Identical springs of stecl and copper `(Y_("Steel") gt Y_("copper"))` are equally stretched.

Options

A.  Less work is done on copper spring.
B. Less work is done on steel spring.
C. Equal work is cone on both the springs.
D. Data is incomplete.

Explanation

To solve the problem, we need to analyze the work done on identical springs made of steel and copper when they are equally stretched. We know that the Young's modulus of steel is greater than that of copper.

### Step-by-step Solution:

1. **Understanding Work Done on Springs**:
The work done (W) on a spring when it is stretched can be expressed as:
\[ W = \frac{F \cdot \Delta L}{2} \]
where \( F \) is the force applied and \( \Delta L \) is the extension of the spring.

2. **Relating Force and Extension to Young's Modulus**:
Young's modulus (Y) is defined as:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
Rearranging this equation gives us:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]

3. **Substituting Extension into Work Done**:
Now, we can substitute \( \Delta L \) into the work done equation:
\[ W = \frac{F \cdot \left( \frac{F \cdot L}{A \cdot Y} \right)}{2} \]
Simplifying this gives:
\[ W = \frac{F^2 \cdot L}{2 \cdot A \cdot Y} \]

4. **Analyzing the Relationship**:
From the equation \( W = \frac{F^2 \cdot L}{2 \cdot A \cdot Y} \), we can see that the work done is inversely proportional to Young's modulus (Y):
\[ W \propto \frac{1}{Y} \]
This means that as Young's modulus increases, the work done decreases.

5. **Comparing Steel and Copper**:
Given that \( Y_{steel} > Y_{copper} \), we can conclude:
\[ W_{steel} < W_{copper} \]
Therefore, less work is done on the steel spring compared to the copper spring.

6. **Final Conclusion**:
Since the work done on the steel spring is less than that on the copper spring, the correct answer is that less work is done on the steel spring.