Question

A wooden cube of side 10 cm floats at the interface between water and oil with its lower surface 3 cm below the interface. If the density of oil is 0.9 g/cm³, the mass of the wooden cube is:

• A. 940 g
• B. 900 g
• C. 1000 g
• D. 930 g \bigskip

Hint:

The equilibrium of the floating cube is determined by the principle of buoyancy. The cube experiences upward buoyant forces due to both water and oil. The total buoyant force equals the weight of the cube. Using the volume submerged in each liquid and their respective densities, we can calculate the cube's mass correctly.

Answer 1

The Correct Option is D

Given the density of oil \(\rho_{\text{oil}} = 0.9 \, \text{g/cm}^3\) and the side of the cube \(s = 10 \, \text{cm}\), the volume of the cube is: \[ V = s^3 = 10^3 = 1000 \, \text{cm}^3 \] Let the mass of the wooden cube be \(M\). The cube floats at the interface between oil and water, and the lower surface is 3 cm below the interface. Since the cube is floating, the weight of the displaced oil and water must be equal to the weight of the cube. The density of water is \( \rho_{\text{water}} = 1 \, \text{g/cm}^3 \). Let the height of the cube submerged in oil be \(h_{\text{oil}} = 3 \, \text{cm}\), and the height submerged in water be \(h_{\text{water}} = 7 \, \text{cm}\). The weight of the displaced oil is: \[ W_{\text{oil}} = \rho_{\text{oil}} \cdot V_{\text{oil}} = 0.9 \cdot 3 \cdot 10^2 = 270 \, \text{g} \] The weight of the displaced water is: \[ W_{\text{water}} = \rho_{\text{water}} \cdot V_{\text{water}} = 1 \cdot 7 \cdot 10^2 = 700 \, \text{g} \] Thus, the total weight of the displaced liquid is \(W_{\text{total}} = 230 + 700 = 930 \, \text{g}\). Since the weight of the cube equals the weight of the displaced liquid: \[ M = 930 \, \text{g} \] Thus, the mass of the cube is \(930 \, \text{g}\).