Question

Metal spheres, each of radius $2\mathrm{cm}$, are packed into a rectangular box of internal dimensions $16\mathrm{cm}\times 8\mathrm{cm}\times 8\mathrm{cm}$. When $16$ spheres are packed the box is filled with preservative liquid. Find the volume of this liquid. Give your answer to the nearest integer. [Use $\mathrm{\pi }=3.14$]

Answer 1

Step 1. Determine the total volume of the given spheres.

Let the radius of the metal spheres be $r$ .

Given that $r=2\mathrm{cm}$.

Since $16$ such spheres are packed into the rectangular box,

As we know that volume of sphere $=\frac{4}{3}{\mathrm{\pi r}}^3$, $r$ is the radius of sphere.

Therefore,

$\begin{array}{rcl}\mathrm{Total}\mathrm{volume}\mathrm{of}16\mathrm{spheres}& =& 16\times \frac{4}{3}\mathrm{\pi }{r}^3\\ & =& 16\times \frac{4}{3}\times 3.14\times {\left(2\right)}^3\\ & =& 16\times \frac{4}{3}\times 3.14\times 8\\ & =& 535.89{\mathrm{cm}}^3\end{array}$

Step 2. Find the volume of the filled liquid.

Given the internal dimensions of the rectangular box are $16\mathrm{cm}\times 8\mathrm{cm}\times 8\mathrm{cm}$.

Therefore,

$\begin{array}{rcl}\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{rectangular}\mathrm{box}& =& \mathrm{L}\times \mathrm{B}\times \mathrm{H}\\ & =& 16\times 8\times 8\\ & =& 1024{\mathrm{cm}}^3\end{array}$

To find the volume of the liquid that is filled in the rectangular box, determine the space left in the rectangular box after the space occupied by the spheres:

Therefore,

$\begin{array}{rcl}\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{liquid}& =& \left(\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{rectangular}\mathrm{box}\right)-\left(\mathrm{Total}\mathrm{volume}\mathrm{of}\mathrm{the}16\mathrm{spheres}\right)\\ & =& \left(1024{\mathrm{cm}}^3\right)-\left(535.89{\mathrm{cm}}^3\right)\\ & =& 488.11{\mathrm{cm}}^3\end{array}$

Hence, the volume of the liquid rounded off to the nearest integer is $488{\mathrm{cm}}^3$.