Question

A 10 cm\(^3\) cube floats in water with a height of 4 cm\(^3\) remaining above the surface. The density of the material from which the cube is made is:

• A. 0.6 g/cm\(^3\)
• B. 1.0 g/cm\(^3\)
• C. 0.4 g/cm\(^3\)
• D. 0.24 g/cm\(^3\)

Hint:

For an object floating in a fluid, the ratio of the volume submerged to the total volume is equal to the ratio of the density of the object to the density of the fluid.

Answer 1

The Correct Option is A

Step 1: Understand the Principle of Buoyancy.
The buoyant force \( F_b \) acting on the cube is equal to the weight of the water displaced by the cube. The volume of the displaced water is equal to the submerged volume of the cube.

Step 2: Volume Submerged and Floating Condition.
The cube has a total volume of 10 cm\(^3\), and 4 cm\(^3\) remains above the surface. Therefore, the submerged volume is: \[ V_{\text{submerged}} = 10 - 4 = 6 \, \text{cm}^3 \]

Step 3: Use Archimedes’ Principle.
The buoyant force is equal to the weight of the displaced water. The weight of the displaced water is: \[ \text{Weight of displaced water} = \rho_{\text{water}} \times V_{\text{submerged}} \times g \] Since the cube floats, the weight of the cube is equal to the buoyant force: \[ \text{Weight of the cube} = \rho_{\text{cube}} \times V_{\text{cube}} \times g \] Equating the two: \[ \rho_{\text{cube}} \times 10 = \rho_{\text{water}} \times 6 \] Given that the density of water is \( 1 \, \text{g/cm}^3 \), we get: \[ \rho_{\text{cube}} = \frac{6}{10} = 0.6 \, \text{g/cm}^3 \]