Chapter 11, Problem 1TP

To determine

The density when the radius of an under-inflated volleyball increased by 10%

Answer to Problem 1TP

The correct answer is option e.

Explanation of Solution

Given info:

The radius of volleyball increases by 10%.

Formula used:

Density is mass per unit volume of the substance and is numerically given as

  $\rho =\frac{m}{V}$

Calculation:

When radius of volleyball is increased by ten percent it means its volume increases.

The volume of sphere is given by

  $V=\frac{4}{3}\pi r^3$

Initially volleyball's density can be given as

  ${\rho }_i=\frac{m}{V_i}$

Neglected mass of air inside the ball

Substitute $\frac{4}{3}\pi r_i^{3}for{V}_i,$

  $\begin{array}{l}{\rho }_i=\frac{1}{\frac{4}{3}\pi r_i^3}\\ =\frac{\begin{array}{l}\\ 0.24\end{array}}{r_i^3}\end{array}$

Now when radius of ball increased by 10%, then new radius of ball is given as

  $\begin{array}{l}{r}_n=r_i+\frac{1}{10}{r}_i\\ =\frac{11}{10}{r}_i\end{array}$

Therefore, new volume of ball is

  $V_n=\frac{4}{3}\pi r_n^3$

Substitute $\frac{11}{10}{r}_{i}for{r}_n,$

  $V_n=\frac{4}{3}\pi {\left(\frac{11}{10}{r}_i\right)}^3$

Final volleyball's density can be given as

  ${\rho }_n=\frac{m}{V_n}$

Neglected mass of air inside the ball

Substitute $\frac{4}{3}\pi {\left(\frac{11}{10}{r}_i\right)}^3\text{}\text{}for{V}_n,$

  $\begin{array}{l}{\rho }_n=\frac{1}{\frac{4}{3}\pi {\left(\frac{11}{10}{r}_i\right)}^3}\\ =\frac{0.18}{r_i^3}\end{array}$

Thus, change in density is,

  $Changeindensity=\left(\frac{0.18}{r_i^3}-\frac{0.24}{r_i^3}\right)$

= $=-\frac{0.06}{r_i^3}$

Negative sign shows that density if volleyball's decreases.

In percentage this decrease can be given as

  $\begin{array}{l}\text{Percentage decrease in density}\text{=}\left(\frac{\left(\frac{0.06}{r_i^3}\right)}{\left(\frac{0.24}{r_j^3}\right)}\times 100\right)\\ =25\%\end{array}$

The density will decrease by approximately 25%.

Hence, the correct answer is option e and the options a, d, c, and d are incorrect.

Conclusion:

The density will decrease by approximately 25%.