A steel wire of mass `3.16kg` is streached to a tensite strain of `1xx10^(-3)`. What is the elastic deforemaiton energy if density `rho = 7.9 g//cc` and `Y = 2xx10^(11) N//m^(2)`

Options

A. `4 KJ`
B. `0.4 KJ`
C. `0.04 KJ`
D. `4J`

Explanation

To find the elastic deformation energy of a steel wire, we can follow these steps:

### Step 1: Understand the formula for elastic deformation energy
The elastic deformation energy (U) stored in a material can be calculated using the formula:
\[ U = \frac{1}{2} Y \cdot \text{strain}^2 \cdot V \]
where:
- \( Y \) is the Young's modulus (in N/m²),
- strain is the tensile strain (dimensionless),
- \( V \) is the volume of the wire (in m³).

### Step 2: Calculate the volume of the wire
The volume \( V \) can be calculated using the formula:
\[ V = \frac{m}{\rho} \]
where:
- \( m \) is the mass of the wire (in kg),
- \( \rho \) is the density (in kg/m³).

Given:
- Mass \( m = 3.16 \, \text{kg} \)
- Density \( \rho = 7.9 \, \text{g/cm}^3 = 7.9 \times 10^3 \, \text{kg/m}^3 \) (conversion from g/cm³ to kg/m³)

Now, calculate the volume:
\[ V = \frac{3.16 \, \text{kg}}{7.9 \times 10^3 \, \text{kg/m}^3} \]

### Step 3: Substitute the values into the volume formula
Calculating the volume:
\[ V = \frac{3.16}{7.9 \times 10^3} \approx 4.00 \times 10^{-4} \, \text{m}^3 \]

### Step 4: Substitute values into the elastic deformation energy formula
Now, we can substitute the values into the elastic deformation energy formula:
Given:
- \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)
- Strain = \( 1 \times 10^{-3} \)

Substituting these values into the formula:
\[ U = \frac{1}{2} \cdot (2 \times 10^{11}) \cdot (1 \times 10^{-3})^2 \cdot (4.00 \times 10^{-4}) \]

### Step 5: Calculate the elastic deformation energy
Calculating \( U \):
\[ U = \frac{1}{2} \cdot 2 \times 10^{11} \cdot 1 \times 10^{-6} \cdot 4.00 \times 10^{-4} \]
\[ U = 1 \times 10^{11} \cdot 4.00 \times 10^{-10} \]
\[ U = 4.00 \times 10^{1} \]
\[ U = 40 \, \text{J} \]

### Final Result
The elastic deformation energy of the steel wire is \( 40 \, \text{J} \) or \( 0.04 \, \text{kJ} \).

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