Question

A steel wire of length 2 m and Young's modulus \( 2.0 \times 10^{11} \, \text{N/m}^2 \) is stretched by a force. If Poisson's ratio and transverse strain for the wire are \( 0.2 \) and \( 10^{-3} \) respectively, then the elastic potential energy density of the wire is \( \times 10^6\), in SI units .

Hint:

In problems involving Young's modulus and strain, the potential energy density can be calculated using the relationship between the strain and the applied stress, considering Poisson's ratio and the transverse strain.

Answer 1

Correct Answer: 25

Given: \[ \ell = 2 \, \text{m}, \quad Y = 2 \times 10^{11} \, \text{N/m}^2 \] The elastic potential energy density \( \mu \) is given by: \[ \mu = \frac{\Delta \varepsilon}{\ell} = \frac{Y \Delta r}{r} \] where \( \Delta r \) is the elongation. Now, for transverse strain \( u \), we use the formula: \[ u = \frac{1}{2} \times \text{Poisson's ratio} \times \left(\frac{\Delta \varepsilon}{\ell}\right) \] Substitute the values to get the energy density: \[ \mu = \frac{5 \times 10^{-3}}{2} \times 2 \times 10^{11} \times \left[ 5 \times 10^{-3} \right]^2 = 25 \] Thus, the elastic potential energy density is \( 25 \times 10^6 \, \text{N/m}^2 \).