Two blocks of masses 10 kg and 20 kg are located on the X -axis....

The position of the centre of mass of a two-particle system is given by $X_{cm} = \dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

For the centre of mass to remain unchanged, the change in its position must be zero, so $\Delta X_{cm} = 0$.

$\Delta X_{cm} = \dfrac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2} = 0$

$m_1 \Delta x_1 + m_2 \Delta x_2 = 0$

Substituting the given values, $m_1 = 10 \text{ kg}$, $\Delta x_1 = 2 \text{ cm}$, and $m_2 = 20 \text{ kg}$:

$10 \times 2 + 20 \times \Delta x_2 = 0$

$20 + 20 \Delta x_2 = 0$

$\Delta x_2 = -1 \text{ cm}$

The negative sign indicates that the second mass must be moved in the opposite direction. The distance it must be moved is $1 \text{ cm}$.