Question

From a solid sphere of mass $M$ and radius $R$, a cube of maximum possible volume is cut. 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 $\frac{4M{R}^2}{9\sqrt{3}\pi }$


When the volume of the cube is maximum, the longest diagonal of cube will be equal to diameter of the sphere.
$FG=GC=L\Rightarrow FC=\sqrt{{\left(FG\right)}^2+{\left(GC\right)}^2}=\sqrt{L^2+L^2}=\sqrt{2}L\&FD=\sqrt{{\left(FC\right)}^2+{\left(CD\right)}^2}=\sqrt{{\left(\sqrt{2}L\right)}^2+L^2}=\sqrt{3}L\therefore \sqrt{3}L=2R\Rightarrow L=\frac{2R}{\sqrt{3}}$
Since $\text{mass}\propto \text{volume}$, we have
$\frac{M_C}{M_S}=\frac{V_C}{V_S}\Rightarrow M_C=\frac{V_C}{V_S}\times M_S\Rightarrow M_C=\frac{{\left(\frac{2R}{\sqrt{3}}\right)}^3}{\frac{4}{3}\pi R^3}\times M\Rightarrow M_C=\frac{2M}{\sqrt{3}\pi }$
And moment of inertia of cube about an axis passing through its center and perpendicular to one of its faces is given by
$I=\frac{1}{6}M{L}^2\Rightarrow I=\frac{1}{6}\times \frac{2M}{\sqrt{3}\pi }\times {\left(\frac{2R}{\sqrt{3}}\right)}^2=\frac{4M{R}^2}{9\sqrt{3}\pi }$