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

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

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

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

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\frac{2 π}{5} \sec$

$\frac{2 π}{3} \sec$

$\frac{π}{3} \sec$

$2 πsec$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} ω_{1} = \frac{v_{1}}{R_{1}} = 2 rad / s \\ ω_{2} = \frac{v_{2}}{R_{2}} = \frac{15}{3} = 5 rad / s \\ θ_{2} = θ_{1} + π \\ ω_{2} t = ω_{1} t + π \Rightarrow t = \frac{π}{ω_{2} - ω_{1}} = \frac{π}{5 - 2} = \frac{π}{3} s \end{array}$