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(B) The moment of inertia of a uniform ring about an axis passing through its center and perpendicular to its plane is $I_{cm} = Mr^2$.According to th

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$\frac{3}{2}Mr^2$

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(B) The moment of inertia of a uniform ring about an axis passing through its center and perpendicular to its plane is $I_{cm} = Mr^2$.According to the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance between the parallel axes.For a tangent in the plane of the ring,the distance $d$ between the center of mass axis and the tangent axis is equal to the radius $r$.Substituting the values,we get $I = Mr^2 + M(r)^2 = Mr^2 + Mr^2 = 2Mr^2$.Wait,the question asks for a tangent in its own plane. The moment of inertia about a diameter is $I_{diameter} = \frac{1}{2}Mr^2$.Using the parallel axis theorem for the tangent in the plane: $I = I_{diameter} + Md^2 = \frac{1}{2}Mr^2 + Mr^2 = \frac{3}{2}Mr^2$.

The moment of inertia of a uniform ring of mass $M$ and radius $r$ about a tangent in its own plane is:

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