A block of wood floats in freshwater with two - third of its volume submerged . In oil , the block floats with one - fourth of its volume submerged. The density of oil is

Options

A. `2666.7 kg m ^(-3)`
B. `5333.3 kg m ^(-3)`
C. `1333.3 kg m ^(-3)`
D. `3333.3 kg m^(-3)`

Answer

To solve the problem, we need to analyze the conditions under which the block of wood floats in both freshwater and oil. We will use the principle of buoyancy, which states that the weight of the fluid displaced by a submerged object is equal to the weight of the object itself.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- The block of wood floats in freshwater with \( \frac{2}{3} \) of its volume submerged.
- The block floats in oil with \( \frac{1}{4} \) of its volume submerged.
- We need to find the density of the oil.

2. **Let’s Define Variables:**
- Let the density of the block be \( \rho_{block} \).
- Let the total volume of the block be \( V \).
- The density of freshwater \( \rho_{water} = 1000 \, \text{kg/m}^3 \).
- Let the density of oil be \( \rho_{oil} \).

3. **Applying the Principle of Buoyancy in Water:**
- The weight of the block is equal to the buoyant force when it is floating.
- The volume of water displaced when the block is in freshwater is \( \frac{2}{3} V \).
- The weight of the displaced water is given by:
\[ \text{Weight of displaced water} = \rho_{water} \cdot \left(\frac{2}{3} V\right) \cdot g \]
- The weight of the block can also be expressed as:
\[ \text{Weight of block} = \rho_{block} \cdot V \cdot g \]
- Setting these equal gives:
\[ \rho_{block} \cdot V \cdot g = \rho_{water} \cdot \left(\frac{2}{3} V\right) \cdot g \]
- Cancelling \( g \) and \( V \) (assuming \( V \neq 0 \)):
\[ \rho_{block} = \rho_{water} \cdot \frac{2}{3} = 1000 \cdot \frac{2}{3} = \frac{2000}{3} \, \text{kg/m}^3 \]

4. **Applying the Principle of Buoyancy in Oil:**
- The volume of oil displaced when the block is in oil is \( \frac{1}{4} V \).
- The weight of the displaced oil is given by:
\[ \text{Weight of displaced oil} = \rho_{oil} \cdot \left(\frac{1}{4} V\right) \cdot g \]
- Setting the weight of the block equal to the buoyant force in oil gives:
\[ \rho_{block} \cdot V \cdot g = \rho_{oil} \cdot \left(\frac{1}{4} V\right) \cdot g \]
- Cancelling \( g \) and \( V \):
\[ \rho_{block} = \rho_{oil} \cdot \frac{1}{4} \]

5. **Equating the Two Expressions for Density of the Block:**
- From the two equations we have:
\[ \frac{2000}{3} = \rho_{oil} \cdot \frac{1}{4} \]
- Rearranging gives:
\[ \rho_{oil} = \frac{2000}{3} \cdot 4 = \frac{8000}{3} \approx 2666.67 \, \text{kg/m}^3 \]

### Final Answer:
The density of oil is approximately \( 2666.67 \, \text{kg/m}^3 \).

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