Question

A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone. [4 MARKS]

Answer 1

Formula: 1 Mark
Application: 1 Mark
Solution: 2 Marks

External radius of the hollow sphere $=\frac{8}{2}cm$
= 4 cm
Internal radius of the hollow sphere $=\frac{4}{2}cm$
= 2 cm
$\therefore$ Volume of the hollow sphere
= Volume of external sphere - Volume of internal sphere
$=\frac{4}{3}\pi (4{)}^3-\frac{4}{3}\pi (2{)}^3=\frac{4}{3}\pi [4^3-2^3]$
$=\frac{4}{3}\pi [64-8]=\frac{4}{3}\pi \times 56c{m}^3\dots \dots \left(i\right)$
Radius of the cone $=\frac{8}{2}cm=4cm$
Let h be the height of the cone.
Then, volume of the cone $=\frac{1}{3}\pi (4{)}^{2}h$
$=\frac{1}{3}\pi \times 16hc{m}^3\dots \dots \left(ii\right)$
According to the question,
$\frac{4}{3}\pi \times 56=\frac{1}{3}\pi \times 16h$ [From (i) and (ii)]
$\Rightarrow h=\frac{4\times 56}{16}cm=14cm$