$A$ one-quarter sector is cut from a uniform circular disc of radius $R$. This sector has mass $M$. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

$A$ thin metal rod of mass $M$ and length $L$ is cut into $4$ equal parts by cutting it perpendicular to its length. If the moment of inertia of the rod about an axis passing through its centre and perpendicular to its axis is $I$,then what is the moment of inertia of each part about a similar axis?

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:

Two identical thin rods of mass $M \ kg$ and length $L \ m$ are connected as shown in the figure. The moment of inertia of the combined rod system about an axis passing through point $P$ and perpendicular to the plane of the rods is $\frac{x}{12} M L^2 \ kg \ m^2$. The value of $x$ is . . . . . . .

If $I_1$ is the moment of inertia of a thin rod of mass $M$ and length $l$ about an axis perpendicular to its length and passing through its centre of mass,and $I_2$ is the moment of inertia of the ring formed by bending the rod about an axis passing through its centre and perpendicular to its plane,then:

Radius of gyration of a thin uniform circular disc about the axis passing through its centre and perpendicular to its plane is $K_c$. Radius of gyration of the same disc about a diameter of the disc is $K_d$. The ratio $K_d : K_c$ is