Two springs have their force constant as `K_(1)` and `K_(2)(K_(1)lt K_(2))`. When they are stretched by the same force:

Options

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

Answer

To solve the problem, we need to analyze the relationship between the force constants of the springs, the forces applied to them, and the work done on each spring.

### Step-by-Step Solution:

1. **Understanding the Force on Springs**:
We are given two springs with force constants \( K_1 \) and \( K_2 \) such that \( K_1 < K_2 \). When the same force \( F \) is applied to both springs, we can denote the forces on the springs as \( F_1 \) for spring 1 and \( F_2 \) for spring 2. Since the same force is applied, we have:
\[ F_1 = F_2 = F \]

2. **Using Hooke's Law**:
According to Hooke's Law, the force exerted by a spring is given by:
\[ F = K \cdot x \]
where \( x \) is the extension or compression of the spring. For the two springs, we can write:
\[ F = K_1 \cdot x_1 \quad \text{(for spring 1)} \]
\[ F = K_2 \cdot x_2 \quad \text{(for spring 2)} \]

3. **Relating Extensions to Force Constants**:
Since both springs are subjected to the same force \( F \), we can express the extensions \( x_1 \) and \( x_2 \) in terms of the force constants:
\[ x_1 = \frac{F}{K_1} \]
\[ x_2 = \frac{F}{K_2} \]

4. **Comparing Extensions**:
Since \( K_1 < K_2 \), it follows that:
\[ x_1 = \frac{F}{K_1} > x_2 = \frac{F}{K_2} \]
This means the extension of spring 1 is greater than that of spring 2.

5. **Calculating Work Done on Each Spring**:
The work done on a spring when it is stretched is given by the formula:
\[ W = \frac{1}{2} K x^2 \]
Therefore, the work done on spring 1 is:
\[ W_1 = \frac{1}{2} K_1 x_1^2 \]
and for spring 2:
\[ W_2 = \frac{1}{2} K_2 x_2^2 \]

6. **Substituting the Extensions into Work Done**:
Substitute the expressions for \( x_1 \) and \( x_2 \):
\[ W_1 = \frac{1}{2} K_1 \left(\frac{F}{K_1}\right)^2 = \frac{1}{2} \frac{F^2}{K_1} \]
\[ W_2 = \frac{1}{2} K_2 \left(\frac{F}{K_2}\right)^2 = \frac{1}{2} \frac{F^2}{K_2} \]

7. **Comparing Work Done**:
Since \( K_1 < K_2 \), it follows that:
\[ \frac{1}{K_1} > \frac{1}{K_2} \]
Therefore:
\[ W_1 > W_2 \]
This means more work is done on spring 1 compared to spring 2.

### Conclusion:
The work done on the first spring (with lower spring constant) is greater than the work done on the second spring (with higher spring constant).

### Final Answer:
More work is done on the first spring than on the second spring.

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