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Physics Question 46 – JEE-MAIN 2026

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A copper wire of length 3 m is stretched by 3 mm by applying an external force. The volume of the wire is 600×10−6 m3. The elastic potential energy stored in the wire in stretched condition would be _______ J. (Given Young modulus of copper =1.1×1011 N/m2)

The energy stored in a deformed elastic material due to its deformation is known as elastic potential energy.

Step 1: Identify Given Parameters and Formula✦ Active

The given parameters are: original length L=3 m, extension ΔL=3 mm =3×10−3 m, volume V=600×10−6 m3, and Young's modulus Y=1.1×1011 N/m2. The formula for elastic potential energy stored in a wire is U=12Y(Strain)2V.

Step 2: Calculate Strain○ Expand

Strain (ϵ) is defined as the ratio of change in length to the original length. Convert ΔL to meters.

ϵ=ΔLL=3×10−3 m3 m=1×10−3
💡 Teacher's Secret Hint

Ensure consistent units (SI) for all quantities before calculation.

Step 3: Calculate Elastic Potential Energy○ Expand

Substitute the calculated strain and the given values into the formula for elastic potential energy.

U=12Yϵ2V U=12(1.1×1011 N/m2)(1×10−3)2(600×10−6 m3) U=12(1.1×1011)(1×10−6)(600×10−6) U=12×1.1×600×10(11−6−6) U=1.1×300×10−1 U=1.1×30 U=33 J
💡 Teacher's Secret Hint

Pay close attention to the exponents when multiplying powers of 10.

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