Three particles each of mass m are placed at the three corners of an equilateral triangle of side a. The work done on the system to increase the sides of the triangle to 2a is:

Options

A. `(3Gm^2)/(a)`
B. `(3Gm^2)/(2a)`
C. `(2Gm^2)/(3a)`
D. `(Gm^2)/(2a)`

Answer

B

Explanation

The initial potential energy of the system is,
`U_(i)=-(Gmm)/(a)-(Gmm)/(a)-(Gmm)/(a)=-(3Gm^2)/(a)`
The final potential energy of the system is,
`U_(f)=-(Gmm)/(2a)-(Gmm)/(2a)-(Gmm)/(2a)=-(3Gm^2)/(2a)`
`therefore` Work done = `U_(f)-U_(i)`
`=-(3Gm^2)/(2a)-((-3Gm^2)/(a))=(3Gm^2)/(2a)`