Given: Mass of the sphere, \( M \)
Radius of the sphere, \( R \)
Step 1: Formula for Moment of Inertia of a Solid Sphere The moment of inertia of a solid sphere about an axis passing through its diameter is given by the formula: \[ I = \frac{2}{5} M R^2 \] where: - \( M \) is the mass of the sphere, - \( R \) is the radius of the sphere.
Step 2: Conclusion Thus, the moment of inertia of the solid sphere about its diameter is \( \frac{2}{5} M R^2 \).
Answer: The correct answer is option (a): \( \frac{2}{5} M R^2 \).
A solid cylinder of radius $\dfrac{R}{3}$ and length $\dfrac{L}{2}$ is removed along the central axis. Find ratio of initial moment of inertia and moment of inertia of removed cylinder. 
A, B and C are disc, solid sphere and spherical shell respectively with the same radii and masses. These masses are placed as shown in the figure. 
The moment of inertia of the given system about PQ is $ \frac{x}{15} I $, where $ I $ is the moment of inertia of the disc about its diameter. The value of $ x $ is: