The length of a wire is increased by `1 mm` on the application, of a given load. In a wire of the same material, but of length and radius twice that of the first, on application of the same load, extension is

Options

A. `0.25mm`
B. `0.5mm`
C. `2mm`
D. `4mm`

Explanation

To solve the problem, we need to find the extension in a wire that has twice the length and radius of another wire, under the same load. We will use the concept of Young's modulus and the relationship between stress, strain, and extension.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Initial wire length (L1) is increased by 1 mm under a given load (F).
- New wire has length (L2) = 2 * L1 and radius (R2) = 2 * R1.

2. **Recall the Formula for Extension**:
The extension (ΔL) in a wire can be expressed using Young's modulus (Y):
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
where:
- \( A \) is the cross-sectional area of the wire, given by \( A = \pi R^2 \).

3. **Substitute the Area**:
For the first wire:
\[ A_1 = \pi R_1^2 \]
Therefore, the extension for the first wire (ΔL1) is:
\[ \Delta L_1 = \frac{F \cdot L_1}{\pi R_1^2 \cdot Y} \]

4. **Calculate the Extension for the Second Wire**:
For the second wire:
\[ A_2 = \pi R_2^2 = \pi (2R_1)^2 = 4\pi R_1^2 \]
The extension for the second wire (ΔL2) is:
\[ \Delta L_2 = \frac{F \cdot L_2}{A_2 \cdot Y} = \frac{F \cdot (2L_1)}{4\pi R_1^2 \cdot Y} \]

5. **Relate the Extensions**:
We can express ΔL2 in terms of ΔL1:
\[ \Delta L_2 = \frac{2F \cdot L_1}{4\pi R_1^2 \cdot Y} = \frac{1}{2} \cdot \frac{F \cdot L_1}{\pi R_1^2 \cdot Y} = \frac{1}{2} \Delta L_1 \]

6. **Substitute the Known Value**:
Since ΔL1 = 1 mm, we find:
\[ \Delta L_2 = \frac{1}{2} \cdot 1 \text{ mm} = 0.5 \text{ mm} \]

### Final Answer:
The extension in the second wire is **0.5 mm**.