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 circular disc has radius \( R_1 \) and thickness \( T_1 \). Another circular disc made of the same material has radius \( R_2 \) and thickness \( T_2 \). If the moments of inertia of both the discs are same and \[ \frac{R_1}{R_2} = 2, \quad \text{then} \quad \frac{T_1}{T_2} = \frac{1}{\alpha}. \] The value of \( \alpha \) is __________.
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. 
Which part of root absorb mineral?