Question

The elastic limit of steel is 8 × 10⁸ N m⁻² and its Young modulus 2 × 10¹¹ N m⁻². Find the maximum elongation of a half-metre steel wire that can be given without exceeding the elastic limit.

Answer 1

Given:
$$\begin{gathered} \mathrm{Elastic}\mathrm{limit}\mathrm{of}\mathrm{steel}\frac{F}{A}=8\times {10}^5\mathrm{N}/{\mathrm{m}}^2 \\ \mathrm{Young}\text{'}\mathrm{s}\mathrm{modulus}\mathrm{of}\mathrm{steel}Y=2\times {10}^{11}\mathrm{N}/{\mathrm{m}}^2 \\ \mathrm{Length}\mathrm{of}\mathrm{steel}\mathrm{wire}L=\frac{1}{2}\mathrm{m}=0.5\mathrm{m} \end{gathered}$$

The elastic limit of steel indicates the maximum pressure that steel can bear.
Let the maximum elongation of steel wire be $∆L$.
$$\begin{gathered} Y=\frac{F}{A}\frac{L}{∆L} \\ \Rightarrow ∆L=\frac{FL}{AY} \\ \Rightarrow ∆L=\frac{8\times {10}^5\times \left(0.5\right)}{2\times {10}^{11}} \\ =2\times {10}^{-3}\mathrm{m}=2\mathrm{mm} \end{gathered}$$

Hence, the required elongation of steel wire is 2 mm.