Chapter 12.8, Problem 4CYP

To determine

The volume of softball is to be estimated.

Answer to Problem 4CYP

The volume of softball is $28.12i{n}^3$ .

Explanation of Solution

Given:

  $\begin{array}{l}1in:\text{}1.3in\\ V_b=12.8i{n}^3\end{array}$

Concept Used:

The volume of sphere is given by the following expression:

  $V=\frac{4}{3}\pi r^3$

Where,

Radius of sphere: $r$

Calculation:

Given that:

Softballs and baseballs are similar in shape that means the ratio value of its dimension is same:

The shape of baseballs and softballs is sphere. So, radius ratio of both balls is given as:

  $\begin{array}{l}1in:\text{}1.3in\\ \frac{r_b}{r_s}=\frac{1}{1.3}\end{array}$

Take the ratio of volume of baseballs and softballs to find the volume of softball:

  $\begin{array}{l}\frac{V_b}{V_s}=\frac{\frac{4}{3}\pi r_b{}^3}{\frac{4}{3}\pi r_s{}^3}\\ \frac{V_b}{V_s}=\frac{r_b{}^3}{r_s{}^3}\\ \frac{V_b}{V_s}={\left(\frac{r_b}{r_s}\right)}^3\\ \frac{V_b}{V_s}={\left(\frac{1}{1.3}\right)}^3\\ \frac{V_b}{V_s}=0.4551\\ V_s=\frac{V_b}{0.4551}\\ V_s=\frac{12.8}{0.4551}\\ V_s=28.12i{n}^3\end{array}$

Conclusion:

Hence, the volume of softball is calculated as $V_s=28.12i{n}^3$ .