Question

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 $2l$ will be

• A. $\frac{3G_{2}m}{2l}$
• B. $\frac{G{m}^2}{2l}$
• C. $\frac{3G{m}^2}{2l}$
• D. $\frac{3G{m}^2}{l}$

Answer 1

The correct option is C $\frac{3G{m}^2}{2l}$

Initial Potential energy $P_i=\frac{-3Gmm}{l}$
Final Potential energy $P_f=\frac{-3Gmm}{2l}$
Work done $=$ change in potential energy $=P_f-P_i=\frac{-3Gmm}{2l}-\frac{-3Gmm}{l}=\frac{3G{m}^2}{2l}$