A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.

Radius of solid sphere (r₁) = 6 cm

Volume of solid sphere (V) = $\dfrac{4}{3}πr_1^3$

$= \dfrac{4}{3}π \times 6^3 \\[1em] = 4π \times 2 \times 6^2 \\[1em] = 288π \text{ cm}^3$

External radius of cylinder (R) = 5 cm, height (h) = 32 cm.

Let r be inner radius of cylinder.

Volume of cylinder = Volume of sphere.

$\therefore V = π(R^2 - r^2)h \\[1em] \Rightarrow 288π = π(5^2 - r^2) \times 32 \\[1em] \Rightarrow 288 = 32(25 - r^2) \\[1em] \Rightarrow \dfrac{288}{32} = 25 - r^2 \\[1em] \Rightarrow 9 = 25 - r^2 \\[1em] \Rightarrow r^2 = 25 - 9 \\[1em] \Rightarrow r^2 = 16 \\[1em] \Rightarrow r = 4 \text{ cm}.$

Thickness of hollow cylinder = R - r = 5 - 4 = 1 cm.

Hence, the thickness of the cylinder = 1 cm.