A) \[2.5\times {{10}^{-5}}J/{{m}^{3}}\]
B) \[5.0\times {{10}^{8}}J/{{m}^{3}}\]
C) \[2.5\times {{10}^{-8}}J/{{m}^{3}}\]
D) \[0.5\times {{10}^{11}}J/{{m}^{3}}\]
Correct Answer: A
Solution :
Energy stored per unit volume \[=\frac{1}{2}\times Y\times {{(strain)}^{2}}\] \[=\frac{1}{2}\times 2\times {{10}^{11}}\times {{\left( \frac{0.5}{100} \right)}^{2}}=2.5\times {{10}^{6}}J/{{m}^{3}}\]You need to login to perform this action.
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