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A thin rod of mass `M` and length `L` is bent into a circular ring. The expression for moment of inertia of ring about an axis passing through its diameter isA. `(ML^(2))/(2pi^(2))`B. `(ML^(2))/(4 pi^(2))`C. `(ML^(2))/(8pi^(2))`D. `(ML^(2))/(pi^(2))` |
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Answer» Correct Answer - C `I=(MR^(2))/2` but `2piR=LrArrR=L/(2pi)` `:. I=(ML^(2))/(8pi^(2))` |
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