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Elastic Potential Energy question from NEET 2019

Q11NEET 2019Easy

Elastic Potential Energy

When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L + $l$ ). The elastic potential energy stored in the extended wire is:

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Answer: (c)

Elastic energy stored $U = \dfrac{1}{2} \times \text{load} \times \text{extension}$ .
Load $= Mg$ , extension $= l$ , so $U = \dfrac{1}{2}Mgl$ .
So the answer is (c).

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