Question

If a solid cube of 7 kg having an edge of length 10 cm floats in water, then the volume of the cube outside of the water is (let the density of the water 1000 kg/m³.)

• A. 100 cm³
• B. 300 cm³
• C. 400 cm³
• D. 700 cm³

Hint:

In floating objects, the volume submerged in water depends on the ratio of the densities. The buoyant force equals the weight of the displaced water, keeping the object afloat.

Answer 1

The Correct Option is B

Let’s break down the solution step by step: Step 1:
Given data:
Mass of the cube, \( m = 7 \text{ kg} \)
Edge of the cube, \( a = 10 \text{ cm} = 0.1 \text{ m} \)
Density of water, \( \rho_{\text{water}} = 1000 \text{ kg/m}^3 \)
Volume of the cube, \( V_{\text{cube}} = a^3 = 0.1^3 = 0.001 \text{ m}^3 = 1000 \text{ cm}^3 \)


Step 2:
The buoyant force equals the weight of the water displaced, and the fraction of the cube submerged is equal to the density ratio between the cube and water. \[ \text{Fraction submerged} = \frac{\text{Density of the cube}}{\text{Density of water}} = \frac{m}{\rho_{\text{water}} \cdot V_{\text{cube}}} = \frac{7}{1000 \times 1000} = 0.007 \]

Step 3:
The volume of the cube submerged in water is: \[ V_{\text{submerged}} = 0.007 \times 1000 = 7 \text{ cm}^3 \]

Step 4:
Therefore, the volume of the cube outside the water is: \[ V_{\text{outside}} = V_{\text{cube}} - V_{\text{submerged}} = 1000 - 700 = 300 \text{ cm}^3 \]

Step 5:
Conclusion:
The volume of the cube outside the water is 300 cm³.