A circular disc of radius \(\mathrm{R}\) and thickness \(\mathrm{R} / 6\) has moment of inertia I about an axis passing through its centre and perpendicular to its plane. It is melted and re-casted in to a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is \(\ldots\) \(\\{\mathrm{A}\\} \mathrm{I}\) \(\\{\mathrm{B}\\}(2 \mathrm{I} / 8)\) \(\\{\mathrm{C}\\}(\mathrm{I} / 5)\) \(\\{\mathrm{D}\\}(\mathrm{I} / 10)\)

The moment of inertia of a circular disc of radius R and mass M about an axis passing through its center and perpendicular to its plane is given by: \( I_{disc} = \frac{1}{2}MR^2 \). We know I = I_{disc}.

We are given the thickness of the disc which is R/6. The volume of the disc can be calculated using the formula \(V_{disc} = \pi R^2 * \frac{R}{6}\). To find the mass of the disc, we can use the formula \(M_{disc} = \rho V_{disc}\), where \(\rho\) is the density of the material.

Since the disc is melted and re-casted into a solid sphere, the mass of the sphere will be the same as the mass of the disc, i.e., \(M_{sphere} = M_{disc}\).

As the mass of the sphere and the disc are the same and the material is the same, we can use the densities to determine the radius of the sphere. The volume of the sphere can be calculated as \(V_{sphere} = \frac{4}{3}\pi R_{sphere}^3\). Using the formula \(M_{sphere} = \rho V_{sphere}\), we can find the radius of the sphere, denoted by R_{sphere}.

The moment of inertia of a solid sphere of mass M and radius R about its diameter as the axis of rotation is given by: \( I_{sphere} = \frac{2}{5}MR_{sphere}^2 \). As we have the mass and radius of the sphere, we can calculate the moment of inertia of the sphere.

We need to find the relationship between the moment of inertia of the sphere and the disc. We can find this by dividing the moment of inertia of the sphere by that of the disc: \[\frac{I_{sphere}}{I_{disc}} = \frac{\frac{2}{5}MR_{sphere}^2}{\frac{1}{2}MR^2} = \frac{2}{5}\frac{R_{sphere}^2}{R^2}\] Now we need to find \(R_{sphere}\) in terms of R. From Step 4, we know that \(\frac{4}{3}\pi R_{sphere}^3 = \pi R^2 * \frac{R}{6}\). Solving for \(R_{sphere}\), we get: \[R_{sphere} = \sqrt[3]{\frac{3}{8}R^3}\] Substituting this value of \(R_{sphere}\) in the above equation, we get: \[\frac{I_{sphere}}{I_{disc}} = \frac{2}{5}\frac{\left(\sqrt[3]{\frac{3}{8}R^3}\right)^2}{R^2}\] Simplifying this, we obtain: \[\frac{I_{sphere}}{I_{disc}} = \frac{1}{10}\] Therefore, the moment of inertia of the sphere about its diameter as axis of rotation is \(\frac{1}{10}\) times the moment of inertia of the disc, which corresponds to option D.