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Question

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 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

Work done in stretching wire

=12YAL2L=12F.l

Where

L= length of wire

l= increase in length

We know that

E=12×F×AL

=12×200×103

=0.1J

Hence (A) option is correct.


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