JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2025
- Medium
- Numerical
A steel wire of length 2 m and Young's modulus \(2.0 \times 10^{11} \mathrm{Nm}^{-2}\) is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and \(10^{-3}\) respectively, then the elastic potential energy density of the wire is _______ \(\times 10^5\) (in SI units)
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Answer & Solution
Correct Answer
(B) 25
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \ell=2 \mathrm{~m} ; \mathrm{Y}=2 \times 10^{11} \frac{\mathrm{~N}}{\mathrm{~m}^2} \\ & \begin{aligned} \mu=-\frac{\left(\frac{\Delta \mathrm{r}}{\mathrm{r}}\right)}{\left(\frac{\Delta \ell}{\ell}\right)} \Rightarrow \frac{\Delta \ell}{\ell} & =\frac{1}{\mu} \times\left(\frac{\Delta \mathrm{r}}{\mathrm{r}}\right) \\ & =\frac{1}{0.2} \times\left(10^{-3}\right)\end{aligned}\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{\Delta \ell}{\ell}=5 \times 10^{-3} \\ & \begin{aligned} \mathrm{u}=\frac{1}{2} \mathrm{y} \varepsilon_{\ell}^2 & =\frac{1}{2} \times 2 \times 10^{11} \times\left[5 \times 10^{-3}\right]^2 \\ & =25\end{aligned}\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{\Delta \ell}{\ell}=5 \times 10^{-3} \\ & \begin{aligned} \mathrm{u}=\frac{1}{2} \mathrm{y} \varepsilon_{\ell}^2 & =\frac{1}{2} \times 2 \times 10^{11} \times\left[5 \times 10^{-3}\right]^2 \\ & =25\end{aligned}\end{aligned}\)
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