When a force is applied on a wire of uniform cross-sectional area `3xx10^-6m^2` and length 4 m, the increase in length is 1mm. Energy stored in it will be `(Y=2xx10^(11)(N)/(m^2)`)

Options

A. 6250 J
B. 0.177 J
C. 0.075 J
D. 0.150 J

Explanation

To find the energy stored in the wire when a force is applied, we can follow these steps:

### Step 1: Understand the given data
- Cross-sectional area \( A = 3 \times 10^{-6} \, \text{m}^2 \)
- Original length \( L = 4 \, \text{m} \)
- Increase in length \( \Delta L = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Young's modulus \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)

### Step 2: Calculate the restoring force
Using the formula for restoring force \( F \):
\[ F = \frac{Y \cdot A \cdot \Delta L}{L} \]
Substituting the values:
\[ F = \frac{(2 \times 10^{11} \, \text{N/m}^2) \cdot (3 \times 10^{-6} \, \text{m}^2) \cdot (1 \times 10^{-3} \, \text{m})}{4 \, \text{m}} \]

### Step 3: Simplify the expression
Calculating the numerator:
\[ 2 \times 10^{11} \cdot 3 \times 10^{-6} \cdot 1 \times 10^{-3} = 6 \times 10^{2} \, \text{N} \]
Now substituting back into the force equation:
\[ F = \frac{6 \times 10^{2}}{4} = 150 \, \text{N} \]

### Step 4: Calculate the energy stored
The energy stored \( U \) in the wire is given by:
\[ U = \frac{1}{2} F \Delta L \]
Substituting the values:
\[ U = \frac{1}{2} \cdot 150 \, \text{N} \cdot (1 \times 10^{-3} \, \text{m}) \]

### Step 5: Calculate the final energy
Calculating the energy:
\[ U = \frac{1}{2} \cdot 150 \cdot 10^{-3} = 0.075 \, \text{J} \]

### Final Answer
The energy stored in the wire is \( 0.075 \, \text{J} \).

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