Chapter 12.6, Problem 14PSC

To determine

To find: The ratio of the volume of a sphere to the volume of the smallest right cylinder that can contain it.

Answer to Problem 14PSC

The ratio of the volume of a sphere to the volume of the cylinder is 2:3.

Explanation of Solution

Given Information:

Right circular cylinder contains the sphere

Formula used:

Volume of a sphere $\left(V\right)=\frac{4}{3}\pi r^3$

  $"r"$ is the radius of a sphere

Volume of a cylinder $\left(V\text{'}\right)=\pi {\left(r_1\right)}^{2}h$

  $r_1\text{and}h$ are the radius and height of the cylinder

Calculation:

Let $r$ be the radius of the sphere

  $\therefore$ Volume of a sphere $\left(V\right)=\frac{4}{3}\pi r^3$

Now, the smallest right cylinder which contain the sphere will have radius same as that the sphere.

Also, the cylinder will have height same as the diameter of the sphere.

We know that, Volume of a cylinder $\left(V\text{'}\right)=\pi {\left(r_1\right)}^{2}h$

  $r_1\text{and}h$ are the radius and height of the cylinder

  $\therefore r_1=r\text{and}h=2r$

  $\begin{array}{l}\therefore V\text{'}=\pi r^2\left(2r\right)\\ \therefore V\text{'}=2\pi r^3\end{array}$

  $\begin{array}{l}\therefore \frac{V}{V\text{'}}=\frac{\frac{4}{3}\pi r^3}{2\pi r^3}\\ \therefore \frac{V}{V\text{'}}=\frac{4}{3}\times \frac{1}{2}\\ \therefore \frac{V}{V\text{'}}=\frac{2}{3}\end{array}$

Hence, the ratio of the volume of a sphere to the volume of the cylinder is 2:3.