InterviewSolution
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InterviewSolution
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if `l_(1)` is te moment of inertia of a thin rod about an axis perpendicular to its length and passing thorugh its centre of mass and `l_(2)` te moment of inertia of the ring formed by the same rod about an axis passing through the centre of mass of the ring and perpendicular tot he plane of the ring. then find the ratio `(l_(1))/(l_(2))`.A. `(pi^(2))/(3)`B. `(3)/(pi^(2))`C. `(2)/(pi^(2))`D. `(pi^(2))/(2)` |
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Answer» Correct Answer - A `I_(1) = (Ml^(2))/(12)` If `r` is radius of ring, then `l = 2 pi r, r = l//2 pi` `l_(2) = M r^(2) = M(l//2pi)^(2) = (Ml^(2))/(4pi^(2))` `(I_(1))/(I_(2)) = (Ml^(2))/(12).(4pi^(2))/(Ml^(2)) = (pi^(2))/(3)` |
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