1.

Two particles of equal mass 'm' go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is :

Answer»

`sqrt((Gm)/(4R))`
`sqrt((Gm)/(3R))`
`sqrt((Gm)/(2R))`
`1/2sqrt((Gm)/(R ))`

Solution :`(Gm^(2))/((2R)^(2))=m omega^(2)R`
`(Gm)/(4R^(3))=omega^(2)`
`omega=sqrt((Gm)/(4R^(3)))`
`v=omega R`
`v= sqrt((Gm)/(4R^(3)))xxR=sqrt((Gm)/(4R))`
So, correct CHOICE is (a).


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