Question

Four particles each of mass are placed at the vertices of a square of side The potential energy of the system is

• A. $- \frac { \sqrt { 2 } \mathrm { Gm } ^ { 2 } } { 1 } \left( 2 - \frac { 1 } { \sqrt { 2 } } \right)$
• B. $- \frac { 2 \mathrm { Gm } ^ { 2 } } { 1 } \left( 2 + \frac { 1 } { \sqrt { 2 } } \right)$
• C. $- \frac { \sqrt { 2 } \mathrm { Gm } ^ { 2 } } { 1 } \left( \sqrt { 2 } + \frac { 1 } { 2 } \right)$
• D. $- \frac { 2 \mathrm { Gm } ^ { 2 } } { 1 } \left( \sqrt { 2 } - \frac { 1 } { \sqrt { 2 } } \right)$

Answer 1

Answer: (B)

$- \frac { 2 \mathrm { Gm } ^ { 2 } } { 1 } \left( 2 + \frac { 1 } { \sqrt { 2 } } \right)$

Answer 2

From figure

AB = BC = CD = AD = l

$\therefore$AC = BD

Total potential energy of the system of four particles each of mass m placed at the vertices A,B,C and D of a square is

$= - \frac { 4 \mathrm { Gm } ^ { 2 } } { 1 } - \frac { 2 \mathrm { Gm } ^ { 2 } } { 1 \sqrt { 2 } } = - \frac { 2 \mathrm { Gm } ^ { 2 } } { 1 } \left( 2 + \frac { 1 } { \sqrt { 2 } } \right)$