# A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about axis passing through its edge and perpendicular to the plane remains I. Then, radius of the disc will be

$\begin {array} {1 1} (a)\;\frac{R}{\sqrt {15}} & \quad (b)\;R \sqrt {\frac{2}{15}} \\ (c)\;\frac{2R}{\sqrt {15}} & \quad (d)\;\frac{R}{4} \end {array}$

The moment of inertia of a solid spherical shell of mass M and radius R as shown in the figure above is $I_{\text{sphere}} = \large\frac{2}{5}$$MR^2 Given that a solid sphere or mass M and radius R is recast into a disc of thickness t, but the moment of inertia remains the same. The moment of inertia of a sphere about its central axis and a solid spherical shell of mass M and radius r as is I = \large\frac{1}{2}$$Mr^2$
Using the theorem of parallel axes, the momentum of inertia of a disc about its edge $I_{\text{disc}} = \large\frac{1}{2} $$Mr^2+Mr^2 = \large\frac{3}{2}$$Mr^2$
Given that $I_{\text{disc}} =I_{\text{sphere}} \rightarrow \large\frac{2}{5}$$MR^2$$ = \large\frac{3}{2} $$Mr^2 \Rightarrow r^2 = \large\frac{4}{15}$$R^2$
$\Rightarrow r= \large\frac{2R}{\sqrt {15}}$
edited Mar 25, 2014
but the axis is perpendicular to the plane then shouldn't we apply the perpendicular theorem to 1/2MR^2 and then we should apply parallel axis theorem