To solve this problem, we need to determine the distance covered by the smaller body before the collision.$\\$Given:$\\$• Mass of the smaller body $= M \\$• Mass of the larger body $= 5M \\$• Radius of the smaller body $= R \\$• Radius of the larger body $= 2R \\$• Initial separation between their centers $= 12R \\$When two bodies attract each other due to gravitational force, they will move towards each other and collide when the distance between their surfaces becomes zero.$\\$The initial separation between the surfaces of the two bodies is:$\\ d_{initial} = 12R - (R + 2R) = 9R \\$Let $s_1$ be the distance covered by the smaller body and $s_2$ be the distance covered by the larger body.$\\$Since the bodies attract each other due to gravitational force, their center of mass remains constant.$\\$The center of mass $x_{cm}$ is given by:$\\ x_{cm} = \frac{M \cdot 0 + 5M \cdot 12R}{M + 5M} = \frac{60MR}{6M} = 10R \\$The center of mass does not move, so the total distance covered by both bodies is $9R. \\$Using the concept of center of mass:$\\ \frac{s_1}{s_2} = \frac{5M}{M} = 5 \\$This implies:$\\ s_1 = 5s_2 \\$Since $s_1 + s_2 = 9R,$ we substitute $s_1 = 5s_2: \\ 5s_2 + s_2 = 9R \\ 6s_2 = 9R \\ s_2 = \frac{9R}{6} = 1.5R \\$Therefore, the distance covered by the smaller body $s_1$ is:$\\ s_1 = 5s_2 = 5 \times 1.5R = 7.5R \\$Thus, the distance covered by the smaller body before collision is $7.5R. \\$This corresponds to Option 2.