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Physics Question 46 – JEE-MAIN 2025

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A steel wire of length 2 m and Young's modulus 2.0×1011 N m−2 is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10−3 respectively, then the elastic potential energy density of the wire is _______ ×105 (in SI units).

Elastic potential energy density is the energy stored per unit volume in a deformed elastic material.

● Video Walkthrough
Step 1: Determine Longitudinal Strain✦ Active

The Poisson's ratio (σ) relates transverse strain (ϵt) to longitudinal strain (ϵL) as σ=ϵtϵL (considering magnitudes). Given σ=0.2 and ϵt=10−3.

ϵL=ϵtσ=10−30.2=10−32×10−1=0.5×10−2=5×10−3
Step 2: Calculate Elastic Potential Energy Density○ Expand

The elastic potential energy density (u) is given by the formula u=12YϵL2, where Y is Young's modulus. Given Y=2.0×1011 N m−2.

u=12×(2.0×1011)×(5×10−3)2 =(1.0×1011)×(25×10−6) =25×1011−6=25×105 J m−3
Step 3: Format the Final Answer○ Expand

The question asks for the value in the format `_______ x 10^5`. Therefore, the value in the blank is 25.

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