The moment of inertia of a circular ring of mass $M$ and diameter r about a tangential axis lying in the…

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 :
  1. $\frac{1}{2} \mathrm{Mr}^2$
  2. $\frac{3}{8} \mathrm{Mr}^2$
  3. $\frac{3}{2} \mathrm{Mr}^2$
  4. $2 \mathrm{Mr}^2$

Solution

Diameter is given as R.
$\begin{aligned}
& \therefore \text { Radius }=\mathrm{R} / 2 \\ & \mathrm{I}_{\text {tan gent }}=\frac{3}{2} \mathrm{~m}\left(\frac{\mathrm{R}}{2}\right)^2=\frac{3}{8} \mathrm{mR}^2
\end{aligned}$

Asked in: JEE Main 2025 (02 Apr Shift 2)

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