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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:

1. Mgl
2. MgL
3. 1/2 Mgl
4. 1/2 MgL

Correct answer: 1/2 Mgl

Solution

The elastic potential energy stored in the wire is given by the work done in stretching it. Since the force increases linearly with extension, the average force is (Mg)/2, and the work done is (1/2) × Mg × l.

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