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JEE Main 2017 (9 Apr, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2017 (9 Apr, Shift 1)

Q172JEE Main 2017 · 9 APR · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

A circular hole of radius $\frac{R}{4}$ is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is:

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Answer: (b)

Hole mass $= \frac{M}{16}$ , centre at $\frac{3R}{4}$ .
$I_{hole} = \frac{1}{2}\frac{M}{16}\frac{R^2}{16} + \frac{M}{16}\frac{9R^2}{16} = \frac{19MR^2}{512}$
$I = \frac{256MR^2}{512} - \frac{19MR^2}{512} = \frac{237MR^2}{512}$
So the answer is (b).

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