The densit of a block of wood which flots on water with 0.1 of its volume above water is:

Options

A. `10^(3)kg//m^(3)`
B. `1.1xx10^(3)kg//m^(3)`
C. `0.9xx10^(3)kg//m^(3)`
D. `0.8xx10^(3)kg//m^(3)`

Explanation

To find the density of a block of wood that floats on water with 0.1 of its volume above the water, we can follow these steps:

### Step 1: Understand the situation
The block of wood floats with 0.1 of its volume above the water. This means that 0.9 of its volume is submerged in the water.

### Step 2: Define variables
Let:
- \( V \) = total volume of the wooden block
- \( \rho_w \) = density of water (approximately \( 1000 \, \text{kg/m}^3 \))
- \( \rho_{wood} \) = density of the wooden block
- The volume submerged = \( 0.9V \)

### Step 3: Apply the principle of flotation
According to Archimedes' principle, the weight of the water displaced by the submerged part of the block is equal to the weight of the block itself.

The weight of the water displaced can be expressed as:
\[ \text{Weight of water displaced} = \text{Volume submerged} \times \text{Density of water} \times g = 0.9V \times \rho_w \times g \]

The weight of the wooden block can be expressed as:
\[ \text{Weight of wood} = \text{Total volume} \times \text{Density of wood} \times g = V \times \rho_{wood} \times g \]

### Step 4: Set the weights equal
Since the weight of the water displaced equals the weight of the wooden block, we can set the two equations equal to each other:
\[ 0.9V \times \rho_w \times g = V \times \rho_{wood} \times g \]

### Step 5: Simplify the equation
We can cancel \( V \) and \( g \) from both sides (assuming \( V \) and \( g \) are not zero):
\[ 0.9 \rho_w = \rho_{wood} \]

### Step 6: Substitute the density of water
Now, substituting the density of water (\( \rho_w = 1000 \, \text{kg/m}^3 \)):
\[ \rho_{wood} = 0.9 \times 1000 \, \text{kg/m}^3 = 900 \, \text{kg/m}^3 \]

### Conclusion
The density of the block of wood is \( 900 \, \text{kg/m}^3 \).

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