JEE Main 2025 — Rotational Motion Question with Solution
JEE Main 2025 (2 Apr Shift 2)
Question
The moment of inertia of a circular ring of mass and diameter r about a tangential axis lying in the plane of the ring is :
Choose an option
Show full solutionCorrect option: B
Correct answer
B
Step-by-step explanation
Diameter is given as R.
$\begin{aligned}
& \therefore \text { Radius }=\mathrm{R} / 2 \\ & \mathrm{I}_{\text {tan gent }}=\frac{3}{2} \mathrm{~m}\left(\frac{\mathrm{R}}{2}\right)^2=\frac{3}{8} \mathrm{mR}^2
\end{aligned}$
$\begin{aligned}
& \therefore \text { Radius }=\mathrm{R} / 2 \\ & \mathrm{I}_{\text {tan gent }}=\frac{3}{2} \mathrm{~m}\left(\frac{\mathrm{R}}{2}\right)^2=\frac{3}{8} \mathrm{mR}^2
\end{aligned}$
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This is a previous-year question from JEE Main 2025, covering the Rotational Motion chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.