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 \[L=2m,\text{ }A=50\text{ }m{{m}^{2}}=50\times {{10}^{-6}}{{m}^{2}}\] \[l=0.5mm=0.5\times {{10}^{-3}}m,M=250kg.\] \[\therefore \] \[Y=\frac{250\times 9.8\times 2}{50\times {{10}^{-6}}\times 0.5\times {{10}^{-3}}}\] \[=19.6\times {{10}^{10}}N/{{m}^{2}}\]You need to login to perform this action.
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