A wire is stretched `1 mm` by a force of `1 kN`. How far would a wire of the same material and length but of four times that diameter be stretched by the same force?

Options

A. `1/2mm`
B. `1/4mm`
C. `1/8mm`
D. `1/16mm`

Explanation

To solve the problem, we need to use the relationship between the extension of a wire, its diameter, and the force applied. The key concept here is Young's modulus, which relates stress and strain in materials.

### Step-by-step Solution:

1. **Understanding the Given Information**:
- The first wire is stretched by a force of \( F = 1 \, \text{kN} = 1000 \, \text{N} \).
- The extension of the first wire, \( \Delta L_1 = 1 \, \text{mm} = 0.001 \, \text{m} \).
- The diameter of the first wire is \( D_1 \).
- The diameter of the second wire is \( D_2 = 4D_1 \).

2. **Using Young's Modulus**:
- Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
- Here, \( A \) is the cross-sectional area of the wire, and for a circular wire, \( A = \frac{\pi}{4} D^2 \).

3. **Relating the Extensions**:
- For the first wire:
\[ Y = \frac{F}{\Delta L_1 \cdot A_1} = \frac{F}{\Delta L_1 \cdot \frac{\pi}{4} D_1^2} \]
- For the second wire:
\[ Y = \frac{F}{\Delta L_2 \cdot A_2} = \frac{F}{\Delta L_2 \cdot \frac{\pi}{4} D_2^2} \]
- Since the Young's modulus is the same for both wires (same material), we can equate the two expressions.

4. **Setting Up the Proportionality**:
- Since \( A_2 = \frac{\pi}{4} D_2^2 = \frac{\pi}{4} (4D_1)^2 = 16 \cdot \frac{\pi}{4} D_1^2 \), we can substitute this into the equation.
- The relationship between the extensions can be derived:
\[ \Delta L_2 = \Delta L_1 \cdot \left(\frac{D_1}{D_2}\right)^2 \]
- Substituting \( D_2 = 4D_1 \):
\[ \Delta L_2 = \Delta L_1 \cdot \left(\frac{D_1}{4D_1}\right)^2 = \Delta L_1 \cdot \left(\frac{1}{4}\right)^2 = \Delta L_1 \cdot \frac{1}{16} \]

5. **Calculating the Extension of the Second Wire**:
- Now substituting \( \Delta L_1 = 1 \, \text{mm} \):
\[ \Delta L_2 = 1 \, \text{mm} \cdot \frac{1}{16} = \frac{1}{16} \, \text{mm} \]

### Final Answer:
The extension of the second wire, \( \Delta L_2 \), is \( \frac{1}{16} \, \text{mm} \).