A) \[2\times {{10}^{11}}N/{{m}^{2}}\]
B) \[2\times {{10}^{-11}}N/{{m}^{2}}\]
C) \[2\times {{10}^{-12}}N/{{m}^{2}}\]
D) \[2\times {{10}^{-13}}N/{{m}^{2}}\]
Correct Answer: A
Solution :
[a] From the graph \[l={{10}^{-4}}m,\,F=20N\] \[A={{10}^{-6}}{{m}^{2}},\,L=1m\] \[\therefore \,\,\,Y=\frac{FL}{Al}=\frac{20\times 1}{{{10}^{-6}}\times {{10}^{-4}}}\] \[20\times {{10}^{10}}=2\times {{10}^{11}}N/{{m}^{2}}\]
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