The moment of inertia of a solid sphere of mass M and radius a, about any tangent is

Correct option is A

Solution:

$$\text{Moment of Inertia about center: } I_{\text{center}} = \frac{2}{5} M a^2 \\[8pt]\text{Using Parallel Axis Theorem: } I_{\text{tangent}} = I_{\text{center}} + M a^2 \\[8pt]= \frac{2}{5} M a^2 + M a^2 = \left( \frac{2}{5} + \frac{5}{5} \right) M a^2 \\[8pt]= \frac{7}{5} M a^2 \\[8pt]\boxed{I_{\text{tangent}} = \frac{7}{5} M a^2}$$​