Physics: Floating Block Submerged Height Calculation

This problem involves the principle of flotation, where an object floats if its weight is balanced by the buoyant force exerted by the fluid. We need to determine the submerged portion of a cubical block.

Principle of Flotation

For an object to float, the upward buoyant force ($F_B$) must equal the downward gravitational force (weight, $W$) acting on the object.

• Weight of the block ($W$) is given by $W = m_b \times g = \rho_b \times V_{block} \times g$.
• Buoyant force ($F_B$) is the weight of the fluid displaced by the submerged part of the object, given by $F_B = m_{fluid} \times g = \rho_l \times V_{submerged} \times g$.

Therefore, for flotation: $ \rho_b \times V_{block} \times g = \rho_l \times V_{submerged} \times g $ $ \rho_b \times V_{block} = \rho_l \times V_{submerged} $

Calculating Submerged Height Ratio

We can rearrange the equation to find the ratio of the submerged volume to the total volume:

$ \frac{V_{submerged}}{V_{block}} = \frac{\rho_b}{\rho_l} $

Since the block is cubical, let its total height be $H$ and its base area be $A$. The total volume is $V_{block} = A \times H$. If the height of the submerged part is $h$, the submerged volume is $V_{submerged} = A \times h$.

Substituting these into the ratio equation:

$ \frac{A \times h}{A \times H} = \frac{\rho_b}{\rho_l} $

The area $A$ cancels out, leaving:

$ \frac{h}{H} = \frac{\rho_b}{\rho_l} $

This shows that the fraction of the block's height that is submerged is equal to the ratio of the block's density to the liquid's density.

Applying Given Values

We are given:

• Density of the block, $\rho_b = 600 \text{ kg/m}^3$.
• Density of the liquid, $\rho_l = 900 \text{ kg/m}^3$.
• Total height of the block, $H = 8.0 \text{ cm}$.

Now, we can calculate the submerged height ($h$):

$ h = H \times \frac{\rho_b}{\rho_l} $

Final Calculation

Substitute the values into the formula:

$ h = 8.0 \text{ cm} \times \frac{600 \text{ kg/m}^3}{900 \text{ kg/m}^3} $

Simplify the density ratio:

$ \frac{600}{900} = \frac{6}{9} = \frac{2}{3} $

Perform the final multiplication:

$ h = 8.0 \text{ cm} \times \frac{2}{3} $

$ h = \frac{16.0}{3} \text{ cm} $

$ h \approx 5.33 \text{ cm} $

The height of the submerged part is approximately 5.33 cm.