Question

The moment of inertia of semicircular ring about an axis which is perpendicular to the plane of the ring and passes through the centre

• A. $M{R}^2$
• B. $\frac{M{R}^2}{2}$
• C. $\frac{M{R}^2}{4}$
• D. None of these

Answer 1

The correct option is A $M{R}^2$

$\begin{array}{c}\text{Let us consider a circular Ring of same Radius(}R\text{)}\\ \text{and mass (2M)}\\ \text{There fore the moment of Inertia of the Ring will be-}\\ I=\left(2M\right)R^2=2M{R}^2\end{array}$

$\begin{array}{c}\text{Therefore tha half portion of the Ring will be a}\\ \text{semicircular ring of Radius}\left(R\right)\text{and mass}\left(M\right)-\\ \Rightarrow \text{Moment of Inertia of the semicircular Ring will}\\ \text{also be halved}\\ I^{\prime }=\frac{I}{2}=\frac{2M{R}^2}{2}=M{R}^2\end{array}$