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Q.

Three point masses each of mass ‘m’ rotate in a circle of radius r with constant angular velocity  ω due to their mutual gravitational attraction. If at any instant, the masses are on the vertex of an equilateral triangle of side ‘a’, then the value of ω  is

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a

Gma3

b

3Gma3

c

Gm3a3

d

Zero

answer is B.

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Detailed Solution

F=F12+F22+2F1F2cosθ ⇒If F1=F2 then Fnet=2Fcosθ2θ=60∴ Fnet=2Fcos602=2Fcos30=2Gmma232=3Gmma2Fnet provides centripetal force3GMma2=mrω23GMa2=a3ω2ω2=3GMa3ω=3GMa3
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Three point masses each of mass ‘m’ rotate in a circle of radius r with constant angular velocity  ω due to their mutual gravitational attraction. If at any instant, the masses are on the vertex of an equilateral triangle of side ‘a’, then the value of ω  is