Two similar springs `P` and `Q` have spring constant `K_(P)` and `K_(Q)` such that `K_(P) gt K_(Q)`. They are stretched, first by the same amount (case a), then the same force (case b). The work done by the spring `W_(P)` and `W_(Q)` are related as, in case (b), respectively

Options

A. `W_(P)=W_(Q),W_(P)gtW_(Q)`
B. `W_(P)=W_(Q),W_(P)=W_(Q)`
C. `W_(P)gtW_(Q),W_(Q)gtW_(P)`
D. `W_(P)ltW_(Q),W_(Q)gtW_(P)`

Explanation

To solve the problem, we will analyze the work done by two springs, P and Q, in two different cases: when they are stretched by the same amount and when they are stretched by the same force.

### Case A: Same Amount of Stretch

1. **Understanding the Work Done Formula**:
The work done by a spring when it is stretched by an amount \( \Delta x \) is given by the formula:
\[ W = \frac{1}{2} k \Delta x^2 \]
where \( k \) is the spring constant.

2. **Work Done by Spring P**:
For spring P, the work done \( W_P \) can be expressed as:
\[ W_P = \frac{1}{2} K_P \Delta x^2 \]

3. **Work Done by Spring Q**:
For spring Q, the work done \( W_Q \) can be expressed as:
\[ W_Q = \frac{1}{2} K_Q \Delta x^2 \]

4. **Taking the Ratio of Work Done**:
Now, we take the ratio of the work done by the two springs:
\[ \frac{W_P}{W_Q} = \frac{\frac{1}{2} K_P \Delta x^2}{\frac{1}{2} K_Q \Delta x^2} \]
The \( \frac{1}{2} \) and \( \Delta x^2 \) terms cancel out:
\[ \frac{W_P}{W_Q} = \frac{K_P}{K_Q} \]

5. **Using the Given Condition**:
Since it is given that \( K_P > K_Q \), we conclude:
\[ W_P > W_Q \]

### Case B: Same Force Applied

1. **Understanding the Force and Work Done**:
When the same force \( F \) is applied to both springs, the work done can be expressed as:
\[ W = \frac{1}{2} F \Delta x \]

2. **Work Done by Spring P**:
For spring P, the work done \( W_P \) is:
\[ W_P = \frac{1}{2} F \Delta x_P \]

3. **Work Done by Spring Q**:
For spring Q, the work done \( W_Q \) is:
\[ W_Q = \frac{1}{2} F \Delta x_Q \]

4. **Taking the Ratio of Work Done**:
The ratio of the work done by the two springs is:
\[ \frac{W_P}{W_Q} = \frac{\frac{1}{2} F \Delta x_P}{\frac{1}{2} F \Delta x_Q} \]
The \( \frac{1}{2} F \) terms cancel out:
\[ \frac{W_P}{W_Q} = \frac{\Delta x_P}{\Delta x_Q} \]

5. **Relating the Displacements**:
Since \( K \) is inversely proportional to the displacement for a given force (Hooke's Law), we have:
\[ K_P \Delta x_P = K_Q \Delta x_Q = F \]
Therefore, we can express:
\[ \Delta x_P = \frac{F}{K_P} \quad \text{and} \quad \Delta x_Q = \frac{F}{K_Q} \]
Since \( K_P > K_Q \), it follows that:
\[ \Delta x_P < \Delta x_Q \]

6. **Final Relation for Work Done**:
Thus, substituting back into the ratio:
\[ \frac{W_P}{W_Q} = \frac{\Delta x_P}{\Delta x_Q} < 1 \]
This implies:
\[ W_P < W_Q \]

### Conclusion
In summary, the relationships for the work done by the springs in both cases are:
- Case A: \( W_P > W_Q \)
- Case B: \( W_P < W_Q \)