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

The energy stored per unit volume in a stretched rod due to the force applied can be calculated using the formula for elastic potential energy density. The formula for energy stored per unit volume \( u \) is given by:

\(u = \frac{1}{2} \times \text{stress} \times \text{strain}\)

Where:

• \(\text{stress} = \frac{\text{Force}}{\text{Area}}\)
• \(\text{strain} = \frac{\text{stress}}{\text{Young's modulus}}\)

Given:

• \(\text{Force} = 9 \times 10^4 \text{ N}\)
• \(\text{Area} = 3 \text{ cm}^2 = 3 \times 10^{-4} \text{ m}^2\) (since \(1 \text{ cm}^2 = 10^{-4} \text{ m}^2))
• \(\text{Young's modulus} = 2 \times 10^{11} \text{ Nm}^{-2}\)

First, calculate the stress:

\(\text{stress} = \frac{9 \times 10^4}{3 \times 10^{-4}} = 3 \times 10^8 \text{ Nm}^{-2}\)

Next, calculate the strain:

\(\text{strain} = \frac{3 \times 10^8}{2 \times 10^{11}} = 1.5 \times 10^{-3}\)

Now, calculate the energy stored per unit volume:

\(u = \frac{1}{2} \times 3 \times 10^8 \times 1.5 \times 10^{-3} = 2.25 \times 10^5 \ \text{Jm}^{-3}\)

Thus, the correct option is \(2.25 \times 10^5 \, \text{Jm}^{-3}\).