A cubical block of density $$\rho_{b}= 600kg/m^{3}$$ floats in a liquid of density $$\rho_{e}= 900kg/m^{3}$$. If the height of block is H = 8.0 cm then height of the submerged part is ________ cm.

We need to find the height of the submerged part of a floating cubical block.

The density of the block is $$\rho_b = 600$$ kg/m³, and the density of the liquid is $$\rho_l = 900$$ kg/m³. The block has height H = 8.0 cm.

By the principle of floatation, the weight of the block equals the weight of the liquid displaced, so

$$\rho_b \times V_{block} \times g = \rho_l \times V_{submerged} \times g$$

Because the block is cubical with base area A, its total volume is A × H and its submerged volume is A × h, giving

$$\rho_b \times A \times H = \rho_l \times A \times h$$

Solving for h yields

$$h = \frac{\rho_b}{\rho_l} \times H = \frac{600}{900} \times 8.0 = \frac{2}{3} \times 8.0 = 5.33 \text{ cm} \approx 5.3 \text{ cm}$$

Therefore, the height of the submerged part is Option 4: 5.3 cm.