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The moment of inertia of ring about an axis passing through its diameter is `I`. Then moment of inertia of that ring about an axis passing through its centre and perpendicular to its plane isA. `(I)/(2)`B. `2I`C. `(I)/(4)`D. `4I` |
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Answer» Correct Answer - B For a ring, `I_(z) =MR^(2)` From perpendicular axis theorem, `I_(x)+I_(y)=I_(z)` Given, `I_(x)=I_(y)=I` From Eq. (i), we get `2I-I_(z)` . |
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