The moment of inertia of a rod of length $l$ about an axis passing through its centre of mass and perpendicular to the rod is $I$. The moment of inertia of a hexagonal shape formed by six such rods,about an axis passing through its centre of mass and perpendicular to its plane,will be (in $I$)

$A$ 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:

Three rings each of mass $M$ and radius $R$ are arranged as shown in the figure. The moment of inertia of the system about the axis $YY'$ will be

$(a)$ Prove the theorem of perpendicular axes. (Hint: Square of the distance of a point $(x, y)$ in the $x-y$ plane from an axis through the origin and perpendicular to the plane is $x^{2}+y^{2}$)
$(b)$ Prove the theorem of parallel axes. (Hint: If the centre of mass of a system of $n$ particles is chosen to be the origin,$\sum m_{i} r_{i}=0$)

What is the moment of inertia of a thin rod of length $L$ and mass $M$ about an axis perpendicular to the rod and passing through a point at a distance of $L/3$ from one of its ends?

Moment of inertia of a body about two perpendicular axes $X$ and $Y$ in the plane of the lamina are $20 \text{ kg m}^2$ and $25 \text{ kg m}^2$ respectively. Its moment of inertia about an axis perpendicular to the plane of the lamina and passing through the point of intersection of $X$ and $Y$ axes is: