If in a wire of Young's moduls `Y`, longitudinal strain `X` is produced then the potential energy stored in its unit volume will be:

Options

A. `0.5YX^(2)`
B. `0.5Y^(2)X`
C. `2YX^(2)`
D. `YX^(2)`

Explanation

To find the potential energy stored in a wire of Young's modulus \( Y \) when a longitudinal strain \( X \) is produced, we can follow these steps:

### Step 1: Understand the formula for potential energy per unit volume
The potential energy (P.E.) stored in a material per unit volume due to deformation can be expressed as:
\[ \text{P.E. per unit volume} = \frac{1}{2} \times \text{Stress} \times \text{Strain} \]

### Step 2: Relate stress and strain using Young's modulus
Young's modulus \( Y \) is defined as the ratio of stress to strain:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
From this, we can express stress in terms of Young's modulus and strain:
\[ \text{Stress} = Y \times \text{Strain} \]

### Step 3: Substitute the strain value
Given that the longitudinal strain is \( X \), we can substitute this into the stress equation:
\[ \text{Stress} = Y \times X \]

### Step 4: Substitute stress into the potential energy formula
Now, we can substitute the expression for stress back into the potential energy formula:
\[ \text{P.E. per unit volume} = \frac{1}{2} \times (Y \times X) \times X \]

### Step 5: Simplify the expression
This simplifies to:
\[ \text{P.E. per unit volume} = \frac{1}{2} Y X^2 \]

### Final Answer
Thus, the potential energy stored in its unit volume is:
\[ \text{P.E. per unit volume} = \frac{1}{2} Y X^2 \]

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