The moment of inertia of a sphere about its diameter is $I$. Four such spheres are arranged as shown in the figure. Calculate the moment of inertia of the system about the axis $XX'$. (in $I$)

The moment of inertia of a uniform circular ring,having a mass $M$ and a radius $R$,about an axis tangential to the ring and perpendicular to its plane,is

The moment of inertia of a solid cylinder of mass $M$, length $L = 2R$ and radius $R$ about an axis passing through the centre of mass and perpendicular to the axis of the cylinder is $I_1$, and about an axis passing through one end of the cylinder and perpendicular to the axis of the cylinder is $I_2$. Then:

Identical rings are arranged in a hexagonal plane pattern such that each ring touches its neighbouring rings. Each ring has mass $M$ and radius $R$. The moment of inertia of the system of seven rings about an axis passing through the centre of the central ring and normal to the plane of all rings is: (in $MR^2$)

The figure shows an isosceles triangular plate of mass $M$ and base $L$. The angle at the apex is $90^o$. The apex lies at the origin and the base is parallel to the $X$-axis. The moment of inertia of the plate about its base parallel to the $x$-axis is

If a solid sphere of mass $5 \, kg$ and a disc of mass $4 \, kg$ have the same radius $R$,then the ratio of the moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent is $\frac{x}{7}$. The value of $x$ is $.........$