Question

An iron rod of length $2m$ and area of cross-section $50m{m}^2$ stretches by $0.5mm$ , when a mass of $250kg$ 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, $=2m,$
$A=50m{m}^2$
$=50\times 10^{-6}{m}^2$
$l=0.5mm$
$=0.5\times 10^{-3}m,$
$m=250kg$
$g=9.8m/s^2$
$Y=\frac{250\times 9.8}{50\times 10^{-6}}\times \frac{2}{0.5\times 10^{-3}}$
$Y=19.6\times 10^{10}N/m^2$
Note : Young's modulus can be determined only for solids and it is the characteristic of the material of a solid.