Question

A cubical block of density \( \rho_b = 600\,\text{kg/m}^3 \) floats in a liquid of density \( \rho_l = 900\,\text{kg/m}^3 \). If the height of block is \(H = 8.0\,\text{cm}\), then height of the submerged part is ________ cm.

• A. 5.3
• B. 6.3
• C. 7.3
• D. 4.3

Hint:

Fraction submerged of a floating body depends only on density ratio.

Answer 1

The Correct Option is A

To determine the height of the submerged part of the cubical block, we can utilize the principle of floatation. According to this principle, the weight of the fluid displaced by the submerged part of the block is equal to the weight of the block. This principle can be expressed mathematically as:

\(W_{\text{fluid displaced}} = W_{\text{block}}\)

Given:

• Density of the block, \(\rho_b = 600\, \text{kg/m}^3\)
• Density of the liquid, \(\rho_l = 900\, \text{kg/m}^3\)
• Height of the block, \(H = 8.0\, \text{cm}\)

The volume of the block is \(H^3\) and the volume of the submerged part is \(hH^2\), where \(h\) is the submerged height. Applying the principle of floatation:

\(\rho_l \cdot g \cdot hH^2 = \rho_b \cdot g \cdot H^3\)

We can simplify this equation by canceling out \(g\) and \(H^2\):

\(\rho_l \cdot h = \rho_b \cdot H\)

Solving for \(h\), we have:

\(h = \frac{\rho_b \cdot H}{\rho_l}\)

Now, substitute the given values:

\(h = \frac{600 \cdot 8}{900}\)

\(h = \frac{4800}{900}\)

\(h = 5.333\ \text{cm}\)

Therefore, the height of the submerged part of the block is approximately 5.3 cm. This matches with the given option, confirming it as the correct answer.