Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by `Deltal` on applying a force F, how much force is needed to stretch the second wire by the same amount?

Options

A. 9 F
B. F
C. 4 F
D. 6 F

Answer

To solve the problem, we need to analyze the relationship between the two wires using the concept of Young's modulus and the properties of the materials involved.

### Step-by-Step Solution:

1. **Understanding Young's Modulus**:
Young's modulus (Y) is defined as the ratio of stress to strain. For a wire, it can be expressed as:
\[ Y = \frac{F/A}{\Delta L/L} \]
where \( F \) is the force applied, \( A \) is the cross-sectional area, \( \Delta L \) is the change in length, and \( L \) is the original length.

2. **For Wire 1**:
Let’s denote the properties of the first wire:
- Cross-sectional area: \( A \)
- Original length: \( L \)
- Force applied: \( F \)
- Change in length: \( \Delta L \)

The Young's modulus for wire 1 can be expressed as:
\[ Y = \frac{F/A}{\Delta L/L} \]

3. **For Wire 2**:
Now, let’s denote the properties of the second wire:
- Cross-sectional area: \( 3A \) (three times that of wire 1)
- Original length: \( L' \)
- Force applied: \( F' \)
- Change in length: \( \Delta L \) (same as wire 1)

The volume of both wires is the same, so we can express their volumes:
\[ V_1 = A \cdot L \quad \text{and} \quad V_2 = 3A \cdot L' \]
Since the volumes are equal:
\[ A \cdot L = 3A \cdot L' \implies L' = \frac{L}{3} \]

4. **Young's Modulus for Wire 2**:
The Young's modulus for wire 2 can be expressed as:
\[ Y = \frac{F'/3A}{\Delta L/(L/3)} \]
Rearranging gives:
\[ Y = \frac{F'}{3A} \cdot \frac{3}{\Delta L} = \frac{F'}{A \cdot \Delta L} \]

5. **Setting the Young's Modulus Equal**:
Since both wires are made of the same material, their Young's moduli are equal:
\[ \frac{F}{A \cdot \Delta L} = \frac{F'}{A \cdot \Delta L} \]

6. **Solving for \( F' \)**:
We can simplify:
\[ F = \frac{F'}{3} \implies F' = 3F \]

7. **Final Result**:
Therefore, the force needed to stretch the second wire by the same amount \( \Delta L \) is:
\[ F' = 9F \]

### Conclusion:
To stretch the second wire by the same amount \( \Delta L \), a force of \( 9F \) is needed.