We have two spheres,one solid and one hollow. Their moments of inertia about their respective diameters are equal. The ratio of their radii will be:

Three thin rods,each of mass $M$ and length $L$,are placed along the $X$,$Y$,and $Z$ axes,which are mutually perpendicular. One end of each rod is at the origin. The moment of inertia of the system about the $Z$ axis is:

If the length of a thin uniform rod is $L$ and the radius of gyration of the rod about an axis perpendicular to its length and passing through one end is $K$, then $K: L=$

Five particles of mass $2 \ kg$ each are attached to the rim of a circular disc of radius $0.1 \ m$ and negligible mass. The moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is ........ $kg \ m^2$.

$A$ straight rod of length $L$ is made of a material having mass per unit length $m(x) = \lambda|x|$, where $x$ is measured from the center of the rod. The moment of inertia about an axis perpendicular to the rod and passing through one end of the rod is to be calculated. Given $L = 1 \ m$ and $\lambda = 16 \ kg/m^2$.

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 . . . . . . .