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 will use the principles of buoyancy and Archimedes' principle. Let's break down the solution step by step.

### Step 1: Understand the Problem
We have a block of wood that floats in two different liquids: freshwater and oil. We know how much of the block is submerged in each liquid, and we need to find the density of the oil.

### Step 2: Define Variables
- Let the volume of the block be \( V \).
- The density of freshwater (water) is \( \rho_{water} = 1000 \, \text{kg/m}^3 \).
- The density of oil is \( \rho_{oil} \) (which we need to find).
- The weight of the block is \( W \).

### Step 3: Apply Archimedes' Principle for Freshwater
When the block is floating in freshwater, two-thirds of its volume is submerged. The buoyant force \( F_B \) acting on the block is equal to the weight of the water displaced by the submerged volume.

\[ F_B = \rho_{water} \cdot \text{Volume submerged} \cdot g = \rho_{water} \cdot \left(\frac{2}{3} V\right) \cdot g \]

Since the block is in equilibrium, the buoyant force equals the weight of the block:

\[ F_B = W \]

Thus, we can write:

\[ \rho_{water} \cdot \left(\frac{2}{3} V\right) \cdot g = W \tag{1} \]

### Step 4: Apply Archimedes' Principle for Oil
When the block is floating in oil, one-fourth of its volume is submerged. The buoyant force in this case is:

\[ F_B = \rho_{oil} \cdot \text{Volume submerged} \cdot g = \rho_{oil} \cdot \left(\frac{1}{4} V\right) \cdot g \]

Again, since the block is in equilibrium, we have:

\[ \rho_{oil} \cdot \left(\frac{1}{4} V\right) \cdot g = W \tag{2} \]

### Step 5: Equate the Two Expressions for Weight
From equations (1) and (2), we can set them equal to each other since both equal \( W \):

\[ \rho_{water} \cdot \left(\frac{2}{3} V\right) \cdot g = \rho_{oil} \cdot \left(\frac{1}{4} V\right) \cdot g \]

### Step 6: Cancel Common Terms
We can cancel \( V \) and \( g \) from both sides:

\[ \rho_{water} \cdot \frac{2}{3} = \rho_{oil} \cdot \frac{1}{4} \]

### Step 7: Solve for the Density of Oil
Now, we can solve for \( \rho_{oil} \):

\[ \rho_{oil} = \frac{\rho_{water} \cdot \frac{2}{3}}{\frac{1}{4}} = \rho_{water} \cdot \frac{2}{3} \cdot 4 = \rho_{water} \cdot \frac{8}{3} \]

Substituting \( \rho_{water} = 1000 \, \text{kg/m}^3 \):

\[ \rho_{oil} = 1000 \cdot \frac{8}{3} = \frac{8000}{3} \approx 2666.67 \, \text{kg/m}^3 \]

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

---