In a system two particles of masses m₁ = 3kg and m₂ = 2kg are placed at certain distance from each other. The particle of mass m₁ is moved towards the center of mass of the system through a distance 2cm. In order to keep the center of mass of the system at the original position, the particle of mass m2 should move towards the center of mass by the distance ____ cm.

To find the distance the second particle must move to keep the center of mass of the system stationary, we can analyze the position of the center of mass of a two-particle system.

The position of the center of mass ($X_{\mathrm{cm}}$) is given by the formula:
$X_{\mathrm{cm}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$
where $m_1$ and $m_2$ are the masses of the two particles, and $x_1$ and $x_2$ are their coordinates.

If the particles are shifted, the change in the position of the center of mass ($\Delta X_{\mathrm{cm}}$) is related to the individual displacements ($\Delta x_1$ and $\Delta x_2$) by:
$\Delta X_{\mathrm{cm}}=\frac{m_1\Delta x_1+m_2\Delta x_2}{m_1+m_2}$

To keep the center of mass of the system at the original position, the change in its position must be zero ($\Delta X_{\mathrm{cm}}=0$). This gives the condition:
$m_1\Delta x_1+m_2\Delta x_2=0$

If we denote the distance moved by the first mass towards the center of mass as $d_1$ and the distance moved by the second mass towards the center of mass as $d_2$, the direction of their movements towards the center of mass are opposite to each other. Therefore, we can write:
$m_1{d}_1=m_2{d}_2$

Given the values from the question:
$m_1=3\text{kg}$
$m_2=2\text{kg}$
$d_1=2\text{cm}$

Substituting these values into the relation:
$3\times 2=2\times d_2$
$6=2d_2$
$d_2=3\text{cm}$

Thus, to keep the center of mass in its original position, the particle of mass $m_2$ must be moved towards the center of mass by a distance of 3 cm.