Question

An iron rod of length 2 m and cross-sectional area of 50 mm2 is stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. Young's modulus of iron rod is

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

Answer 1

Answer: (C)

Young's modulus of material of the body is $Y=\frac{\text{longitudial}\text{stress}}{\text{longitudinal}\text{strain}}=\frac{MgL}{Al}$ Putting the numerical values, we have $L=2m,A=50m{m}^2=50\times {10}^{-6}{m}^2$ $l=0.5mm=0.5\times {10}^{-3}m,M=250kg$ $\therefore$ $Y=\frac{250\times 9.8\times 2}{50\times {10}^{-6}\times 0.5\times {10}^{-3}}=19.6\times {10}^{10}N{m}^{-2}$