The disc of mass $M$ with uniform surface mass density $\sigma$ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position $\left(\frac{xa}{3\pi }, \frac{xa}{3\pi }\right)$ where $x$ is _______ . (Round off to the Nearest Integer) [$a$ is an area as shown in the figure]

Moment of inertia $\left(\mathrm{M}.\mathrm{I}.\right)$ of four bodies, having same mass and radius, are reported as;

$I_1=\mathrm{M}.\mathrm{I}.$ of thin circular ring about its diameter,

$I_2=\mathrm{M}.\mathrm{I}.$ of circular disc about an axis perpendicular to disc and going through the centre,

$I_3=\mathrm{M}.\mathrm{I}.$ of solid cylinder about its axis and

$I_4=\mathrm{M}.\mathrm{I}.$ of solid sphere about its diameter.

Then:

A circular hole of radius $\left(\frac{a}{2}\right)$ is cut out of a circular disc of radius $a$ as shown in figure. The centroid of the remaining circular portion with respect to point $O$ will be:

A sphere of radius $a$ and mass $m$ rolls along a horizontal plane with constant speed $v_0$. It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?