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

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Question

(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 $\

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 $\frac{\Delta l}{l} = \frac{F}{AY} = \frac{22}{(0.02 	imes 10^{-4}) 	imes (1.1 	imes 10^{11})} = \frac{22}{2.2 	imes 10^4} = 10^{-4}$.Poisson's ratio is defined as $\sigma = -\frac{\Delta r/r}{\Delta l/l}$. The lateral strain is $\frac{\Delta r}{r} = -\sigma \frac{\Delta l}{l} = -0.32 	imes 10^{-4}$.The area is $A = \pi r^2$,so $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.Substituting the values,$\frac{\Delta A}{A} = 2 	imes (-0.32 	imes 10^{-4}) = -0.64 	imes 10^{-4}$.The decrease in area is $\Delta A = |\frac{\Delta A}{A}| 	imes A = (0.64 	imes 10^{-4}) 	imes (0.02 \,cm^2) = 1.28 	imes 10^{-6} \,cm^2$.

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

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