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

Question

Accepted Answer

Correct answer: C. 1/2 Mgl — When a wire stretches under load, elastic potential energy is stored due to the deformation of the material, which can be calculated using the relationship between stress, strain, and the extension. **Step 1: Identify the Elastic Potential Energy Formula** Elastic potential energy stored in a stretched wire is given by the work done against the restoring force during extension. U = \frac{1}{2} 	imes 	ext{Force} 	imes 	ext{Extension} **Step 2: Apply Forces and Calculate** The force stretching the wire is the weight of the block: F = Mg The extension of the wire is: \Delta L = l Substituting into the elastic potential energy formula: U = \frac{1}{2} 	imes Mg 	imes l U = \frac{1}{2}Mgl **The elastic potential energy stored in the extended wire is** \boxed{\frac{1}{2}Mgl}

Answer

A. Mgl

Answer

B. MgL

Answer

D. 1/2MgL

NEET Question Paper 2019 - MCQ Practice Questions