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

Answer

To solve the problem, we will use the relationship between stress, strain, and Young's modulus. Let's break down the solution step by step.

### Step 1: Understand the relationship between stress, strain, and Young's modulus
Young's modulus (Y) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Where:
- Stress = \( \frac{F}{A} \) (Force per unit area)
- Strain = \( \frac{\Delta x}{L} \) (Change in length per original length)

### Step 2: Write the Young's modulus equation for wire 1
For wire 1, with cross-sectional area \( A \) and original length \( L \):
\[ Y = \frac{F}{A} \cdot \frac{L}{\Delta x} \]
Rearranging gives:
\[ F = Y \cdot \frac{A \cdot \Delta x}{L} \]

### Step 3: Determine the dimensions of wire 2
Since both wires have the same volume, we can express the volume of both wires:
- Volume of wire 1: \( V_1 = A \cdot L \)
- Volume of wire 2: \( V_2 = 3A \cdot L_2 \)

Setting the volumes equal:
\[ A \cdot L = 3A \cdot L_2 \]
This simplifies to:
\[ L_2 = \frac{L}{3} \]

### Step 4: Write the Young's modulus equation for wire 2
For wire 2, with cross-sectional area \( 3A \) and original length \( L_2 = \frac{L}{3} \):
\[ Y = \frac{F'}{3A} \cdot \frac{L/3}{\Delta x} \]
Rearranging gives:
\[ F' = Y \cdot \frac{3A \cdot \Delta x}{L/3} \]
This simplifies to:
\[ F' = Y \cdot \frac{9A \cdot \Delta x}{L} \]

### Step 5: Relate the forces for both wires
From the equation for wire 1:
\[ F = Y \cdot \frac{A \cdot \Delta x}{L} \]
Now, substituting this into the equation for wire 2:
\[ F' = 9 \cdot \frac{F}{3} \]
Thus:
\[ F' = 9F \]

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