A point `P` lies on the axis of a fixed ring of mass `M` and radius `a`, at a distance `a` from its centre `C`. A small particle starts from `P` and reaches `C` under gravitational attraction only. Its speed at `C` will be.

Answer

`sqrt((2GM)/(a)(1-(1)/(sqrt(2))))`

Explanation

Potential at the axis of ring `V_(p) = -(GM)/(sqrt(R^(2)+x^(2)))`
`V_(p)=-(GM)/(sqrt(a^(2)+a^(2))) rArr V_(p) = -(GM)/(sqrt(2)a) rArr V_(0) = -(GM)/(a)`
From energy conservation Loss of potential of particle = Gain of kinetic energy.
`U_(i)-U_(f)=K_(f)-K_(i)`
`-(GMm)/(sqrt(2)a)-(-(GMm)/(a))-(1)/(2)mv^(2) rArr (GM)/(a) (1-(1)/(sqrt(2)))=(v^(2))/(2)`
`v^(2) = (2GM)/(a) (1-(1)/(sqrt(2))) rArr v= sqrt((2GM)/(a)(1-(1)/(sqrt(2))))`