The total weight of a piece of wood is 6 kg in the floating state in water its 1/3 part remains inside the water on this floating solid what maximum weight is to be put such that the whole of the piece of wood is to be drowned in the water

Options

A. 12 kg
B. 10 kg
C. 14 kg
D. 15 kg

Answer

To solve the problem step by step, we will analyze the situation of the piece of wood floating in water and determine the maximum additional weight that can be added before the wood is completely submerged.

### Step-by-Step Solution:

1. **Understanding the Floating Condition**:
The piece of wood has a total weight of 6 kg and is floating in water. According to the problem, 1/3 of the wood's volume is submerged in water.

2. **Applying Archimedes' Principle**:
According to Archimedes' principle, the weight of the water displaced by the submerged part of the wood is equal to the weight of the wood. Since 1/3 of the wood is submerged, the buoyant force (upthrust) acting on the wood is equal to the weight of the wood.

3. **Calculating the Buoyant Force**:
The weight of the wood (W_wood) is given as 6 kg. Therefore, the buoyant force (F_b) can be calculated as:
\[ F_b = W_{wood} = 6 \, \text{kg} \cdot g \]
where \( g \) is the acceleration due to gravity.

4. **Weight of the Displaced Water**:
The weight of the displaced water is equal to the volume of the submerged part of the wood multiplied by the density of water (ρ) and gravity (g). Since 1/3 of the wood is submerged, we can express this as:
\[ F_b = \frac{1}{3} V \cdot \rho \cdot g \]
where \( V \) is the total volume of the wood.

5. **Setting the Forces Equal**:
Since the buoyant force equals the weight of the wood, we can write:
\[ 6 \, \text{kg} \cdot g = \frac{1}{3} V \cdot \rho \cdot g \]
Canceling \( g \) from both sides gives us:
\[ 6 \, \text{kg} = \frac{1}{3} V \cdot \rho \]

6. **Finding the Total Volume of the Wood**:
Rearranging the equation to find the total volume \( V \):
\[ V = 18 \, \text{kg} / \rho \]

7. **Calculating the Maximum Additional Weight**:
To find the maximum additional weight \( W_{max} \) that can be added before the wood is fully submerged, we need to consider that when the wood is fully submerged, the buoyant force will equal the total weight (weight of the wood plus the additional weight):
\[ F_b = W_{wood} + W_{max} \]
The maximum buoyant force when the wood is fully submerged is:
\[ F_b = V \cdot \rho \cdot g \]
Since \( V = 18 \, \text{kg} / \rho \), we have:
\[ F_b = 18 \, \text{kg} \]
Therefore, we can write:
\[ 18 \, \text{kg} = 6 \, \text{kg} + W_{max} \]
Solving for \( W_{max} \):
\[ W_{max} = 18 \, \text{kg} - 6 \, \text{kg} = 12 \, \text{kg} \]

8. **Conclusion**:
The maximum weight that can be added to the piece of wood without it being completely submerged is 12 kg.

### Final Answer:
The maximum weight that can be added is **12 kg**.