The weight of a balloon is `W_(1)` when empty and `W_(2)` when filled with air. Both are weighed in air by the same sensitive spring balance and under identical conditions.

Options

A. `W_(1) = W_(2)`, as the weight of air in the balloon is offset by the force of buoyancy on it.
B. `W_(2) lt W_(1)` due to the force of buoyancy acting on the filled balloon.
C. `W_(2) gt W_(1)`, as the air inside is at a greater pressure and hence hs greater density than the air outside.
D. `W_(2) = W_(1)+` weight of the air inside it.

Answer

To solve the problem, we need to analyze the weights of the balloon in two different states: when it is empty and when it is filled with air. We will derive the relationship between the two weights, \( W_1 \) and \( W_2 \).

### Step-by-Step Solution:

1. **Define the Weights**:
- Let \( W_1 \) be the weight of the empty balloon.
- Let \( W_2 \) be the weight of the balloon when it is filled with air.

2. **Weight of the Empty Balloon**:
- The weight of the empty balloon can be expressed as:
\[ W_1 = M_1 \cdot g \]
where \( M_1 \) is the mass of the empty balloon and \( g \) is the acceleration due to gravity.

3. **Weight of the Filled Balloon**:
- When the balloon is filled with air, the weight \( W_2 \) can be expressed as:
\[ W_2 = M_1 \cdot g + \text{Weight of the air inside the balloon} \]
- The weight of the air inside the balloon can be calculated as:
\[ \text{Weight of air} = V \cdot \rho_{\text{inside}} \cdot g \]
where \( V \) is the volume of the balloon and \( \rho_{\text{inside}} \) is the density of the air inside the balloon.

4. **Buoyant Force**:
- The buoyant force acting on the balloon when it is filled with air is given by:
\[ F_B = V \cdot \rho_{\text{outside}} \cdot g \]
where \( \rho_{\text{outside}} \) is the density of the air outside the balloon.

5. **Net Weight Measurement**:
- The spring balance measures the apparent weight of the balloon, which is the actual weight minus the buoyant force:
\[ W_2 = (M_1 \cdot g + V \cdot \rho_{\text{inside}} \cdot g) - (V \cdot \rho_{\text{outside}} \cdot g) \]

6. **Combine the Equations**:
- Substituting the expressions for \( W_1 \) and the weights into the equation for \( W_2 \):
\[ W_2 = W_1 + V \cdot g \cdot (\rho_{\text{inside}} - \rho_{\text{outside}}) \]

7. **Conclusion**:
- Since the density of the air inside the balloon (\( \rho_{\text{inside}} \)) is greater than the density of the air outside (\( \rho_{\text{outside}} \)), we conclude that:
\[ W_2 > W_1 \]
- Therefore, the relationship between the weights is:
\[ W_2 = W_1 + \text{(additional weight due to air inside)} \]