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}} $