Question

A metal rod of area of cross-section \(3 \text{cm}^2\) is stretched along its length by applying a force of \(9 \times 10^4 \text{N}\). If the Young's modulus of the material of the rod is \(2 \times 10^{11} \text{Nm}^{-2}\), the energy stored per unit volume in the stretched rod is

• A. \(2.25 \times 10^5 \, \text{Jm}^{-3}\)
• B. \(9 \times 10^5 \, \text{Jm}^{-3}\)
• C. \(13.5 \times 10^5 \, \text{Jm}^{-3}\)
• D. \(4.5 \times 10^5 \, \text{Jm}^{-3}\)

Answer 1

The Correct Option is A

Step 1: Formula for Elastic Potential Energy Density: Energy density (u) = Energy / Volume. \[ u = \frac{1}{2} \times \text{stress} \times \text{strain} \] Since \( Y = \frac{\text{stress}}{\text{strain}} \implies \text{strain} = \frac{\text{stress}}{Y} \). Substitute strain: \[ u = \frac{1}{2} \times \text{stress} \times \frac{\text{stress}}{Y} = \frac{(\text{stress})^2}{2Y} \] 
Step 2: Calculate Stress: Force \( F = 9 \times 10^4 \, \text{N} \). Area \( A = 3 \, \text{cm}^2 = 3 \times 10^{-4} \, \text{m}^2 \). Stress \( \sigma = \frac{F}{A} = \frac{9 \times 10^4}{3 \times 10^{-4}} = 3 \times 10^8 \, \text{Nm}^{-2} \). 
Step 3: Calculate Energy Density: Young's Modulus \( Y = 2 \times 10^{11} \, \text{Nm}^{-2} \). \[ u = \frac{(3 \times 10^8)^2}{2 \times (2 \times 10^{11})} \] \[ u = \frac{9 \times 10^{16}}{4 \times 10^{11}} \] \[ u = \frac{9}{4} \times 10^5 = 2.25 \times 10^5 \, \text{Jm}^{-3} \]