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 \( 25 \times 10^6 \, \text{N/m}^2 \).

Answer 2

We need to find the elastic potential energy density of a steel wire. We are given its Young's modulus, Poisson's ratio, and the transverse strain it experiences when stretched.

Concept Used:

The solution involves two key concepts from the properties of matter:

1. Poisson's Ratio (\( \sigma \)): It is defined as the ratio of the magnitude of the transverse strain to the magnitude of the longitudinal (or axial) strain.

\[ \sigma = \frac{|\text{Transverse Strain}|}{|\text{Longitudinal Strain}|} = \frac{|\epsilon_t|}{|\epsilon_l|} \]

2. Elastic Potential Energy Density (\( U_d \)): This is the elastic potential energy stored per unit volume of the material. It is given by the formula:

\[ U_d = \frac{1}{2} \times \text{Stress} \times \text{Strain} \]

Using the relation Stress = Young's Modulus (\(Y\)) \( \times \) Longitudinal Strain (\(\epsilon_l\)), the formula can be expressed as:

\[ U_d = \frac{1}{2} \times (Y \cdot \epsilon_l) \times \epsilon_l = \frac{1}{2} Y \epsilon_l^2 \]

Step-by-Step Solution:

Step 1: Identify the given values.

• Young's Modulus, \( Y = 2.0 \times 10^{11} \, \text{N/m}^2 \)
• Poisson's ratio, \( \sigma = 0.2 \)
• Transverse strain, \( \epsilon_t = 10^{-3} \)

Note: The length of the wire (2 m) is not required for calculating the energy density.

Step 2: Calculate the longitudinal strain (\(\epsilon_l\)) using Poisson's ratio.

From the definition of Poisson's ratio:

\[ \epsilon_l = \frac{\epsilon_t}{\sigma} \]

Substituting the given values:

\[ \epsilon_l = \frac{10^{-3}}{0.2} = \frac{1}{0.2} \times 10^{-3} = 5 \times 10^{-3} \]

Step 3: Calculate the elastic potential energy density (\( U_d \)) using the formula involving Young's modulus and longitudinal strain.

\[ U_d = \frac{1}{2} Y \epsilon_l^2 \]

Substitute the values of \( Y \) and \( \epsilon_l \):

\[ U_d = \frac{1}{2} \times (2.0 \times 10^{11} \, \text{N/m}^2) \times (5 \times 10^{-3})^2 \]

Step 4: Perform the final computation.

\[ U_d = \frac{1}{2} \times (2.0 \times 10^{11}) \times (25 \times 10^{-6}) \] \[ U_d = (1.0 \times 10^{11}) \times (25 \times 10^{-6}) \] \[ U_d = 25 \times 10^{11-6} = 25 \times 10^5 \, \text{J/m}^3 \]

To express this in the form \( \times 10^6 \), we can write:

\[ U_d = 2.5 \times 10^6 \, \text{J/m}^3 \]

The elastic potential energy density of the wire is \( 2.5 \times 10^6 \) in SI units. The value to be filled in the blank is 2.5.