Two identical rods each of mass $M$ and length $l$ are joined in a crossed position as shown in the figure. The moment of inertia of this system about a bisector ($B_1$ or $B_2$) 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.

$A$ thin rod of length $L$ and mass $M$ is bent at the middle point $O$ at an angle of $60^{\circ}$ as shown in the figure. The moment of inertia of the rod about an axis passing through $O$ and perpendicular to the plane of the rod will be:

Five masses,each of $2\, kg$,are placed on a horizontal circular disc,which can be rotated about a vertical axis passing through its centre. All the masses are equidistant from the axis at a distance of $10\, cm$. The moment of inertia of the whole system (in $g\cdot cm^2$) is: (Assume the disc is of negligible mass)

Two discs of same mass and different radii are made of different materials such that their thicknesses are $1\,cm$ and $0.5\,cm$ respectively. The densities of materials are in the ratio $3:5$. The moment of inertia of these discs respectively about their diameters will be in the ratio of $\frac{x}{6}$. The value of $x$ is $.......$.

The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is