A boat carrying steel balls is floating on the surface of water in a tank. If the balls are thrown into the tank one by one, how will it affect the level of water ?

Options

A. it will remain unchanged
B. it will rise
C. It will fall
D. First it will rise and then fall

Explanation

To solve the problem of how the water level in the tank is affected when steel balls are thrown from a boat into the tank, we can break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- The boat is floating on the surface of the water, carrying steel balls. The total weight of the boat and the steel balls is balanced by the buoyant force exerted by the water.
- Let the mass of the steel balls be \( m \) and the mass of the boat be \( M \).

2. **Buoyant Force and Displacement**:
- According to Archimedes' principle, the buoyant force is equal to the weight of the water displaced by the boat.
- The initial buoyant force can be expressed as:
\[ F_b = (M + m)g \]
- The volume of water displaced, \( V_1 \), can be calculated using the density of water \( \rho \):
\[ V_1 = \frac{(M + m)}{\rho} \]

3. **Throwing the Steel Balls into the Tank**:
- When a steel ball of mass \( m \) is thrown into the tank, it sinks because the density of steel is greater than that of water.
- The volume of the steel ball that displaces water when submerged is given by:
\[ V_{ball} = \frac{m}{\rho_{steel}} \]
- Here, \( \rho_{steel} \) is the density of the steel, which is greater than \( \rho \).

4. **Final Condition After Throwing the Ball**:
- After throwing one steel ball into the tank, the new buoyant force is now only acting on the remaining mass of the boat \( M \):
\[ F_b' = Mg \]
- The volume of water displaced by the boat now is:
\[ V_2 = \frac{M}{\rho} \]
- The total volume of water displaced after the ball is thrown in is:
\[ V_{total} = V_2 + V_{ball} = \frac{M}{\rho} + \frac{m}{\rho_{steel}} \]

5. **Comparison of Volumes**:
- Since the density of steel is greater than that of water, the volume displaced by the steel ball when it is in the water is less than the volume that was displaced when it was in the boat.
- Therefore, we can conclude that:
\[ V_1 > V_{total} \]
- This means that the total volume of water displaced after throwing the ball is less than the volume displaced when the ball was in the boat.

6. **Conclusion**:
- As a result, the water level in the tank will **fall** when the steel balls are thrown into the tank one by one.

### Final Answer:
The level of water in the tank will decrease when the steel balls are thrown into the tank.