A block of wood floats in water with `2/3` of its volume submerged. Its relative density is

Options

A. `1/3`
B. `2/3`
C. `4/3`
D. `1/9`

Answer

To find the relative density of a block of wood that floats in water with \( \frac{2}{3} \) of its volume submerged, we can follow these steps:

### Step 1: Understand the problem
We know that the block is floating, which means the weight of the block is equal to the buoyant force acting on it. The buoyant force is equal to the weight of the water displaced by the submerged part of the block.

### Step 2: Define the variables
Let:
- \( V \) = total volume of the wooden block
- \( V' \) = volume of the block submerged in water = \( \frac{2}{3} V \)
- \( \sigma \) = density of the wooden block
- \( \rho \) = density of water (approximately \( 1000 \, \text{kg/m}^3 \))

### Step 3: Write the equations
According to Archimedes' principle, the buoyant force \( F_B \) is given by:
\[ F_B = \text{Weight of the water displaced} = V' \cdot \rho \cdot g \]
Since \( V' = \frac{2}{3} V \), we can write:
\[ F_B = \left(\frac{2}{3} V\right) \cdot \rho \cdot g \]

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

### Step 4: Set the forces equal
For the block to float, the buoyant force must equal the weight of the block:
\[ \left(\frac{2}{3} V\right) \cdot \rho \cdot g = V \cdot \sigma \cdot g \]

### Step 5: Simplify the equation
We can cancel \( g \) and \( V \) (assuming \( V \neq 0 \)):
\[ \frac{2}{3} \rho = \sigma \]

### Step 6: Find the relative density
Relative density (specific gravity) is defined as the ratio of the density of the substance to the density of water:
\[ \text{Relative Density} = \frac{\sigma}{\rho} \]

Substituting \( \sigma \) from the previous step:
\[ \text{Relative Density} = \frac{\frac{2}{3} \rho}{\rho} = \frac{2}{3} \]

### Conclusion
The relative density of the block of wood is \( \frac{2}{3} \).