1.

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))`

Answer» Correct Answer - C
`I=(MR^(2))/2` but `2piR=LrArrR=L/(2pi)`
`:. I=(ML^(2))/(8pi^(2))`


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