JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 :
- A \(\frac{1}{2} \mathrm{Mr}^2\)
- B \(\frac{3}{8} \mathrm{Mr}^2\)
- C \(\frac{3}{2} \mathrm{Mr}^2\)
- D \(2 \mathrm{Mr}^2\)
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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