From a solid sphere of mass $\mathrm{M}$ and radius $\mathrm{R}$ a cube of maximum possible volume is cut.…

From a solid sphere of mass $\mathrm{M}$ and radius $\mathrm{R}$ 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 faces is :
  1. $\frac{4 \mathrm{MR}^{2}}{9 \sqrt{3} \pi}$
  2. $\frac{4 \mathrm{MR}^{2}}{3 \sqrt{3} \pi}$
  3. $\frac{\mathrm{MR}^{2}}{32 \sqrt{2} \pi}$
  4. $\frac{\mathrm{MR}^{2}}{16 \sqrt{2} \pi}$

Solution

Here a
$=\frac{\frac{4}{3} \pi \mathrm{R}^{3}}{\left(\frac{2}{\sqrt{3}} \mathrm{R}\right)^{3}}=\frac{\sqrt{3}}{2} \pi \cdot \quad \mathrm{M}^{\prime}=\frac{2 \mathrm{M}}{\sqrt{3} \pi}$
Moment of inertia of the cube about the given axis, $I=\frac{M^{\prime} a^{2}}{6}=\frac{\frac{2 M}{\sqrt{3} \pi} \times\left(\frac{2}{\sqrt{3}} R\right)^{2}}{6}=\frac{4 M R^{2}}{9 \sqrt{3} \pi}$
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Asked in: JEE Mains - Rotational Motion - Test 1

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