From a solid sphere of mass M and radius R a cube of maximum possible… | JEE Main 2015

Q180JEE Main 2015Difficult

Moment of Inertia & Radius of Gyration

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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Answer: (a)

Largest cube: body diagonal $= 2R \Rightarrow a = \frac{2R}{\sqrt{3}}$ .
Mass $m = M\frac{a^3}{\frac{4}{3}\pi R^3} = M \cdot \frac{8R^3}{3\sqrt{3}} \cdot \frac{3}{4\pi R^3} = \frac{2M}{\sqrt{3}\pi}$
$I = \frac{ma^2}{6} = \frac{2M}{\sqrt{3}\pi} \cdot \frac{4R^2}{3 \times 6} = \frac{4MR^2}{9\sqrt{3}\pi}$
So the answer is (a).

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