To find the center of mass of the system, we use the formula:$\\ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \\$Given:$\\$• $m_1 = 10$ kg (mass of the first object)$\\$• $m_2 = 20$ kg (mass of the second object)$\\$• Length of the rod $= 10$ m$\\$Assume the 10 kg mass is at position $x_1 = 0$ m (left end of the rod).$\\$Then the 20 kg mass is at position $x_2 = 10$ m (right end of the rod).$\\$Substitute the values into the center of mass formula:$\\ x_{cm} = \frac{10 \times 0 + 20 \times 10}{10 + 20} \\ x_{cm} = \frac{0 + 200}{30} \\ x_{cm} = \frac{200}{30} \\ x_{cm} = \frac{20}{3}$ m$\\$Therefore, the distance of the center of mass from the 10 kg mass is $\frac{20}{3}$ m.$\\$This corresponds to Option 2.