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JEE Advanced2011PhysicsOscillationsActual

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

Options

  1. ARestoring torque in case A= Restoring torque in case B
  2. BRestoring torque in case A < Restoring torque in case B
  3. CAngular frequency for case A> Angular frequency for case B
  4. DAngular frequency for case A < Angular frequency for case B

Correct answer

A. Restoring torque in case A= Restoring torque in case B

Step-by-step solution

_A= _B=m g L 2 + MgL = Restoring torque about point O . In case A , moment of inertia will be more. Hence, angular acceleration ( = / I) will be less. Therefore, angular frequency will be less. Correct options are (a) and (d). Analysis of Question Question is difficult from my point of view. Because this type of SHM is rarely taught in the class and questions of this type are not given in standard books.

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