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A thin wire of mass m and length l is bent in the form of a ring . Moment of inertia of that ring about an axis passing through its centre and perpendicular to its through its centre and perpendicular to its plane isA. `ml^(2)//4pi`B. `ml^(2)//2pi`C. `ml^(2)//4pi^(2)`D. `ml^(2)//2pi^(2)` |
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Answer» Correct Answer - C `2pi r = l " " therefore r = (l)/(2 pi)` `I = mr^(2)` `= (ml^(2))/(4pi^(2))`. `I = mr^(2)` `= (ml^(2))/(4pi^(2))`. |
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