Two springs have their force constant as `k_(1)` and `k_(2) (k_(1) gt k_(2))`. When they are streched by the same force.

Options

A. No work is done in case of both the springs
B. Equal work is done in case of both the spriongs
C. More work is done in case of second spring
D. more work is done in case of first spring

Explanation

To solve the problem, we need to analyze the relationship between the work done on two springs with different spring constants when they are stretched by the same force.

### Step-by-Step Solution:

1. **Understanding the Work Done on a Spring**:
The work done (W) on a spring when it is stretched by a force (F) can be expressed using the formula:
\[ W = \frac{F^2}{2k} \]
where \( k \) is the spring constant.

2. **Identifying the Given Information**:
We have two springs with spring constants \( k_1 \) and \( k_2 \) such that \( k_1 > k_2 \). This means the first spring is stiffer than the second spring.

3. **Analyzing the Work Done on Each Spring**:
- For the first spring (with spring constant \( k_1 \)):
\[ W_1 = \frac{F^2}{2k_1} \]
- For the second spring (with spring constant \( k_2 \)):
\[ W_2 = \frac{F^2}{2k_2} \]

4. **Comparing the Work Done**:
Since \( k_1 > k_2 \), it follows that:
\[ \frac{1}{k_1} < \frac{1}{k_2} \]
Therefore, we can conclude:
\[ W_1 < W_2 \]
This indicates that more work is done on the second spring compared to the first spring.

5. **Conclusion**:
Since \( k_1 \) is greater than \( k_2 \), the work done on the first spring is less than the work done on the second spring. Thus, the correct answer to the question is that more work is done in the case of the second spring.

### Final Answer:
More work is done in the case of the second spring.

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