Two blocks of masses 10 kg and 30 kg are situated along a vertical...

The displacement of the centre of mass of a two-particle system is given by the formula:

$\Delta y_{cm} = \dfrac{m_1 \Delta y_1 + m_2 \Delta y_2}{m_1 + m_2}$

Given $m_1 = 10 \text{ kg}$, $m_2 = 30 \text{ kg}$, $\Delta y_1 = 7 \text{ cm}$ (upward), and $\Delta y_{cm} = 1 \text{ cm}$ (upward).

Substituting the values into the formula:

$1 = \dfrac{10 \times 7 + 30 \times \Delta y_2}{10 + 30}$

$1 = \dfrac{70 + 30 \Delta y_2}{40}$

$40 = 70 + 30 \Delta y_2$

$30 \Delta y_2 = -30$

$\Delta y_2 = -1 \text{ cm}$

The negative sign indicates that the displacement is in the downward direction. Thus, the second block must be displaced by $1 \text{ cm}$ downward.