Question
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.
$$\frac{{M{R^2}}}{{16\sqrt 2 \pi }}$$
Answer :
$$\frac{{4M{R^2}}}{{9\sqrt 3 \pi }}$$
Solution :
Here $$a = \frac{2}{{\sqrt 3 }}R$$ Now, $$\frac{M}{{M'}} = \frac{{\frac{4}{3}\pi {R^3}}}{{{a^3}}}$$
$$ = \frac{{\frac{4}{3}\pi {R^3}}}{{{{\left( {\frac{2}{3}R} \right)}^3}}} = \frac{{\sqrt 3 }}{2}\pi .$$
$$M' = \frac{{2M}}{{\sqrt 3 \pi }}$$
Moment of inertia of the cube about the given axis, $$I = \frac{{M'{a^2}}}{6}$$
$$\eqalign{ & = \frac{{\frac{{2M}}{{\sqrt 3 \pi }} \times {{\left( {\frac{2}{{\sqrt 3 }}R} \right)}^2}}}{6} \cr & = \frac{{4M{R^2}}}{{9\sqrt 3 \pi }} \cr} $$
Here $$a = \frac{2}{{\sqrt 3 }}R$$ Now, $$\frac{M}{{M'}} = \frac{{\frac{4}{3}\pi {R^3}}}{{{a^3}}}$$
$$ = \frac{{\frac{4}{3}\pi {R^3}}}{{{{\left( {\frac{2}{3}R} \right)}^3}}} = \frac{{\sqrt 3 }}{2}\pi .$$
$$M' = \frac{{2M}}{{\sqrt 3 \pi }}$$
Moment of inertia of the cube about the given axis, $$I = \frac{{M'{a^2}}}{6}$$
$$\eqalign{ & = \frac{{\frac{{2M}}{{\sqrt 3 \pi }} \times {{\left( {\frac{2}{{\sqrt 3 }}R} \right)}^2}}}{6} \cr & = \frac{{4M{R^2}}}{{9\sqrt 3 \pi }} \cr} $$
