Question

The elastic energy stored per unit volume in a stretched wire is: \[ \left( Y = \text{Young's modulus of the material of the wire}; \quad S = \text{stress acting on the wire} \right) \]

• A. \( \dfrac{1}{2} \left( \dfrac{S}{Y} \right) \)
• B. \( \dfrac{1}{2} \left( \dfrac{S}{Y^2} \right) \)
• C. \( \dfrac{1}{2} \left( \dfrac{S^2}{Y} \right) \)
• D. \( \dfrac{1}{2} \left( \dfrac{S^2}{Y^2} \right) \)
• E. \( \dfrac{1}{2} (SY) \)

Hint:

The strain energy per unit volume in a material is directly related to the square of the stress and inversely related to the Young's modulus of the material.

Answer 1

The Correct Option is C

The elastic energy per unit volume (also called strain energy density) stored in a stretched wire is given by: \[ U = \frac{1}{2} \cdot {stress} \cdot {strain} \] The stress \( S \) is defined as: \[ S = Y \cdot {strain} \] where \( Y \) is the Young's modulus of the material. Thus, the strain is \( \frac{S}{Y} \), and the strain energy per unit volume becomes: \[ U = \frac{1}{2} \cdot S \cdot \left( \frac{S}{Y} \right) = \frac{1}{2} \cdot \frac{S^2}{Y} \] Hence, the correct answer is (C).