Two wires one of copper and other of iron having same cross-section area and lengths 2 m and 1.5 m respectively are fastened end and stretched by a load 10 N. If copper wire is stretched by 2mm then find the total extension of combined wire, `Y_(Cu)=1xx10^(11)N//m^(2)` and `Y_(Fe)=3xx10^(11)N//m^(2)`.

Answer

To solve the problem, we need to find the total extension of two wires (one copper and one iron) when they are stretched by a load of 10 N. We are given the Young's modulus for both materials, their lengths, and the extension of the copper wire.

### Step-by-Step Solution:

1. **Understand Young's Modulus**:
Young's modulus (Y) is defined as the ratio of stress to strain. Mathematically, it can be expressed as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L} \]
where:
- \( F \) = Force applied
- \( A \) = Cross-sectional area
- \( \Delta L \) = Change in length (extension)
- \( L \) = Original length

2. **Set Up the Equations**:
For the copper wire:
\[ Y_{Cu} = \frac{F \cdot L_{Cu}}{A \cdot \Delta L_{Cu}} \]
For the iron wire:
\[ Y_{Fe} = \frac{F \cdot L_{Fe}}{A \cdot \Delta L_{Fe}} \]

3. **Given Values**:
- \( Y_{Cu} = 1 \times 10^{11} \, \text{N/m}^2 \)
- \( Y_{Fe} = 3 \times 10^{11} \, \text{N/m}^2 \)
- \( L_{Cu} = 2 \, \text{m} \)
- \( L_{Fe} = 1.5 \, \text{m} \)
- \( F = 10 \, \text{N} \)
- \( \Delta L_{Cu} = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

4. **Calculate the Extension in Iron Wire**:
From the equations for Young's modulus, we can express the extensions in terms of the known values. Rearranging the equation for iron:
\[ \Delta L_{Fe} = \frac{F \cdot L_{Fe}}{Y_{Fe} \cdot A} \]
We can also relate the two equations by dividing them:
\[ \frac{Y_{Cu}}{Y_{Fe}} = \frac{L_{Cu} \cdot \Delta L_{Fe}}{L_{Fe} \cdot \Delta L_{Cu}} \]
Rearranging gives:
\[ \Delta L_{Fe} = \frac{Y_{Cu}}{Y_{Fe}} \cdot \frac{L_{Fe}}{L_{Cu}} \cdot \Delta L_{Cu} \]

5. **Substituting Known Values**:
Substituting the known values into the equation:
\[ \Delta L_{Fe} = \frac{1 \times 10^{11}}{3 \times 10^{11}} \cdot \frac{1.5}{2} \cdot (2 \times 10^{-3}) \]
Simplifying:
\[ \Delta L_{Fe} = \frac{1}{3} \cdot \frac{1.5}{2} \cdot (2 \times 10^{-3}) \]
\[ \Delta L_{Fe} = \frac{1.5 \times 2 \times 10^{-3}}{6} = \frac{3 \times 10^{-3}}{6} = 0.5 \times 10^{-3} \, \text{m} = 0.5 \, \text{mm} \]

6. **Calculate Total Extension**:
Now, we can find the total extension:
\[ \Delta L_{\text{total}} = \Delta L_{Cu} + \Delta L_{Fe} = 2 \, \text{mm} + 0.5 \, \text{mm} = 2.5 \, \text{mm} \]

### Final Answer:
The total extension of the combined wire is \( 2.5 \, \text{mm} \).