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?

Two loops $A$ and $B$ of radii $R_1$ and $R_2$ are made from a uniform wire. If the moment of inertia of $A$ is $I_A$ and that of $B$ is $I_B$,then find the ratio $R_2 / R_1$ given that $I_A / I_B = 27$.

For the same total mass,which of the following will have the largest moment of inertia about an axis passing through its centre of mass and perpendicular to the plane of the body?

Two discs having moment of inertia $I_1$ and $I_2$ are made from the same material and have the same mass. Their thicknesses and radii are $t_1, t_2$ and $R_1, R_2$ respectively. The relation between the moment of inertia of each disc about an axis passing through its centre and perpendicular to its plane and its thickness is:

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

We have two spheres,one of which is hollow and the other solid. They have identical masses and moments of inertia about their respective diameters. The ratio of their radii is given by: