Question

A copper wire of length 3 m is stretched by 3 mm by applying an external force. The volume of the wire is \( 600 \times 10^{-6} \text{ m}^3 \). The elastic potential energy stored in the wire in stretched condition would be _____J. (Given Young's modulus of copper = \( 1.1 \times 10^{11} \text{ N/m}^2 \))}

Answer 1

Correct Answer: 33

Step 1: Understanding the Concept:
Elastic potential energy is the energy stored in a body when it is deformed within its elastic limit. For a stretched wire, this energy is calculated using Young's modulus, strain, and volume.
: Key Formula or Approach:
Energy density (\( u \)) = \( \frac{1}{2} \cdot Y \cdot (\text{strain})^2 \).
Total Potential Energy (\( U \)) = \( u \cdot \text{Volume} \).
Strain = \( \frac{\Delta L}{L} \).
Step 2: Detailed Explanation:
Given:
\( L = 3 \text{ m} \).
\( \Delta L = 3 \text{ mm} = 3 \times 10^{-3} \text{ m} \).
\( V = 600 \times 10^{-6} \text{ m}^3 \).
\( Y = 1.1 \times 10^{11} \text{ N/m}^2 \).
Calculate Strain:
\[ \text{Strain} = \frac{\Delta L}{L} = \frac{3 \times 10^{-3}}{3} = 10^{-3} \].
Calculate Potential Energy:
\[ U = \frac{1}{2} \cdot Y \cdot (\text{Strain})^2 \cdot V \]
\[ U = \frac{1}{2} \times 1.1 \times 10^{11} \times (10^{-3})^2 \times 600 \times 10^{-6} \]
\[ U = \frac{1}{2} \times 1.1 \times 10^{11} \times 10^{-6} \times 600 \times 10^{-6} \]
\[ U = 0.5 \times 1.1 \times 600 \times 10^{-1} \]
\[ U = 0.5 \times 1.1 \times 60 = 33 \text{ J} \].
Step 3: Final Answer:
The elastic potential energy stored in the wire is 33 J.