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NEET Question Paper 2019

NEET Question Paper 2019

Practice NEET Question Paper 2019 MCQ questions with detailed answers and step-by-step explanations. 1+ free questions available with instant solutions — perfect for competitive exam preparation.
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Difficulty: All Easy Medium Hard 1–1 of 1
Q.1 Easy
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:
A Mgl
B MgL
C 1/2 Mgl
D 1/2MgL
Correct Answer:  C. 1/2 Mgl
EXPLANATION

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.

\[U = \frac{1}{2} \times \text{Force} \times \text{Extension}\]

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:

\[U = \frac{1}{2} \times Mg \times l\]
\[U = \frac{1}{2}Mgl\]

The elastic potential energy stored in the extended wire is \(\boxed{\frac{1}{2}Mgl}\)

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