Question

If three particles each of mass $m$ are placed at the corners of an equilateral triangle of side $a$, the potential energy of the system and the work done if the side of the triangle is changed from $a$ to $2a$, are

• A. $\frac{3GM}{a^2},\frac{3GM}{2a}$
• B. $-\frac{3G{M}^2}{a},\frac{3G{M}^2}{2a}$
• C. $-\frac{3G{M}^2}{a^2},\frac{3G{M}^2}{4a^2}$
• D. $-\frac{3G{M}^2}{a},\frac{3GM}{2a}$

Answer 1

The correct option is B $-\frac{3G{M}^2}{a},\frac{3G{M}^2}{2a}$

Potential energy of the system will be

$U=3(-\frac{G{M}^2}{a})$

In the latter case, the potential energy will be

$U^{\prime }=3(-\frac{G{m}^2}{2a})=-\frac{3G{m}^2}{2a}$

Thus, work done is

$W=-\Delta U=\frac{G{M}^2}{a}(3-\frac{3}{2})=\frac{3G{M}^2}{2a}$

Answer is option B.