The moment of inertia $(M.I.)$ of four bodies,having the same mass $M$ and radius $R$,are reported as follows:
$I_{1} = M.I.$ of a thin circular ring about its diameter.
$I_{2} = M.I.$ of a circular disc about an axis perpendicular to the disc and passing through the centre.
$I_{3} = M.I.$ of a solid cylinder about its axis.
$I_{4} = M.I.$ of a solid sphere about its diameter.
Then:

The moment of inertia of a solid disc made of thin metal of radius $R$ and mass $M$ about one of its diameters is given by $\frac{M R^2}{4}$. What will be the moment of inertia about this axis,if the disc is folded in half about this diameter?

Two discs have the same thickness. Their radii are $R_1$ and $R_2$ and their densities are $d_1$ and $d_2$. The moment of inertia of the second disc is greater than the first if:

$A$ non-uniform rod $OM$ (of length $l$) is kept along the $x$-axis and is rotating about an axis $AB$,which is perpendicular to the rod as shown in the figure. The rod has a linear mass density that varies with the distance $r$ from the left end $O$ of the rod according to $\lambda = \lambda_0 \left( \frac{r^3}{l^3} \right)$,where $\lambda_0$ is a constant. If the axis $AB$ is at a distance $x$ from the end $O$,what is the value of $x$ such that the moment of inertia of the rod about the axis $AB$ $(I_{AB})$ is minimum?

$A$ circular disc of radius $R$ and thickness $\frac{R}{6}$ has moment of inertia $I$ about an axis passing through its centre and perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is :

Two spheres of equal masses,one of which is a thin spherical shell and the other a solid sphere,have the same moment of inertia about their respective diameters. The ratio of their radii is