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A ring of radius `R` is first rotated with an angular velocity `omega` and then carefully placed on a rough horizontal surface. The coefficient of friction between the surface and the ring is `mu`. Time after which its angular speed is reduced to half isA. `(omega_(0)muR)/(2g)`B. `(2omega_(0)R)/(mug)`C. `(omega_(0)R)/(2mug)`D. `(omega_(0)g)/(2mug)` |
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Answer» Correct Answer - 3 `alpha= (tau)/(I) = (mug)/(R)` `omega_(t) =(omega_(0))/(2) = omega_(0) - alphat` So`t=(omega_(0)R)/(2 mug)` |
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