The length of a wire is `1.0 m` and the area of cross-section is `1.0 xx 10^(-2) cm^(2)`. If the work done for increase in length by `0.2 cm` is 0.4 joule, then Young's modulus of the material of the wire is

Options

A. `2.0xx10^(10)N//m^(2)`
B. `4xx10^(10)N//m^(2)`
C. `2.0xx10^(11) N//m^(2)`
D. `2xx10^(10)N//m^(2)`

Answer

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

### Step 1: Understand the given data
- Length of the wire, \( L = 1.0 \, \text{m} \)
- Area of cross-section, \( A = 1.0 \times 10^{-2} \, \text{cm}^2 \)
- Increase in length, \( \Delta L = 0.2 \, \text{cm} \)
- Work done, \( W = 0.4 \, \text{J} \)

### Step 2: Convert units
Convert the area of cross-section from cm² to m² and the increase in length from cm to m:
- \( A = 1.0 \times 10^{-2} \, \text{cm}^2 = 1.0 \times 10^{-2} \times 10^{-4} \, \text{m}^2 = 1.0 \times 10^{-6} \, \text{m}^2 \)
- \( \Delta L = 0.2 \, \text{cm} = 0.2 \times 10^{-2} \, \text{m} = 0.002 \, \text{m} \)

### Step 3: Relate work done to Young's modulus
The work done on the wire can be expressed in terms of Young's modulus \( Y \):
\[ W = \frac{1}{2} \cdot \frac{Y \cdot A}{L} \cdot (\Delta L)^2 \]

### Step 4: Substitute the known values
Substituting the values into the equation:
\[ 0.4 = \frac{1}{2} \cdot \frac{Y \cdot (1.0 \times 10^{-6})}{1.0} \cdot (0.002)^2 \]

### Step 5: Simplify the equation
Calculating \( (0.002)^2 \):
\[ (0.002)^2 = 0.000004 = 4 \times 10^{-6} \]
Now substituting this back into the equation:
\[ 0.4 = \frac{1}{2} \cdot Y \cdot (1.0 \times 10^{-6}) \cdot (4 \times 10^{-6}) \]
\[ 0.4 = \frac{1}{2} \cdot Y \cdot 4 \times 10^{-12} \]
\[ 0.4 = 2 \times 10^{-12} Y \]

### Step 6: Solve for Young's modulus \( Y \)
Rearranging the equation to solve for \( Y \):
\[ Y = \frac{0.4}{2 \times 10^{-12}} = 2 \times 10^{11} \, \text{N/m}^2 \]

### Final Answer
Thus, the Young's modulus of the material of the wire is:
\[ Y = 2 \times 10^{11} \, \text{N/m}^2 \]