Moment of Inertia of Ring JEE Main 2025| Rigid Body Dynamics

JEE MAIN 2025

02-04-2025 SHIFT-2

Question

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 :

Select the correct option:

$\frac{3}{8} M r^{2}$

$\frac{3}{2} M r^{2}$

$2 M r^{2}$

$\frac{1}{2} M r^{2}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Diameter is given as R.
∴ Radius $=R / 2$
$I_{\text {tan gent }}=\frac{3}{2} m\left(\frac{R}{2}\right)^{2}=\frac{3}{8} m R^{2}$

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