 # Air enclosed in a sphere has density 1.4 kg/m^3. What will the density be if the radius of the sphere is halved, compressing the air within? The main purpose of this question is to find the density of the air enclosed in the sphere if the radius of the sphere is halved.

A sphere is a  $3-$dimensional body with a circular shape. It is divided into three $x-$axis, the $y-$axis, and the $z-$axis. This is the primary distinction between a sphere and a circle. A sphere, unlike other $3-$dimensional shapes, has no vertices or edges. All the points present on the sphere’s surface are equally spaced from the center. More generally, any point on the sphere’s surface is equidistant from its center.

The radius of the sphere is regarded as the length of a line segment from the sphere’s center to a point on the surface of the sphere. Also, the diameter of the sphere is defined as the length of a line segment from one point to another and which passes through its center. Moreover, a sphere’s circumference can be measured using the length of the largest possible circle drawn around a sphere usually known as a great circle. Being a $3-$dimensional shape, a sphere possesses a space usually known as volume which is measured in cubic units. Similarly, the surface of a sphere also requires an area to be occupied, which is known as its surface area and is expressed in square units.

Let $\rho$ be the density of air enclosed in the sphere, $V_1=\dfrac{4}{3}\pi r^3$ and $m_1$, be the volume and mass of the sphere respectively, then:

$\rho=\dfrac{m_1}{V_1}$

Let $V$ be the volume of the sphere when the radius is halved, then:

$V=\dfrac{4}{3}\pi \left(\dfrac{r}{2}\right)^3$

$V=\dfrac{4}{3}\cdot \dfrac{1}{8}\pi r^3$

$V=\dfrac{1}{8}\cdot \dfrac{4}{3}\pi r^3$

Or   $V=\dfrac{1}{8}V_1$

Let $\rho_1$ be the new density when the radius is halved, then:

$\rho_1=\dfrac{m_1}{V}$

$\rho_1=\dfrac{m_1}{\dfrac{1}{8}V_1}$

$\rho_1=8\dfrac{m_1}{V_1}$

$\rho_1=8\rho$

Since $\rho=1.4\,kg/m^3$

$\rho=8( 1.4\,kg/m^3)=11.2\,kg/m^3$

## Example 1

Find the volume of the sphere having the diameter of $6\,cm$.

### Solution

Let $V$ be the volume of the sphere, then:

$V=\dfrac{4}{3}\pi r^3$

Since Diameter $(d)=2r$

Therefore, $r=\dfrac{d}{2}$

$r=\dfrac{6}{2}=3\,cm$

$V=\dfrac{4}{3}\pi (3\,cm)^3$

$V=\dfrac{4}{3}\cdot 27\pi$

$V=36\pi cm^3$

Or use $\pi=\dfrac{22}{7}$ to get:

$V=36\left(\dfrac{22}{7}\right)\,cm^3$

$V=113\,cm^3$

## Example 2

The volume of a sphere is $200\,cm^3$, find its radius in centimeters.

### Solution

Since $V=\dfrac{4}{3}\pi r^3$

Given that $V=200\,cm^3$, therefore:

$200\,cm^3=\dfrac{4}{3}\pi r^3$

Use $\pi=\dfrac{22}{7}$:

$\dfrac{200\cdot 3}{4}\cdot \dfrac{7}{22}\,cm^3=r^3$

$r^3=\dfrac{600}{4}\cdot \dfrac{7}{22}\,cm^3$

$r^3=47.73\,cm^3$

$r=3.63\,cm$

Hence, the radius of the sphere having the volume $200\,cm^3$ is $3.63\,cm$.