Elastic limit of a particular steel wire is ` 2.5 xx 10^(10) N//m^(2)` maximum strain to which the wire be subjected without losing elasticity is `(Y_("steel") = 2 xx 10^(11) N//m^(2))`

Options

A. `0.5`
B. `0.25`
C. `0.125`
D. `1.25`

Explanation

To find the maximum strain to which the wire can be subjected without losing elasticity, we can use the relationship between stress, strain, and Young's modulus. The formula is given by:

\[ Y = \frac{\text{Stress}}{\text{Strain}} \]

Where:
- \( Y \) is Young's modulus,
- Stress is the force per unit area,
- Strain is the deformation per unit length.

### Step 1: Identify the given values
- Elastic limit (Stress) = \( 2.5 \times 10^{10} \, \text{N/m}^2 \)
- Young's modulus for steel \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)

### Step 2: Rearrange the formula to find strain
From the formula \( Y = \frac{\text{Stress}}{\text{Strain}} \), we can rearrange it to find strain:

\[ \text{Strain} = \frac{\text{Stress}}{Y} \]

### Step 3: Substitute the values into the equation
Now, substitute the values for Stress and Young's modulus into the equation:

\[ \text{Strain} = \frac{2.5 \times 10^{10} \, \text{N/m}^2}{2 \times 10^{11} \, \text{N/m}^2} \]

### Step 4: Perform the calculation
Calculating the strain:

\[ \text{Strain} = \frac{2.5}{2} \times \frac{10^{10}}{10^{11}} = 1.25 \times 10^{-1} = 0.125 \]

### Step 5: Conclusion
The maximum strain to which the wire can be subjected without losing elasticity is \( 0.125 \) or \( 12.5\% \).