A metal ball immersed in water weighs `w_(1)` at `0^(@)C` and `w_(2)` at `50^(@)C`. The coefficient of cubical expansion of metal is less than that of water. Then

Options

A. `w_(1)gtw_(2)`
B. `w_(1)ltw_(2)`
C. `w_(1)=w_(?)`
D. Data is insufficient

Answer

To solve the problem, we need to analyze the situation of the metal ball immersed in water at two different temperatures, 0°C and 50°C. We will use the concept of buoyancy and the coefficients of cubical expansion for both the metal and water.

### Step-by-Step Solution:

1. **Understanding the Weights**:
- At 0°C, the weight of the metal ball in water is \( W_1 \).
- At 50°C, the weight of the metal ball in water is \( W_2 \).
- The weight of the ball in water is affected by the buoyant force, which is equal to the weight of the water displaced by the ball.

2. **Buoyant Force at 0°C**:
- The buoyant force \( F_b \) at 0°C can be expressed as:
\[ W_1 = mg - F_b \]
- Here, \( mg \) is the weight of the metal ball in air, and \( F_b \) is the buoyant force at 0°C.

3. **Buoyant Force at 50°C**:
- At 50°C, the buoyant force changes due to the expansion of water. The weight of the ball in water can be expressed as:
\[ W_2 = mg - F_b' \]
- Where \( F_b' \) is the buoyant force at 50°C.

4. **Effect of Temperature on Buoyant Force**:
- The buoyant force depends on the volume of water displaced, which changes with temperature due to the coefficient of cubical expansion.
- The formula for the change in buoyant force due to temperature is:
\[ F_b' = F_b \left(1 + \frac{\gamma_m \Delta T}{1 + \gamma_w \Delta T}\right) \]
- Here, \( \gamma_m \) is the coefficient of cubical expansion of the metal, \( \gamma_w \) is the coefficient of cubical expansion of water, and \( \Delta T \) is the change in temperature (50°C).

5. **Comparing the Two Weights**:
- Since it is given that the coefficient of cubical expansion of metal is less than that of water (\( \gamma_w > \gamma_m \)), this means that the buoyant force at 50°C will be greater than at 0°C.
- Therefore, the weight \( W_2 \) will be less than \( W_1 \) because the buoyant force has increased more significantly due to the higher expansion of water compared to the metal.

6. **Conclusion**:
- Since the buoyant force increases more at 50°C than at 0°C, we conclude that:
\[ W_2 < W_1 \]
- Thus, the correct answer is that \( W_2 \) is less than \( W_1 \).

### Final Answer:
- The relationship between the weights is \( W_2 < W_1 \).

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