To find the center of mass of the system, we use the formula for the center of mass of a system of particles:$\\ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \\$where:$\\$• $m_1 = 5$ kg (mass of the first particle)$\\$• $m_2 = 10$ kg (mass of the second particle)$\\$• $x_1 = 0$ m (position of the first particle)$\\$• $x_2 = 1$ m (position of the second particle, since the rod is 1 m long)$\\$Substitute these values into the formula:$\\ x_{cm} = \frac{5 \cdot 0 + 10 \cdot 1}{5 + 10} \\ x_{cm} = \frac{0 + 10}{15} \\ x_{cm} = \frac{10}{15} \\ x_{cm} = \frac{2}{3}$ m$\\$Convert $x_{cm}$ from meters to centimeters:$\\ x_{cm} = \frac{2}{3} \times 100$ cm$\\ x_{cm} \approx 66.67$ cm$\\$Therefore, the center of mass of the system from the 5 kg particle is nearly at a distance of $67$ cm.$\\$This corresponds to Option 2.