$A 3 \mathrm{~m}$ long steel wire is stretched to increase its length by $0.3 \mathrm{~cm}$. Poisson's ratio for steel is 0.26 . The lateral strain produced in the wire is

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Aicharya · Accepted Answer

Length of steel wire, $l=3 \mathrm{~m}$ Increased length, $\Delta l=0.3 \mathrm{~cm}=3 \times 10^{-3} \mathrm{~m}$ Poisson's ratio $=0.26$ $\Rightarrow \quad \frac{\text { Lateral strain }}{\text { Longitudinal strain }}=0.26$ $\Rightarrow$ Lateral strain $=0.26 \times$ longitudinal strain $\begin{aligned} & =0.26 \times \frac{\Delta l}{l} \ & =0.26 \times \frac{3 \times 10^{-3}}{3}=0.26 \times 10^{-3} \end{aligned}$

\(A 3 \mathrm{~m}\) long steel wire is stretched to increase its length by \(0.3 \mathrm{~cm}\). Poisson's…

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