Two particles start moving on circles of radius R1 = 1m and R2 = 3m shown with velocity v1 = 2 m/s, v2 = 15 m/s, respectively. The minimum time after which the particles will again be collinear with centre is equal to

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2π3sec

2π5sec

2πsec

π3sec

answer is D.

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Detailed Solution

ω1=v1R1=2rad/sω2=v2R2=153=5rad/sθ2=θ1+πω2t=ω1t+π⇒t=πω2−ω1=π5−2=π3s

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