Two wires of the same material have lengths in the ratio `1:2` and their radii are in the ratio `1:sqrt(2)` If they are stretched by applying equal forces, the increase in their lengths will be in the ratio

Options

A. `sqrt(2):2`
B. `2:sqrt(2)`
C. `1:1`
D. `1:2`

Explanation

To solve the problem, we need to find the ratio of the increase in lengths of two wires that are stretched by equal forces. We will use the relationship between stress, strain, and Young's modulus.

### Step-by-Step Solution:

1. **Identify Given Ratios**:
- Lengths of the wires: \( L_1 : L_2 = 1 : 2 \)
- Radii of the wires: \( r_1 : r_2 = 1 : \sqrt{2} \)

2. **Understand Young's Modulus**:
- Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
- Rearranging gives:
\[ \Delta L = \frac{F L}{A Y} \]
- Where \( A \) is the cross-sectional area of the wire.

3. **Calculate the Cross-Sectional Area**:
- The cross-sectional area \( A \) of a wire is given by:
\[ A = \pi r^2 \]
- Therefore, for the two wires:
\[ A_1 = \pi r_1^2 \quad \text{and} \quad A_2 = \pi r_2^2 \]

4. **Express the Increase in Length**:
- For wire 1:
\[ \Delta L_1 = \frac{F L_1}{\pi r_1^2 Y} \]
- For wire 2:
\[ \Delta L_2 = \frac{F L_2}{\pi r_2^2 Y} \]

5. **Find the Ratio of Increases in Length**:
- The ratio of the increases in lengths is:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{F L_1}{\pi r_1^2 Y} \cdot \frac{\pi r_2^2 Y}{F L_2} = \frac{L_1 r_2^2}{L_2 r_1^2} \]

6. **Substitute the Ratios**:
- Substitute \( L_1 = 1 \), \( L_2 = 2 \), \( r_1 = 1 \), and \( r_2 = \sqrt{2} \):
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{1 \cdot (\sqrt{2})^2}{2 \cdot 1^2} = \frac{1 \cdot 2}{2 \cdot 1} = \frac{2}{2} = 1 \]

7. **Conclusion**:
- The ratio of the increase in lengths of the two wires is:
\[ \Delta L_1 : \Delta L_2 = 1 : 1 \]

### Final Answer:
The increase in their lengths will be in the ratio \( 1 : 1 \).