A 5 m long aluminium wire `(Y=7xx10^10(N)/(m^2)`) of diameter 3 mm supprts a 40 kg mass. In order to have the same elongation in a copper wire `(Y=12xx10^10(N)/(m^2)`) of the same length under the same weight, the diameter should now, in mm

Options

A. 1.75
B. 1.5
C. 2.5
D. 5

Explanation

To solve the problem, we need to find the diameter of a copper wire that will have the same elongation as a 5 m long aluminum wire under the same load.

### Step-by-step Solution:

1. **Identify Given Data:**
- Length of the aluminum wire, \( L = 5 \, \text{m} \)
- Diameter of the aluminum wire, \( d_{Al} = 3 \, \text{mm} \) (which gives a radius \( r_{Al} = \frac{3}{2} = 1.5 \, \text{mm} \))
- Mass supported by the wire, \( m = 40 \, \text{kg} \)
- Young's modulus of aluminum, \( Y_{Al} = 7 \times 10^{10} \, \text{N/m}^2 \)
- Young's modulus of copper, \( Y_{Cu} = 12 \times 10^{10} \, \text{N/m}^2 \)

2. **Calculate the Force (Weight) Acting on the Wire:**
\[ F = m \cdot g = 40 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 392.4 \, \text{N} \]

3. **Use the Formula for Stress and Strain:**
- Stress, \( \sigma = \frac{F}{A} = \frac{F}{\pi r^2} \)
- Strain, \( \epsilon = \frac{\Delta L}{L} \)
- From the definition of Young's modulus:
\[ Y = \frac{\sigma}{\epsilon} \]
Therefore, we can express strain in terms of stress and Young's modulus:
\[ \epsilon = \frac{\sigma}{Y} \]

4. **Set Up the Equation for Both Wires:**
Since both wires have the same elongation (strain), we can equate the strains:
\[ \frac{F}{\pi r_{Al}^2 Y_{Al}} = \frac{F}{\pi r_{Cu}^2 Y_{Cu}} \]
The force \( F \) cancels out:
\[ \frac{1}{r_{Al}^2 Y_{Al}} = \frac{1}{r_{Cu}^2 Y_{Cu}} \]

5. **Rearranging the Equation:**
\[ r_{Cu}^2 = r_{Al}^2 \cdot \frac{Y_{Cu}}{Y_{Al}} \]

6. **Substituting the Values:**
- First, calculate \( r_{Al}^2 \):
\[ r_{Al} = 1.5 \, \text{mm} = 0.0015 \, \text{m} \]
\[ r_{Al}^2 = (0.0015)^2 = 2.25 \times 10^{-6} \, \text{m}^2 \]
- Now substitute the values of Young's modulus:
\[ r_{Cu}^2 = 2.25 \times 10^{-6} \cdot \frac{12 \times 10^{10}}{7 \times 10^{10}} \]
\[ r_{Cu}^2 = 2.25 \times 10^{-6} \cdot \frac{12}{7} \]
\[ r_{Cu}^2 = 2.25 \times 10^{-6} \cdot 1.714285714 \approx 3.86 \times 10^{-6} \, \text{m}^2 \]

7. **Calculating the Radius of the Copper Wire:**
\[ r_{Cu} = \sqrt{3.86 \times 10^{-6}} \approx 0.00196 \, \text{m} = 1.96 \, \text{mm} \]

8. **Finding the Diameter of the Copper Wire:**
\[ d_{Cu} = 2 \cdot r_{Cu} = 2 \cdot 1.96 \approx 3.92 \, \text{mm} \]

### Final Answer:
The diameter of the copper wire should be approximately **3.92 mm**.