The maximum elongation of a steel wire of $1$ m length if the elastic limit of steel and its Young's modulus, respectively, are $8 \times 10^8\ \text{N m}^{-2}$ and $2 \times 10^{11}\ \text{N m}^{-2}$, is :
Answer: (C) $4$ mm
The maximum stress is the elastic limit. Maximum strain:
$$\frac{\Delta l}{l} = \frac{\text{stress}}{Y} = \frac{8 \times 10^8}{2 \times 10^{11}} = 4 \times 10^{-3}$$
$$\Delta l = 4 \times 10^{-3} \times 1\ \text{m} = 4\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics