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. 0.5
B. 0.8
C. 0.6
D. 0.7

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the initial conditions
The wooden block is floating in water with \( \frac{4}{5} \) of its volume submerged. This indicates that the buoyant force acting on the block is equal to the weight of the block.

### Step 2: Establish the relationship between the densities
Let:
- \( V \) = Volume of the block
- \( \rho_b \) = Density of the block
- \( \rho_w \) = Density of water

From the first condition, we can write:
\[ \text{Buoyant force} = \text{Weight of the block} \]
\[ \rho_w \cdot \frac{4}{5}Vg = \rho_b \cdot Vg \]
Dividing both sides by \( Vg \):
\[ \rho_w \cdot \frac{4}{5} = \rho_b \]
Thus, we have:
\[ \rho_b = \frac{4}{5} \rho_w \]

### Step 3: Analyze the situation after pouring oil
When oil is poured into the bucket, the block is found to have half of its volume submerged in water and half in oil. This means:
- Volume submerged in water = \( \frac{1}{2}V \)
- Volume submerged in oil = \( \frac{1}{2}V \)

### Step 4: Set up the buoyancy equation for the new situation
The weight of the block must equal the total buoyant force from both the water and the oil:
\[ \rho_b \cdot Vg = \text{Buoyant force from oil} + \text{Buoyant force from water} \]
This can be expressed as:
\[ \rho_b \cdot Vg = \rho_o \cdot \frac{1}{2}Vg + \rho_w \cdot \frac{1}{2}Vg \]
Dividing through by \( Vg \):
\[ \rho_b = \frac{1}{2} \rho_o + \frac{1}{2} \rho_w \]

### Step 5: Substitute the density of the block
Now, substituting \( \rho_b = \frac{4}{5} \rho_w \) into the equation:
\[ \frac{4}{5} \rho_w = \frac{1}{2} \rho_o + \frac{1}{2} \rho_w \]

### Step 6: Rearranging the equation
Multiply through by 2 to eliminate the fraction:
\[ \frac{8}{5} \rho_w = \rho_o + \rho_w \]
Rearranging gives:
\[ \rho_o = \frac{8}{5} \rho_w - \rho_w \]
\[ \rho_o = \left(\frac{8}{5} - 1\right) \rho_w \]
\[ \rho_o = \frac{3}{5} \rho_w \]

### Step 7: Determine the density of oil relative to water
To find the density of oil relative to water:
\[ \frac{\rho_o}{\rho_w} = \frac{3}{5} = 0.6 \]

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

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