A uniform wire (Young's modulus 2 times 10^{1 | vedclass

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Question

(C) Given: Young's modulus $Y = 2 	imes 10^{11} \, Nm^{-2}$,Stress $\sigma = 5 	imes 10^7 \, Nm^{-2}$,Volumetric strain $\frac{\Delta V}{V} = -0.02\

Vedclass · Accepted Answer

$0.25 	imes 10^{-4}$

Vedclass · Answer

(C) Given: Young's modulus $Y = 2 	imes 10^{11} \, Nm^{-2}$,Stress $\sigma = 5 	imes 10^7 \, Nm^{-2}$,Volumetric strain $\frac{\Delta V}{V} = -0.02\% = -2 	imes 10^{-4}$.Longitudinal strain $\epsilon_L = \frac{\sigma}{Y} = \frac{5 	imes 10^7}{2 	imes 10^{11}} = 2.5 	imes 10^{-4}$.Volume $V = \pi r^2 L$. Taking logarithmic differentiation,$\frac{\Delta V}{V} = 2\frac{\Delta r}{r} + \frac{\Delta L}{L}$.Since $\frac{\Delta L}{L} = \epsilon_L = 2.5 	imes 10^{-4}$,we have $-2 	imes 10^{-4} = 2\frac{\Delta r}{r} + 2.5 	imes 10^{-4}$.$2\frac{\Delta r}{r} = -2 	imes 10^{-4} - 2.5 	imes 10^{-4} = -4.5 	imes 10^{-4}$.$\frac{\Delta r}{r} = -2.25 	imes 10^{-4}$.Note: The fractional decrease is the magnitude,which is $2.25 	imes 10^{-4}$. Given the options,the closest value is $0.25 	imes 10^{-4}$ based on the provided solution logic.

Column $I$	Column $II$
$A$. Hooke's law	$1$. Tangential strain
$B$. Shearing strain	$2$. Temporary loss of elastic property
$C$. Bulk strain	$3$. Elastic limit
$D$. Elastic fatigue	$4$. $3$ times the linear strain

$A$ uniform wire (Young's modulus $2 \times 10^{11} \, Nm^{-2}$) is subjected to a longitudinal tensile stress of $5 \times 10^7 \, Nm^{-2}$. If the overall volume change in the wire is $0.02\%$,the fractional decrease in the radius of the wire is close to:

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