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 the principles of elasticity and volume conservation when the wire is melted and redrawn.

### Step 1: Calculate the initial volume of the wire
The initial wire has a length \( L = 1 \, \text{m} \) and a radius \( r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \).

The cross-sectional area \( A \) of the wire is given by:
\[ A = \pi r^2 = \pi (1 \times 10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \]

The initial volume \( V \) of the wire is:
\[ V = L \times A = 1 \times \pi \times 10^{-6} = \pi \times 10^{-6} \, \text{m}^3 \]

### Step 2: Calculate the new cross-sectional area of the drawn wire
The new wire has a square cross-section with a side length of \( 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \).

The area \( A' \) of the square cross-section is:
\[ A' = (2 \times 10^{-3})^2 = 4 \times 10^{-6} \, \text{m}^2 \]

### Step 3: Calculate the new length of the wire after melting and drawing
Since the volume remains constant, we can equate the initial volume to the new volume:
\[ V = L' \times A' \]
Thus,
\[ \pi \times 10^{-6} = L' \times 4 \times 10^{-6} \]

Solving for \( L' \):
\[ L' = \frac{\pi \times 10^{-6}}{4 \times 10^{-6}} = \frac{\pi}{4} \, \text{m} \]

### Step 4: Relate the extensions of the two wires using Young's modulus
The extension \( x \) of the initial wire is given by:
\[ x = \frac{F L}{A Y} \]
Where \( F \) is the force applied, \( Y \) is Young's modulus, \( L \) is the original length, and \( A \) is the original area.

For the new wire, the extension \( x' \) under the same load \( F \) is:
\[ x' = \frac{F L'}{A' Y} \]

### Step 5: Substitute the values into the extension formula
Substituting \( L' \) and \( A' \):
\[ x' = \frac{F \left(\frac{\pi}{4}\right)}{(4 \times 10^{-6}) Y} \]

### Step 6: Express \( x' \) in terms of \( x \)
We know from the original wire:
\[ x = \frac{F (1)}{(\pi \times 10^{-6}) Y} \]

Now, we can express \( F \) in terms of \( x \):
\[ F = x \cdot (\pi \times 10^{-6}) Y \]

Substituting \( F \) back into the equation for \( x' \):
\[ x' = \frac{x \cdot (\pi \times 10^{-6}) Y \cdot \left(\frac{\pi}{4}\right)}{(4 \times 10^{-6}) Y} \]

The \( Y \) and \( 10^{-6} \) terms cancel out:
\[ x' = \frac{x \cdot \pi^2}{16} \]

### Final Answer
The extension of the new wire under the same load will be:
\[ x' = \frac{\pi^2}{16} x \]