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 figure shows a uniform rod of length $L$ and mass $M$ lying along the $x$-axis with one end at the origin $O$. The locus of all points $(x, y)$ in the $xy$-plane,about which the moment of inertia of the rod is the same as that about $O$,is:

$A$ circular disc of radius $b$ has a hole of radius $a$ at its centre (see figure). If the mass per unit area of the disc varies as $\sigma(r) = \frac{\sigma_0}{r}$,then the radius of gyration of the disc about its axis passing through the centre is

$A$ uniform solid cylinder with radius $R$ and length $L$ has a moment of inertia $I_1$ about its central axis. $A$ concentric solid cylinder of radius $R' = \frac{R}{2}$ and length $L' = \frac{L}{2}$ is carved out of the original cylinder. If $I_2$ is the moment of inertia of the carved-out portion of the cylinder about the same axis,then $\frac{I_1}{I_2} = ..........$

The moment of inertia of a solid cylinder of mass $2.5 \ kg$ and radius $10 \ cm$ about its axis is (in $kg \ m^2$)

The figure shows two solid discs with radius $R$ and $r$ respectively. If the mass per unit area is the same for both,what is the ratio of the moment of inertia $(MI)$ of the bigger disc about axis $AB$ (which is perpendicular to the plane of the disc and passing through its centre) to the $MI$ of the smaller disc about one of its diameters lying on its plane? Given $M$ is the mass of the larger disc.