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 : [JEE Main Solved Paper-2015 ]

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 :

\[2R=\sqrt{3}a\] \[\frac{2R}{\sqrt{3}}=a\] \[I=\frac{mass{{\left( side \right)}^{2}}}{6}\] \[=\left( \frac{M}{\frac{4}{3}\pi {{R}^{3}}} \right){{a}^{3}}\frac{4{{R}^{2}}}{3}\frac{1}{6}\] \[=\frac{3M}{4\pi {{R}^{3}}}\frac{8{{R}^{3}}}{3\sqrt{3}}\frac{4{{R}^{2}}}{3\times 6}\] \[=\frac{8M{{R}^{2}}}{18\sqrt{3}\pi }=\frac{4M{{R}^{2}}}{\pi 9\sqrt{3}}\]