Calculate the energy stored in a wire given its length, cross-sectional area, elongation, and Young's modulus. Assume A = 5x10⁻⁶ m2.

To find the energy stored in the wire, we use the formula for elastic potential energy:

$$U = \frac{1}{2} \frac{Y A (\Delta L)^2}{L}$$

where:

• $Y$ is Young's modulus
• $A$ is the cross-sectional area
• $\Delta L$ is the elongation
• $L$ is the original length of the wire

Given values from the problem:

• Length of wire ($L$) = $1.0 \text{ m}$
• Cross-sectional area ($A$) = $5 \times 10^{-5} \text{ m}^2$
• Elongation ($\Delta L$) = $3/2 \text{ mm} = 1.5 \text{ mm}$
• Young's modulus ($Y$) = $2.0 \times 10^{11} \text{ Pa}$

Self-correction/Assumption: Based on the provided options, it appears there might be a typo in the cross-sectional area. If we use $A = 5 \times 10^{-5} \text{ m}^2$, the calculated energy is $11.25 \text{ J}$. However, if we assume the cross-sectional area is $5 \times 10^{-6} \text{ m}^2$, the result matches one of the options. We will proceed with the assumption that $A = 5 \times 10^{-6} \text{ m}^2$ to match the given correct option.

Step 1: Convert all given values to SI units.

• $L = 1.0