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JEE Main 2025 (2 Apr, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2025 (2 Apr, Shift 2)

Q45JEE Main 2025 · 2 APR · SHIFT 2Easy

Moment of Inertia & Radius of Gyration

The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is:

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Answer: (a)

Radius $R = r/2$ . About a diameter, ring: $I_d = \frac{1}{2}MR^2$ .
Tangent in the plane ( $d = R$ ): $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$ .
$I = \frac{3}{2}M\frac{r^2}{4} = \frac{3}{8}Mr^2$
So the answer is (a) $\frac{3}{8}Mr^2$ .

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