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

Explanation

To solve the problem, we need to understand the relationship between the forces, lengths, and areas of the two wires made of the same material and having the same volume.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Wire 1 has a cross-sectional area \( A \) and stretches by \( \Delta x \) under a force \( F \).
- Wire 2 has a cross-sectional area \( 3A \) and we need to find the force \( F_1 \) required to stretch it by the same amount \( \Delta x \).
- Both wires have the same volume.

2. **Volume of the Wires**:
- The volume \( V \) of a wire is given by the formula:
\[ V = \text{Area} \times \text{Length} \]
- For Wire 1:
\[ V = A \times L_1 \]
- For Wire 2:
\[ V = 3A \times L_2 \]
- Since the volumes are equal:
\[ A \times L_1 = 3A \times L_2 \]
- Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ L_1 = 3L_2 \]

3. **Using Hooke’s Law**:
- According to Hooke's Law, the extension \( \Delta L \) in a wire is given by:
\[ \Delta L = \frac{F L}{A Y} \]
- Where \( F \) is the force applied, \( L \) is the original length, \( A \) is the cross-sectional area, and \( Y \) is Young's modulus (which is the same for both wires since they are made of the same material).

4. **Setting Up the Equations**:
- For Wire 1:
\[ \Delta x = \frac{F L_1}{A Y} \]
- For Wire 2:
\[ \Delta x = \frac{F_1 L_2}{3A Y} \]

5. **Equating the Extensions**:
- Since both wires stretch by the same amount \( \Delta x \):
\[ \frac{F L_1}{A Y} = \frac{F_1 L_2}{3A Y} \]
- Canceling \( A \) and \( Y \) from both sides:
\[ F L_1 = \frac{F_1 L_2}{3} \]
- Rearranging gives:
\[ F_1 = 3F \frac{L_1}{L_2} \]

6. **Substituting Lengths**:
- From our earlier result, we know \( L_1 = 3L_2 \):
\[ F_1 = 3F \frac{3L_2}{L_2} = 9F \]

### Final Answer:
The force needed to stretch wire 2 by the same amount \( \Delta x \) is \( F_1 = 9F \).

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