There are two massless springs A and B spring constant `K_(A)` and `K_(B)` respectively and `K_(A)ltK_(B)` if `W_(A)` and `W_(B)` be denoted as work done on A and work done on B respectvely. Then

Options

A. if they are are compressed to same distance `W_(A)gtW_(B)`
B. if they are compressed by same force (upto equilibrium state) `W_(A)ltW_(B)`
C. if they are compressed by ame distance `W_(A)=W_(B)`
D. if they are compressed by same force (upto equilibrium state) `W_(A)gtW_(B)`

Explanation

To solve the problem regarding the work done on two massless springs A and B with spring constants \( K_A \) and \( K_B \) respectively, where \( K_A < K_B \), we will follow these steps:

### Step 1: Understand the Work Done on a Spring
The work done \( W \) on a spring when it is compressed or stretched by a distance \( x \) is given by the formula:
\[ W = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant.

### Step 2: Set Up the Work Done for Each Spring
For spring A:
\[ W_A = \frac{1}{2} K_A x_A^2 \]
For spring B:
\[ W_B = \frac{1}{2} K_B x_B^2 \]

### Step 3: Analyze the Compression of the Springs
If both springs are compressed by the same distance \( x \), we can set \( x_A = x_B = x \). Thus, the work done on each spring becomes:
\[ W_A = \frac{1}{2} K_A x^2 \]
\[ W_B = \frac{1}{2} K_B x^2 \]

### Step 4: Compare the Work Done
Since \( K_A < K_B \), we can compare the work done:
\[ W_A = \frac{1}{2} K_A x^2 < \frac{1}{2} K_B x^2 = W_B \]
This shows that the work done on spring A is less than the work done on spring B.

### Step 5: Conclusion
Thus, we conclude that:
\[ W_A < W_B \]