Three particles of equal mass 'm' are situated at the vertices of an equilateral triangle of side `L`. The work done in increasing the side of the triangle to `2L` is

Options

A. `(2G^(2)m)/(2L)`
B. `(Gm^(2))/(2L)`
C. `(3Gm^(2))/2L`
D. `(3Gm^(2))/(L)`

Answer

C

Explanation

Initial potential energy `U_(i)=(-3Gm^(2))/L`
Final potential energy `U_(f)=-(3Gm^(2))/(2L)`
`W=U_(f)-U_(i)=-(3Gm^(2))/(2L)-(-(3Gm^(2))/L)=(3Gm^(2))/(2L)`