Chapter 12.6, Problem 13PSC

a.

To determine

To find: The volume of the largest sphere that can be inscribed in the cube of volume $\text{1000}{\text{m}}^3.$

Answer to Problem 13PSC

The volume of the largest sphere that can be inscribed in the cube is $524{\text{m}}^3.$

Explanation of Solution

Given Information:

Volume of a cube $\left(V\right)=1000{\text{m}}^3$

Formula used:

Volume of a cube $\left(V\right)=a^3$

  $"a"$ is the edge of a cube

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

  $"r"$ is the radius of a sphere

Calculation:

Let $a$ be the edge of a cube

We know that, Volume of a cube $\left(V\right)=a^3$

As $V=1000{\text{m}}^3$

  $\begin{array}{l}\therefore 1000{\text{m}}^3=a^3\\ \therefore a=10\text{m}\end{array}$

Now, the largest sphere which fits into a cube will be a sphere of whose diameter is equal to the edge of cube.

Let $d$ be the diameter of the sphere

  $\begin{array}{l}\therefore d=10\text{m}\\ \therefore r=\frac{d}{2}=5\text{m}\end{array}$

Now we know that, Volume of a sphere $\left(V\text{'}\right)=\frac{4}{3}\pi r^3$

  $\begin{array}{l}\therefore V\text{'}=\frac{4}{3}\times \frac{22}{7}\times 5^3\\ \therefore V\text{'}=523.80{\text{m}}^3\\ \therefore V\text{'}\approx 524{\text{m}}^3\end{array}$

Hence, the volume of the largest sphere that can be inscribed in the cube is $524{\text{m}}^3.$

b.

To determine

To find: The volume of the smallest sphere that can be circumscribed about the cube.

Answer to Problem 13PSC

The volume of the smallest sphere is $2722{\text{m}}^3.$

Explanation of Solution

Given Information:

Volume of a cube $\left(V\right)=1000{\text{m}}^3$

Formula used:

Volume of a cube $\left(V\right)=a^3$

  $"a"$ is the edge of a cube

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

  $"r"$ is the radius of a sphere

Diagonal of a square $=a_1\sqrt{2}$

  $"a_1"$ is the side of a square

In a right triangle, ${\text{Hypotenuse}}^2={\text{Base}}^2+{\text{Altitude}}^2$

Calculation:

Let $a$ be the edge of a cube

We know that, Volume of a cube $\left(V\right)=a^3$

As $V=1000{\text{m}}^3$

  $\begin{array}{l}\therefore 1000{\text{m}}^3=a^3\\ \therefore a=10\text{m}\end{array}$

So, each side of every square base of a cube will be 10 m.

Therefore, diagonal of each surface of a cube will be $\left(d\right)=10\sqrt{2}\text{m}$

Now, the diagonal of a cube $\left(D\right)$ , diagonal of a square base and edge of a cube forms a right triangle.

Here, $D$ is the hypotenuse, $d$ is the base and $a$ is the altitude.

Using Pythagoras theorem, we have

  ${\text{Hypotenuse}}^2={\text{Base}}^2+{\text{Altitude}}^2$

  $\begin{array}{l}\therefore D^2=d^2+a^2\\ \therefore D^2={\left(10\sqrt{2}\right)}^2+{10}^2\\ \therefore D^2=200+100\\ \therefore D=10\sqrt{3}\text{m}\end{array}$

Now, $D$ becomes diameter of a sphere which circumscribes the cube.

  $\therefore$ Radius of the sphere $r=\frac{D}{2}=5\sqrt{3}\text{m}$

Now we know that, Volume of a sphere $\left(V\text{'}\right)=\frac{4}{3}\pi r^3$

  $\begin{array}{l}\therefore V\text{'}=\frac{4}{3}\times \frac{22}{7}\times {\left(5\sqrt{3}\right)}^3\\ \therefore V\text{'}=\frac{88}{21}\times 125\times 3\sqrt{3}{\text{m}}^3\\ \therefore V\text{'}=2721.79{\text{m}}^3\\ \therefore V\text{'}\approx 2722{\text{m}}^3\end{array}$

Hence, the volume of the smallest sphere that can circumscribe the cube is $2722{\text{m}}^3.$