Consider two configurations of a system of three particles of masses $m, 2 m$ and $3 m$ . The work done by gravity in changing the configuration of the system from figure (i) to figure (ii) is

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$\frac{6 G m^{2}}{a} \{1 + \frac{1}{\sqrt{2}}\}$

$\frac{6 G m^{2}}{a} \{1 - \frac{1}{\sqrt{2}}\}$

$\frac{6 G m^{2}}{a} \{2 - \frac{1}{\sqrt{2}}\}$

answer is C.

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Detailed Solution

$W = - Δ U = U_{i} - U_{f}$

$= - \frac{G m m}{a} [\{\frac{(1) (2)}{1} + \frac{(1) (3)}{1} + \frac{(2) (3)}{\sqrt{2}}\}]$ $- \{\frac{(1 \times 2)}{1} + \frac{(1 \times 3)}{1} + \frac{(2 \times 3)}{1}\}$

$= \frac{6 G m^{2}}{a} (1 - \frac{1}{\sqrt{2}})$

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