The energy stored per unit volume in copper wire, which produces longitudinal strain of 0.1% is
`(Y = 1.1 xx 10^(11) N//m^(2))`

Options

A. `11 xx 10^(3) Jm^(-3)`
B. `5.5 xx 10^(3) Jm^(-3)`
C. `5.5 xx 10^(4) Jm^(-3)`
D. `11 xx 10^(4) Jm^(-3)`

Explanation

To solve the problem of calculating the energy stored per unit volume in a copper wire producing a longitudinal strain of 0.1%, we can follow these steps:

### Step 1: Understanding the Formula
The energy stored per unit volume (u) in a material under stress is given by the formula:
\[ u = \frac{1}{2} \times \text{stress} \times \text{strain} \]

### Step 2: Relating Stress to Young's Modulus
Stress (σ) can be related to Young's Modulus (Y) and strain (ε) by the formula:
\[ \text{stress} = Y \times \text{strain} \]
Thus, we can rewrite the energy stored per unit volume as:
\[ u = \frac{1}{2} \times Y \times \epsilon^2 \]

### Step 3: Converting Strain to Decimal
The given longitudinal strain is 0.1%. To use it in calculations, we convert this percentage to a decimal:
\[ \epsilon = \frac{0.1}{100} = 0.001 \]

### Step 4: Substituting Values
Now we can substitute the values into the formula. We know:
- Young's Modulus (Y) for copper = \( 1.1 \times 10^{11} \, \text{N/m}^2 \)
- Strain (ε) = 0.001

Substituting these values into the energy formula:
\[ u = \frac{1}{2} \times (1.1 \times 10^{11}) \times (0.001)^2 \]

### Step 5: Calculating the Energy Stored
Calculating \( (0.001)^2 \):
\[ (0.001)^2 = 0.000001 = 10^{-6} \]

Now substituting this back into the equation:
\[ u = \frac{1}{2} \times (1.1 \times 10^{11}) \times (10^{-6}) \]
\[ u = \frac{1.1 \times 10^{11}}{2} \times 10^{-6} \]
\[ u = 0.55 \times 10^{5} \]

### Step 6: Final Result
Converting \( 0.55 \times 10^{5} \) to standard form gives:
\[ u = 5.5 \times 10^{4} \, \text{J/m}^3 \]

### Conclusion
The energy stored per unit volume in the copper wire is:
\[ \boxed{5.5 \times 10^{4} \, \text{J/m}^3} \]

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