Two balls of masses 5m and m have radii 2R and R. Their centre of masses are separated by 12R. They move towards each other under their gravitational force. The distance moved by the centre of smaller sphere when the spheres touch each other is

Given,

effective distance = 9R

let distance traveled by smaller mass be x

$\mathrm{x}=\left(\frac{{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right)9\mathrm{R}$

Now the gravitational force between m and 5m be,

${\mathrm{F}}_{\mathrm{G}}=\frac{\mathrm{G}\times \mathrm{m}\times 5\mathrm{m}}{{(12\mathrm{R}-\mathrm{x})}^2}=\frac{5{\mathrm{Gm}}^2}{{(12\mathrm{R}-\mathrm{x})}^2}$

$$\begin{gathered} \mathrm{acceleration}, \\ {\mathrm{a}}_{\mathrm{m}}=\frac{{\mathrm{F}}_{\mathrm{G}}}{\mathrm{m}}=\frac{5\mathrm{Gm}}{{(12\mathrm{R}-\mathrm{x})}^2} \end{gathered}$$

${\mathrm{a}}_{5\mathrm{m}}=\frac{{\mathrm{F}}_{\mathrm{G}}}{5\mathrm{m}}=\frac{\mathrm{Gm}}{{(12\mathrm{R}-\mathrm{x})}^2}$.

$\mathrm{Distance} \mathrm{travelled} \mathrm{by} \mathrm{mass} \mathrm{m},$

$\mathrm{x}=\frac{1}{2}{\mathrm{a}}_{\mathrm{m}}{\mathrm{t}}^2=\frac{1}{2}\frac{5\mathrm{Gm}}{{(12\mathrm{R}-\mathrm{x})}^2}{\mathrm{t}}^2\mathrm{\_\_\_\_\_\_\_}\left(1\right)$

$\mathrm{Distance} \mathrm{travelled} \mathrm{by} \mathrm{mass} 5\mathrm{m},$

$9\mathrm{R}-\mathrm{x}=\frac{1}{2}{\mathrm{a}}_{5\mathrm{m}}{\mathrm{t}}^2=\frac{1}{2}\frac{\mathrm{Gm}}{{(12\mathrm{R}-\mathrm{x})}^2}\mathrm{\_\_\_\_\_\_}\left(2\right)$

$\mathrm{dividing} 1 \mathrm{by} 2$

$\frac{\mathrm{x}}{9\mathrm{R}-\mathrm{x}}=\frac{\frac{1}{2}\frac{\mathrm{Gm}}{{(12\mathrm{R}-\mathrm{x})}^2}{\mathrm{t}}^2}{\frac{1}{2}\frac{\mathrm{Gm}}{{(12\mathrm{R}-\mathrm{x})}^2}{\mathrm{t}}^2}=5$

$\mathrm{x}=45\mathrm{R}-5\mathrm{x}\Rightarrow \mathrm{x}=7.5\mathrm{R}$

Hence the correct answer is 7.5R.