Three particles of mass M each are lying on the comers of an equilateral triangle of side l . If the length of each side of the triangle is increased to twice, the work done is given by

Options

A. `(-3 GM^(2))/(2l)`
B. `(3 DM^(2))/(2l)`
C. `(GM^(2))/(l)`
D. `(-1.5 GM)/(l)`

Answer

B

Explanation

Using ` U = - (Gm_(1) m_(2)) /(r) ` we get
`U_(1) = 3 (- (GM ^(2))/( l)) = - (3GM ^(2))/(l)`
`U_(2) = 3 (- (GM ^(2))/( 2l)) = - (3)/(2) (GM^(2))/( l)`
`:.` Work done ` = Delta ` in P.E. ` = U_(2) = U_(1) =- (3)/(2) (GM^(2))/( l) + (3 GM^(2))/( l) = (3)/(2) (GM^(2))/(l)`