In a system two particles of masses $$m_1 = 3$$ kg and $$m_2 = 2$$ kg are placed at certain distance from each other. The particle of mass $$m_1$$ is moved towards the center of mass of the system through a distance $$2$$ cm. In order to keep the center of mass of the system at the original position, the particle of mass $$m_2$$ should move towards the center of mass by the distance _____ cm.

We need to find how far $$m_2$$ must move to keep the centre of mass at its original position. The position of the centre of mass is $$x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$, and for $$x_{\text{cm}}$$ to remain unchanged the net displacement must satisfy $$m_1 \Delta x_1 + m_2 \Delta x_2 = 0$$.

Here $$m_1 = 3$$ kg moves towards the centre of mass by $$\Delta x_1 = 2$$ cm, so we have $$3 \times 2 = 2 \times d_2$$, giving $$d_2 = \frac{6}{2} = 3 \text{ cm}$$. Both particles move towards the CM, so both displacements reduce the distance to the CM.

The correct answer is 3 cm.