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(D) Let the handle be a rod of length $L = 6r$ and the ring be a circular hoop of radius $R = r$. The mass of both is $M$.$1$. Moment of inertia of th

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$52$

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(D) Let the handle be a rod of length $L = 6r$ and the ring be a circular hoop of radius $R = r$. The mass of both is $M$.$1$. Moment of inertia of the handle about an axis through its center of mass (perpendicular to it) is $I_{cm, rod} = \frac{M(6r)^2}{12} = 3Mr^2$.The distance of this axis from the rotation axis is $d_1 = (6r/2) - r/2 = 5r/2$.Using the parallel axis theorem,$I_{rod} = I_{cm, rod} + M(d_1)^2 = 3Mr^2 + M(5r/2)^2 = 3Mr^2 + 6.25Mr^2 = 9.25Mr^2$.$2$. Moment of inertia of the ring about its diameter is $I_{cm, ring} = \frac{MR^2}{2} = \frac{Mr^2}{2} = 0.5Mr^2$.The distance of the center of the ring from the rotation axis is $d_2 = 6r - r/2 + r = 6.5r = 13r/2$.Using the parallel axis theorem,$I_{ring} = I_{cm, ring} + M(d_2)^2 = 0.5Mr^2 + M(13r/2)^2 = 0.5Mr^2 + 42.25Mr^2 = 42.75Mr^2$.Total moment of inertia $I = I_{rod} + I_{ring} = 9.25Mr^2 + 42.75Mr^2 = 52Mr^2$.

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