Two wires of the same length `l` and radius are joined end to end and loaded. If the Young's moduli of the materials of the wires are `Y_(1)` and `Y_(2)`, the combination behaves as a single wire of Young's modulus will be

Options

A. `(Y_(1)+Y_(2))`
B. `sqrt(Y_(1)Y_(2))`
C. `(2Y_(1)Y_(2))/((Y_(1)+Y_(2)))`
D. `(Y_(1)Y_(2))/((Y_(1)+Y_(2)))`

Explanation

To find the equivalent Young's modulus of two wires of the same length and radius joined end to end, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**: We have two wires of the same length \( l \) and radius \( r \) joined end to end. The Young's moduli of the materials of the wires are \( Y_1 \) and \( Y_2 \).

2. **Identify Forces and Areas**: Since the wires are in series, the same force \( F \) (tension) acts on both wires. The cross-sectional area \( A \) of both wires is the same, given by:
\[ A = \pi r^2 \]

3. **Define Extensions**: Let the extensions of the first and second wires be \( \Delta L_1 \) and \( \Delta L_2 \) respectively. The total extension \( \Delta L \) of the combined wire is:
\[ \Delta L = \Delta L_1 + \Delta L_2 \]

4. **Use Young's Modulus Definition**: Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / l} \]
Rearranging gives us:
\[ \Delta L = \frac{F \cdot l}{A \cdot Y} \]

5. **Calculate Extensions for Each Wire**:
- For wire 1:
\[ \Delta L_1 = \frac{F \cdot l}{A \cdot Y_1} \]
- For wire 2:
\[ \Delta L_2 = \frac{F \cdot l}{A \cdot Y_2} \]

6. **Substitute Extensions into Total Extension**:
\[ \Delta L = \Delta L_1 + \Delta L_2 = \frac{F \cdot l}{A \cdot Y_1} + \frac{F \cdot l}{A \cdot Y_2} \]

7. **Factor Out Common Terms**:
\[ \Delta L = \frac{F \cdot l}{A} \left( \frac{1}{Y_1} + \frac{1}{Y_2} \right) \]

8. **Express Total Extension in Terms of Equivalent Young's Modulus**:
The equivalent Young's modulus \( Y_{\text{eq}} \) for the combined wire can be defined as:
\[ \Delta L = \frac{F \cdot (2l)}{A \cdot Y_{\text{eq}}} \]

9. **Set the Two Expressions for Total Extension Equal**:
\[ \frac{F \cdot (2l)}{A \cdot Y_{\text{eq}}} = \frac{F \cdot l}{A} \left( \frac{1}{Y_1} + \frac{1}{Y_2} \right) \]

10. **Cancel Common Terms**:
Cancel \( F \) and \( A \) from both sides:
\[ \frac{2l}{Y_{\text{eq}}} = \frac{l}{Y_1} + \frac{l}{Y_2} \]

11. **Simplify**:
Dividing through by \( l \):
\[ \frac{2}{Y_{\text{eq}}} = \frac{1}{Y_1} + \frac{1}{Y_2} \]

12. **Find Equivalent Young's Modulus**:
Rearranging gives:
\[ Y_{\text{eq}} = \frac{2 Y_1 Y_2}{Y_1 + Y_2} \]

### Final Answer:
The equivalent Young's modulus \( Y_{\text{eq}} \) of the combination of the two wires is:
\[ Y_{\text{eq}} = \frac{2 Y_1 Y_2}{Y_1 + Y_2} \]