Four point masses (each of mass $m$) are arranged in the $X-Y$ plane at coordinates $(a, 0)$,$(0, a)$,$(-a, 0)$,and $(0, 2a)$. What is the moment of inertia of this system about the $Y$-axis?

The moment of inertia of a semicircular ring of mass $M$ and radius $R$ about its centre is:

Moment of inertia of a thin uniform rod of length $L$ and mass $M$ rotating about the perpendicular axis passing through its centre is $I$. If the same rod is bent in the form of a ring,its moment of inertia about the diameter is $I_1$. If $I_1 = xI$,then the value of $x$ is:

$A$ 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 :

From a uniform wire,two circular loops are made: $(i)\ P$ of radius $r$ and $(ii)\ Q$ of radius $nr$. If the moment of inertia of $Q$ about an axis passing through its center and perpendicular to its plane is $8$ times that of $P$ about a similar axis,the value of $n$ is (diameter of the wire is very much smaller than $r$ or $nr$).

The radius of gyration of a uniform rod of length $l$ about an axis passing through one of its ends and perpendicular to its length is