Explanation

(c) Let at O there will be a collision. If smaller sphere moves $x$ distance to reach at O, then bigger sphere will move a distance of $(9R - x)$
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$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{small} = \frac{F}{M} = \frac{G \times 5M}{(12R - x)^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{big} = \frac{F}{5M} = \frac{GM}{(12R - x)^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \frac{1}{2}a_{small}t^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\frac{G \times 5M}{(12R - x)^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
$\ \ \ \ \ \ \ (9R - x) = \frac{1}{2}a_{big}t^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\frac{GM}{{(12R - x)}^{2}}t^{2}\ \ \ \ \ \ \ \ \ \ldots(ii)$
Thus, dividing Eq. (i) by Eq. (ii), we get
$\therefore\ \ \ \ \ \ \ \ \frac{x}{9R - x} = 5$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 45R - 5x$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6x = 45R$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 7.5\ R$