A) Ö (GM/2L)
B) Ö (GM/L)
C) Ö (2GM/L)
D) Ö (GM/3L)
Correct Answer: B
Solution :
The resultant gravitational force on each particle provides the necessary centripetal force. \[\therefore \]\[m{{v}^{2}}/r=\]Ö\[({{F}^{2}}+{{F}^{2}}+2{{F}^{2}}\cos {{60}^{o}})=\]Ö3F But (Ö3L/2)x (2/3)=1/Ö3 \[\therefore \]\[m{{v}^{2}}\]Ö3/L=Ö3GM2/L2\[\Rightarrow v=\]Ö(GM/L)You need to login to perform this action.
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