Two half rings are joined as shown in the fig | vedclass

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(A) The moment of inertia of a complete ring of mass $2M$ and radius $R$ about its diameter is $I = \frac{1}{2}(2M)R^2 = MR^2$.Since the system consis

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$MR^2$

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(A) The moment of inertia of a complete ring of mass $2M$ and radius $R$ about its diameter is $I = \frac{1}{2}(2M)R^2 = MR^2$.Since the system consists of two half-rings of mass $M$ each,joined such that the axis $XX'$ passes through their common diameter,we can treat the system as a single ring of mass $2M$ and radius $R$ rotating about its diameter.The moment of inertia of a thin ring of mass $m$ and radius $R$ about its diameter is $\frac{1}{2}mR^2$.For the first half-ring of mass $M$,the moment of inertia about the axis $XX'$ is $I_1 = \frac{1}{2}MR^2$.For the second half-ring of mass $M$,the moment of inertia about the axis $XX'$ is $I_2 = \frac{1}{2}MR^2$.Using the principle of superposition,the total moment of inertia of the system is $I = I_1 + I_2 = \frac{1}{2}MR^2 + \frac{1}{2}MR^2 = MR^2$.

List-$I$	List-$II$
$(a)$ $MI$ of the rod (length $L$,mass $M$,about an axis $\perp$ to the rod passing through the midpoint)	$(i) \frac{8ML^2}{3}$
$(b)$ $MI$ of the rod (length $L$,mass $2M$,about an axis $\perp$ to the rod passing through one of its ends)	$(ii) \frac{ML^2}{3}$
$(c)$ $MI$ of the rod (length $2L$,mass $M$,about an axis $\perp$ to the rod passing through its midpoint)	$(iii) \frac{ML^2}{12}$
$(d)$ $MI$ of the rod (length $2L$,mass $2M$,about an axis $\perp$ to the rod passing through one of its ends)	$(iv) \frac{2ML^2}{3}$

Two half rings are joined as shown in the figure. Each half ring has a radius $R$ and mass $M$. The moment of inertia of the system about the axis $XX'$ is

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