A steel wire of length 2 m and cross‐sectional area 1.0 cm² has a Young’s modulus of 200 GPa. If a force of 1000 N is applied, what is its extension in millimetres?

Step-by-step solution

1. Write Hooke’s law for extension: $\Delta L=\frac{FL}{AE}$.
2. Convert the area to SI units: $1.0{\text{cm}}^2=1.0\times 10^{-4}{\text{m}}^2$.
3. Substitute the given values into the formula: $\Delta L=\frac{1000\times 2}{(1.0\times 10^{-4})\times (2\times 10^{11})}$.
4. Calculate the numerator: $1000\times 2=2000$.
5. Calculate the denominator: $(1.0\times 10^{-4})\times (2\times 10^{11})=2\times 10^7$.
6. Divide to get $\Delta L=\frac{2000}{2\times 10^7}=1\times 10^{-4}\text{m}$.
7. Convert to millimetres: $1\times 10^{-4}\text{m}=0.1\text{mm}$.

Concept tested: Extension due to Hooke’s law

Published 10 October 2026 · How this solution was prepared and checked