Question

Metallic spheres of radii $6\mathrm{cm}$, $8\mathrm{cm}$ and $10\mathrm{cm}$ respectively are melted to form a single solid sphere. Find the radius of the resulting sphere.

Answer 1

Step 1. Determine the volumes of the given three spheres.

Let the radii of the given spheres be $r_1$, $r_2$ and $r_{\mathit{3}}$ respectively.

Given: $r_1=6\mathrm{cm}$, $r_2=8\mathrm{cm}$ and $r_{\mathit{3}}=10\mathrm{cm}$.

Now, Volume of a sphere $=\frac{4}{3}\pi r^3$

The volume of the first sphere:

$\begin{array}{rcl}{V}_{\mathit{1}}& \mathit{=}& \frac{\mathit{4}}{\mathit{3}}\pi {\left(r_1\right)}^3\\ & =& \frac{\mathit{4}}{\mathit{3}}\pi {\left(6\right)}^3\\ & =& \frac{\mathit{4}}{\mathit{3}}\pi \times 216\end{array}$

The volume of the second sphere:

$\begin{array}{rcl}{V}_2& \mathit{=}& \frac{\mathit{4}}{\mathit{3}}\pi {\left(r_2\right)}^3\\ & =& \frac{\mathit{4}}{\mathit{3}}\pi {\left(8\right)}^3\\ & =& \frac{\mathit{4}}{\mathit{3}}\pi \times 512\end{array}$

The volume of the third sphere:

$\begin{array}{rcl}{V}_3& \mathit{=}& \frac{\mathit{4}}{\mathit{3}}\pi {\left(r_3\right)}^3\\ & =& \frac{\mathit{4}}{\mathit{3}}\pi {\left(10\right)}^3\\ & =& \frac{\mathit{4}}{\mathit{3}}\pi \times 1000\end{array}$

Step 2: Find the radius of the resulting sphere.

Let $r$ be the radius of the resulting sphere.

As the three spheres are melted to form a single new sphere.

Therefore,

$\begin{array}{rcl}\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{new}\mathrm{sphere}& =& \mathrm{Total}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{three}\mathrm{spheres}\\ & =& V_1+V_2+V_3\end{array}$

Hence,

$\begin{array}{rcl}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\frac{\mathit{4}}{\mathit{3}}\pi r^{\mathit{3}}& \mathit{=}& \left(\frac{\mathit{4}}{\mathit{3}}\pi \times 216\right)+\left(\frac{\mathit{4}}{\mathit{3}}\pi \times 512\right)+\left(\frac{\mathit{4}}{\mathit{3}}\pi \times 1000\right)\\ & \Rightarrow & \frac{\mathit{4}}{\mathit{3}}\pi r^{\mathit{3}}\mathit{=}\frac{\mathit{4}}{\mathit{3}}\pi \times \left(216+512+1000\right)\\ & \Rightarrow & \frac{\mathit{4}}{\mathit{3}}\pi r^3\mathit{=}\frac{\mathit{4}}{\mathit{3}}\pi \times \left(1728\right)\\ & \Rightarrow & r^{\mathit{3}}\mathit{=}\mathit{1728}\\ & \Rightarrow & r\mathit{=}\sqrt[3]{1728}\\ & \Rightarrow & \mathit{}r\mathit{=}\sqrt[\mathit{3}]{\mathit{12}\mathit{\times }\mathit{12}\mathit{\times }\mathit{12}}\\ & \Rightarrow & r\mathit{=}12\mathrm{cm}\end{array}$

Hence, the radius of the resulting sphere is $12\mathrm{cm}$.