Three identical particles are joined together by a thread as shown in figure. All the three particles are moving in a horizontal plane. 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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(d) Let $\omega $ is the angular speed of revolution
${T_3} = m{\omega ^2}3l$
${T_2} - {T_3} = m{\omega ^2}2l$ ==> ${T_2} = m{\omega ^2}5l$
${T_1} - {T_2} = m{\omega ^2}l \Rightarrow {T_1} = m{\omega ^2}6l$
${T_3}:{T_2}:{T_1} = 3:5:6$
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