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 step by step, we will use the concepts of Young's modulus, stress, and strain.

### Step 1: Understand the problem
We have a copper wire and a steel wire connected end to end. The lengths of the wires and their diameters are given. The total elongation when a load is applied is also provided.

### Step 2: Identify the given values
- Length of copper wire, \( L_{Cu} = 2.4 \, \text{m} \)
- Length of steel wire, \( L_{steel} = 1.6 \, \text{m} \)
- Diameter of both wires, \( D = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m} \)
- Radius, \( r = \frac{D}{2} = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m} \)
- Total elongation, \( \Delta L = 0.7 \, \text{mm} = 0.7 \times 10^{-3} \, \text{m} \)
- Young's modulus for copper, \( Y_{Cu} = 1.2 \times 10^{11} \, \text{N/m}^2 \)
- Young's modulus for steel, \( Y_{steel} = 2 \times 10^{11} \, \text{N/m}^2 \)

### Step 3: Calculate the cross-sectional area of the wires
The cross-sectional area \( A \) of the wires can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 = \pi (1.5 \times 10^{-3})^2 = \pi \times 2.25 \times 10^{-6} \approx 7.06858 \times 10^{-6} \, \text{m}^2 \]

### Step 4: Relate the elongations of the wires
Since both wires are in series, the total elongation is the sum of the elongations of each wire:
\[ \Delta L_{Cu} + \Delta L_{steel} = \Delta L \]
Let \( \Delta L_{Cu} = \Delta L_1 \) and \( \Delta L_{steel} = \Delta L_2 \).

### Step 5: Use Young's modulus to relate stress and strain
For the copper wire:
\[ \frac{F}{A} = Y_{Cu} \cdot \frac{\Delta L_{Cu}}{L_{Cu}} \implies \Delta L_{Cu} = \frac{F L_{Cu}}{A Y_{Cu}} \]
For the steel wire:
\[ \frac{F}{A} = Y_{steel} \cdot \frac{\Delta L_{steel}}{L_{steel}} \implies \Delta L_{steel} = \frac{F L_{steel}}{A Y_{steel}} \]

### Step 6: Substitute the elongation equations into the total elongation equation
Substituting the expressions for \( \Delta L_{Cu} \) and \( \Delta L_{steel} \):
\[ \frac{F L_{Cu}}{A Y_{Cu}} + \frac{F L_{steel}}{A Y_{steel}} = \Delta L \]
Factoring out \( F \):
\[ F \left( \frac{L_{Cu}}{A Y_{Cu}} + \frac{L_{steel}}{A Y_{steel}} \right) = \Delta L \]
\[ F = \frac{\Delta L}{\frac{L_{Cu}}{A Y_{Cu}} + \frac{L_{steel}}{A Y_{steel}}} \]

### Step 7: Substitute the known values
Substituting the known values into the equation:
\[ F = \frac{0.7 \times 10^{-3}}{\frac{2.4}{7.06858 \times 10^{-6} \times 1.2 \times 10^{11}} + \frac{1.6}{7.06858 \times 10^{-6} \times 2 \times 10^{11}}} \]

Calculating the denominators:
1. For copper:
\[ \frac{2.4}{7.06858 \times 10^{-6} \times 1.2 \times 10^{11}} \approx 2.84 \times 10^{-3} \]
2. For steel:
\[ \frac{1.6}{7.06858 \times 10^{-6} \times 2 \times 10^{11}} \approx 1.14 \times 10^{-3} \]

Adding these:
\[ \frac{L_{Cu}}{A Y_{Cu}} + \frac{L_{steel}}{A Y_{steel}} \approx 2.84 \times 10^{-3} + 1.14 \times 10^{-3} = 3.98 \times 10^{-3} \]

### Step 8: Calculate the force
Now substituting back to find \( F \):
\[ F = \frac{0.7 \times 10^{-3}}{3.98 \times 10^{-3}} \approx 176.4 \, \text{N} \]

### Final Answer
The load applied is approximately \( F \approx 176.4 \, \text{N} \).

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