$A$ long slender rod is welded to a thin circular disc of diameter $0.5 \ m$ at a point on its circumference. The rod is in the same plane as that of the disc and forms a tangent to the disc. The radius of gyration of the disc about the rod (in $m$) is

Moment of Inertia $(M.I.)$ of four bodies having same mass $M$ and radius $2R$ are as follows:
$I_{1} =$ $M.I.$ of solid sphere about its diameter
$I_{2} =$ $M.I.$ of solid cylinder about its axis
$I_{3} =$ $M.I.$ of solid circular disc about its diameter
$I_{4} =$ $M.I.$ of thin circular ring about its diameter
If $2(I_{2} + I_{3}) + I_{4} = x I_{1}$,then the value of $x$ will be...

$A$ ring and a solid sphere rotating about an axis passing through their centers have same radii of gyration. The axis of rotation is perpendicular to the plane of the ring. The ratio of the radius of the ring to that of the sphere is $\sqrt{\frac{2}{x}}$. The value of $x$ is $.......$

From a circular ring of mass $M$ and radius $R$,an arc corresponding to a $90^{\circ}$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is $K$ times $MR^{2}$. Then the value of $K$ is

Two rings of the same radius and mass are placed such that their centers are at the same point and their planes are mutually perpendicular. The moment of inertia of this system about an axis passing through the center and perpendicular to the plane of one of the rings is (mass of ring = $m$,radius = $r$):

Two discs of the same thickness but different radii are made of two different materials such that their masses are equal. The densities of the materials are in the ratio $1:3$. The ratio of the moments of inertia of the discs about the axes passing through their centers and perpendicular to their planes is: