A solid sphere of mass M and a disc of mass $\dfrac{M}{2}$ have the same radius. The ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be

Let the radius of the solid sphere and the disc be $R$.

Mass of the solid sphere, $m_{s} = M$

Mass of the disc, $m_{d} = \dfrac{M}{2}$

Moment of inertia of the disc about its diameter is $\dfrac{1}{4}m_{d}R^2$. Using the parallel axis theorem, the moment of inertia of the disc about a tangent in its plane is:

$I_{disc} = \dfrac{1}{4}m_{d}R^2 + m_{d}R^2 = \dfrac{5}{4}m_{d}R^2$

Substituting $m_{d} = \dfrac{M}{2}$:

$I_{disc} = \dfrac{5}{4}\left(\dfrac{M}{2}\right)R^2 = \dfrac{5}{8}MR^2$

Moment of inertia of the solid sphere about its diameter is $\dfrac{2}{5}m_{s}R^2$. Using the parallel axis theorem, the moment of inertia of the solid sphere about its tangent is:

$I_{sphere} = \dfrac{2}{5}m_{s}R^2 + m_{s}R^2 = \dfrac{7}{5}m_{s}R^2$

Substituting $m_{s} = M$:

$I_{sphere} = \dfrac{7}{5}MR^2$

The required ratio is:

$\dfrac{I_{disc}}{I_{sphere}} = \dfrac{\dfrac{5}{8}MR^2}{\dfrac{7}{5}MR^2} = \dfrac{5}{8} \times \dfrac{5}{7} = \dfrac{25}{56}$

Answer: $25 : 56$