Question

From a solid sphere of mass M and radius R, a cube of maximum possible volume is cut. The 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. $\frac{M{R}^2}{16\sqrt{2}\pi }$
• 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 $\frac{4M{R}^2}{9\sqrt{3}\pi }$

d = 2R = $a\sqrt{3}$

$\Rightarrow a=\frac{2}{\sqrt{3}}R$
$\frac{M}{M^{\prime }}=\frac{\frac{4}{3}\pi R^3}{{\left(\frac{2}{\sqrt{3}}R\right)}^3}=\frac{\sqrt{3}}{2}\pi$
$\Rightarrow M^{\prime }=\frac{2M}{\sqrt{3}\pi }$
$I=\frac{M^{\prime }{a}^2}{6}=\frac{2M}{\sqrt{3}\pi }\times \frac{4}{3}{R}^2\times \frac{1}{6}$
$I=\frac{4M{R}^2}{9\sqrt{3}\pi }$