A) 0.2 J
B) 10 J
C) 20 J
D) 0.1 J
Correct Answer: D
Solution :
Elastic energy stored in the wire is \[U=\frac{1}{2}\]stress\[\times \]strain\[\times \]volume \[=\frac{1}{2}\frac{F}{A}\times \frac{\Delta L}{L}\times AL\] \[=\frac{1}{2}F\Delta L\] \[=\frac{1}{2}\times 200\times 1\times {{10}^{-3}}=0.1\,\,J\]
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