A wire is stretched through 2mm by a certain load. The extension produced in a wire of the same material with double the length and radius with the same load will be

Options

A. 2mm
B. 4mm
C. 1mm
D. 0.5mm

Explanation

To solve the problem, we need to determine the extension produced in a wire of the same material when its length and radius are doubled while keeping the same load applied.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- Let the original length of the wire be \( L \) and the radius be \( R \).
- The extension produced in this wire when a load \( F \) is applied is given as \( \Delta L = 2 \, \text{mm} \).

2. **Formulating the Extension**:
- The formula for extension \( \Delta L \) in a wire is given by:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
- Where:
- \( F \) = applied force (load)
- \( L \) = original length of the wire
- \( A \) = cross-sectional area of the wire
- \( Y \) = Young's modulus of the material

3. **Calculating the Cross-Sectional Area**:
- The cross-sectional area \( A \) of the wire with radius \( R \) is:
\[ A = \pi R^2 \]

4. **Applying the Formula for the Original Wire**:
- Substituting the area into the extension formula:
\[ \Delta L = \frac{F \cdot L}{\pi R^2 \cdot Y} \]

5. **Considering the New Wire**:
- For the new wire, the length is doubled (\( 2L \)) and the radius is also doubled (\( 2R \)).
- The new cross-sectional area \( A' \) becomes:
\[ A' = \pi (2R)^2 = \pi \cdot 4R^2 = 4\pi R^2 \]

6. **Calculating the Extension for the New Wire**:
- The extension \( \Delta L' \) in the new wire can be expressed as:
\[ \Delta L' = \frac{F \cdot (2L)}{A' \cdot Y} = \frac{F \cdot (2L)}{4\pi R^2 \cdot Y} \]

7. **Simplifying the Expression**:
- Substitute \( A' \) into the extension formula:
\[ \Delta L' = \frac{2F \cdot L}{4\pi R^2 \cdot Y} = \frac{1}{2} \cdot \frac{F \cdot L}{\pi R^2 \cdot Y} \]
- Recognizing that \( \frac{F \cdot L}{\pi R^2 \cdot Y} \) is equal to the original extension \( \Delta L \):
\[ \Delta L' = \frac{1}{2} \Delta L \]

8. **Substituting the Known Value**:
- Since \( \Delta L = 2 \, \text{mm} \):
\[ \Delta L' = \frac{1}{2} \cdot 2 \, \text{mm} = 1 \, \text{mm} \]

### Final Answer:
The extension produced in the new wire is **1 mm**.