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 centre and perpendicular to one of its faces is

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

Answer 1

Answer: (B)

$I=\frac{M{x}^2}{6}$
edge length : (x)
$2R=\sqrt{3}x$
$x=\frac{2R}{\sqrt{3}}$
Now,
mass of cube :
$m=\frac{M}{\left(\frac{4}{3}\pi R^3\right)}{\left(\frac{2R}{\sqrt{3}}\right)}^3$
$\left(\frac{3M}{4\pi R^3}\right)\left(\frac{8R^3}{3\sqrt{3}}\right)$
$m=\frac{2M}{\sqrt{3}\pi }$
$I=\frac{1}{3}\left(\frac{2M}{\sqrt{3}\pi }\right)\left[\frac{4R^2}{3}\right]$
$=\frac{4M{R}^2}{9\sqrt{3}\pi }$