A wire of length `1 m` and radius `1mm` is subjected to a load. The extension is `x`.The wire is melted and then drawn into a wire of square cross - section of side `2 mm` Its extension under the same load will be

Options

A. A) `(pi^(2))x/8`
B. B) `(pi^(2))x/16`
C. C) ` (pi^(2))x/2`
D. D) `x/(2pi)`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given data
- Length of the first wire (L1) = 1 m
- Radius of the first wire (r1) = 1 mm = \(1 \times 10^{-3}\) m
- Extension of the first wire = x
- Side of the square cross-section of the second wire = 2 mm = \(2 \times 10^{-3}\) m

### Step 2: Calculate the cross-sectional area of the first wire (A1)
The cross-sectional area of a cylindrical wire is given by the formula:
\[ A_1 = \pi r_1^2 \]
Substituting the value of \(r_1\):
\[ A_1 = \pi (1 \times 10^{-3})^2 = \pi \times 10^{-6} \text{ m}^2 \]

### Step 3: Calculate the cross-sectional area of the second wire (A2)
The cross-sectional area of a square wire is given by the formula:
\[ A_2 = \text{side}^2 \]
Substituting the value of the side:
\[ A_2 = (2 \times 10^{-3})^2 = 4 \times 10^{-6} \text{ m}^2 \]

### Step 4: Use the conservation of volume
Since the wire is melted and reformed, the volume before and after must be equal:
\[ \text{Volume of first wire} = \text{Volume of second wire} \]
The volume of the first wire (cylinder) is:
\[ V_1 = A_1 \times L_1 = \pi \times 10^{-6} \times 1 \]
The volume of the second wire (cuboid) is:
\[ V_2 = A_2 \times L_2 = 4 \times 10^{-6} \times L_2 \]
Setting these equal:
\[ \pi \times 10^{-6} = 4 \times 10^{-6} \times L_2 \]

### Step 5: Solve for L2
Rearranging the equation to find \(L_2\):
\[ L_2 = \frac{\pi \times 10^{-6}}{4 \times 10^{-6}} = \frac{\pi}{4} \text{ m} \]

### Step 6: Relate the extensions using Young's modulus
Using the formula for Young's modulus:
\[ Y = \frac{F L}{A \delta L} \]
For the first wire:
\[ Y = \frac{F L_1}{A_1 x} \]
For the second wire:
\[ Y = \frac{F L_2}{A_2 \delta L_2} \]
Since the material is the same, we can set these two equal:
\[ \frac{F L_1}{A_1 x} = \frac{F L_2}{A_2 \delta L_2} \]
Canceling \(F\) from both sides:
\[ \frac{L_1}{A_1 x} = \frac{L_2}{A_2 \delta L_2} \]

### Step 7: Solve for \(\delta L_2\)
Rearranging gives:
\[ \delta L_2 = \frac{L_2 A_2 x}{L_1 A_1} \]
Substituting the values:
\[ \delta L_2 = \frac{\left(\frac{\pi}{4}\right) (4 \times 10^{-6}) x}{(1) (\pi \times 10^{-6})} \]
Simplifying:
\[ \delta L_2 = \frac{\pi \times 10^{-6} x}{\pi \times 10^{-6}} = \frac{\pi}{4} x \]

### Final Result
Thus, the extension of the second wire under the same load is:
\[ \delta L_2 = \frac{\pi}{4} x \]