Question

Gravitational potential energy of the mass at the centre of a square as shown in figure is {Potential energy at infinite is taken to be zero}

• A. $- \frac {4Gm^2}{l\sqrt{3}}$
• B. $- \frac {2\sqrt{2} Gm^2}{l}$
• C. $- \frac {-4 Gm^2}{l}$
• D. $- \frac {4\sqrt{2} Gm^2}{l}$

Answer 1

Answer: (D)

U = $- \Delta\,V × m$
= $(V_o-V_\infty)m$
$ = \frac{-Gm\times4\times m}{\frac{l}{\sqrt{2}}}=-\frac{4\sqrt{2Gmm}}{l}=-\frac{4\sqrt{2}Gm^{2}}{l}$