Question 22 of 280
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 :
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Answer: (B)
Diameter is given as R.
$\begin{aligned}
& \therefore \text { Radius }=\mathrm{R} / 2 \\
& \mathrm{I}_{\text {tan gent }}=\frac{3}{2} \mathrm{~m}\left(\frac{\mathrm{R}}{2}\right)^2=\frac{3}{8} \mathrm{mR}^2
\end{aligned}$
$\begin{aligned}
& \therefore \text { Radius }=\mathrm{R} / 2 \\
& \mathrm{I}_{\text {tan gent }}=\frac{3}{2} \mathrm{~m}\left(\frac{\mathrm{R}}{2}\right)^2=\frac{3}{8} \mathrm{mR}^2
\end{aligned}$