A particle of mass `1kg` is placed at a distance of `4m` from the centre and on the axis of a uniform ring mass `5kg` and radius `3m`. The work done to increase the distance of the particle from `4m` to `3sqrt(3)`m is

Answer

`1.11xx10^(-11)"joule"`

Explanation

`V_(1)=(GM)/((R^(2)+x^(2))^(1//2))=(G(5))/((16+9)^(1//2))=G`
`V_(2)=(GM)/((R^(2)+x^(2))^(1//2))`
`implies(G(5))/((9+27)^(1//2))implies(G5)/(6)`
work done `=m[V_(2)-V_(1)]=(G)/(6)=1.11xx10^(-11)` joule.