The volume of the largest possible cube that can be inscribed in a hollow spherical ball of radius $r$ cm is

### Concept & Cube Inscribed in a Sphere When the largest possible cube is inscribed inside a sphere, the longest interior distance of the cube (its main body diagonal) perfectly spans the maximum width of the sphere (its diameter). $$\text{Body Diagonal of Cube} = \text{Diameter of Sphere}$$ $$\sqrt{3}a = 2r$$ Where $a$ is the side of the cube and $r$ is the radius of the sphere. ### Step-by-Step Solution * **Given:** Radius of the sphere is $r$. Let the side of the inscribed cube be $a$. * Equate the body diagonal of the cube to the sphere's diameter: $\sqrt{3}a = 2r$. * Isolate the side length of the cube, $a$: $a = \frac{2r}{\sqrt{3}}$. * The volume of the cube $V = a^3$. * Substitute the value of $a$ into the volume formula: $V = \left(\frac{2r}{\sqrt{3}}\right)^3$ * Cube the numerator and the denominator separately: Numerator $= (2r)^3 = 8r^3$. Denominator $= (\sqrt{3})^3 = \sqrt{3} \times \sqrt{3} \times \sqrt{3} = 3\sqrt{3}$. * Resulting volume $V = \frac{8r^3}{3\sqrt{3}}$. * This matches the format $\frac{8}{3\sqrt{3}}r^3$. ### Exam Strategy & Shortcut Use dimensional analysis. Volume must be a cubic unit ($r^3$). Options (a) and (b) are squared units ($r^2$), which represent area, not volume. They can be instantly eliminated. Between the remaining options, knowing the diagonal relates via $\sqrt{3}$ directly leads to $3\sqrt{3}$ in the denominator when cubed. ### Common Pitfall Confusing an inscribed cube with an inscribed sphere. If the sphere is inscribed, $a=2r$. If the cube is inscribed, $\sqrt{3}a=2r$. Mixing these up leads to entirely incorrect structural equations. ### Final Answer Therefore, the correct answer is **$\frac{8}{3\sqrt{3}}r^3$**.