The elastic potential energy stored in a steel wire of length $20\,m$ stretched through $2 \,m$ is $80\,J$. The cross sectional area of the wire is $.........\,mm ^2$ (Given, $y =2.0 \times 10^{11}\,Nm ^{-2}$ )
JEE MAIN 2023, Medium
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Energy per unit volume $=\frac{1}{2} \text { stress } \times \text { strain }$
Energy $=\frac{1}{2} \text { stress } \times \text { strain } \times \text { volume }$
$80=\frac{1}{2} \times Y \times \text { strain }^2 A \times \ell$
$80=\frac{1}{2} \times 2 \times 10^{11} \times \frac{\left(2 \times 10^{-2}\right)^2}{400} \times A \times 20$
$20=\frac{10^{+\prime}}{20} \times A$
$40 \times 10^{-6} m ^2=A$
$A=40\,mm ^2$
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