A soft plastic ball filled with air at atmospheric pressure weighs `W_1`. If weight becomes `W_2` when air is removed from it. Then

Options

A. `W_1 gt W_2`
B. `W_1=W_2`
C. `W_1 lt W_2`
D. `W_1 = 2W_2`

Answer

To solve the question, we need to analyze the weights of the soft plastic ball filled with air and the ball after the air has been removed. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Weights**:
- Let \( W_1 \) be the weight of the soft plastic ball filled with air at atmospheric pressure.
- Let \( W_2 \) be the weight of the soft plastic ball after the air has been removed.

2. **Components of the Weights**:
- The weight \( W_1 \) consists of two components:
- The weight of the plastic ball itself (let's call this \( W_b \)).
- The weight of the air inside the ball (let's call this \( W_a \)).
- Therefore, we can express \( W_1 \) as:
\[ W_1 = W_b + W_a \]

3. **Weight After Air Removal**:
- When the air is removed from the ball, the weight \( W_2 \) only includes the weight of the plastic ball:
\[ W_2 = W_b \]

4. **Comparing the Weights**:
- Now, we can compare \( W_1 \) and \( W_2 \):
- Since \( W_1 = W_b + W_a \) and \( W_2 = W_b \), we can see that:
\[ W_1 = W_2 + W_a \]
- Here, \( W_a \) (the weight of the air) is a positive value because air has weight. Therefore:
\[ W_1 > W_2 \]

5. **Conclusion**:
- We conclude that the weight of the soft plastic ball filled with air (\( W_1 \)) is greater than the weight of the ball without air (\( W_2 \)).
- Thus, the correct relationship is:
\[ W_1 > W_2 \]

### Final Answer:
The correct option is: **W1 is greater than W2**.

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