1) An iron rod of length 2 cm and cross-sectional area of 50 cm2 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) $19.6\times 10^{20}N/m^{2}$ B) $19.6\times 10^{18}N/m^{2}$ C) $19.6\times 10^{10}N/m^{2}$ D) $19.6\times 10^{15}N/m^{2}$ Answer: Option CExplanation:$Y= \frac{F/A}{\triangle l/l}=\frac{\frac{250\times 9.8}{50\times 10^{-6}}}{\frac{0.5\times 10^{-3}}{2}}$ $\frac{250\times 9.8}{50\times 10^{-6}}\times\frac{2}{0.5 \times 10^{-3}}\Rightarrow19.6 \times10^{10} N/m^{2}$
