Question

From a uniform solid sphere of mass $M$ and radius $R$, a cube of a maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre 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 1

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

Given: From a uniform solid sphere of mass M and radius R, a cube of a maximum possible volume is cut.

To find the moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces

Solution:

Consider the cross-sectional view of a diametric plane as shown in fig(i), and cross-sectional view of the cube and sphere in fig(ii),

Let the side of the cube be $A$

Using geometry of the cube

$PQ=2R=\sqrt{3}s⟹a=\frac{2R}{\sqrt{3}}.......\left(i\right)$

Volume densityof the solid sphere is

$\rho =\frac{M}{V}⟹\rho =\frac{M}{\frac{4}{3}\pi R^3}⟹\rho =\frac{3M}{4\pi R^3}......\left(ii\right)$

Now mass of the cube is

$m=\rho \times a^2$

$⟹m=\frac{3M}{4\pi R^3}\times \frac{(2R{)}^2}{(\sqrt{3}{)}^2}$ (from eqn(i) and (ii))

$⟹m=\frac{3m}{4\pi R^3}\times \frac{8R^3}{3\sqrt{3}}⟹m=\frac{2M}{\sqrt{3}\pi }........\left(iii\right)$

Moment of inertia of the cube about the given axis is

$I_Y=\frac{m{a}^2}{12}(a^2+a^2)⟹I_Y=\frac{m{a}^2}{6}$

Substituting the values of m and a we get

$⟹I_Y=\frac{2M}{\sqrt{3}\pi }\times \frac{1}{6}\times \frac{(2R{)}^2}{(\sqrt{3}{)}^2}⟹I_Y=\frac{4M{R}^2}{9\sqrt{3}\pi }$

is the the moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces