A) \[2\times {{10}^{11\,}}N/{{m}^{2}}\]
B) \[2\times {{10}^{-11}}N/{{m}^{2}}\]
C) \[3\times {{10}^{-12}}N/{{m}^{2}}\]
D) \[2\times {{10}^{-13}}N/{{m}^{2}}\]
Correct Answer: A
Solution :
From the graph \[l={{10}^{-4}}m,\ F=20N\] \[A={{10}^{-6}}{{m}^{2}},\ L=1m\] \ \[Y=\frac{FL}{Al}\]\[=\frac{20\times 1}{{{10}^{-6}}\times {{10}^{-4}}}=20\times {{10}^{10}}=2\times {{10}^{11}}N/{{m}^{2}}\]
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