The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young’s modulus, respectively, are 8 × 10⁸ N m^–2 and 2 × 10¹¹ N m^–2 , is:

To find the maximum elongation of the steel wire before it exceeds its elastic limit, we can use Hooke's Law and the relationship between stress, strain, and Young's modulus.

Young's modulus ($Y$) is defined as the ratio of tensile stress to tensile strain:
$Y=\frac{\text{Stress}}{\text{Strain}}$

The elastic limit represents the maximum stress the material can withstand without undergoing permanent deformation. Therefore, the maximum strain (${\epsilon }_{\text{max}}$) is corresponding to the stress at the elastic limit (${\sigma }_{\text{max}}$):
${\sigma }_{\text{max}}=8\times {10}^8{\text{N m}}^{-2}$
$Y=2\times {10}^{11}{\text{N m}}^{-2}$

Strain is defined as the change in length ($\Delta L$) divided by the original length ($L$):
$\text{Strain}=\frac{\Delta L}{L}$

Substituting this definition into the formula for Young's modulus gives:
$Y=\frac{{\sigma }_{\text{max}}}{\Delta L_{\text{max}}/L}$

Rearranging the equation to solve for the maximum elongation ($\Delta L_{\text{max}}$):
$\Delta L_{\text{max}}=\frac{{\sigma }_{\text{max}}\times L}{Y}$

Given that the original length of the wire is $L=1\text{m}$, we substitute the values into the equation:
$\Delta L_{\text{max}}=\frac{(8\times {10}^8{\text{N m}}^{-2})\times 1\text{m}}{2\times {10}^{11}{\text{N m}}^{-2}}$

Let's simplify the expression:
$\Delta L_{\text{max}}=4\times {10}^{-3}\text{m}$

Since $1{\text{mm}}^{}={10}^{-3}\text{m}$, we convert the result to millimeters:
$\Delta L_{\text{max}}=4\text{mm}$

Thus, the maximum elongation of the steel wire is 4 mm.