The maximum elongation of a steel wire of 1m 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 :

The problem asks for the maximum elongation of a steel wire of length 1 m when the stress reaches the elastic limit.

Given data:

Original length L₀ = 1 m

Elastic‑limit stress σₑ = 8 × 10⁸ N·m^–2

Young’s modulus E = 2 × 10¹¹ N·m^–2

First, compute the strain ε at the elastic limit using Hooke’s law:

$\epsilon \; =\; \backslash frac\{\sigma ₑ\}\{E\}$

Substituting the numbers:

$\epsilon \; =\; \backslash frac\{8 \times 10^\{8\}\}\{2 \times 10^\{11\}\}\; =\; 4 \times 10^\{-3\}$

The strain is dimensionless; it represents the fractional change in length.

Now calculate the actual elongation ΔL:

$\Delta L\; =\; \epsilon · L₀$

Since L₀ = 1 m:

$\Delta L\; =\; 4 \times 10^\{-3\} · 1 m\; =\; 4 \times 10^\{-3\} m$

Convert metres to millimetres (1 m = 1000 mm):

$\Delta L\; =\; 4 \times 10^\{-3\} m �� 1000 mm/m\; =\; 4 mm$

Therefore, the maximum elongation of the wire before it exceeds the elastic limit is 4 mm.