When a spring is stretched by 2cm, it stores 100J of energy. If it is stretched further by 2cm, the stored energy will be increased by
A
100J
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B
200J
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C
300J
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D
400J
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Solution
The correct option is C300J x=2cm energy stored=12kx2 100=0.5×k×22 k=1002=50 now.x=4cm energy stored =12kx2 =0.5×50×42 =400J so stored energy increases by=400−100=300