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Four particles, each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

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a

GMR

b

22GMR

c

GMR(22+1)

d

GMR22+14

answer is D.

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Detailed Solution

F12=GM22R2F14=GM22R2The resultant of these two forces is2GM2/2R2. Now, F13=GM2/4R2The combined resultant of all the forces is2GM22R2+GM24R2 or GM2R222+14Equating this with centripetal force, we getMv2R=GM2R222+14or v2=GMR22+14 or v=GMR22+14
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