A wire of area of cross-section `10^(-6) m^(2)` is increased in length by 0.1 %. The tension produced is 1000 N. The Young's modulus of wire is

Options

A. `10^(12) N//m^(2)`
B. `10^(11) N//m^(2)`
C. `10^(10) N//m^(2)`
D. `10^(9) N//m^(2)`

Answer

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

### Step 1: Understand the given data
- Area of cross-section (A) = \(10^{-6} \, m^2\)
- Increase in length (ΔL) = 0.1% of original length (L)
- Tension (Force, F) = 1000 N

### Step 2: Convert the percentage increase in length to a decimal
The increase in length as a decimal can be calculated as:
\[ \text{Strain} = \frac{\Delta L}{L} = \frac{0.1}{100} = 0.001 \]

### Step 3: Calculate the stress in the wire
Stress (σ) is defined as the force applied per unit area. It can be calculated using the formula:
\[ \sigma = \frac{F}{A} \]
Substituting the values:
\[ \sigma = \frac{1000 \, N}{10^{-6} \, m^2} = 1000 \times 10^6 \, N/m^2 = 10^9 \, N/m^2 \]

### Step 4: Use the definition of Young's modulus
Young's modulus (E) is defined as the ratio of stress to strain:
\[ E = \frac{\sigma}{\text{Strain}} \]
Substituting the values we calculated:
\[ E = \frac{10^9 \, N/m^2}{0.001} = 10^9 \times 10^3 \, N/m^2 = 10^{12} \, N/m^2 \]

### Step 5: Conclusion
The Young's modulus of the wire is:
\[ E = 10^{12} \, N/m^2 \]

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