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(C) Let the angular velocity be $\Omega = 2\omega$. For particle $B$ of mass $m$ at distance $2r$ from $O$,the tension $T_{AB}$ provides the necessary

Vedclass · Accepted Answer

$2 : 1$

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(C) Let the angular velocity be $\Omega = 2\omega$. For particle $B$ of mass $m$ at distance $2r$ from $O$,the tension $T_{AB}$ provides the necessary centripetal force: $T_{AB} = m \Omega^2 (2r) = m (2\omega)^2 (2r) = m (4\omega^2) (2r) = 8m\omega^2 r$. For particle $A$ of mass $2m$ at distance $r$ from $O$,the net force towards the center is $T_{OA} - T_{AB}$: $T_{OA} - T_{AB} = (2m) \Omega^2 r = (2m) (2\omega)^2 r = (2m) (4\omega^2) r = 8m\omega^2 r$. Substituting $T_{AB} = 8m\omega^2 r$: $T_{OA} = 8m\omega^2 r + 8m\omega^2 r = 16m\omega^2 r$. Therefore,the ratio is: $T_{OA} / T_{AB} = (16m\omega^2 r) / (8m\omega^2 r) = 2 / 1 = 2 : 1$.

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