Two wires of equal dimensions and young's modulus `Y_1` and `Y_2` are connected end to end. What is the equivalent young's modulus for the combination?

Options

A. `(Y_1+Y_2)/2`
B. `sqrt(Y_1 + Y_2)`
C. `(2 Y_1 Y_2 )/ (Y_1 + Y_2)`
D. `sqrt((Y_1^2 y_2^2)/2)`

Answer

To find the equivalent Young's modulus \( Y \) for two wires of equal dimensions and Young's moduli \( Y_1 \) and \( Y_2 \) connected end to end, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have two wires, both with length \( L \) and cross-sectional area \( A \).
- The Young's moduli of the wires are \( Y_1 \) and \( Y_2 \).
- When connected end to end, the total length of the combination becomes \( L + L = 2L \).

2. **Define Young's Modulus**:
- Young's modulus \( Y \) is defined as the ratio of stress to strain:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
- Here, \( F \) is the force applied, \( A \) is the cross-sectional area, \( \Delta L \) is the change in length, and \( L \) is the original length.

3. **Calculate Change in Length for Each Wire**:
- For the first wire with Young's modulus \( Y_1 \):
\[ \Delta L_1 = \frac{F L}{A Y_1} \]
- For the second wire with Young's modulus \( Y_2 \):
\[ \Delta L_2 = \frac{F L}{A Y_2} \]

4. **Total Change in Length**:
- Since the wires are connected in series, the total change in length \( \Delta L \) is the sum of the changes in length of both wires:
\[ \Delta L = \Delta L_1 + \Delta L_2 = \frac{F L}{A Y_1} + \frac{F L}{A Y_2} \]

5. **Combine the Expressions**:
- Factor out common terms:
\[ \Delta L = \frac{F L}{A} \left( \frac{1}{Y_1} + \frac{1}{Y_2} \right) \]

6. **Express the Equivalent Young's Modulus**:
- The equivalent Young's modulus \( Y \) for the combined wire of length \( 2L \) is given by:
\[ Y = \frac{F/A}{\Delta L/(2L)} \]
- Substituting the expression for \( \Delta L \):
\[ Y = \frac{F/A}{\frac{F L}{A} \left( \frac{1}{Y_1} + \frac{1}{Y_2} \right) / (2L)} \]
- Simplifying this gives:
\[ Y = \frac{2L}{L} \cdot \frac{1}{\frac{1}{Y_1} + \frac{1}{Y_2}} = \frac{2 Y_1 Y_2}{Y_1 + Y_2} \]

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