$A$ rod has a length of $1 \ m$ and a mass of $0.12 \ kg$. The moment of inertia about an axis passing through its center and perpendicular to its length is ...... $kg \cdot m^2$.

The moment of inertia of a thin uniform rod about a perpendicular axis passing through one of its ends is $I$. Now,the rod is bent into a ring and its moment of inertia about its diameter is $I_{1}$. Then $\frac{I}{I_{1}}$ is:

Four point masses,each of value $m$,are placed at the corners of a square $ABCD$ of side $l$. The moment of inertia of this system about an axis passing through $A$ and parallel to $BD$ is

Five solid spheres,each of mass $m$ and radius $r$,are arranged as shown in the figure. The axis of rotation $A-A'$ passes through the centers of three spheres. Calculate the moment of inertia of the system about the axis of rotation $A-A'$. (in $m r^2$)

Two spheres $A$ and $B$,each of mass $5 \ kg$,are attached to the ends of a light rod of length $1 \ m$. Treating the spheres as point masses,find the ratio of the moment of inertia of the system about an axis passing through $A$ to that about an axis passing through the center of the rod,both axes being perpendicular to the rod.

$A$ thin disc of mass $M$ and radius $R$ has mass per unit area $\sigma (r) = kr^2$,where $r$ is the distance from its centre. Its moment of inertia about an axis passing through its centre of mass and perpendicular to its plane is