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(D) Let $\omega$ be the angular speed of revolution of the particles.The distances of the particles $A, B,$ and $C$ from the center $O$ are $l, 2l,$ a

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$3:5:6$

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(D) Let $\omega$ be the angular speed of revolution of the particles.The distances of the particles $A, B,$ and $C$ from the center $O$ are $l, 2l,$ and $3l$ respectively.The centripetal force required for each particle is provided by the tension in the string.For particle $C$ (outermost): The tension $T_3$ provides the centripetal force: $T_3 = m\omega^2(3l) = 3m\omega^2l$.For particle $B$: The net centripetal force is $T_2 - T_3 = m\omega^2(2l)$. Substituting $T_3$,we get $T_2 = 2m\omega^2l + 3m\omega^2l = 5m\omega^2l$.For particle $A$: The net centripetal force is $T_1 - T_2 = m\omega^2(l)$. Substituting $T_2$,we get $T_1 = m\omega^2l + 5m\omega^2l = 6m\omega^2l$.The ratio of tensions $T_3 : T_2 : T_1$ is $3m\omega^2l : 5m\omega^2l : 6m\omega^2l = 3:5:6$.

Three identical particles of mass $m$ are joined together by a thread as shown in the figure. All three particles are moving in a horizontal plane in a circle about the point $O$. If the velocity of the outermost particle is $v_0$,then the ratio of tensions in the three sections of the string is:

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