Hence, the amount of liquid is 0.80 L.