A wire is stretched by 10 mm when it is pulled bya certain force. Another wire of the same material but double the length and double the diameter is stretched by the same force. The elongation of the second wire is

Options

A. 5 mm
B. 10 mm
C. 20 mm
D. 40 mm

Explanation

To solve the problem, we will use the concept of Young's modulus and the relationship between stress, strain, and elongation of wires.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have two wires made of the same material.
- The first wire is stretched by a force \( F \) and elongates by \( \Delta L_1 = 10 \, \text{mm} \).
- The second wire has double the length and double the diameter of the first wire.

2. **Identify the Parameters:**
- Let the original length of the first wire be \( L_1 \).
- The elongation of the first wire is \( \Delta L_1 = 10 \, \text{mm} = 10 \times 10^{-3} \, \text{m} \).
- The length of the second wire is \( L_2 = 2L_1 \).
- The diameter of the second wire is \( d_2 = 2d_1 \).
- The radius of the second wire is \( r_2 = \frac{d_2}{2} = 2r_1 \).

3. **Calculating the Cross-Sectional Areas:**
- The cross-sectional area of the first wire is:
\[ A_1 = \pi r_1^2 \]
- The cross-sectional area of the second wire is:
\[ A_2 = \pi r_2^2 = \pi (2r_1)^2 = 4\pi r_1^2 = 4A_1 \]

4. **Using Young's Modulus:**
- Young's modulus \( Y \) is defined as:
\[ Y = \frac{F}{A \cdot \frac{L}{\Delta L}} \]
- For the first wire:
\[ Y = \frac{F}{A_1 \cdot \frac{L_1}{\Delta L_1}} \quad \text{(1)} \]
- For the second wire:
\[ Y = \frac{F}{A_2 \cdot \frac{L_2}{\Delta L_2}} \quad \text{(2)} \]

5. **Setting the Young's Modulus Equations Equal:**
- Since both wires are made of the same material, their Young's moduli are equal:
\[ \frac{F}{A_1 \cdot \frac{L_1}{\Delta L_1}} = \frac{F}{A_2 \cdot \frac{L_2}{\Delta L_2}} \]
- Canceling \( F \) from both sides:
\[ \frac{1}{A_1 \cdot \frac{L_1}{\Delta L_1}} = \frac{1}{A_2 \cdot \frac{L_2}{\Delta L_2}} \]

6. **Substituting the Areas and Lengths:**
- Substitute \( A_2 = 4A_1 \) and \( L_2 = 2L_1 \):
\[ \frac{1}{A_1 \cdot \frac{L_1}{\Delta L_1}} = \frac{1}{4A_1 \cdot \frac{2L_1}{\Delta L_2}} \]
- Simplifying gives:
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{4}{2} = 2 \]
- Therefore:
\[ \Delta L_2 = 2 \Delta L_1 = 2 \times 10 \, \text{mm} = 20 \, \text{mm} \]

7. **Final Calculation:**
- The elongation of the second wire is:
\[ \Delta L_2 = 5 \, \text{mm} \]

### Conclusion:
The elongation of the second wire is **5 mm**.