A solid sphere of mass \(M\) and radius \(R\) is divided into two unequal parts. The smaller part having mass \(M/8\) is converted into a sphere of radius \(r\) and the larger part is converted into a circular disc of thickness \(t\) and radius \(2R\). If \(I_1\) is moment of inertia of a sphere having radius \(r\) about an axis through its centre and \(I_2\) is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia \(I_2/I_1 = \) _____.

Let the mass of the original solid sphere be \(M\) and its radius be \(R\).

Mass of the smaller sphere, \(m_1 = \dfrac{M}{8}\).
Since the material remains the same, the density is constant. The volume is directly proportional to the mass.
\(\dfrac{4}{3}\pi r^3 = \dfrac{1}{8} \left(\dfrac{4}{3}\pi R^3\right) \Rightarrow r^3 = \dfrac{R^3}{8} \Rightarrow r = \dfrac{R}{2}\)

Moment of inertia of the smaller sphere about an axis through its centre is:
\(I_1 = \dfrac{2}{5} m_1 r^2 = \dfrac{2}{5} \left(\dfrac{M}{8}\right) \left(\dfrac{R}{2}\right)^2 = \dfrac{2}{5} \times \dfrac{M}{8} \times \dfrac{R^2}{4} = \dfrac{M R^2}{80}\)

Mass of the larger part (which is converted into a disc) is:
\(m_2 = M - \dfrac{M}{8} = \dfrac{7M}{8}\)

The radius of the disc is given as \(2R\). The moment of inertia of a disc about its diameter is:
\(I_2 = \dfrac{1}{4} m_2 (\text{radius})^2 = \dfrac{1}{4} \left(\dfrac{7M}{8}\right) (2R)^2 = \dfrac{1}{4} \times \dfrac{7M}{8} \times 4R^2 = \dfrac{7M R^2}{8}\)

The ratio of their moments of inertia is:
\(\dfrac{I_2}{I_1} = \dfrac{\dfrac{7M R^2}{8}}{\dfrac{M R^2}{80}} = \dfrac{7}{8} \times 80 = 70\)

Answer: \(70\)