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

Answer: 3. 1/2Mgl The energy stored in a stretched wire equals the work done by the stretching force, which grows linearly from 0 to Mg . U=1/2 F l=1/2 Mg l . Option 1 would be true only if the force stayed constant at Mg during the stretch, which is not so.

Topic: Elastic behaviour, stress-strain, Hooke's law, Young's, bulk and rigidity moduli· Properties of Solids and Liquids

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. \(\frac{1}{2}Mgl\)

4. \(\frac{1}{2}MgL\)

Doubtly Asked 30 Oct 2026 NEET UG 2019 Practice test neet Save WhatsApp Report

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Answer: 3. \(\frac{1}{2}Mgl\)

The energy stored in a stretched wire equals the work done by the stretching force, which grows linearly from 0 to \(Mg\).

\(U=\tfrac12 F\,l=\tfrac12 Mg\,l\).

Option 1 would be true only if the force stayed constant at \(Mg\) during the stretch, which is not so.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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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 :

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