The figure shows a thin metallic triangular sheet $ABC.$ The mass of the sheet is $M.$ The moment of inertia of the sheet about side $AC$ is

Three point masses,each of mass $m$,are placed at the corners of an equilateral triangle of side $\ell$. The moment of inertia of the system about an axis along any one side of the triangle is:

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:

If the length of a thin uniform rod is $L$ and the radius of gyration of the rod about an axis perpendicular to its length and passing through one end is $K$,then $K: L=$

The moment of inertia $I$ of a solid sphere having fixed volume depends upon its volume $V$ as

The ratio of the radius of gyration of a hollow sphere to that of a solid cylinder of equal mass,for the moment of inertia about their diameter axis $AB$ as shown in the figure,is $\sqrt{\frac{8}{x}}$. The value of $x$ is: