The moment of inertia of a uniform circular disc about its diameter is $I$. What is its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim (in $, I$)?

The moment of inertia of a uniform cylinder of length $l$ and radius $R$ about its perpendicular bisector is $I$. What is the ratio $l/R$ such that the moment of inertia is minimum?

$A$ disc and a ring both have the same mass and radius. The ratio of the moment of inertia of the disc about its diameter to that of a ring about a tangent in its plane is

Seven identical discs each of mass $M$ and radius $R$ are arranged in a hexagonal plane pattern so as to touch each neighbour disc as shown in the figure. The moment of inertia of the system of seven discs about an axis passing through the centre of the central disc and normal to the plane of all discs is

$A$ lamina is made by removing a small disc of diameter $2R$ from a bigger disc of uniform mass density and radius $2R$,as shown in the figure. The moment of inertia of this lamina about axes passing through $O$ and $P$ is $I_0$ and $I_P$,respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $\frac{I_P}{I_0}$ to the nearest integer is:

$A$ solid sphere of mass $5 \text{ kg}$ and a disc of mass $4 \text{ kg}$ have the same radius. The ratio of the moment of inertia of the sphere about its tangent to the moment of inertia of the disc about a tangent in its plane will be $x : y$. The value of $x$ and $y$ respectively is