A steel wire and copper wire of the same diameter and of length 1m and 2m respectively are connected end to end and a force is applied. The increase in length of the combination is 0.01m.Calculate the elongation produced in the individual wires. Given that Y of copper is `12 xx 10^(10) Nm^(2)` and that of steel is `20 xx 10^(10) N m^(-2)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have two wires: a steel wire of length \( L_1 = 1 \, \text{m} \) and a copper wire of length \( L_2 = 2 \, \text{m} \). They are connected end to end, and the total elongation of the combination when a force is applied is given as \( \Delta L = 0.01 \, \text{m} \). We need to find the individual elongations \( \Delta L_1 \) (for steel) and \( \Delta L_2 \) (for copper).

### Step 2: Apply the Concept of Young's Modulus
Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L_0} \]
Where:
- \( F \) is the force applied,
- \( A \) is the cross-sectional area,
- \( \Delta L \) is the change in length,
- \( L_0 \) is the original length.

Since both wires have the same diameter and are subjected to the same force, we can express the elongations in terms of Young's modulus and original lengths.

### Step 3: Set Up the Equations
For the steel wire:
\[ \Delta L_1 = \frac{F L_1}{A Y_{\text{steel}}} \]
For the copper wire:
\[ \Delta L_2 = \frac{F L_2}{A Y_{\text{copper}}} \]

### Step 4: Relate the Elongations
From the total elongation:
\[ \Delta L = \Delta L_1 + \Delta L_2 = 0.01 \, \text{m} \]

### Step 5: Express \( \Delta L_1 \) and \( \Delta L_2 \) in Terms of Each Other
Using the relationships from Young's modulus, we can express \( \Delta L_1 \) and \( \Delta L_2 \):
- For steel:
\[ \Delta L_1 = \frac{F \cdot 1}{A \cdot (20 \times 10^{10})} \]
- For copper:
\[ \Delta L_2 = \frac{F \cdot 2}{A \cdot (12 \times 10^{10})} \]

### Step 6: Find the Ratio of Elongations
Taking the ratio of the elongations:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{L_1/Y_{\text{steel}}}{L_2/Y_{\text{copper}}} \]
Substituting the values:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{1/20 \times 10^{10}}{2/12 \times 10^{10}} = \frac{12}{40} = \frac{3}{10} \]
Thus, we can write:
\[ \Delta L_1 = 0.3 \Delta L_2 \]

### Step 7: Substitute into the Total Elongation Equation
Now substituting \( \Delta L_1 \) in the total elongation equation:
\[ 0.3 \Delta L_2 + \Delta L_2 = 0.01 \]
\[ 1.3 \Delta L_2 = 0.01 \]
\[ \Delta L_2 = \frac{0.01}{1.3} \approx 0.0077 \, \text{m} \]

### Step 8: Calculate \( \Delta L_1 \)
Now, substituting \( \Delta L_2 \) back to find \( \Delta L_1 \):
\[ \Delta L_1 = 0.3 \times 0.0077 \approx 0.0023 \, \text{m} \]

### Final Results
- Elongation in steel wire \( \Delta L_1 \approx 0.0023 \, \text{m} \)
- Elongation in copper wire \( \Delta L_2 \approx 0.0077 \, \text{m} \)