A block of wood floats in water with `(4//5)th` of its volume submerged. If the same block just floats in a liquid, the density of liquid in `(kg m^(-3))` is

Options

A. 1250
B. 600
C. 400
D. 800

Explanation

To solve the problem, we need to determine the density of the liquid in which the wooden block floats just at the surface. We start with the information given about the block floating in water.

### Step-by-Step Solution:

1. **Understanding the Floating Condition**:
When an object floats, the weight of the object is equal to the weight of the fluid displaced by the object. This can be expressed as:
\[ \text{Weight of block} = \text{Weight of displaced fluid} \]

2. **Weight of the Block**:
Let the volume of the block be \( V \) and the density of the block be \( \rho_s \). The weight of the block can be expressed as:
\[ \text{Weight of block} = m \cdot g = \rho_s \cdot V \cdot g \]
where \( g \) is the acceleration due to gravity.

3. **Weight of Displaced Water**:
The problem states that \( \frac{4}{5} \) of the block's volume is submerged in water. Therefore, the volume of water displaced is \( \frac{4}{5} V \). The density of water \( \rho_w \) is given as \( 1000 \, \text{kg/m}^3 \). The weight of the displaced water is:
\[ \text{Weight of displaced water} = \rho_w \cdot \left(\frac{4}{5} V\right) \cdot g \]

4. **Setting Up the Equation**:
From the floating condition, we set the weight of the block equal to the weight of the displaced water:
\[ \rho_s \cdot V \cdot g = \rho_w \cdot \left(\frac{4}{5} V\right) \cdot g \]
We can cancel \( g \) and \( V \) (assuming \( V \neq 0 \)):
\[ \rho_s = \rho_w \cdot \frac{4}{5} \]

5. **Calculating the Density of the Block**:
Substituting \( \rho_w = 1000 \, \text{kg/m}^3 \):
\[ \rho_s = 1000 \cdot \frac{4}{5} = 800 \, \text{kg/m}^3 \]

6. **Finding the Density of the Liquid**:
When the block just floats in another liquid, the condition remains the same. Thus, the density of the liquid \( \rho_l \) must equal the density of the block:
\[ \rho_l = \rho_s = 800 \, \text{kg/m}^3 \]

### Conclusion:
The density of the liquid in which the block just floats is:
\[ \boxed{800 \, \text{kg/m}^3} \]