A tension of 20 ,N is applied to a copper wir | vedclass

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(A) Given: Tension $F = 20 \,N$, Area $A = 0.01 \,cm^2 = 10^{-6} \,m^2$, Young's modulus $Y = 1.1 	imes 10^{11} \,N/m^2$, Poisson's ratio $\sigma = 0

Vedclass · Accepted Answer

$1.16 	imes 10^{-6} \,cm^2$

Vedclass · Answer

(A) Given: Tension $F = 20 \,N$, Area $A = 0.01 \,cm^2 = 10^{-6} \,m^2$, Young's modulus $Y = 1.1 	imes 10^{11} \,N/m^2$, Poisson's ratio $\sigma = 0.32$.The longitudinal strain is given by $\frac{\Delta l}{l} = \frac{F}{AY} = \frac{20}{10^{-6} 	imes 1.1 	imes 10^{11}} = \frac{20}{1.1 	imes 10^5} \approx 1.818 	imes 10^{-4}$.Poisson's ratio is defined as $\sigma = -\frac{\Delta r/r}{\Delta l/l}$, so the lateral strain is $\frac{\Delta r}{r} = -\sigma \frac{\Delta l}{l} = -0.32 	imes 1.818 	imes 10^{-4} \approx -0.5818 	imes 10^{-4}$.The area is $A = \pi r^2$, so $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.The decrease in area is $\Delta A = 2 	imes A 	imes \sigma 	imes \frac{\Delta l}{l} = 2 	imes 10^{-6} 	imes 0.32 	imes 1.818 	imes 10^{-4} \approx 1.16 	imes 10^{-10} \,m^2$.Converting to $cm^2$: $1.16 	imes 10^{-10} 	imes (10^2 \,cm)^2 = 1.16 	imes 10^{-6} \,cm^2$.

$A$ tension of $20 \,N$ is applied to a copper wire of cross-sectional area $0.01 \,cm^2$. 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 the cross-sectional area of the wire is:

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