The elastic potential energy stored in a steel wire of length \(20\,m\) stretched through \(2 \,cm\) is \(80\,J\). The cross sectional area of the wire is \(.........\,mm ^2\) (Given, \(y =2.0 \times 10^{11}\,Nm ^{-2}\) )

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Answer & Solution

Correct Answer

(C) \(40\)

Detailed explanation

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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