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Question

A ball is placed inside a box into which it will fit tightly. If the radius of the ball is 10 cm, calculate: a) the volume of the ball, b) the volume of the box, c) the percentage volume of the box not occupied by the ball.

Solution

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$\textbf{a)}$

The ball is spherical in shape. The volume of a sphere with radius $r$ is given by:

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

Substitute $r=10$ cm:

$V=\dfrac{4}{3}\pi (10)^3$

$V=\dfrac{4000}{3}\pi$

$V\approx \color{#c34632}4188.8\text{ cm}^3$

$\textbf{b)}$

If the ball fits tightly inside the box, then the box is a cube with side length equal to the diameter of the ball which is $2(10)=20$ cm. So, its volume is:

$V=s^3=20^3$

$V=\color{#c34632}8000\text{ cm}^3$

$\textbf{c)}$

The percentage volume of the box not occupied by the ball is the ratio of the difference of the volumes of the box and the ball to the volume of the box:

$\dfrac{V_{box}- V_{ball}}{ V_{box}}=\dfrac{8000-\dfrac{4000}{3}\pi }{8000}\approx 0.476\to \color{#c34632}47.6\%$

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