question_answer

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 faces is

A) \[\frac{4M{{R}^{2}}}{9\sqrt{3\pi }}\]

B) \[\frac{4M{{R}^{2}}}{3\sqrt{3\pi }}\]

C) \[\frac{M{{R}^{2}}}{32\sqrt{2\pi }}\]

D) \[\frac{M{{R}^{2}}}{16\sqrt{2\pi }}\]

Correct Answer: A

Solution :

[a] Here \[a=\frac{2}{\sqrt{3}}R\]  Now, \[\frac{M}{M'}=\frac{\frac{4}{3}\pi {{R}^{3}}}{{{a}^{3}}}\] \[=\frac{\frac{4}{3}\pi {{R}^{3}}}{{{\left( \frac{2}{\sqrt{3}}R \right)}^{3}}}=\frac{\sqrt{3}}{2}\pi .\] \[M'=\frac{2M}{\sqrt{3}\pi }\] Moment of inertia of the cube about the given axis, \[I=\frac{M'{{a}^{2}}}{6}\] \[=\frac{\frac{2M}{\sqrt{3}\pi }\times {{\left( \frac{2}{\sqrt{3}}R \right)}^{2}}}{6}=\frac{4M{{R}^{2}}}{9\sqrt{3}\pi }\]