A force of one newton doubles the length of a cord having cross-sectional area `1mrh^(2)`. The Young's modulus of the material of the cord is

Options

A. `1Nm^(-2)`
B. `5xx10^(5)Nm^(-2)`
C. `10^(6)Nm^(-2)`
D. `2xx10^(6)Nm^(-2)`

Answer

To find the Young's modulus of the material of the cord, we can follow these steps:

### Step 1: Identify the given values
- Force (F) = 1 N
- Change in length (ΔL) = L (since the length doubles, ΔL = L)
- Cross-sectional area (A) = 1 mm² = 1 × 10⁻⁶ m²

### Step 2: Understand the formula for Young's modulus
Young's modulus (Y) is defined as the ratio of stress to strain:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]

Where:
- Stress = \( \frac{F}{A} \)
- Strain = \( \frac{\Delta L}{L} \)

### Step 3: Calculate stress
Using the formula for stress:
\[ \text{Stress} = \frac{F}{A} = \frac{1 \text{ N}}{1 \times 10^{-6} \text{ m}^2} = 1 \times 10^{6} \text{ N/m}^2 \]

### Step 4: Calculate strain
Since the length doubles, the change in length (ΔL) is equal to the original length (L):
\[ \text{Strain} = \frac{\Delta L}{L} = \frac{L}{L} = 1 \]

### Step 5: Substitute values into the Young's modulus formula
Now, substituting the values of stress and strain into the Young's modulus formula:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{1 \times 10^{6} \text{ N/m}^2}{1} = 1 \times 10^{6} \text{ N/m}^2 \]

### Step 6: Conclusion
Thus, the Young's modulus of the material of the cord is:
\[ Y = 10^{6} \text{ N/m}^2 \]

### Final Answer:
The Young's modulus of the material of the cord is \( 10^{6} \text{ N/m}^2 \).

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