A uniform circular disc of radius ' R ' and mass ' M ' is rotating… | JEE Main 2025 (22 Jan, Shift 1)

Q61JEE Main 2025 · 22 JAN · SHIFT 1Difficult

Moment of Inertia & Radius of Gyration

A uniform circular disc of radius ' $R$ ' and mass ' $M$ ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius $R/2$ is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.

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Answer: (b)

Removed disc: radius $\frac{R}{2}$ , mass $\frac{M}{4}$ , centre at $\frac{R}{2}$ from the axis.
$I_{hole} = \frac{1}{2}\frac{M}{4}\frac{R^2}{4} + \frac{M}{4}\frac{R^2}{4} = \frac{3MR^2}{32}$
$I = \frac{MR^2}{2} - \frac{3MR^2}{32} = \frac{13}{32}MR^2$
So the answer is (b).

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