In CGS system, the Young's modulusm of a steel wire is `2 xx 10^(12)`. To double the length of a wire of unit cross-section area, the force

Options

A. `4 xx 10^(6)` dynes
B. `2 xx 10^(12)` dynes
C. `2 xx 10^(12)` newtons
D. `2 xx 10^(8)` dynes

Answer

To solve the problem, we will use the formula for Young's modulus (Y), which relates stress and strain in a material. The steps are as follows:

### Step 1: Understand the given data
- Young's modulus (Y) of the steel wire = \(2 \times 10^{12} \, \text{dyn/cm}^2\)
- Cross-sectional area (A) = 1 cm² (unit cross-section area)
- We want to double the length of the wire, so if the original length is L, the change in length (\(\Delta L\)) will be equal to L.

### Step 2: Write the formula for Young's modulus
Young's modulus is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Where:
- Stress = \(\frac{F}{A}\) (Force per unit area)
- Strain = \(\frac{\Delta L}{L}\) (Change in length per original length)

### Step 3: Substitute the expressions for stress and strain into the Young's modulus formula
Substituting these into the Young's modulus equation gives:
\[ Y = \frac{\frac{F}{A}}{\frac{\Delta L}{L}} \]

### Step 4: Rearrange the equation to solve for force (F)
Rearranging the equation, we have:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
From this, we can express force (F) as:
\[ F = Y \cdot \frac{A \cdot \Delta L}{L} \]

### Step 5: Substitute the known values
We know:
- \(Y = 2 \times 10^{12} \, \text{dyn/cm}^2\)
- \(A = 1 \, \text{cm}^2\)
- \(\Delta L = L\)

Substituting these values into the equation:
\[ F = 2 \times 10^{12} \cdot \frac{1 \cdot L}{L} = 2 \times 10^{12} \, \text{dyn} \]

### Step 6: Conclusion
The force required to double the length of the wire is:
\[ F = 2 \times 10^{12} \, \text{dyn} \]