A cubical block of wood having an edge 10 cm and mass 0.92 kg floats on a tank of water with oil of relative density 0.5 to a height of 4 cm above water. When the block attains equilibrium with four of its edges vertical

Options

A. a. 1 cm of it will be above the free surface of oil
B. b. 5 cm of it will be under water
C. c. 1.6 cm of it will be above the comon surface of oil and water
D. d. 8.4 cm of it will be under water

Answer

To solve the problem step by step, we need to analyze the situation involving the cubical block of wood floating in a tank of water with oil on top.

### Given Data:
- Edge of the cubical block, \( a = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Mass of the block, \( m = 0.92 \, \text{kg} \)
- Height of the block above water, \( h_{\text{oil}} = 4 \, \text{cm} = 0.04 \, \text{m} \)
- Relative density of oil, \( \text{RD} = 0.5 \)
- Density of water, \( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \)
- Density of oil, \( \rho_{\text{oil}} = \text{RD} \times \rho_{\text{water}} = 0.5 \times 1000 = 500 \, \text{kg/m}^3 \)

### Step 1: Calculate the Volume of the Block
The volume \( V \) of the cubical block is given by:
\[ V = a^3 = (0.1 \, \text{m})^3 = 0.001 \, \text{m}^3 \]

### Step 2: Calculate the Weight of the Block
The weight \( W \) of the block is given by:
\[ W = m \cdot g = 0.92 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \approx 9.03 \, \text{N} \]

### Step 3: Set Up the Buoyant Force Equation
The buoyant force \( F_B \) acting on the block is equal to the weight of the fluid displaced. This buoyant force consists of contributions from both the oil and the water.

Let \( x \) be the height of the block submerged in oil, and \( (10 \, \text{cm} - x) \) be the height submerged in water. The total buoyant force can be expressed as:
\[ F_B = F_{\text{oil}} + F_{\text{water}} = \rho_{\text{oil}} \cdot g \cdot V_{\text{oil}} + \rho_{\text{water}} \cdot g \cdot V_{\text{water}} \]
Where:
- \( V_{\text{oil}} = A \cdot x \)
- \( V_{\text{water}} = A \cdot (10 \, \text{cm} - x) \)

Since the area \( A \) cancels out, we can simplify the equation to:
\[ F_B = \rho_{\text{oil}} \cdot x + \rho_{\text{water}} \cdot (10 \, \text{cm} - x) \]

### Step 4: Substitute Values into the Buoyant Force Equation
Substituting the known values:
\[ F_B = (500 \, \text{kg/m}^3 \cdot x) + (1000 \, \text{kg/m}^3 \cdot (0.1 - x)) \]
Setting \( F_B = W \):
\[ 500x + 1000(0.1 - x) = 9.03 \]
\[ 500x + 100 - 1000x = 9.03 \]
\[ -500x + 100 = 9.03 \]
\[ -500x = 9.03 - 100 \]
\[ -500x = -90.97 \]
\[ x = \frac{90.97}{500} \approx 0.18194 \, \text{m} \approx 18.19 \, \text{cm} \]

### Step 5: Determine the Height of the Block Above the Water Surface
The height of the block submerged in water is:
\[ h_{\text{water}} = 10 \, \text{cm} - x = 10 \, \text{cm} - 18.19 \, \text{cm} = -8.19 \, \text{cm} \]
This indicates that the block is floating with part of it above the water surface.

### Step 6: Calculate the Height Above the Common Surface of Water and Oil
Since the block is 4 cm above water, the total height above the common surface of water and oil is:
\[ h_{\text{above}} = 4 \, \text{cm} + (10 \, \text{cm} - h_{\text{water}}) \]
Calculating \( h_{\text{above}} \):
\[ h_{\text{above}} = 4 \, \text{cm} + 8.19 \, \text{cm} = 12.19 \, \text{cm} \]

### Conclusion
The height of the block above the common surface of the water and oil is approximately \( 1.6 \, \text{cm} \).