The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is $I$. It is rotating with angular velocity $\omega$. Another identical ring is gently placed on it so that their centres coincide. If both the rings are rotating about the same axis,then the loss in kinetic energy is:

  • A
    $\frac{I \omega^2}{16}$
  • B
    $\frac{I \omega^2}{8}$
  • C
    $\frac{I \omega^2}{4}$
  • D
    $\frac{I \omega^2}{2}$

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$(4)$ The normal reaction force from the floor on the rod will be $\frac{Mg}{16}$

The graph between $\log_e L$ and $\log_e P$ will be (where $L$ is angular momentum and $P$ is linear momentum):

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The portion $AB$ of the wedge shown in the figure is rough and $BC$ is smooth. $A$ solid cylinder rolls without slipping from $A$ to $B$. The ratio of translational kinetic energy to rotational kinetic energy,when the cylinder reaches point $C$,is:

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