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

Answer

B

Explanation

`V_(B)=` Potential at B due to A + Potential at A due to B
`V_(B)=(-Gm_(2))/(R)-(-Gm_(1))/(sqrt(2)R)rArrV_(A)=(-Gm_(1))/(R)-(-Gm_(2))/(sqrt(2)R)`
`W_(ArarrB)=m(V_(B)-V_(A))=(Gm(m_(1)-m_(2)))/(sqrt(2)R)(sqrt(2)-1)`.