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

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(D) Young's modulus is given by $Y = \frac{F/A}{\Delta l/l}$, so the longitudinal strain is $\frac{\Delta l}{l} = \frac{F}{YA}$.Given $F = 22 \,N$, $Y

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$1.28 	imes 10^{-6} \,cm^2$

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(D) Young's modulus is given by $Y = \frac{F/A}{\Delta l/l}$, so the longitudinal strain is $\frac{\Delta l}{l} = \frac{F}{YA}$.Given $F = 22 \,N$, $Y = 1.1 	imes 10^{11} \,N/m^2$, and $A = 0.01 \,cm^2 = 10^{-6} \,m^2$.Calculating longitudinal strain: $\frac{\Delta l}{l} = \frac{22}{1.1 	imes 10^{11} 	imes 10^{-6}} = 2 	imes 10^{-4}$.Poisson's ratio $\sigma$ is defined as $\sigma = \frac{	ext{Lateral strain}}{	ext{Longitudinal strain}} = \frac{\Delta r/r}{\Delta l/l}$.Thus, $\frac{\Delta r}{r} = \sigma \cdot \frac{\Delta l}{l} = 0.32 	imes 2 	imes 10^{-4} = 6.4 	imes 10^{-5}$.The area of the cross-section is $A = \pi r^2$. Taking the derivative, the fractional change in area is $\frac{\Delta A}{A} = 2 \frac{\Delta r}{r}$.Substituting the values: $\frac{\Delta A}{A} = 2 	imes 6.4 	imes 10^{-5} = 12.8 	imes 10^{-5}$.Therefore, the decrease in area is $\Delta A = (12.8 	imes 10^{-5}) 	imes (0.01 \,cm^2) = 1.28 	imes 10^{-6} \,cm^2$.

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

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