Chapter 1, Problem 20MCQ

(a)

To determine

To find: the volume and surface area of a sphere with radius 2 inches.

Answer to Problem 20MCQ

The volume is $V=\frac{32}{3}\pi$ and surface of the area is $SA=16\pi$

Explanation of Solution

Formula Used:

The formula for the surface area of a sphere is $SA=4\pi r^2$ and the formula for the volume of a sphere is $V=\frac{4}{3}\pi r^3$ .

Calculation:

The formula for the volume of a sphere is

  $V=\frac{4}{3}\pi r^3$

Now, substitute the value of $r=2$ in the formula,

  $\begin{array}{c}V=\frac{4}{3}\pi r^3\\ 3V=4\pi \cdot 8\\ 3V=32\pi \\ V=\frac{32\pi }{3}\end{array}$

Now, Substitute the value $r=2$ in the surface area formula,

  $\begin{array}{l}SA=4\pi r^2\\ SA=4\pi \cdot 2^2\\ SA=16\pi \end{array}$

So, the volume is $V=\frac{32}{3}\pi$ and surface of the area is $SA=16\pi$

(b)

To determine

Whether it is possible for a sphere to have the same numerical value for the surface area and volume, and to find the radius of such a sphere.

Answer to Problem 20MCQ

The radius of the sphere is 3.

Explanation of Solution

Formula Used:

The formula for the surface area of a sphere is $SA=4\pi r^2$ and the formula for the volume of a sphere is $V=\frac{4}{3}\pi r^3$ .

Calculation:

The surface area is equal to the volume, $V=SA$

So,

  $\begin{array}{c}\frac{4}{3}\pi \times r^3=4\pi \times r^2\\ \frac{4}{3}\times r^3=4\times r^2\\ r=\frac{12}{4}\\ r=3\end{array}$

So, the radius of the sphere is 3.