The internal and external diameters of a hollow hemispherical shell are 6 cm and 10 cm, respectively.

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.

Solution:

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

Radius of cone = 7 cm
Let the height of the cone be h cm.

Volume of cone $=\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h}=\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \mathrm{~h}=\frac{154 \mathrm{~h}}{3} \mathrm{~cm}^{3}$

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

$\frac{154 h}{3}=\frac{616}{3}$

$\Rightarrow h=\frac{616}{154}=4 \mathrm{~cm}$

 

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