A meta cube is plced in an empty vessel. When water is filed in the vessel so that the cubeis completely immersed in the water, the fore on the bottom of the vessel in contact with the cube

Options

A. will increase
B. will decrease
C. will remain the same
D. will become zero

Answer

To solve the problem, we need to analyze the forces acting on the metal cube when it is fully immersed in water. Let's break it down step by step.

### Step-by-step Solution:

1. **Identify the Forces Acting on the Cube:**
- When the metal cube is placed in the vessel, it has a weight (W) acting downwards due to gravity. This weight can be expressed as:
\[ W = mg \]
where \( m \) is the mass of the cube and \( g \) is the acceleration due to gravity.

2. **Understanding the Buoyant Force:**
- When the cube is completely immersed in water, it displaces a volume of water equal to its own volume. According to Archimedes' principle, the buoyant force (B) acting on the cube is equal to the weight of the water displaced:
\[ B = \text{Weight of displaced water} = \rho V g \]
where \( \rho \) is the density of water, \( V \) is the volume of the cube, and \( g \) is the acceleration due to gravity.

3. **Forces on the Bottom of the Vessel:**
- The force exerted on the bottom of the vessel due to the cube is the sum of the weight of the cube and the buoyant force acting on it. However, the buoyant force acts upwards and balances the weight of the cube. Therefore, the net force on the bottom of the vessel is simply the weight of the cube:
\[ F_{\text{bottom}} = W = mg \]

4. **Conclusion:**
- The force on the bottom of the vessel in contact with the cube remains equal to the weight of the cube, regardless of the buoyant force acting on it. Thus, the force on the bottom of the vessel is:
\[ F_{\text{bottom}} = mg \]

### Final Answer:
The force on the bottom of the vessel in contact with the cube is \( mg \).

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