The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length 100 cm to stretch it by 1 mm is (if Young's modulus of the wire = 2.0 × 10¹¹ Nm⁻²): (a) 10¹¹ (b) 10¹⁷ (c) 10⁷ (d) 10⁵

Answer: (d). Energy per unit volume u = 1/2Y(strain)². Strain = (1 mm)/(100 cm) = 10⁻³ u = 1/2 × 2 × 10¹¹ × 10⁻⁶ = 10⁵ J/m³ So the answer is (d) 10⁵.

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Elastic Potential Energy

The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length $100\text{ cm}$ to stretch it by $1\text{ mm}$ is (if Young's modulus of the wire $= 2.0 \times 10^{11}\text{ Nm}^{-2}$ ):

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Answer: (d)

Energy per unit volume $u = \dfrac{1}{2}Y(\text{strain})^2$ .
Strain $= \dfrac{1\text{ mm}}{100\text{ cm}} = 10^{-3}$
$u = \dfrac{1}{2} \times 2 \times 10^{11} \times 10^{-6} = 10^5\text{ J/m}^3$
So the answer is (d) $10^5$ .

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