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JEE Main 2022 · 26 Jun · Shift 1, Physics question 53

The elastic behaviour of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of 5×10−45\times10^{-4} is ______ kJ/m3^3. Assume that material is elastic upto the linear strain of 5×10−45\times10^{-4}.

Strain against stress (N/m²): a straight line through the origin with marked points (20, 1×10⁻¹⁰), (40, 2×10⁻¹⁰), (60, 3×10⁻¹⁰) and (80, 4×10⁻¹⁰).
Answer and worked solution

Official answer: 25

Young's modulus from the line: Y=201×10−10=2×1011Y=\dfrac{20}{1\times10^{-10}}=2\times10^{11} N m−2^{-2}.

Energy density. 12Yε2=12(2×1011)(5×10−4)2=2.5×104\frac12Y\varepsilon^2=\frac12(2\times10^{11})(5\times10^{-4})^2=2.5\times10^4 J m−3=25^{-3}=25 kJ m−3^{-3}.

Answer: 25

Topics:Elastic potential energyHooke's law and elastic moduli
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