A metal ball immersed in water weighs `w_(1)` at `5^(@)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) gt w_(2)`
B. `w_(1) lt w_(2)`
C. `w_(1) = w_(2)`
D. data is insufficient

Explanation

To solve the problem, we need to analyze the weights of the metal ball at two different temperatures and the relationship between the coefficients of cubical expansion of the metal and water.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a metal ball immersed in water. The weight of the ball is measured at two different temperatures: \( w_1 \) at \( 5^\circ C \) and \( w_2 \) at \( 50^\circ C \).

2. **Buoyancy Force**:
- The weight of the ball in water is affected by the buoyancy force, which is given by:
\[ F_B = V \cdot \rho \cdot g \]
- Here, \( V \) is the volume of the ball, \( \rho \) is the density of the water, and \( g \) is the acceleration due to gravity.

3. **Weight Equations**:
- At \( 5^\circ C \):
\[ w_1 = F_B(5) = V(5) \cdot \rho_w(5) \cdot g \]
- At \( 50^\circ C \):
\[ w_2 = F_B(50) = V(50) \cdot \rho_w(50) \cdot g \]

4. **Dividing the Weight Equations**:
- We can divide the two equations to find the relationship between \( w_1 \) and \( w_2 \):
\[ \frac{w_2}{w_1} = \frac{V(50) \cdot \rho_w(50)}{V(5) \cdot \rho_w(5)} \]

5. **Considering Volume Changes**:
- The volume of the metal ball will change with temperature according to its coefficient of cubical expansion \( \gamma_m \):
\[ V(50) = V(5) \cdot (1 + \gamma_m \Delta T) \]
- Similarly, the volume of water will change according to its coefficient of cubical expansion \( \gamma_w \):
\[ \rho_w(50) = \rho_w(5) \cdot (1 - \gamma_w \Delta T) \]

6. **Substituting Volume and Density**:
- Substitute these expressions back into the weight ratio:
\[ \frac{w_2}{w_1} = \frac{(V(5) \cdot (1 + \gamma_m \Delta T)) \cdot (\rho_w(5) \cdot (1 - \gamma_w \Delta T))}{V(5) \cdot \rho_w(5)} \]
- Simplifying this gives:
\[ \frac{w_2}{w_1} = \frac{(1 + \gamma_m \Delta T)(1 - \gamma_w \Delta T)}{1} \]

7. **Analyzing the Coefficients**:
- Given that \( \gamma_w > \gamma_m \), we can conclude that:
\[ (1 + \gamma_m \Delta T)(1 - \gamma_w \Delta T) < 1 \]
- This implies that \( \frac{w_2}{w_1} < 1 \), or equivalently, \( w_2 < w_1 \).

8. **Final Conclusion**:
- Therefore, we conclude that:
\[ w_2 > w_1 \]

### Answer:
The weight \( w_2 \) at \( 50^\circ C \) is greater than the weight \( w_1 \) at \( 5^\circ C \).