Two wires are made of the same material and have the same volume. However wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by `Deltax` on applying force F, how much force is needed to stretch wire 2 by the same amount?

Options

A. (a) 4F
B. (b) 6F
C. (c) 9F
D. (d) F

Explanation

To solve the problem, we need to analyze the relationship between the forces applied to the two wires and their respective elongations. We will use the concept of Young's modulus, which relates stress and strain in a material.

### Step-by-Step Solution:

1. **Understanding the Wires**:
- Wire 1 has a cross-sectional area \( A \) and length \( L \).
- Wire 2 has a cross-sectional area \( 3A \) and since both wires have the same volume, we can find the length of wire 2.

2. **Volume of Wires**:
- The volume \( V \) of a wire is given by the formula:
\[ V = \text{Cross-sectional Area} \times \text{Length} \]
- For wire 1:
\[ V_1 = A \times L \]
- For wire 2:
\[ V_2 = 3A \times L_2 \]
- Since \( V_1 = V_2 \), we have:
\[ A \times L = 3A \times L_2 \]
- Simplifying gives:
\[ L_2 = \frac{L}{3} \]

3. **Applying Young's Modulus**:
- Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta x / L} \]
- For wire 1, applying force \( F \):
\[ Y = \frac{F/A}{\Delta x / L} \]
- Rearranging gives:
\[ Y = \frac{F \cdot L}{A \cdot \Delta x} \]

4. **For Wire 2**:
- Let \( F' \) be the force required to stretch wire 2 by the same amount \( \Delta x \):
\[ Y = \frac{F'}{3A} \cdot \frac{L/3}{\Delta x} \]
- Rearranging gives:
\[ Y = \frac{F' \cdot (L/3)}{3A \cdot \Delta x} \]

5. **Setting the Young's Modulus Equations Equal**:
- Since both wires are made of the same material, their Young's moduli are equal:
\[ \frac{F \cdot L}{A \cdot \Delta x} = \frac{F' \cdot (L/3)}{3A \cdot \Delta x} \]
- Canceling \( \Delta x \) from both sides and simplifying gives:
\[ \frac{F \cdot L}{A} = \frac{F' \cdot (L/3)}{3A} \]
- Further simplifying leads to:
\[ F \cdot L = \frac{F' \cdot L}{3} \]
- Cancelling \( L \) from both sides (assuming \( L \neq 0 \)):
\[ F = \frac{F'}{3} \]
- Rearranging gives:
\[ F' = 3F \]

6. **Final Calculation**:
- To find the force \( F' \) required to stretch wire 2 by the same amount \( \Delta x \):
\[ F' = 9F \]

### Conclusion:
The force needed to stretch wire 2 by the same amount \( \Delta x \) is \( 9F \).