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)-M_(1)))/(a)`
B. `(Gm(M_(2)-M_(1)))/(a sqrt(2)) (sqrt(2) + 1)`
C. `(Gm(M_(2)-M_(1)))/(a sqrt(2)) (sqrt(2) - 1)`
D. `(Gm(M_(2)-M_(1)))/( sqrt(2)) a`

Answer

C

Explanation

w = m `(V_(2) - V_(1))`
When, ` " "V_(1) = - [ (GM_(1))/(a) + (GM_(2))/(sqrt(2a)) ] ` ,
` V_(2) = - [ (GM_(2))/(a) + (GM_(1))/(sqrt(2a)) ] `
`therefore " "` W = `(Gm(M_(2) - M))/(a sqrt(2)) (sqrt(2) - 1)`.