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A point P lies on the axis of a fixed ri...

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.

Text Solution

Verified by Experts

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