A steel rod of length $1.5\text{m}$ and cross‑sectional area $0.8{\text{cm}}^2$ has Young’s modulus $210\text{GPa}$. If a tensile force of $800\text{N}$ is applied, what is the extension of the rod in millimetres?

Step-by-step solution

1. Convert area to SI units: $A=0.8{\text{cm}}^2=0.8\times 10^{-4}{\text{m}}^2=8\times 10^{-5}{\text{m}}^2$.
2. Write the formula for extension: $\Delta L=\frac{FL}{AE}$.
3. Insert numerical values: $F=800\text{N}$, $L=1.5\text{m}$, $A=8\times 10^{-5}{\text{m}}^2$, $E=210\times 10^9\text{Pa}$.
4. Compute the denominator: $AE=8\times 10^{-5}\times 210\times 10^9=1.68\times 10^7$.
5. Compute the numerator: $FL=800\times 1.5=1200$.
6. Find $\Delta L=1200/1.68\times 10^7=7.14\times 10^{-5}\text{m}$.
7. Convert to millimetres: $7.14\times 10^{-5}\text{m}=0.0714\text{mm}$.
8. Answer is $0.071\text{mm}$ (rounded to three significant figures).

Concept tested: Young’s modulus and linear deformation

Published 16 October 2026 · How this solution was prepared and checked