A wire `(Y=2xx10^(11)N//m` has length 1m and area `1mm^(2)`. The work required to increase its length by 2mm is

Options

A. 400J
B. 40J
C. 0.4J
D. 0.04J

Answer

To solve the problem of calculating the work required to increase the length of a wire by 2 mm, we can follow these steps:

### Step 1: Understand the Given Data
- Young's modulus (Y) = \(2 \times 10^{11} \, \text{N/m}^2\)
- Original length of the wire (L) = 1 m
- Cross-sectional area (A) = \(1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2\)
- Change in length (ΔL) = 2 mm = \(2 \times 10^{-3} \, \text{m}\)

### Step 2: Use the Formula for Young's Modulus
Young's modulus is defined as:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
Where:
- F = Force applied
- L = Original length
- A = Cross-sectional area
- ΔL = Change in length

### Step 3: Rearranging the Formula to Find Force (F)
We can rearrange the formula to solve for the force (F):
\[ F = \frac{Y \cdot A \cdot \Delta L}{L} \]

### Step 4: Substitute the Values into the Formula
Now, substituting the known values:
\[ F = \frac{(2 \times 10^{11} \, \text{N/m}^2) \cdot (1 \times 10^{-6} \, \text{m}^2) \cdot (2 \times 10^{-3} \, \text{m})}{1 \, \text{m}} \]

### Step 5: Calculate the Force (F)
Calculating the above expression:
\[ F = \frac{(2 \times 10^{11}) \cdot (1 \times 10^{-6}) \cdot (2 \times 10^{-3})}{1} \]
\[ F = 2 \times 10^{2} \, \text{N} = 200 \, \text{N} \]

### Step 6: Calculate the Work Done (W)
The work done (W) to stretch the wire can be calculated using the formula:
\[ W = F \cdot \Delta L \]
Substituting the values we have:
\[ W = 200 \, \text{N} \cdot (2 \times 10^{-3} \, \text{m}) \]

### Step 7: Calculate the Work Done (W)
Calculating the above expression:
\[ W = 200 \cdot 2 \times 10^{-3} = 400 \times 10^{-3} \, \text{J} = 0.4 \, \text{J} \]

### Final Answer
The work required to increase the length of the wire by 2 mm is **0.4 Joules**.

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