question_answer

An iron rod of length 2 m and cross-sectional area of\[50\text{ }m{{m}^{2}}\]is stretched by 0.5 mm, 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) \[19.6\times {{10}^{15}}N/{{m}^{2}}\]

Correct Answer: C

Solution :

                 Key Idea: Ratio of stress to strain is constant for the material of the given body and is the Young's modulus. Let the length of wire be L, weight Mg is applied to the other end. Within elastic limit, \[Longitudinal\text{ }stress=\frac{force\text{ }(weight\text{ }suspended)}{area}\]\[=\frac{Mg}{A}\] \[Longitudinal\text{ }strain=\frac{increase\text{ }in\text{ }length}{original\text{ }length}\] Young's modulus of material of the body is \[Y=\frac{longitudinal\text{ }stress}{longitudinal\text{ }strain}=\frac{MgL}{Al}\] Putting the numerical values, we have \[L=2m,\text{ }A=50\text{ }m{{m}^{2}}=50\times {{10}^{-6}}{{m}^{2}}\] \[l=0.5mm=0.5\times {{10}^{-3}}m,M=250kg.\] \[\therefore \]  \[Y=\frac{250\times 9.8\times 2}{50\times {{10}^{-6}}\times 0.5\times {{10}^{-3}}}\]                 \[=19.6\times {{10}^{10}}N/{{m}^{2}}\]