Asked in: March 2026 board paperSurface area of a sphere and total surface area of a cube are the same. Show that the ratio of the volume of the sphere to that of the cube is \(\sqrt6 : \sqrt\pi\).

Answer: Shown: \(V_{\text{sphere}} : V_{\text{cube}} = \sqrt6 : \sqrt\pi\)

Let the radius of the sphere be r and the side of the cube be a.

Given \(4\pi r^2 = 6a^2\) \(\therefore \dfrac{r^2}{a^2} = \dfrac{6}{4\pi} = \dfrac{3}{2\pi}\) \(\therefore \dfrac{r}{a} = \dfrac{\sqrt3}{\sqrt{2\pi}}\)

\(\dfrac{V_{\text{sphere}}}{V_{\text{cube}}} = \dfrac{\dfrac{4}{3}\pi r^3}{a^3} = \dfrac{4\pi}{3} \times \left(\dfrac{r}{a}\right)^3 = \dfrac{4\pi}{3} \times \dfrac{3\sqrt3}{2\pi\sqrt{2\pi}}\)

\(= \dfrac{2\sqrt3}{\sqrt{2\pi}} = \dfrac{\sqrt4 \times \sqrt3}{\sqrt2 \times \sqrt\pi} = \dfrac{\sqrt{12}}{\sqrt2\,\sqrt\pi} = \dfrac{\sqrt6}{\sqrt\pi}\)

\(\therefore\) Volume of sphere : volume of cube \(= \sqrt6 : \sqrt\pi\).