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