A) 0.1 J
B) 0.2 J
C) 0.3 J
D) 0.4 J
Correct Answer: A
Solution :
When a wire is stretched, work is done against the interatomic forces. This work is stored in the wire in the form of elastic potential energy. If length of wire is L, and area of cross-section is A, suppose on applying a force F along length of wire, the length increases by I. Then Youngs modulus \[Y=\frac{F/A}{L/L}\] \[\Rightarrow \] \[F=\frac{YA}{L}l\] Work done \[dW=F\times dl=\int_{0}^{l}{\,\,\frac{YA}{L}}l\,dl=\frac{1}{2}YA\frac{{{l}^{2}}}{L}\] Given, \[A=1\,m{{m}^{2}}={{10}^{-6}}\], \[l=1\,mm={{10}^{-3}}m,\,Y=2\times {{10}^{11}}N/{{m}^{2}},L=1m\] \[\therefore \] \[W=\frac{1}{2}\times 2\times {{10}^{11}}\times {{10}^{-6}}\times {{({{10}^{-3}})}^{2}}\] \[W=0.1\,J\]
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