Question

The elastic potential energy stored in a copper rod of length one meter and area of cross-section \(1 \, \text{mm}^2\) when stretched by 1 mm is:

• A. \( 6 \times 10^{-2} \, \text{J} \)
• B. \( 3 \times 10^{-2} \, \text{J} \)
• C. \( 60 \, \text{J} \)
• D. \( 3 \, \text{J} \)

Hint:

The elastic potential energy stored is proportional to the square of the force and inversely proportional to the Young's modulus.

Answer 1

The Correct Option is A

The elastic potential energy stored in a stretched material is given by: \[ U = \frac{1}{2} \frac{F^2 L}{A Y}, \] where \(F\) is the force, \(L\) is the length of the rod, \(A\) is the area of cross-section, and \(Y\) is the Young’s modulus. First, calculate the force using the elongation formula: \[ \Delta L = \frac{F L}{A Y}, \] where \(\Delta L = 1 \, \text{mm}\). After solving for the energy, we get: \[ U = 6 \times 10^{-2} \, \text{J}. \]