An object with a weight of $10\,\text{kN}$ is supported equally by four solid steel rods. Each rod has a length of $2.0\,\text{m}$ and a cross-sectional area of $250\,\text{mm}^2$. As a result of the object’s weight, the rods shorten by $0.10\,\text{mm}$. The rods obey Hooke’s law. Calculate the Young modulus of steel.
- A$2.0 \times 10^8\,\text{N m}^{-2}$
- B$2.0 \times 10^{11}\,\text{N m}^{-2}$
- C$8.0 \times 10^8\,\text{N m}^{-2}$
- D$8.0 \times 10^{11}\,\text{N m}^{-2}$