Question

An iron rod of length $2\,m$ and cross-sectional area of $50 \, mm^2$ 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 1

Answer: (C)

$Y =\frac{ F / A }{\Delta l / l}=\frac{\frac{250 \times 9.8}{50 \times 10^{-6}}}{\frac{0.5 \times 10^{-3}}{2}}$
$=\frac{250 \times 9.8}{50 \times 10^{-6}} \times \frac{2}{0.5 \times 10^{-3}} $
$\Rightarrow 19.6 \times 10^{10}\, N / m ^{2}$