A) \[1.32\times {{10}^{-4}}m,66\times {{10}^{3}}J\]
B) \[~132\times {{10}^{-4}}m,6.6\times {{10}^{2}}J\]
C) \[13.2\times {{10}^{-4}}6.6\times {{10}^{3}}J\]
D) \[0.132\times {{10}^{-4}}m,66\times {{10}^{4}}J\]
Correct Answer: C
Solution :
Given, \[A=1.2\,c{{m}^{2}}\] \[=1.2\times {{10}^{-4}}{{m}^{2}},\] \[F=4.8\times {{10}^{3}}N,\] \[Y=1.2\times {{10}^{11}}N/{{m}^{2}}.\] length \[l=\frac{F\times L}{AY}\] \[=\frac{4.8\times {{10}^{3}}\times 4}{1.2\times {{10}^{-4}}\times 1.2\times {{10}^{11}}}\] \[=13.3\times {{10}^{-4}}m\] The energy stored per unit volume \[E=\frac{1}{2}\times \frac{F}{A}\times \frac{l}{L}\] \[=\frac{1}{2}\times \frac{4.8\times {{10}^{3}}}{1.2\times {{10}^{-4}}}\times \frac{13.3\times {{10}^{-4}}}{4}\] \[E=6.6\times {{10}^{3}}\,J\]
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