A) 0.2 J
B) 10 J
C) 20 J
D) 0.1 J
Correct Answer: D
Solution :
[d] Elastic energy \[\frac{1}{2}\times F\times x\] \[F=200\text{ }N,\text{ }x=1mm={{10}^{-3}}m\] \[\therefore \,\,\,E=\frac{1}{2}\times 200\times 1\times {{10}^{-3}}=0.1J\]
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