Two discs of the same material and thickness have radii $0.2\, m$ and $0.6\, m.$ Their moments of inertia about their axes will be in the ratio

The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to it is $\frac{1}{12} ML^2$. The rod is bent at the center such that the two halves make an angle of $60^\circ$ with each other. Find the moment of inertia of the bent rod about the same axis passing through the center of the rod (the point of bending) and perpendicular to the plane containing the two halves.

Let $M$ and $L$ be the mass and length of a thin uniform rod, respectively. In the first case, the axis of rotation passes through the centre and is perpendicular to the length of the rod. In the second case, the axis of rotation passes through one end and is perpendicular to the length of the rod. The ratio of the radius of gyration in the first case to the second case is:

The moment of inertia of a solid disc rotating about its diameter is $2.5$ times higher than the moment of inertia of a ring rotating in a similar way. The moment of inertia of a solid sphere which has the same radius as the disc and is rotating in a similar way,is $n$ times higher than the moment of inertia of the given ring. Here,$n=$ . . . . . . . Consider all the bodies have equal masses.

The moment of inertia of a circular ring of radius $r$ and mass $M$ about its diameter is:

The analogue of mass in rotational motion is: