An elongation of `0.2%` in a wire of cross section `10^(-4)m^(2)` causes tension 1000 N. Then its young's modulus is

Options

A. `5xx10^(8)N//m^(2)`
B. `5xx10^(9)N//m^(2)`
C. `10^(8)N//m^(2)`
D. `10^(7)N//m^(2)`

Explanation

To find the Young's modulus of the wire given the elongation, cross-sectional area, and tension, we can follow these steps:

### Step 1: Understand the given values
- Elongation (strain) = 0.2% = 0.2/100 = 0.002
- Cross-sectional area (A) = \(10^{-4} \, m^2\)
- Tension (Force, F) = 1000 N

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

### Step 3: Calculate the stress
Using the formula for stress:
\[ \text{Stress} = \frac{F}{A} = \frac{1000 \, N}{10^{-4} \, m^2} \]
Calculating this gives:
\[ \text{Stress} = 1000 \times 10^4 = 10^7 \, N/m^2 \]

### Step 4: Substitute the values into the Young's modulus formula
Now we substitute the values of stress and strain into the Young's modulus formula:
\[ Y = \frac{10^7 \, N/m^2}{0.002} \]

### Step 5: Calculate Young's modulus
Calculating this gives:
\[ Y = 10^7 \div 0.002 = 10^7 \times 5000 = 5 \times 10^9 \, N/m^2 \]

### Conclusion
Thus, the Young's modulus of the wire is:
\[ Y = 5 \times 10^9 \, N/m^2 \]

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