Question

The internal and external diameters of a hollow hemispherical shell are 6 cm and 10 cm, respectively. It is melted and recast into a solid cone of base diameter 14 cm. Find the height of the cone so formed.

Answer 1

Internal diameter of the hemispherical shell = 6 cm
Therefore, internal radius of the hemispherical shell = 3 cm

External diameter of the hemispherical shell = 10 cm
External radius of the hemispherical shell = 5 cm

Volume of hemispherical shell$=\frac{2}{3}\mathrm{\pi }\left(5^3-3^3\right)=\frac{196}{3}\times \frac{22}{7}=\frac{616}{3}{\mathrm{cm}}^3$

Radius of cone = 7 cm
Let the height of the cone be h cm.
Volume of cone $=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}=\frac{1}{3}\times \frac{22}{7}\times 7\times 7\mathrm{h}=\frac{154\mathrm{h}}{3}{\mathrm{cm}}^3$

The volume of the hemispherical shell must be equal to the volume of the cone. Therefore,

$$\begin{gathered} \frac{154h}{3}=\frac{616}{3} \\ \Rightarrow h=\frac{616}{154}=4cm \end{gathered}$$