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

Answer 1

Answer: (B)

$\frac{ 4MR^2}{ 9 \sqrt 3 \pi }$

Answer 2

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