Three rings each of mass M and radius R are a | vedclass

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(C) The total moment of inertia of the system about the axis $YY'$ is $I = I_1 + I_2 + I_3$.For ring $1$,the axis $YY'$ passes through its diameter. T

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$\frac{7}{2}MR^2$

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(C) The total moment of inertia of the system about the axis $YY'$ is $I = I_1 + I_2 + I_3$.For ring $1$,the axis $YY'$ passes through its diameter. Thus,the moment of inertia is $I_1 = \frac{1}{2}MR^2$.For rings $2$ and $3$,the axis $YY'$ is parallel to their diameters at a distance $R$. Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $I_{cm} = \frac{1}{2}MR^2$ and $d = R$.So,$I_2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.Similarly,$I_3 = \frac{3}{2}MR^2$.Therefore,the total moment of inertia is $I = \frac{1}{2}MR^2 + \frac{3}{2}MR^2 + \frac{3}{2}MR^2 = \frac{7}{2}MR^2$.

Three rings each of mass $M$ and radius $R$ are arranged as shown in the figure. The moment of inertia of the system about the axis $YY'$ is:

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