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

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a

0. 2J

b

10 J

c

20J

d

0.1 J

answer is D.

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Detailed Solution

The elastic energy stored in the wire is given byU=12× stress × strain × volume =12×FA×lL×AL=12Fl=12×200×1×10−3=0⋅1J

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