A steel wire of length 2 m and Young's modulus is stretched by a force. If Poisson ratio and transverse strain for for the wire are 0.2 and respectively, then the elastic potential energy density of the wire is _____ (in SI units) [2025]

Question

A steel wire of length 2 m and Young's modulus $2.0 \times 10^{11} {Nm}^{- 2}$ is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and $10^{- 3}$ respectively, then the elastic potential energy density of the wire is _____ $\times 10^{5}$ (in SI units) [2025]

Smart Achievers · Accepted Answer

(25)

$U = \frac{1}{2} \times Y \times {(longitudinal strain)}^{2}$

$σ =$ $| \frac{transverse strain}{longitudinal strain} |$

Longitudinal strain = $\frac{10^{- 3}}{0.2} = 5 \times 10^{- 3}$

$U = \frac{1}{2} \times 2 \times 10^{11} \times 25 \times 10^{- 6}$

$= 25 \times 10^{5} J / m^{3}$

Solution

Q.
A steel wire of length 2 m and Young's modulus $2.0 \times 10^{11} {Nm}^{- 2}$ is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and $10^{- 3}$ respectively, then the elastic potential energy density of the wire is _____ $\times 10^{5}$ (in SI units) [2025]

Ans.
(25)

$U = \frac{1}{2} \times Y \times {(longitudinal strain)}^{2}$

$σ =$ $| \frac{transverse strain}{longitudinal strain} |$

Longitudinal strain = $\frac{10^{- 3}}{0.2} = 5 \times 10^{- 3}$

$U = \frac{1}{2} \times 2 \times 10^{11} \times 25 \times 10^{- 6}$

$= 25 \times 10^{5} J / m^{3}$

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Solution

Q. A steel wire of length 2 m and Young's modulus 2.0×1011Nm–2 is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10–3 respectively, then the elastic potential energy density of the wire is _____ ×105 (in SI units) [2025]

Ans. (25) U=12×Y×(longitudinal strain)2 σ=|transverse strainlongitudinal strain| Longitudinal strain = 10–30.2=5×10–3 U=12×2×1011×25×10–6 =25×105 J/m3

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Q.
A steel wire of length 2 m and Young's modulus $2.0 \times 10^{11} {Nm}^{- 2}$ is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and $10^{- 3}$ respectively, then the elastic potential energy density of the wire is _____ $\times 10^{5}$ (in SI units) [2025]

Ans.
(25)

$U = \frac{1}{2} \times Y \times {(longitudinal strain)}^{2}$

$σ =$ $| \frac{transverse strain}{longitudinal strain} |$

Longitudinal strain = $\frac{10^{- 3}}{0.2} = 5 \times 10^{- 3}$

$U = \frac{1}{2} \times 2 \times 10^{11} \times 25 \times 10^{- 6}$

$= 25 \times 10^{5} J / m^{3}$