Question

The M.I. of a circular ring with mass m and radius R about an axis passing through its center and per-pendicular to its plane is

• A. $\frac{M{R}^2}{4}$
• B. $M{R}^2$
• C. $\frac{M{R}^2}{2}$
• D. $\frac{3}{4}M{R}^2$

Answer 1

The correct option is C $\frac{M{R}^2}{2}$

$\begin{array}{c}\text{Let us take a small element mass}\left(dm\right)\\ \Rightarrow \text{moment of inertia of this element is}\\ dI=dm\cdot R^2\\ \Rightarrow \text{Moment of Inertia of whole Ring is}\end{array}$

$\begin{array}{cc}{\int }_0^{I}dI& ={\int }_0^{M}dm\cdot R^2=M{R}^2\\ I& =M{R}^2\\ \text{Hence option (B) is correct.}& \end{array}$