Question

A stretched wire of a material whose Young's modulus Y = 2 × 10¹¹ Nm^-2 has Poisson's ratio of 0.25. Its lateral strain ε_l = 10^-3. The elastic energy density of the wire is

• A. 1 × 10⁵ Jm^-3
• B. 4 × 10⁵ Jm^-3
• C. 8 × 10⁵ Jm^-3
• D. 16 × 10⁵ Jm^-3

Answer 1

The Correct Option is D

Given: 

• Young's modulus: \( Y = 2 \times 10^{11} \) N/m\(^2\)
• Poisson's ratio: \( \nu = 0.25 \)
• Lateral strain: \( \varepsilon_l = 10^{-3} \)

Step 1: Relationship Between Lateral and Longitudinal Strain

Poisson’s ratio is given by:

\[ \nu = \frac{\text{lateral strain}}{\text{longitudinal strain}} \]

Rearranging for longitudinal strain \( \varepsilon \):

\[ \varepsilon = \frac{\varepsilon_l}{\nu} = \frac{10^{-3}}{0.25} = 4 \times 10^{-3} \]

Step 2: Elastic Energy Density Formula

The elastic energy density \( U \) is given by:

\[ U = \frac{1}{2} Y \varepsilon^2 \]

Substituting values:

\[ U = \frac{1}{2} \times (2 \times 10^{11}) \times (4 \times 10^{-3})^2 \]

\[ U = \frac{1}{2} \times (2 \times 10^{11}) \times (16 \times 10^{-6}) \]

\[ U = \frac{32 \times 10^5}{2} = 16 \times 10^5 \text{ J/m}^3 \]

Answer: The correct option is D (16 × 10⁵ J/m³).