An elongation of `0.1%` in a wire of cross-sectional `10^(-6) m^(2)` casues a tension of `100N.Y` for the wire is

Options

A. `10^(12) N//m^(2)`
B. `10^(11) N//m^(2)`
C. `10^(10) N//m^(2)`
D. `100 N//m^(2)`

Explanation

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

### Step 1: Understand the given data
- Elongation percentage: \(0.1\%\)
- Cross-sectional area \(A = 10^{-6} \, m^2\)
- Tension (Force) \(F = 100 \, N\)

### Step 2: Convert elongation percentage to a decimal
The elongation percentage is given as \(0.1\%\). To convert this to a decimal form:
\[ \frac{\Delta L}{L} = \frac{0.1}{100} = 0.001 = 1 \times 10^{-3} \]

### Step 3: Use the formula for Young's modulus
The formula for Young's modulus \(Y\) is:
\[ Y = \frac{F}{A \cdot \frac{\Delta L}{L}} \]
Where:
- \(F\) is the force applied,
- \(A\) is the cross-sectional area,
- \(\frac{\Delta L}{L}\) is the fractional elongation.

### Step 4: Substitute the values into the formula
Substituting the known values into the Young's modulus formula:
\[ Y = \frac{100 \, N}{10^{-6} \, m^2 \cdot 1 \times 10^{-3}} \]

### Step 5: Simplify the expression
Calculating the denominator:
\[ 10^{-6} \, m^2 \cdot 1 \times 10^{-3} = 10^{-9} \, m^2 \]
Now substituting this back into the equation for \(Y\):
\[ Y = \frac{100}{10^{-9}} = 100 \times 10^{9} \, N/m^2 \]

### Step 6: Final result
Thus, the Young's modulus \(Y\) is:
\[ Y = 10^{11} \, N/m^2 \]

### Summary
The Young's modulus of the wire is \(Y = 10^{11} \, N/m^2\).

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