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(C) Let the mass of each particle be $m$ and the angular velocity be $\omega$. The distance of particles $C$,$B$,and $A$ from the center $O$ are $3l$,

Vedclass · Accepted Answer

$6 : 5 : 3$

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(C) Let the mass of each particle be $m$ and the angular velocity be $\omega$. The distance of particles $C$,$B$,and $A$ from the center $O$ are $3l$,$2l$,and $l$ respectively.For particle $C$: The tension $T_3$ provides the necessary centripetal force. $T_3 = m\omega^2(3l) = 3m\omega^2l$.For particle $B$: The net force is $T_2 - T_3 = m\omega^2(2l)$. Substituting $T_3$,we get $T_2 = m\omega^2(2l) + 3m\omega^2l = 5m\omega^2l$.For particle $A$: The net force is $T_1 - T_2 = m\omega^2(l)$. Substituting $T_2$,we get $T_1 = m\omega^2l + 5m\omega^2l = 6m\omega^2l$.Thus,the ratio of tensions $T_1 : T_2 : T_3 = 6m\omega^2l : 5m\omega^2l : 3m\omega^2l = 6 : 5 : 3$.

Three particles of equal mass are tied to a string and rotated in a horizontal plane as shown. What is the ratio of tensions in the three parts of the string?

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