The Young's modulus of the material of a wire is `2xx10^(10) N m^(-2)` If the elongation strain is 1% then the energy stored in the wire per unit volume in `J m^(-3) ` is

Options

A. `10^(6)Jm^(-3)`
B. `10^(8)Jm^(-3)`
C. `2xx10^(6)Jm^(-3)`
D. `2xx10^(8)Jm^(-3)`

Answer

To find the energy stored in the wire per unit volume, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Young's modulus (Y) = \(2 \times 10^{10} \, \text{N/m}^2\)
- Elongation strain (ε) = 1% = \( \frac{1}{100} = 0.01\)

2. **Formula for Energy Stored per Unit Volume**:
The energy stored in a wire per unit volume (U) can be calculated using the formula:
\[ U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \]

3. **Relating Stress and Young's Modulus**:
Stress (σ) can be expressed in terms of Young's modulus and strain:
\[ \text{Stress} = Y \times \text{Strain} \]
Therefore, we can rewrite the energy formula as:
\[ U = \frac{1}{2} \times (Y \times \text{Strain}) \times \text{Strain} \]
This simplifies to:
\[ U = \frac{1}{2} \times Y \times \text{Strain}^2 \]

4. **Substituting the Values**:
Now we can substitute the values of Young's modulus and strain into the formula:
\[ U = \frac{1}{2} \times (2 \times 10^{10}) \times (0.01)^2 \]

5. **Calculating the Strain Squared**:
Calculate \( (0.01)^2 \):
\[ (0.01)^2 = 0.0001 = 10^{-4} \]

6. **Final Calculation**:
Now substitute this back into the equation:
\[ U = \frac{1}{2} \times (2 \times 10^{10}) \times (10^{-4}) \]
Simplifying this:
\[ U = \frac{1}{2} \times 2 \times 10^{10} \times 10^{-4} = 10^{10} \times 10^{-4} = 10^{6} \, \text{J/m}^3 \]

7. **Conclusion**:
The energy stored in the wire per unit volume is:
\[ U = 10^{6} \, \text{J/m}^3 \]

### Final Answer:
The energy stored in the wire per unit volume is \(10^{6} \, \text{J/m}^3\).

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