A tension of 22 ~N is applied to a copper wir | vedclass

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(A) Given: Tension $F = 22 ~N$, Area $A = 0.02 ~cm^2 = 0.02 	imes 10^{-4} ~m^2$, Young's modulus $Y = 1.1 	imes 10^{11} ~N/m^2$, Poisson's ratio $\s

Vedclass · Accepted Answer

$1.28 	imes 10^{-6} ~cm^2$

Vedclass · Answer

(A) Given: Tension $F = 22 ~N$, Area $A = 0.02 ~cm^2 = 0.02 	imes 10^{-4} ~m^2$, Young's modulus $Y = 1.1 	imes 10^{11} ~N/m^2$, Poisson's ratio $\sigma = 0.32$.Longitudinal strain is given by $\frac{\Delta l}{l} = \frac{F}{AY} = \frac{22}{0.02 	imes 10^{-4} 	imes 1.1 	imes 10^{11}} = \frac{22}{2.2 	imes 10^5} = 10^{-4}$.Poisson's ratio is defined as $\sigma = - \frac{\Delta r/r}{\Delta l/l}$. The lateral strain is $\frac{\Delta r}{r} = -\sigma \frac{\Delta l}{l}$.The area $A = \pi r^2$, so $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.Substituting the lateral strain, $\frac{\Delta A}{A} = -2 \sigma \frac{\Delta l}{l}$.The magnitude of decrease in area is $\Delta A = 2 \sigma A \frac{\Delta l}{l}$.$\Delta A = 2 	imes 0.32 	imes 0.02 ~cm^2 	imes 10^{-4} = 0.64 	imes 0.02 	imes 10^{-4} ~cm^2 = 0.0128 	imes 10^{-4} ~cm^2 = 1.28 	imes 10^{-6} ~cm^2$.

$A$ tension of $22 ~N$ is applied to a copper wire of cross-sectional area $0.02 ~cm^2$. The Young's modulus of copper is $1.1 \times 10^{11} ~N/m^2$ and the Poisson's ratio is $0.32$. The decrease in cross-sectional area will be:

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