A sphere is tightly enclosed inside a cube of side length a. If the cube is then inscribed inside another sphere, what is the ratio of the inner to outer sphere volumes?
\(1:3\sqrt{3}\)
Given: a sphere fits exactly inside a cube of side \(a\) (inner sphere), and the same cube is inscribed inside a larger sphere (outer sphere). We need the ratio of their volumes.
The inner sphere touches all faces of the cube, so its diameter equals the cube's side: \(2r_1 = a\), giving \(r_1 = \dfrac{a}{2}\).
The outer sphere passes through all vertices of the cube, so its diameter equals the cube's space diagonal: \(2r_2 = a\sqrt{3}\), giving \(r_2 = \dfrac{a\sqrt{3}}{2}\).
Since volume is proportional to the cube of the radius, \(\dfrac{V_1}{V_2} = \left(\dfrac{r_1}{r_2}\right)^3\).
The radius ratio is \(\dfrac{r_1}{r_2} = \dfrac{a/2}{a\sqrt{3}/2} = \dfrac{1}{\sqrt{3}}\).
Cubing gives \(\dfrac{V_1}{V_2} = \dfrac{1}{(\sqrt{3})^3} = \dfrac{1}{3\sqrt{3}}\).
Hence the ratio of inner to outer sphere volumes is 1 : 3√3.
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