Question:
A solid sphere of radius 'r' is melted and recast into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 4 cm, its height 24 cm and thickness 2 cm, find the value of 'r'.
Solution:
Volume of sphere $=\frac{4}{3} \pi r^{3}$ ......(i)
Since,
The sphere is recast in to a hollow cylinder of uniform thickness 2 cm.
The external radius of hollow cylinder r1 = 4 cm
The internal radius of hollow cylinder $r_{2}=4-2=2 \mathrm{~cm}$
and height, h = 24 cm
Clearly,
The volume of hollow cylinder = volume of sphere
$\left(4^{2}-2^{2}\right) \times 24=\frac{4}{3} r^{3}$
$12 \times 24=\frac{4}{3} r^{3}$
$r=\sqrt{12 \times 6 \times 3}$
$=\sqrt[3]{3 \times 2 \times 2 \times 2 \times 3 \times 3}$
$r=6 \mathrm{~cm}$