A wire of length `2m` is made from `10 cm^(2)` of copper. A force F is applied so that its length increases by `2mm`. Another wire of length `8m` is made from the same volume of copper. If the force F is applied to it, its length will increase by

Options

A. 0.8 cm
B. 1.6 cm
C. 2.4 cm
D. 3.2 cm

Answer

To solve the problem, we will use the relationship between stress, strain, and Young's modulus, which is given by Hooke's Law.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- For the first wire:
- Length \( L_1 = 2 \, \text{m} = 200 \, \text{cm} \)
- Cross-sectional area \( A = 10 \, \text{cm}^2 \)
- Increase in length \( \Delta L_1 = 2 \, \text{mm} = 0.2 \, \text{cm} \)

- For the second wire:
- Length \( L_2 = 8 \, \text{m} = 800 \, \text{cm} \)
- Cross-sectional area \( A \) (same volume of copper, so it will be calculated later)

2. **Calculate the Volume of the First Wire**:
\[ V = A \times L_1 = 10 \, \text{cm}^2 \times 200 \, \text{cm} = 2000 \, \text{cm}^3 \]

3. **Calculate the Cross-Sectional Area of the Second Wire**:
Since the volume of the second wire is the same as the first wire:
\[ V = A \times L_2 \implies A = \frac{V}{L_2} = \frac{2000 \, \text{cm}^3}{800 \, \text{cm}} = 2.5 \, \text{cm}^2 \]

4. **Use Hooke's Law**:
The relationship between the increase in length, force, length, area, and Young's modulus is given by:
\[ \Delta L = \frac{F L}{A Y} \]
where \( Y \) is the Young's modulus.

5. **Set Up the Ratios for the Two Wires**:
For the first wire:
\[ \Delta L_1 = \frac{F L_1}{A_1 Y} = \frac{F \cdot 200}{10 \cdot Y} \]
For the second wire:
\[ \Delta L_2 = \frac{F L_2}{A_2 Y} = \frac{F \cdot 800}{2.5 \cdot Y} \]

6. **Find the Ratio of Increments in Length**:
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{\frac{F \cdot 800}{2.5 \cdot Y}}{\frac{F \cdot 200}{10 \cdot Y}} = \frac{800 \cdot 10}{200 \cdot 2.5} \]
Simplifying this gives:
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{8000}{500} = 16 \]

7. **Calculate \(\Delta L_2\)**:
Since \(\Delta L_1 = 0.2 \, \text{cm}\):
\[ \Delta L_2 = 16 \cdot \Delta L_1 = 16 \cdot 0.2 \, \text{cm} = 3.2 \, \text{cm} \]

### Final Answer:
The length increase in the second wire, \(\Delta L_2\), will be **3.2 cm**.