Question

A jeweler has three small solid spheres made of gold, of radius 2 mm, 3 mm, and 4 mm. He decides to melt these down and make just one sphere out of them. What will the radius of this larger sphere be?

Short answer

The radius of the larger sphere will be 4.63mm approximately.

Step-by-step solution

Step-1. Volume of sphere.

The volume of the sphere is$V=\frac{4}{3}\pi r^3$ where, r is the radius.

Step-2. Volume of the larger sphere.

The radius of the small solid spheres are 2 mm, 3 mm and 4 mm.

First find the volume of the lager sphere.

$\begin{array}{l}V=\frac{4}{3}\pi {r_1}^3+\frac{4}{3}\pi {r_2}^3+\frac{4}{3}\pi {r_3}^3\\ =\frac{4}{3}\pi (2{)}^3+\frac{4}{3}\pi (3{)}^3+\frac{4}{3}\pi (4{)}^3\\ =\frac{4}{3}\pi (8+27+64)\\ =132\pi \end{array}$

Step-3. Determine the radius of larger sphere.

The volume of larger sphere is $132\pi$.

Put$132\pi$ for V into the volume of sphere.

$\begin{array}{l}132\pi =\frac{4}{3}\pi r^3\\ 132\pi \times \frac{3}{4\pi }=r^3\\ 99=r^3\\ r\approx 4.63\end{array}$

The radius of the larger sphere will be 4.63 mm approximately.