Moment of inertia of a disc about a diameter is $I$. Find the moment of inertia of the disc about an axis perpendicular to its plane and passing through its rim. (in $I$)

The figure shows an isosceles triangular plate of mass $M$ and base $L$. The angle at the apex is $90^o$. The apex lies at the origin and the base is parallel to the $X$-axis. The moment of inertia of the plate about its base parallel to the $x$-axis is

$A$ rigid body of mass $m$ is rotating with angular velocity $\omega$ about an axis at a distance '$a$' from the center of mass $G$. The radius of gyration about an axis passing through $G$ is $K$. The rotational kinetic energy of this body about the new parallel axis will be:

$A$ solid sphere of radius $R$ has a moment of inertia $I$ about its geometrical axis. It is melted into a disc of radius $r$ and thickness $t$. If its moment of inertia about the tangential axis (which is perpendicular to the plane of the disc) is also equal to $I$,then the value of $r$ is equal to:

$I_{CM}$ is the moment of inertia of a circular disc about an axis $(CM)$ passing through its center and perpendicular to the plane of the disc. $I_{AB}$ is its moment of inertia about an axis $AB$ perpendicular to the plane and parallel to the axis $CM$ at a distance $\frac{2}{3}R$ from the center,where $R$ is the radius of the disc. The ratio of $I_{AB}$ and $I_{CM}$ is $x:9$. The value of $x$ is $........$

Four identical discs each of mass $M$ and diameter $a$ are arranged in a plane as shown in the figure. If the moment of inertia of the system about $OO'$ is $\frac{x}{4} Ma^2$,then the value of $x$ will be: