Moment of Inertia $(M.I.)$ of four bodies having same mass $M$ and radius $2R$ are as follows:
$I_{1} =$ $M.I.$ of solid sphere about its diameter
$I_{2} =$ $M.I.$ of solid cylinder about its axis
$I_{3} =$ $M.I.$ of solid circular disc about its diameter
$I_{4} =$ $M.I.$ of thin circular ring about its diameter
If $2(I_{2} + I_{3}) + I_{4} = x I_{1}$,then the value of $x$ will be...

$A$ solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The first part has a mass of $\frac{7M}{8}$ and is converted into a uniform disc of radius $2R$. The second part is converted into a uniform solid sphere. Let $I_1$ be the moment of inertia of the disc about its axis and $I_2$ be the moment of inertia of the new sphere about its axis. The ratio of $I_1/I_2$ is given by

$A$ solid spherical ball of density $\rho_{1}$ and a hollow spherical ball of density $\rho_{2}$ have the same outer radius $R$ and the same mass $M$. What is the ratio of the moment of inertia of the hollow sphere to that of the solid sphere about an axis passing through their centers?

$A$ solid sphere $A$ of radius $R$ and mass $M$ is attached at a point to a smaller solid sphere $B$ of radius $r < R$ and mass $m < M$. Assume that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of $A$ is $I_A$ and that calculated about a vertical axis passing through the centre of $B$ is $I_B$. The difference $I_A - I_B$ is :

Four point masses $m, 2m, 3m$ and $4m$ are kept at the corners $A, B, C$ and $D$ respectively of a square $ABCD$ of side '$b$'. The moment of inertia of the system about an axis perpendicular to the plane of the square and passing through the point $D$ is (in $mb^2$)

Two rods,each of mass $m$ and length $l$,are joined at their centers to form a cross. What is the moment of inertia of the cross about an axis passing through their common center and perpendicular to the plane of the rods?