$(a)$ Find the moment of inertia of a sphere about a tangent to the sphere,given the moment of inertia of the sphere about any of its diameters to be $2 M R^{2} / 5,$ where $M$ is the mass of the sphere and $R$ is the radius of the sphere.
$(b)$ Given the moment of inertia of a disc of mass $M$ and radius $R$ about any of its diameters to be $M R^{2} / 4,$ find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

(N/A) The moment of inertia $(M.I.)$ of a sphere about its diameter is $I_{cm} = \frac{2}{5} M R^{2}$.
According to the theorem of parallel axes,the moment of inertia about any axis is $I = I_{cm} + M d^{2}$,where $d$ is the distance between the parallel axes. For a tangent,$d = R$.
Therefore,the $M.I.$ about a tangent is $I = \frac{2}{5} M R^{2} + M R^{2} = \frac{7}{5} M R^{2}$.
$(b)$ The moment of inertia of a disc about its diameter is $I_{d} = \frac{1}{4} M R^{2}$.
According to the theorem of perpendicular axes,the moment of inertia about an axis perpendicular to the plane and passing through the centre is $I_{z} = I_{x} + I_{y} = \frac{1}{4} M R^{2} + \frac{1}{4} M R^{2} = \frac{1}{2} M R^{2}$.
Applying the theorem of parallel axes to find the $M.I.$ about an axis normal to the disc and passing through a point on its edge (distance $d = R$ from the centre):
$I = I_{z} + M R^{2} = \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2}$.