A wooden block floating in a bucket of water has 4/5 of its volume submerged. When certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is:

Options

A. 0.7
B. 0.8
C. 0.5
D. 0.6

Explanation

To solve the problem, we need to analyze the situation step by step using the principles of buoyancy and density.

### Step 1: Understand the Initial Condition
The wooden block is floating in water with 4/5 of its volume submerged. Let the volume of the block be \( V \).

- Submerged volume in water: \( V_{sub} = \frac{4}{5}V \)
- The weight of the block is equal to the buoyant force acting on it, which is given by Archimedes' principle.

### Step 2: Apply Archimedes' Principle
The buoyant force \( F_b \) is equal to the weight of the water displaced by the submerged part of the block:
\[ F_b = \rho_{water} \cdot V_{sub} \cdot g = \rho_{water} \cdot \left(\frac{4}{5}V\right) \cdot g \]
Where \( \rho_{water} \) is the density of water and \( g \) is the acceleration due to gravity.

The weight of the block \( W \) is given by:
\[ W = \rho_{block} \cdot V \cdot g \]

Setting the two forces equal gives:
\[ \rho_{water} \cdot \left(\frac{4}{5}V\right) \cdot g = \rho_{block} \cdot V \cdot g \]
Cancelling \( V \) and \( g \) from both sides, we have:
\[ \rho_{water} \cdot \frac{4}{5} = \rho_{block} \]
Thus, we can express the density of the block in terms of the density of water:
\[ \rho_{block} = \frac{4}{5} \rho_{water} \quad \text{(Equation 1)} \]

### Step 3: Analyze the New Condition with Oil
When oil is added, the block floats with half of its volume submerged in water and half in oil. This means:
- Volume submerged in water: \( V_{water} = \frac{1}{2}V \)
- Volume submerged in oil: \( V_{oil} = \frac{1}{2}V \)

### Step 4: Write the Buoyant Force in the New Condition
The buoyant force now consists of contributions from both the water and the oil:
\[ F_b = \rho_{water} \cdot V_{water} \cdot g + \rho_{oil} \cdot V_{oil} \cdot g \]
Substituting the volumes:
\[ F_b = \rho_{water} \cdot \left(\frac{1}{2}V\right) \cdot g + \rho_{oil} \cdot \left(\frac{1}{2}V\right) \cdot g \]
Factoring out common terms:
\[ F_b = \frac{1}{2}Vg \left(\rho_{water} + \rho_{oil}\right) \]

### Step 5: Set the Buoyant Force Equal to the Weight of the Block
The weight of the block remains the same:
\[ W = \rho_{block} \cdot V \cdot g \]
Setting the buoyant force equal to the weight of the block:
\[ \frac{1}{2}Vg \left(\rho_{water} + \rho_{oil}\right) = \rho_{block} \cdot V \cdot g \]
Cancelling \( V \) and \( g \):
\[ \frac{1}{2} \left(\rho_{water} + \rho_{oil}\right) = \rho_{block} \]

### Step 6: Substitute the Density of the Block
Using Equation 1:
\[ \frac{1}{2} \left(\rho_{water} + \rho_{oil}\right) = \frac{4}{5} \rho_{water} \]

### Step 7: Solve for the Density of Oil
Multiplying through by 2:
\[ \rho_{water} + \rho_{oil} = \frac{8}{5} \rho_{water} \]
Rearranging gives:
\[ \rho_{oil} = \frac{8}{5} \rho_{water} - \rho_{water} = \frac{8}{5} \rho_{water} - \frac{5}{5} \rho_{water} = \frac{3}{5} \rho_{water} \]

### Step 8: Find the Relative Density of Oil
The relative density of oil with respect to water is:
\[ \frac{\rho_{oil}}{\rho_{water}} = \frac{3/5 \rho_{water}}{\rho_{water}} = \frac{3}{5} = 0.6 \]

### Final Answer
The density of oil relative to that of water is \( 0.6 \).

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