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Question

(C) Length of steel wire,$l = 3 \ m$. Increased length,$\Delta l = 0.3 \ cm = 0.3 	imes 10^{-2} \ m = 3 	imes 10^{-3} \ m$. Poisson's ratio $(\sigma

Vedclass · Accepted Answer

$0.26 	imes 10^{-3}$

Vedclass · Answer

(C) Length of steel wire,$l = 3 \ m$. Increased length,$\Delta l = 0.3 \ cm = 0.3 	imes 10^{-2} \ m = 3 	imes 10^{-3} \ m$. Poisson's ratio $(\sigma)$ $= 0.26$. By definition,Poisson's ratio is the ratio of lateral strain to longitudinal strain: $\sigma = \frac{	ext{Lateral strain}}{	ext{Longitudinal strain}} = 0.26$. Longitudinal strain $= \frac{\Delta l}{l} = \frac{3 	imes 10^{-3} \ m}{3 \ m} = 10^{-3}$. Therefore,Lateral strain $= 0.26 	imes 	ext{Longitudinal strain} = 0.26 	imes 10^{-3}$.

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

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