A copper wire and a steel wire having the same cross-sectional area are fastened end to end and stretched by a fore F. The lengths of cpper and steel wires are in the ratio of `2:1` and their moduli of elasticity are in the ratio of `1:2`. What is the ratio `e_(c)/e_(s)` of their extensions?

Options

A. `1:2`
B. `4:1`
C. `2:1`
D. `1:4`

Explanation

To solve the problem, we need to find the ratio of extensions \( e_c \) (extension in copper wire) to \( e_s \) (extension in steel wire) when both wires are subjected to the same force \( F \).

### Given Data:
1. Length ratio of copper wire to steel wire: \( \frac{L_C}{L_S} = \frac{2}{1} \)
2. Young's modulus ratio of copper to steel: \( \frac{Y_C}{Y_S} = \frac{1}{2} \)

### Step-by-Step Solution:

1. **Understanding Young's Modulus**:
Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \frac{F L}{A \Delta L} \]
where \( F \) is the force applied, \( A \) is the cross-sectional area, \( \Delta L \) is the extension, and \( L \) is the original length.

2. **Expressing Extensions**:
Rearranging the formula for Young's modulus gives us:
\[ \Delta L = \frac{F L}{A Y} \]
Therefore, the extensions for copper and steel wires can be expressed as:
\[ \Delta L_C = \frac{F L_C}{A Y_C} \quad \text{and} \quad \Delta L_S = \frac{F L_S}{A Y_S} \]

3. **Finding the Ratio of Extensions**:
The ratio of extensions \( \frac{e_C}{e_S} \) can be written as:
\[ \frac{e_C}{e_S} = \frac{\Delta L_C}{\Delta L_S} = \frac{\frac{F L_C}{A Y_C}}{\frac{F L_S}{A Y_S}} \]
The forces \( F \) and areas \( A \) cancel out:
\[ \frac{e_C}{e_S} = \frac{L_C Y_S}{L_S Y_C} \]

4. **Substituting the Known Ratios**:
Now, substituting the known ratios:
- \( \frac{L_C}{L_S} = \frac{2}{1} \)
- \( \frac{Y_C}{Y_S} = \frac{1}{2} \) implies \( Y_S = 2Y_C \)

Thus, we have:
\[ \frac{e_C}{e_S} = \frac{L_C Y_S}{L_S Y_C} = \frac{L_C}{L_S} \cdot \frac{Y_S}{Y_C} = \frac{2}{1} \cdot \frac{2}{1} = 4 \]

5. **Final Result**:
Therefore, the ratio of extensions is:
\[ \frac{e_C}{e_S} = 4 \]

### Conclusion:
The ratio of extensions \( \frac{e_C}{e_S} \) is \( 4:1 \).

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