Question

An iron rod of length 2m and cross- sectional area of $50m{m}^2$ stretched by 0.5mm, 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

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

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