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

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 ratio of the radius of gyration of a solid sphere of mass $M$ and radius $R$ about its own axis to the radius of gyration of a thin hollow sphere of the same mass and radius about its axis is:

In the given figure,about which axis will the moment of inertia of the triangular frame be maximum? [Given: $AB < BC < AC$]

The densities of two solid spheres $A$ and $B$ of the same radii $R$ vary with radial distance $r$ as $\rho_A(r) = k \left(\frac{r}{R}\right)$ and $\rho_B(r) = k \left(\frac{r}{R}\right)^5$,respectively,where $k$ is a constant. The moments of inertia of the individual spheres about axes passing through their centres are $I_A$ and $I_B$,respectively. If $\frac{I_B}{I_A} = \frac{n}{10}$,the value of $n$ is

Moment of Inertia $(M.I.)$ of four bodies having same mass $M$ and radius $2R$ are as follows:
$I_{1} =$ $M.I.$ of solid sphere about its diameter
$I_{2} =$ $M.I.$ of solid cylinder about its axis
$I_{3} =$ $M.I.$ of solid circular disc about its diameter
$I_{4} =$ $M.I.$ of thin circular ring about its diameter
If $2(I_{2} + I_{3}) + I_{4} = x I_{1}$,then the value of $x$ will be...