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NEET 2010 (AIPMT, Mains) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from NEET 2010 (AIPMT, Mains)

Q184NEET 2010 · AIPMT · MAINSMedium

Moment of Inertia & Radius of Gyration

From a circular disc of radius R and mass 9M, a small disc of mass M and radius $\frac{R}{3}$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

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Answer: (a)

Both discs are concentric, so no parallel-axis term.
$I = \frac{9MR^2}{2} - \frac{M}{2}\left(\frac{R}{3}\right)^2 = \frac{81MR^2 - MR^2}{18}$
$I = \frac{40}{9}MR^2$ , so the answer is (a).

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