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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2025
  • Easy
  • MCQ

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 & Solution

Correct Answer

(B) \(\frac{3}{8} \mathrm{Mr}^2\)

Step-by-step Solution

Detailed explanation

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