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(B) Let the angular positions of $A, B$ and $C$ at $t=0$ be $	heta_A(0) = 0$,$	heta_B(0) = \pi/2$,and $	heta_C(0) = \pi$ respectively.The angular v

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$5 : 4$

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(B) Let the angular positions of $A, B$ and $C$ at $t=0$ be $	heta_A(0) = 0$,$	heta_B(0) = \pi/2$,and $	heta_C(0) = \pi$ respectively.The angular velocities are $\omega_A = v_A/r = \pi \, rad/s$,$\omega_B = v_B/r = 2.5\pi \, rad/s$,and $\omega_C = v_C/r = 2\pi \, rad/s$.At time $t$,the angular positions are $	heta_A(t) = \pi t$,$	heta_B(t) = \pi/2 + 2.5\pi t$,and $	heta_C(t) = \pi + 2\pi t$.For the particles to meet,their angular positions must be equal modulo $2\pi$. Let $	heta_A(t) = 	heta_B(t) + 2n\pi$ and $	heta_B(t) = 	heta_C(t) + 2m\pi$.$\pi t = \pi/2 + 2.5\pi t + 2n\pi \implies -1.5t = 0.5 + 2n \implies t = -(2n + 0.5)/1.5$.$\pi/2 + 2.5\pi t = \pi + 2\pi t + 2m\pi \implies 0.5t = 0.5 + 2m \implies t = 1 + 4m$.Equating $t$: $1 + 4m = -(2n + 0.5)/1.5 \implies 1.5 + 6m = -2n - 0.5 \implies 2n + 6m = -2 \implies n + 3m = -1$.For the first meeting,let $m = -1$,then $n = 2$. Thus $t = 1 + 4(-1) = -3$ (taking magnitude $t=3s$).Distance $s_B = v_B t = 2.5 	imes 3 = 7.5 \, m$.Distance $s_C = v_C t = 2 	imes 3 = 6 \, m$.Ratio $s_B : s_C = 7.5 : 6 = 5 : 4$.

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