A) \[2.0\times {{10}^{10}}\,N/{{m}^{2}}\]
B) \[4\times {{10}^{10}}\,N/{{m}^{2}}\]
C) \[2.0\times {{10}^{11}}\,N/{{m}^{2}}\]
D) \[2\times {{10}^{10}}\,N/{{m}^{2}}\]
Correct Answer: C
Solution :
\[W=\frac{1}{2}\frac{YA{{l}^{2}}}{L}\]Þ\[0.4=\frac{1}{2}\times \frac{Y\times {{1}^{-6}}\times {{(0.2\times {{10}^{-2}})}^{2}}}{1}\] Y\[=2\times {{10}^{11}}N/{{m}^{2}}\]
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