A wooden block floats in water with two third of its volume submerged. A. Calculate the density of wood.b. When the same block is placed in oil, three - quarter of its volume in immersed in oil. Calculate the density of oil.

Answer

To solve the problem step by step, we will first calculate the density of the wooden block when it is floating in water and then calculate the density of the oil when the same block is submerged in it.

### Step 1: Calculate the Density of the Wooden Block

1. **Understanding the Problem**: The wooden block floats in water with two-thirds of its volume submerged. This means that the weight of the water displaced by the submerged part of the block equals the weight of the block itself.

2. **Using Archimedes' Principle**: According to Archimedes' Principle, the buoyant force (which is equal to the weight of the fluid displaced) is given by:
\[ \text{Buoyant Force} = \text{Weight of the Block} \]

3. **Volume of the Block**: Let the volume of the wooden block be \( V \). Since two-thirds of it is submerged, the volume of water displaced is:
\[ V_{\text{submerged}} = \frac{2}{3}V \]

4. **Weight of the Water Displaced**: The weight of the water displaced can be calculated using the density of water (\( \rho_w = 1000 \, \text{kg/m}^3 \)):
\[ \text{Weight of water displaced} = \rho_w \cdot V_{\text{submerged}} \cdot g = 1000 \cdot \frac{2}{3}V \cdot g \]

5. **Weight of the Wooden Block**: The weight of the wooden block is given by:
\[ \text{Weight of the block} = \rho \cdot V \cdot g \]
where \( \rho \) is the density of the wooden block.

6. **Setting the Two Weights Equal**: Since the buoyant force equals the weight of the block:
\[ 1000 \cdot \frac{2}{3}V \cdot g = \rho \cdot V \cdot g \]

7. **Cancelling \( V \) and \( g \)**: We can cancel \( V \) and \( g \) from both sides:
\[ 1000 \cdot \frac{2}{3} = \rho \]

8. **Calculating the Density**:
\[ \rho = \frac{2000}{3} \approx 667 \, \text{kg/m}^3 \]

### Step 2: Calculate the Density of the Oil

1. **Understanding the New Situation**: When the same block is placed in oil, three-quarters of its volume is submerged. This means that the weight of the oil displaced equals the weight of the block.

2. **Volume of Oil Displaced**: The volume of oil displaced is:
\[ V_{\text{oil}} = \frac{3}{4}V \]

3. **Weight of the Oil Displaced**: The weight of the oil displaced can be calculated using the density of oil (\( \rho_o \)):
\[ \text{Weight of oil displaced} = \rho_o \cdot V_{\text{oil}} \cdot g = \rho_o \cdot \frac{3}{4}V \cdot g \]

4. **Setting the Weights Equal Again**: The weight of the block remains the same:
\[ \rho_o \cdot \frac{3}{4}V \cdot g = \rho \cdot V \cdot g \]

5. **Cancelling \( V \) and \( g \)**: Again, we can cancel \( V \) and \( g \):
\[ \rho_o \cdot \frac{3}{4} = \rho \]

6. **Calculating the Density of Oil**:
\[ \rho_o = \frac{\rho \cdot 4}{3} = \frac{667 \cdot 4}{3} \approx 889.3 \, \text{kg/m}^3 \]

### Final Answers:
- The density of the wooden block is approximately **667 kg/m³**.
- The density of the oil is approximately **889.3 kg/m³**.