Chapter 12.6, Problem 6PSB

To determine

To calculate: The volume of rubber used to make the ball.

Answer to Problem 6PSB

The volume of rubber used to make the ball is 19.113 cm³.

Explanation of Solution

Given information: The rubber shell’s outer diameter $\left(d_1\right)=48\text{ mm}$ and inner diameter $\left(d_2\right)=42\text{ mm}\text{.}$

Formula used:

Volume of spherical shell $\left(V\right)=\left(\frac{4}{3}\right)\pi \left(r_1{}^3-r_2{}^3\right)$ …………………. (i)

where $r_1$ = outer radius of the spherical shell

  $r_2$ = inner radius of the spherical shell

Calculation: Here required volume of rubber is the volume of spherical shell with given outer and inner diameter.

Now rewriting equation (i) in terms of outer diameter $\left(d_1\right)$ and inner diameter $\left(d_2\right)$ ,

  $\begin{array}{l}V=\left(\frac{4}{3}\right)\times \pi \times \left\{{\left(\frac{d_1}{2}\right)}^3-{\left(\frac{d_2}{2}\right)}^3\right\}\\ V=\left(\frac{4}{3}\right)\times \pi \times \left\{\left(\frac{d_1{}^3}{8}\right)-\left(\frac{d_2{}^3}{8}\right)\right\}\\ V=\left(\frac{4}{3}\right)\times \pi \times \left(\frac{1}{8}\right)\left(d_1{}^3-d_2{}^3\right)\\ V=\left(\frac{1}{6}\right)\times \pi \times \left(d_1{}^3-d_2{}^3\right)\mathrm{......................................}\text{(ii)}\end{array}$

Using $d_1=48\text{ mm}=4.8\text{ cm}$ and $d_2=42\text{ mm}=4.2\text{ cm}$ in equation (ii)

  $\begin{array}{l}V=\left(\frac{1}{6}\right)\times \pi \times \left({4.8}^3-{4.2}^3\right){\text{ cm}}^{\text{3}}\\ V=\left(\frac{1}{6}\right)\times \pi \times 36.504{\text{ cm}}^{\text{3}}\\ V=19.113{\text{ cm}}^{\text{3}}\end{array}$

Hence, the required volume of rubber used to make the ball is 19.113 cm³.