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Question

A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to its lower end. If the weight stretches the wire by 1 mm, then elastic energy stored in the wire is

A
0.1 J
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B
0.2 J
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C
10 J
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D
20 J
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Solution

The correct option is A 0.1 J
W=200 N,
Elongation, x=103 m
We know that,
Elastic energy, U=12Fx
U=12×200×103=0.1 J

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