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 \[\text{=}\frac{\text{force}\,\text{(weight}\,\text{suspended)}}{\text{area}}\] \[=\frac{\Mu g}{A}\] Longitudinal strain \[=\frac{\text{increase}\,\text{in}\,\text{length}}{\text{original}\,\text{length}}\] Youngs modulus of material of the body is \[\text{Y}\,\text{=}\,\,\frac{\text{longitudinal}\,\text{stress}}{\text{longitudinal}\,\text{strain}}\,\,\text{=}\,\,\frac{\text{MgL}}{\text{Al}}\] Putting the numerical values, we have \[L=2m,A=560\,m{{m}^{2}}=50\times {{10}^{-6}}{{m}^{2}}\] \[l=0.5\,mm=0.5\times {{10}^{-3}}m,\,M=250\,kg\] \[\therefore \] \[Y=\frac{250\times 9.8\times 2}{50\times {{10}^{-6}}\times 0.5\times {{10}^{-3}}}\] \[=19.6\times {{10}^{10}}\,N/{{m}^{2}}\]
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