Question

From a solid sphere of mass $M$ and radius $R$, a cube of maximum possible volume is cut. Then, moment of inertia of the cube about an axis passing through its centre and perpendicular to one of its faces is:

• A. $\frac{M{R}^2}{32\sqrt{2}\pi }$
• B. $\frac{M{R}^2}{16\sqrt{2}\pi }$
• C. $\frac{4M{R}^2}{9\sqrt{3}\pi }$
• D. $\frac{4M{R}^2}{3\sqrt{3}\pi }$

Answer 1

The correct option is C $\frac{4M{R}^2}{9\sqrt{3}\pi }$

Assume $L$ is the side of the cube. For maximum volume, it must be circumscribed by the sphere.


Hence, the longest diagonal of the cube will be equal to the diameter of the sphere.
From figure,
$DB=\sqrt{D{C}^2+C{B}^2}=\sqrt{2}L$
$DH=\sqrt{D{B}^2+(HB{)}^2}=\sqrt{(\sqrt{2}L{)}^2+L^2}=\sqrt{3}L$

$\Rightarrow \sqrt{3}L=2R$
$\therefore L=\frac{2R}{\sqrt{3}}...\left(i\right)$

Since mass $\propto$ volume,
$\frac{M_{\text{cube}}}{M_{\text{sphere}}}=\frac{V_{\text{cube}}}{V_{\text{sphere}}}$
Here $M\&V$ stands for the mass and volume respectively.
$\Rightarrow M_{\text{cube}}=\frac{V_{\text{cube}}}{V_{\text{sphere}}}\times M_{\text{sphere}}$
$\Rightarrow M_{\text{cube}}=\frac{{\left(\frac{2R}{\sqrt{3}}\right)}^3}{\frac{4}{3}\pi R^3}\times M_{\text{sphere}}\Rightarrow M_{\text{cube}}=\frac{2M}{\sqrt{3}\pi }$
($\because M_{\text{sphere}}=M$)

Moment of inertia of cube having side ($L$), about an axis passing through its centre and perpendicular to one of its faces is given by:
$I=\frac{1}{6}{M}_{\text{cube}}{L}^2$
$I=\frac{1}{6}\times \frac{2M}{\sqrt{3}\pi }\times {\left(\frac{2R}{\sqrt{3}}\right)}^2$
$\therefore I=\frac{4M{R}^2}{9\sqrt{3}\pi }$