Mechanical Properties of Solids

JEE Main level, numerical answer, medium

110 of 118

Question 110 of 118

A steel wire of length 2 m and Young's modulus 2.0×1011Nm−22.0 \times 10^{11} \mathrm{Nm}^{-2} is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10−310^{-3} respectively, then the elastic potential energy density of the wire is _______ ×105\times 10^5 (in SI units)

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Answer: 25

ℓ=2 m;Y=2×1011 N m2μ=−(Δrr)(Δℓℓ)⇒Δℓℓ=1μ×(Δrr)=10.2×(10−3)\begin{aligned} & \ell=2 \mathrm{~m} ; \mathrm{Y}=2 \times 10^{11} \frac{\mathrm{~N}}{\mathrm{~m}^2} \\ & \begin{aligned} \mu=-\frac{\left(\frac{\Delta \mathrm{r}}{\mathrm{r}}\right)}{\left(\frac{\Delta \ell}{\ell}\right)} \Rightarrow \frac{\Delta \ell}{\ell} & =\frac{1}{\mu} \times\left(\frac{\Delta \mathrm{r}}{\mathrm{r}}\right) \\ & =\frac{1}{0.2} \times\left(10^{-3}\right)\end{aligned}\end{aligned}
⇒Δℓℓ=5×10−3u=12yεℓ2=12×2×1011×[5×10−3]2=25\begin{aligned} & \Rightarrow \frac{\Delta \ell}{\ell}=5 \times 10^{-3} \\ & \begin{aligned} \mathrm{u}=\frac{1}{2} \mathrm{y} \varepsilon_{\ell}^2 & =\frac{1}{2} \times 2 \times 10^{11} \times\left[5 \times 10^{-3}\right]^2 \\ & =25\end{aligned}\end{aligned}

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