Two exactly similar wires of steel and copper are stretched by equal forces. If the total elogation is 1cm. Find by how much is each wire elongated ? Given Y for steel ` = 20 xx 10^(11) dyn e//cm^2 and Y for copper = 12xx10^(11) dyn e//cm^2`

Explanation

To solve the problem, we need to find the elongation of each wire (steel and copper) when they are stretched by equal forces, given that the total elongation is 1 cm.

### Step-by-step Solution:

1. **Understanding the Problem**:
- We have two wires: one made of steel and the other made of copper.
- The total elongation of both wires combined is 1 cm.
- We need to find the individual elongations of the steel wire (\( \Delta L_s \)) and the copper wire (\( \Delta L_c \)).

2. **Using Young's Modulus**:
- Young's modulus (\( Y \)) is defined as:
\[ Y = \frac{F \cdot L_0}{A \cdot \Delta L} \]
where \( F \) is the force applied, \( L_0 \) is the original length, \( A \) is the cross-sectional area, and \( \Delta L \) is the elongation.

3. **Setting Up the Equations**:
- For steel:
\[ Y_s = \frac{F \cdot L_0}{A_s \cdot \Delta L_s} \]
- For copper:
\[ Y_c = \frac{F \cdot L_0}{A_c \cdot \Delta L_c} \]

4. **Taking the Ratio**:
- Since both wires are similar and stretched by the same force, we can set up the ratio of elongations:
\[ \frac{\Delta L_c}{\Delta L_s} = \frac{Y_s}{Y_c} \]
- Plugging in the values of Young's modulus:
\[ \frac{\Delta L_c}{\Delta L_s} = \frac{20 \times 10^{11}}{12 \times 10^{11}} = \frac{20}{12} = \frac{5}{3} \]
- Therefore, we can express \( \Delta L_c \) in terms of \( \Delta L_s \):
\[ \Delta L_c = \frac{5}{3} \Delta L_s \]

5. **Using the Total Elongation**:
- We know that:
\[ \Delta L_s + \Delta L_c = 1 \text{ cm} \]
- Substituting \( \Delta L_c \):
\[ \Delta L_s + \frac{5}{3} \Delta L_s = 1 \]
- Combining the terms:
\[ \frac{8}{3} \Delta L_s = 1 \]

6. **Solving for \( \Delta L_s \)**:
- Rearranging gives:
\[ \Delta L_s = \frac{3}{8} \text{ cm} = 0.375 \text{ cm} \]

7. **Finding \( \Delta L_c \)**:
- Now substituting back to find \( \Delta L_c \):
\[ \Delta L_c = \frac{5}{3} \Delta L_s = \frac{5}{3} \times 0.375 = 0.625 \text{ cm} \]

### Final Answer:
- The elongation of the steel wire (\( \Delta L_s \)) is **0.375 cm**.
- The elongation of the copper wire (\( \Delta L_c \)) is **0.625 cm**.