The radius of gyration of a rod of length $L$ and mass $M$ about an axis perpendicular to its length and passing through a point at a distance $L/3$ from one of its ends,is :-

Two rods of equal mass $m$ and length $l$ lie along the $x$-axis and $y$-axis with their centers at the origin. What is the moment of inertia of both about the line $x=y$?

Two iron solid discs of negligible thickness have radii $R_1$ and $R_2$ and moments of inertia $I_1$ and $I_2$,respectively. For $R_2 = 2 R_1$,the ratio of $I_1$ and $I_2$ would be $1 / x$,where $x = $ . . . . . . .

The moment of inertia of a solid disc made of thin metal of radius $R$ and mass $M$ about one of its diameters is given by $\frac{M R^2}{4}$. What will be the moment of inertia about this axis,if the disc is folded in half about this diameter?

Let $M$ and $L$ be the mass and length of a thin uniform rod,respectively. In the $1^{\text{st}}$ case,the axis of rotation passes through the center and is perpendicular to its length. In the $2^{\text{nd}}$ case,the axis of rotation passes through one end and is perpendicular to its length. The ratio of the radius of gyration in the first case to the second case 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