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Two identical rings $A$ and $B$ are acted upon by torques $ζ_{A}$ and $ζ_{B}$ respectively. $A$ is rotating about an axis passing through the centre of mass and perpendicular to the plane of the ring. $B$ is rotating about a chord at a distance $\frac{1}{\sqrt{2}}$ times the radius from the centre of the ring. if the angular acceleration of the rings is the same, then

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$ζ_{A} = ζ_{B}$

$ζ_{A} > ζ_{B}$

$ζ_{A} < ζ_{B}$

Nothing can be said about $ζ_{A}$ and $ζ_{B}$ as data are insufficient

answer is A.

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Detailed Solution

$I_{B}'$ ( in new given condition )

$= \frac{1}{2} M R^{2} + M {(\frac{R}{\sqrt{2}})}^{2} = M R^{2} = I_{A}$

thus $ζ_{A} = ζ_{B} = ζ_{B}' = I α$

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