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A metal rod of length 'L' and maws 'm' is pivoted at one end A thin disc of mass 'M' and radius 'R' (

Answer» <html><body><p> Restoring torque in case A = Restoring torque in case B <br/> Restorings torque in case A lt Restoring torque in case B <br/> Angular frequency for case A gt Angular frequency for case B. <br/> Angular frquency for case A lt Angular frequency for case B. </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :The fact that disc is free to rotate or not does not affect the forces <a href="https://interviewquestions.tuteehub.com/tag/acting-269" style="font-weight:bold;" target="_blank" title="Click to know more about ACTING">ACTING</a> on the system and hence restoring torque remains same in both the casesso option (a) is correct. <br/> In case A disc is not free to rotate about itscentre hence it has to change its orientation during the oscillation and its moment of inertia as a disc will be counted. But in second case disc will oscillate without changing its orientation andit will be just like a <a href="https://interviewquestions.tuteehub.com/tag/point-1157106" style="font-weight:bold;" target="_blank" title="Click to know more about POINT">POINT</a> mass attached at theend of rod. We can understand that moment of inertia in case A will be more than in case B. Frequency is <a href="https://interviewquestions.tuteehub.com/tag/inversely-7688775" style="font-weight:bold;" target="_blank" title="Click to know more about INVERSELY">INVERSELY</a> proportional to square root of moment of inertia, hence we can understand that option (d) is <a href="https://interviewquestions.tuteehub.com/tag/also-373387" style="font-weight:bold;" target="_blank" title="Click to know more about ALSO">ALSO</a> correct.</body></html>


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