Two solid pieces, one of steel and the other of aluminium when immersed completely in water have equal weights. When the solid pieces are weighed in air

Options

A. the weight of aluminium is half the weight of steel
B. steel piece will weigh more
C. they have the same weight
D. aluminium piece will weigh more

Explanation

To solve the problem, we need to analyze the relationship between the weights of the two solid pieces (steel and aluminum) when they are weighed in air, given that they have equal weights when immersed in water.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two solid pieces: one made of steel and the other made of aluminum.
- When immersed in water, their weights are equal.

2. **Weight in Water**:
- The weight of an object in water is given by the formula:
\[ W_{\text{water}} = W_{\text{air}} - F_b \]
where \( F_b \) is the buoyant force.
- The buoyant force is equal to the weight of the water displaced by the object, which can be calculated using Archimedes' principle:
\[ F_b = V \cdot \rho_{\text{water}} \cdot g \]
where \( V \) is the volume of the object, \( \rho_{\text{water}} \) is the density of water, and \( g \) is the acceleration due to gravity.

3. **Weight Relationship in Water**:
- For the steel piece:
\[ W_{\text{steel, water}} = W_{\text{steel, air}} - V_{\text{steel}} \cdot \rho_{\text{water}} \cdot g \]
- For the aluminum piece:
\[ W_{\text{aluminum, water}} = W_{\text{aluminum, air}} - V_{\text{aluminum}} \cdot \rho_{\text{water}} \cdot g \]
- Since the weights in water are equal:
\[ W_{\text{steel, air}} - V_{\text{steel}} \cdot \rho_{\text{water}} \cdot g = W_{\text{aluminum, air}} - V_{\text{aluminum}} \cdot \rho_{\text{water}} \cdot g \]

4. **Density Considerations**:
- The density of steel (\( \rho_{\text{steel}} \)) is greater than that of aluminum (\( \rho_{\text{aluminum}} \)).
- Therefore, for the same weight, the volume of aluminum must be greater than the volume of steel:
\[ V_{\text{aluminum}} > V_{\text{steel}} \]

5. **Weight in Air**:
- When the objects are weighed in air, we have:
\[ W_{\text{steel, air}} = m_{\text{steel}} \cdot g \]
\[ W_{\text{aluminum, air}} = m_{\text{aluminum}} \cdot g \]
- Since the volume of aluminum is greater, but the mass of aluminum is less than that of steel due to its lower density, we can conclude:
\[ m_{\text{steel}} > m_{\text{aluminum}} \]

6. **Conclusion**:
- Since the weight of the aluminum piece in air is less than that of the steel piece, we conclude:
\[ W_{\text{aluminum, air}} < W_{\text{steel, air}} \]

### Final Answer:
The weight of aluminum is less than the weight of steel when both are weighed in air.