Question

An iron rod of length $2\text{m}$ and cross-sectional area of $50{\text{mm}}^2$ is stretched by $0.5\text{mm}$, when a mass of $250\text{kg}$ is hung from its lower end. Young's modulus of iron rod is

• A. $19.6\times {10}^{20}{\text{N/m}}^2$
• B. $19.6\times {10}^{18}{\text{N/m}}^2$
• C. $19.6\times {10}^{10}{\text{N/m}}^2$
• D. $19.6\times {10}^{15}{\text{N/m}}^2$

Answer 1

The correct option is C $19.6\times {10}^{10}{\text{N/m}}^2$

$Y=\frac{F/A}{\Delta l/l}=\frac{\frac{250\times 9.8}{50\times {10}^{-6}}}{\frac{0.5\times {10}^{-3}}{2}}$

$=\frac{250\times 9.8}{50\times {10}^{-6}}\times \frac{2}{0.5\times {10}^{-3}}\Rightarrow 19.6\times {10}^{10}{\text{N/m}}^2$