Answer

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**Hint:**We have to assume the moment of inertia at first about an axis perpendicular to it and through its center. Now take three concurrent axes, and by the theory of perpendicular axes form an equation.

**Complete step by step answer:**

In the above diagram, we see a circular disc and ab is the diameter of the disc and we are asked to find the moment of inertia by one of its diameter.

We know that the instant of inertia of the disc about an axis perpendicular thereto and thru its center to be known; it's $\dfrac{M{{R}^{2}}}{2}$, where M is that the mass of the disc and R is its radius.

The disc is taken into account to be a planar body. Hence the idea of perpendicular axes is applicable thereto .We take three concurrent axes through the center of the disc, O because the x,y,z axes;x and y-axes dwell the plane of the disc and z is perpendicular thereto. By the theory of perpendicular axes,

\[{{I}_{z}}={{I}_{x}}+{{I}_{y}}\]

And

Now, the axes x and y are along two diameters of the disc, and by design, the moment of inertia of the disc is the same at any diameter. Hence

${{I}_{x}}={{I}_{y}}$

${{I}_{z}}=2{{I}_{x}}$

But,

${{I}_{z}}=\dfrac{M{{R}^{2}}}{2}$

On solving,

${{I}_{x}}=\dfrac{{{I}_{z}}}{2}$

${{I}_{x}}=\dfrac{M{{R}^{2}}}{4}$

Therefore the instant of inertia of a disc about any of its diameter is MR$^{2}$ /4.

**Note:**The axes x and y are along two diameters of the disc. Assume the moment of inertia at the right place. The moment of inertia of the disk about any of its diameter will be the same. In order to find the moment of inertia about any axis, we can use the parallel axis theorem or the perpendicular axis theorem.

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