Question:medium

The work to be done to produce a strain of \(10^{-3}\) in a steel wire of mass 2.96 kg and density 7.4 gcm\(^{-3}\) is (Young's modulus of steel = \(2 \times 10^{11}\) Nm\(^{-2}\))

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Remember the formulas for elastic potential energy. The energy per unit volume is \( \frac{1}{2}(\text{Stress})(\text{Strain}) \), which can also be written as \( \frac{1}{2} Y (\text{Strain})^2 \) or \( \frac{(\text{Stress})^2}{2Y} \). To get the total energy (work done), multiply the energy density by the total volume of the material.

Updated On: Mar 30, 2026

0.04 kJ
0.04 J
100 kJ
400 J

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The Correct Option is A

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