Three particles of mass $m$ each are placed at the vertices of an equilateral triangle $ABC$ of side length $\ell$. What is the moment of inertia of the system about the axis $AX$ passing through vertex $A$ and perpendicular to side $AB$ in the plane of the triangle?

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...

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 solid spheres $A$ and $B$ each of radius $R$ are made of materials of densities $\rho_A$ and $\rho_B$ respectively. Their moments of inertia about a diameter are $I_A$ and $I_B$ respectively. The value of $\frac{I_A}{I_B}$ 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?

The moment of inertia of a solid disc made of thin metal of radius $R$ and mass $M$ about one of its diameters is given by $\frac{M R^2}{4}$. What will be the moment of inertia about this axis,if the disc is folded in half about this diameter?