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)

$19.6\times {{10}^{10}}\,N/{{m}^{2}}$

Answer 2

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.