Q 11-06-020NEETJEE MainEAMCET 2008 (Engineering)Easy
The moment of inertia of a thin circular disc about an axis passing through its centre and perpendicular to its plane is $I$. Then, the moment of inertia of the disc about an axis parallel to its diameter and touching the edge of the rim is
Answer: (B) $\dfrac{5}{2}I$
$I = \dfrac{MR^2}{2}$, so $MR^2 = 2I$.
About a diameter: $\dfrac{MR^2}{4}$. About a parallel tangent (parallel axis theorem):
$$I' = \frac{MR^2}{4} + MR^2 = \frac{5}{4}MR^2 = \frac{5}{4}(2I) = \frac{5}{2}I$$
Solution by Sreeraj P, M.Sc Physics