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Q.

Three identical particles each of mass ‘m’ are arranged at the corner of an equilateral triangle of side “a”. If they are to be in equilibrium, the speed with which they must revolve under the influence  of one another’s gravity in a circular orbit circumscribing the triangle is

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a

3GMa

b

GMa

c

GM3a

d

3GMa2

answer is B.

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Detailed Solution

When F1=F2 then Fnet=2Fcosθ2=2Fcos602=2Fcos30=2F32=F3=3Gmma2 Fnet provides necessary centripetal force ⇒Fnet=mv2r ⇒3Gmma2=mv2a3 ∵r=a3 ⇒3Gma=3v2 ⇒v=Gma
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Three identical particles each of mass ‘m’ are arranged at the corner of an equilateral triangle of side “a”. If they are to be in equilibrium, the speed with which they must revolve under the influence  of one another’s gravity in a circular orbit circumscribing the triangle is