question_answer

An iron rod of length 2 m and area of cross-section \[50\,\,m{{m}^{2}}\] 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)  \[19.6\times {{10}^{10}}\,\,N/{{m}^{2}}\]

Correct Answer: D

Solution :

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