When a block of mass $M$ is suspended by a long wire of length $L$, the length of the wire become $(L+l) .$ The elastic potential energy stoped in the extended wire is :
NEET 2019, Easy
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$\mathrm{U}=\frac{1}{2}$ (force) (elongation)
$=\frac{1}{2}(M g) \ell=\frac{1}{2} M g \ell$
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