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 .
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 .
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Correct Answer: 25
Approach Solution - 1
Approach Solution -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.
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