Question:
From a solid sphere of mass \(M\) and radius \(R\) a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its face is:
From a solid sphere of mass \(M\) and radius \(R\) a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and perpendicular to one of its face is:
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The largest cube has its body diagonal equal to the sphere diameter, so side \(a = 2R/\sqrt{3}\). Find its mass from the density, then use \(I = ma^2/6\).
Updated On: Nov 11, 2026
- \(\frac{4MR^2}{9\sqrt{3\pi}}\)
- \(\frac{4MR^2}{3\sqrt{3\pi}}\)
- \(\frac{MR^2}{32\sqrt{2\pi}}\)
- \(\frac{MR^2}{16\sqrt{2\pi}}\)
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The Correct Option is A
Solution and Explanation
Largest cube inside the sphere: body diagonal \(= 2R\), so \(\sqrt{3}a = 2R\) and \(a = \frac{2R}{\sqrt{3}}\).
Volume of cube \(a^3 = \frac{8R^3}{3\sqrt{3}}\); volume of sphere \(= \frac{4}{3}\pi R^3\).
Mass of cube \(m = M\cdot\frac{8R^3/(3\sqrt{3})}{4\pi R^3/3} = \frac{2M}{\sqrt{3}\,\pi}\).
Moment of inertia of a cube about an axis through its centre perpendicular to a face: \(I = \frac{ma^2}{6}\).
\(I = \frac{1}{6}\cdot\frac{2M}{\sqrt{3}\pi}\cdot\frac{4R^2}{3} = \frac{8MR^2}{18\sqrt{3}\pi} = \frac{4MR^2}{9\sqrt{3}\pi}\).
This corresponds to option (1) (only option (1) has the factors 4 and 9).
Answer: (1) \(\frac{4MR^2}{9\sqrt{3\pi}}\)
Volume of cube \(a^3 = \frac{8R^3}{3\sqrt{3}}\); volume of sphere \(= \frac{4}{3}\pi R^3\).
Mass of cube \(m = M\cdot\frac{8R^3/(3\sqrt{3})}{4\pi R^3/3} = \frac{2M}{\sqrt{3}\,\pi}\).
Moment of inertia of a cube about an axis through its centre perpendicular to a face: \(I = \frac{ma^2}{6}\).
\(I = \frac{1}{6}\cdot\frac{2M}{\sqrt{3}\pi}\cdot\frac{4R^2}{3} = \frac{8MR^2}{18\sqrt{3}\pi} = \frac{4MR^2}{9\sqrt{3}\pi}\).
This corresponds to option (1) (only option (1) has the factors 4 and 9).
Answer: (1) \(\frac{4MR^2}{9\sqrt{3\pi}}\)
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