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. 1.65
C. 2.5
D. `5.0`

Explanation

To solve the problem step by step, we will follow the principles of elasticity and the relationship between Young's modulus, stress, strain, and dimensions of the wire.

### Step 1: Understand the given values
- Length of the aluminum wire, \( L = 5 \, \text{m} \)
- Young's modulus of aluminum, \( Y_1 = 7 \times 10^{10} \, \text{N/m}^2 \)
- Diameter of aluminum wire, \( d_1 = 3 \, \text{mm} = 0.003 \, \text{m} \)
- Mass supported by the wire, \( m = 40 \, \text{kg} \)
- Young's modulus of copper, \( Y_2 = 12 \times 10^{10} \, \text{N/m}^2 \)

### Step 2: Calculate the force acting on the wire
The force \( F \) acting on the wire due to the mass is given by:
\[ F = m \cdot g \]
where \( g \approx 9.81 \, \text{m/s}^2 \). Thus,
\[ F = 40 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 392.4 \, \text{N} \]

### Step 3: Write the formula for Young's modulus
Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L_0} \]
where:
- \( A \) is the cross-sectional area of the wire,
- \( \Delta L \) is the change in length (elongation),
- \( L_0 \) is the original length of the wire.

### Step 4: Calculate the cross-sectional area of the aluminum wire
The cross-sectional area \( A \) of the wire is given by:
\[ A = \frac{\pi d^2}{4} \]
Substituting \( d_1 = 0.003 \, \text{m} \):
\[ A_1 = \frac{\pi (0.003)^2}{4} = \frac{\pi \cdot 0.000009}{4} \approx 7.06858 \times 10^{-6} \, \text{m}^2 \]

### Step 5: Set up the equation for elongation
Using the Young's modulus formula for aluminum:
\[ Y_1 = \frac{F/A_1}{\Delta L/L_0} \]
Rearranging gives:
\[ \Delta L = \frac{F \cdot L_0}{Y_1 \cdot A_1} \]

### Step 6: Set up the equation for the copper wire
For the copper wire, we want the same elongation \( \Delta L \):
\[ Y_2 = \frac{F/A_2}{\Delta L/L_0} \]
Rearranging gives:
\[ \Delta L = \frac{F \cdot L_0}{Y_2 \cdot A_2} \]

### Step 7: Equate the elongations
Since \( \Delta L \) is the same for both wires, we can set the equations equal to each other:
\[ \frac{F \cdot L_0}{Y_1 \cdot A_1} = \frac{F \cdot L_0}{Y_2 \cdot A_2} \]
Cancelling \( F \) and \( L_0 \) from both sides gives:
\[ \frac{1}{Y_1 \cdot A_1} = \frac{1}{Y_2 \cdot A_2} \]

### Step 8: Relate the areas
This simplifies to:
\[ A_2 = A_1 \cdot \frac{Y_2}{Y_1} \]

### Step 9: Substitute for areas
Substituting the area formula:
\[ \frac{\pi d_2^2}{4} = \frac{\pi d_1^2}{4} \cdot \frac{Y_2}{Y_1} \]
Cancelling \( \frac{\pi}{4} \) gives:
\[ d_2^2 = d_1^2 \cdot \frac{Y_2}{Y_1} \]

### Step 10: Solve for \( d_2 \)
Taking the square root:
\[ d_2 = d_1 \cdot \sqrt{\frac{Y_2}{Y_1}} \]
Substituting the values:
\[ d_2 = 0.003 \cdot \sqrt{\frac{12 \times 10^{10}}{7 \times 10^{10}}} \]
Calculating the ratio:
\[ d_2 = 0.003 \cdot \sqrt{\frac{12}{7}} \approx 0.003 \cdot 1.4142 \approx 0.00424 \, \text{m} = 4.24 \, \text{mm} \]

### Final Answer
The diameter of the copper wire should be approximately \( 4.24 \, \text{mm} \).