Two identical thin ring each of radius `R` are co-axially placed at a distance `R`. If the ring have a uniform mass distribution and each has mass `m_(1)` and `m_(2)` respectively, then the work done in moving a mass `m` from the centre of one ring to that of the other is :

Options

A. zero
B. `(Gm(m_(1)-m_(2))(sqrt(2)-1))/(sqrt(2)R)`
C. `(Gmsqrt(2)(m_(1)+m_(2)))/R`
D. `(Gm_(1)m(sqrt(2)+1))/(m_(2)R)`

Answer

B

Explanation

`V_(1)=(-Gm_(1))/R-(Gm_(2))/(sqrt(2)R)` and `V_(2)=(-Gm_(2))/R-(Gm_(1))/(sqrt(2)R)`
`DeltaV=V_(2)-V_(1)=(-Gm_(2))/R-(Gm_(1))/(sqrt(2)R)+(-Gm_(1))/R-(Gm_(2))/(sqrt(2)R)`
`=R(m_(1)-m_(2))(1/R-1/(sqrt(2)R))`
Hence `W=m(DeltaV)=(mG(m_(1)-m_(2))(sqrt(2)-1))/(sqrt(2)R)`