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EEJ MAIN Physics QUESTION #7066
Question 1
A cube of maximum possible volume is cut from a solid sphere of mass $M$ and radius $R$. Find the moment of inertia of this cube about an axis that passes through its center and is perpendicular to one of its faces.
  • $\frac{MR^2}{32\sqrt{2}\pi}$
  • $\frac{MR^2}{16\sqrt{2}\pi}$
  • $\frac{4 MR^2}{9\sqrt{3}\pi}$✔️
  • $\frac{4 MR^2}{3\sqrt{3}\pi}$
Correct Answer Explanation

1. Find Cube Side ($a$): The body diagonal of the cube equals the diameter of the sphere. $\sqrt{3}a = 2R \implies a = \frac{2R}{\sqrt{3}}$.

2. Find Cube Mass ($m$): Density $\rho = \frac{M}{\frac{4}{3}\pi R^3}$. Mass of cube $m = \rho \times a^3 = \left(\frac{3M}{4\pi R^3}\right) \times \left(\frac{8R^3}{3\sqrt{3}}\right) = \frac{2M}{\pi \sqrt{3}}$.

3. Moment of Inertia ($I$): For a cube about the center-face axis, $I = \frac{ma^2}{6}$.
$I = \frac{1}{6} \left( \frac{2M}{\pi \sqrt{3}} \right) \left( \frac{4R^2}{3} \right) = \frac{8MR^2}{18\pi \sqrt{3}} = \frac{4MR^2}{9\sqrt{3}\pi}$.