Question

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is $2cm$ and the diameter of the base is $4cm$. If a right circular cylinder circumscribes the solid. Find how much more space it will cover.

Answer 1

Let $BPC$ be the hemisphere and $ABC$ be the cone mounted on the base of the hemisphere.

Let $EFGH$ be the right circular cylinder circumscribing.
We have
Height of the cone$=OA+2cm$
Diameter of the base of the cone$=BC=4cm$

Radius of the hemisphere $=BO=\frac{1}{2}BC=\frac{4}{2}cm=2cm$
$OP=2cm$
$AP=OP+OA=2+2=4cm$

Volume of the right circular cylinder
$\Rightarrow \pi r^{2}h=\pi \times 2^2\times 4{cm}^3=16\pi {cm}^3$

volume of the solid toy $=\{\frac{2}{3}\pi \times 2^3+\frac{1}{3}\pi \times 2^2\times 2\}{cm}^3=8\pi {cm}^3$

Required space $=$ volume of the right circular cylinder - volume of the toy

$=16\pi {cm}^3-8\pi {cm}^3$

$=8\pi {cm}^3$