Question

An iron rod of length $2 \,m $ and area of cross-section $ 50\,mm^{2}$ stretches by $ 0.5\,mm $ , when a mass of $250\, kg$ is hung from its lower end. The Young's 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}^{15}}\,N/{{m}^{2}} $
• D. $ 19.6\times {{10}^{10}}\,N/{{m}^{2}} $

Answer 1

Answer: (D)

When strain is small, the ratio of the longitudinal stress to the corresponding longitudinal strain is called the Young's moduls of the material of the body.

$Y =\frac{\text { longitudinal stress }}{\text { longitudinal strain }} $
$=\frac{F / A}{l / L} $
Given, $=2\, m , $
$A=50 \,mm ^{2} $
$=50 \times 10^{-6} m ^{2} $
$l=0.5 \,mm $
$=0.5 \times 10^{-3}\, m , $
$m=250 \,kg$
$g=9.8\, m / s ^{2}$
$Y=\frac{250 \times 9.8}{50 \times 10^{-6}} \times \frac{2}{0.5 \times 10^{-3}}$
$Y=19.6 \times 10^{10}\, N / m ^{2}$
Note : Young's modulus can be determined only for solids and it is the characteristic of the material of a solid.