A copper wire of cross-sectional area 0.01 c | vedclass

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(A) Given:Area of cross-section $A = 0.01 \ cm^2 = 10^{-6} \ m^2$.Tension $T = 22 \ N$.Young's modulus $Y = 1.1 	imes 10^{11} \ N \ m^{-2}$.Poisson r

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$12.8 	imes 10^{-3}$

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(A) Given:Area of cross-section $A = 0.01 \ cm^2 = 10^{-6} \ m^2$.Tension $T = 22 \ N$.Young's modulus $Y = 1.1 	imes 10^{11} \ N \ m^{-2}$.Poisson ratio $\sigma = 0.32$.Longitudinal strain $\frac{\Delta l}{l} = \frac{T}{AY} = \frac{22}{10^{-6} 	imes 1.1 	imes 10^{11}} = \frac{22}{1.1 	imes 10^5} = 20 	imes 10^{-5}$.Lateral strain $\frac{\Delta r}{r} = -\sigma \frac{\Delta l}{l} = -0.32 	imes 20 	imes 10^{-5} = -6.4 	imes 10^{-5}$.Since $A = \pi r^2$,the change in area is $\Delta A = 2\pi r \Delta r$,so $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.Percentage change in area $= |2 	imes (-6.4 	imes 10^{-5})| 	imes 100 = 12.8 	imes 10^{-3} \%$.

$A$ copper wire of cross-sectional area $0.01 \ cm^2$ is under a tension of $22 \ N$. Find the percentage change in the cross-sectional area (Young's modulus of copper $= 1.1 \times 10^{11} \ N \ m^{-2}$ and Poisson ratio $= 0.32$).

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