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By **parallel-axis theorem**, the moment of inertia of a thin ring of mass $M$, radius $R$, about a tangent in its plane is:
A$MR^2$
B$\tfrac{1}{2}MR^2$
C$\tfrac{3}{2}MR^2$
D$2MR^2$
Answer & Solution
Correct answer: C. $\tfrac{3}{2}MR^2$
About diameter (in plane), $I_d = MR^2/2$ (perpendicular-axis theorem: $I_z = 2I_d = MR^2$ gives $I_d = MR^2/2$). Tangent in plane: $I = MR^2/2 + MR^2 = 3MR^2/2$.
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