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JEE Main 2023 (1 Feb, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2023 (1 Feb, Shift 2)

Q90JEE Main 2023 · 1 FEB · SHIFT 2Easy

Moment of Inertia & Radius of Gyration

Moment of inertia of a disc of mass ' $M$ ' and radius ' $R$ ' about any of its diameter is $\frac{MR^2}{4}$ . The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, $\frac{x}{2}$ MR $2$ . The value of $x$ is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Perpendicular axis theorem: $I_{cm} = 2\times\frac{MR^2}{4} = \frac{MR^2}{2}$ .
Parallel axis to the edge: $I = \frac{MR^2}{2} + MR^2 = \frac{3}{2}MR^2$ .
Comparing with $\frac{x}{2}MR^2$ : $x = 3$ .

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