# A spherical cannon ball, 28 cm in diameter, is melted and recast into a right circular conical mould with base diameter of 35 cm.

Question:

A spherical cannon ball, 28 cm in diameter, is melted and recast into a right circular conical mould with base diameter of 35 cm. Find the height of the cone.

Solution:

Diameter of cannon ball $=28 \mathrm{~cm}$

Radius of the cannon ball $=14 \mathrm{~cm}$

Volume of ball $=\frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \times(14)^{3} \mathrm{~cm}^{3}$

Diameter of base of cone $=35 \mathrm{~cm}$

Radius of base of cone $=\frac{35}{2} \mathrm{~cm}$

Let the height of the cone be $\mathrm{h} \mathrm{cm}$.

Volume of cone $=\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h}=\frac{1}{3} \pi \times\left(\frac{35}{2}\right)^{2} \times \mathrm{h} \mathrm{cm}^{3}$

From the above results and from the given conditions,
Volume of ball = Volume of cone

Or, $\frac{4}{3} \pi \times(14)^{3}=\frac{1}{3} \pi \times\left(\frac{35}{2}\right)^{2} \times \mathrm{h}$

$\Rightarrow \mathrm{h}=\frac{\frac{4}{3} \pi \times(14)^{3}}{\frac{1}{3} \pi \times\left(\frac{35}{2}\right)^{2}}=\frac{4 \times 14 \times 14 \times 14 \times 2 \times 2}{35 \times 35}=35.84 \mathrm{~cm}$