1. MgL₁/AL 2. Mg(L₁ - L)/AL 3. MgL/AL₁ 4. MgL/A(L₁ - L)

Answer: 4. MgL/A(L₁ - L) Young's modulus Y=stress/strain . Stress =Mg/A Strain =L₁-L/L , since the extension is L₁-L and the original length is L Y=MgL/A(L₁-L) Answer: option 4.

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

A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changes to L_1 when mass M is suspended from its free end. The expression for Young's modulus is :

1. $\frac{MgL_1}{AL}$

2. $\frac{Mg(L_1 - L)}{AL}$

3. $\frac{MgL}{AL_1}$

4. $\frac{MgL}{A(L_1 - L)}$

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Answer: 4. $\frac{MgL}{A(L_1 - L)}$

Young's modulus $Y=\dfrac{\text{stress}}{\text{strain}}$.

Stress $=\dfrac{Mg}{A}$
Strain $=\dfrac{L_1-L}{L}$, since the extension is $L_1-L$ and the original length is $L$

$$Y=\frac{MgL}{A(L_1-L)}$$

Answer: option 4.

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

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A wire of length L, area of cross section A is hanging from a fixed support. The length of the wire changes to L_1 when mass M is suspended from its free end. The expression for Young's modulus is :

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