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A ring of mass $M = 10 k g$ and radius $R = 2 m$ is taken as shown in the figure. Calculate the moment of inertia of ring about the diametric axis

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$25 k g m^{2}$

$30 k g m^{2}$

$40 k g m^{2}$

$20 k g m^{2}$

answer is D.

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Detailed Solution

Using perpendicular axis theorem

$\begin{array}{l} I_{0} = I_{A A^{1}} + I_{B B^{1}} a n d I_{A A^{1}} = I_{B B^{1}} \\ 2 I_{A A^{1}} = M R^{2} \\ ∴ I_{A A^{1}} = \frac{M R^{2}}{2} = \frac{10 \times 2^{2}}{2} = 20 k g m^{2} \\ ∴ I_{A A^{1}} = 20 k g m^{2} \end{array}$

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