Question

Moment of inertia of a solid sphere of mass m and radius R about its tangent is:

\((a)\ \frac{7}{5}MR^2\)

\((b)\ \frac{4}{5}MR^2\)

\((c)\ \frac{2}{5}MR^2\)

\((d)\ \frac{1}{2}MR^2\)

Answer 1

Correct option is : \((a)\ \frac{7}{5}MR^2\)

I_AB = I_d = I_cm = \(\frac{2}{5}MR^2\)

Applying therorem of parallel axes,

I_CD = I_cm + \(\frac{2}{5}\) MR² + MR²

or I_CD = \(\frac{7}{5}\)MR²

∴ Option (a) is correct.