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JEE MAIN 2022
26-06-2022 S1
Question
The elastic behaviour of material for linear streass and linear strain, is shown in the figure. The energy density for a linear strain of $5 \times 10^{-4}$ is ............. $\mathrm{kJ} / \mathrm{m}^3$. Assume that material is elastic upto the linear strain of $5 \times 10^{-4}$.
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Solution
$$ y=\frac{\text { stress }}{\text { strain }}=2.0 \times 10^{10} $$ Energy density $=\frac{1}{2}$ stress × strain $$ \begin{aligned} & =\frac{1}{2}(\text { strain })^2 y=\frac{1}{2}\left(5 \times 10^{-4}\right)^2 \times 20 \times 10^{10} \\ & =25 \times 10^2 \times 10=25 \frac{\mathrm{~kJ}}{\mathrm{~m}^3} \end{aligned} $$ Ans. 25
Question Tags
JEE Main
Physics
Medium
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