From a solid sphere of mass $M$ and radius $R$,a cube of maximum possible volume is cut. The moment of inertia of the cube about an axis passing through its center and perpendicular to one of its faces is

$A$ particle falls freely near the surface of the earth. Consider a fixed point $O$ (not vertically below the particle) on the ground.

Statement-$1$: $A$ body is rotating about an axis with angular velocity $\omega$ and moment of inertia $I$. Its angular momentum $L$ remains constant,but its rotational kinetic energy $K$ decreases,provided no external torque is applied.
Statement-$2$: $L = I\omega$ and $K = \frac{L^2}{2I} = \frac{1}{2} I\omega^2$.

$A$ thin rod of length $L$ is placed vertically on a frictionless horizontal floor and released with a negligible push to allow it to fall. At any moment,the rod makes an angle $\theta$ with the vertical. If the center of mass has acceleration $= A$,and the rod has an angular acceleration $= \alpha$ at that moment,then:

Seven identical circular planar disks,each of mass $M$ and radius $R$,are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point $P$ is:

$A$,$B$,and $C$ are a disc,a solid sphere,and a spherical shell respectively,with the same radii $(R)$ and masses $(M)$. These bodies are placed as shown in the figure. The moment of inertia of the given system about the axis $PQ$ is $\frac{x}{15} I$,where $I$ is the moment of inertia of the disc about its diameter. The value of $x$ is . . . . . . .