Question

Three solid spheres of iron whose diameters are $\mathbf{2}\mathbf{}\mathbf{cm}\mathbf{,}\mathbf{}\mathbf{12}\mathbf{}\mathbf{cm}$ and $\mathbf{16}\mathbf{}\mathbf{cm}$ respectively, are melted to form a single solid sphere. Find the radius of the solid sphere.

Answer 1

Step 1: Given information.

Diameter of first sphere$=2\mathrm{cm}$

Diameter of second sphere$=12\mathrm{cm}$

Diameter of third sphere$=16\mathrm{cm}$

Step 2: Calculate the radius of single solid sphere.

Let the radius of first, second and third spheres are $r_1,r_2$ and $r_3$ respectively and radius of single solid sphere is $R$.

So that,

$$\begin{gathered} \mathrm{Radius}=\frac{\mathrm{Diameter}}{2} \\ \Rightarrow r_1=\frac{2}{2}=1\mathrm{cm} \\ \Rightarrow r_2=\frac{12}{2}=6\mathrm{cm} \\ \Rightarrow r_3=\frac{16}{2}=8\mathrm{cm} \end{gathered}$$

Since, all three sphere are melted and converted into single solid sphere.

Therefore, the volume of single solid sphere will be equal to the sum of volumes of all three spheres.

$$\begin{gathered} \therefore \frac{4}{3}\pi R^3=\frac{4}{3}\pi {r_1}^3+\frac{4}{3}\pi {r_2}^3+\frac{4}{3}\pi {r_3}^3 \\ \Rightarrow \frac{4}{3}\pi R^3=\frac{4}{3}\pi \times \left({r_1}^3+{r_2}^3+{r_3}^3\right) \\ \Rightarrow R^3={r_1}^3+{r_2}^3+{r_3}^3 \\ \Rightarrow R^3=1^3+6^3+8^3 \\ \Rightarrow R^3=1+216+512 \\ \Rightarrow R^3=729 \\ \Rightarrow R=\sqrt[3]{729} \\ \Rightarrow R=9\mathrm{cm} \end{gathered}$$

Hence, the radius of single solid sphere is $9\mathrm{cm}$.