A copper wire of length 2.4 m and a steel wire of length 1.6 m, both of diameter 3 mm, are connected end to end. When stretched by a load, the net elongation found to be 0.7 mm. The load appled is
`(Y_("Copper") = 1.2 xx 10^(11) N m^(-2) , Y_("steel") = 2 xx 10^(11) N m^(-2))`

Options

A. `1.2xx10^(2)N`
B. `1.8xx10^(2)N`
C. `2.4xx10^(2)N`
D. `3.2 xx 10^(2)N`

Explanation

To solve the problem, we need to determine the force experienced by the copper and steel wires when they are stretched by a load. The elongation of both wires combined is given, and we will use the relationship between stress, strain, and Young's modulus to find the force.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Length of copper wire, \( L_c = 2.4 \, \text{m} \)
- Length of steel wire, \( L_s = 1.6 \, \text{m} \)
- Diameter of both wires, \( d = 3 \, \text{mm} = 0.003 \, \text{m} \)
- Young's modulus for copper, \( Y_c = 1.2 \times 10^{11} \, \text{N/m}^2 \)
- Young's modulus for steel, \( Y_s = 2.0 \times 10^{11} \, \text{N/m}^2 \)
- Total elongation, \( \Delta L = 0.7 \, \text{mm} = 0.0007 \, \text{m} \)

2. **Calculate the Cross-Sectional Area:**
The cross-sectional area \( A \) of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{0.003}{2} \right)^2 = \pi \times (0.0015)^2 \approx 7.06858 \times 10^{-6} \, \text{m}^2 \]

3. **Set Up the Relationship Between Elongations:**
Using the relationship between elongation, Young's modulus, and lengths, we can write:
\[ \frac{Y_c}{Y_s} = \frac{L_c \cdot \Delta L_s}{L_s \cdot \Delta L_c} \]
where \( \Delta L_c \) and \( \Delta L_s \) are the elongations of the copper and steel wires, respectively.

4. **Express the Total Elongation:**
Since the total elongation is the sum of the elongations of both wires:
\[ \Delta L_c + \Delta L_s = 0.0007 \, \text{m} \]

5. **Substituting Values:**
From the above relationship, we can express \( \Delta L_c \) in terms of \( \Delta L_s \):
\[ \Delta L_c = \frac{Y_c}{Y_s} \cdot \frac{L_c}{L_s} \cdot \Delta L_s \]
Substituting the values:
\[ \Delta L_c = \frac{1.2 \times 10^{11}}{2.0 \times 10^{11}} \cdot \frac{2.4}{1.6} \cdot \Delta L_s = 0.6 \cdot 1.5 \cdot \Delta L_s = 0.9 \Delta L_s \]

6. **Total Elongation Equation:**
Now substituting \( \Delta L_c \) into the total elongation equation:
\[ 0.9 \Delta L_s + \Delta L_s = 0.0007 \]
\[ 1.9 \Delta L_s = 0.0007 \implies \Delta L_s = \frac{0.0007}{1.9} \approx 0.000368 \, \text{m} \approx 0.368 \, \text{mm} \]

7. **Calculate Elongation of Copper Wire:**
Now, substituting \( \Delta L_s \) back to find \( \Delta L_c \):
\[ \Delta L_c = 0.9 \cdot 0.000368 \approx 0.000331 \, \text{m} \approx 0.331 \, \text{mm} \]

8. **Calculate the Force:**
Using Young's modulus to find the force in the copper wire:
\[ Y_c = \frac{F}{A \cdot \frac{\Delta L_c}{L_c}} \implies F = Y_c \cdot A \cdot \frac{\Delta L_c}{L_c} \]
Substituting the values:
\[ F = 1.2 \times 10^{11} \cdot 7.06858 \times 10^{-6} \cdot \frac{0.000331}{2.4} \]
\[ F \approx 1.2 \times 10^{11} \cdot 7.06858 \times 10^{-6} \cdot 0.000137917 \approx 1.8 \times 10^2 \, \text{N} \]

### Final Answer:
The force applied on both wires is approximately **180 N**.