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(D) Given that stress $\sigma_{s} = \frac{1}{100} Y$, where $Y$ is the Young's modulus.We know that $Y = \frac{	ext{stress}}{	ext{longitudinal strai

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$0.4$

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(D) Given that stress $\sigma_{s} = \frac{1}{100} Y$, where $Y$ is the Young's modulus.We know that $Y = \frac{	ext{stress}}{	ext{longitudinal strain}} = \frac{\sigma_{s}}{\Delta l / l}$.Substituting the value of stress: $Y = \frac{Y / 100}{\Delta l / l} \implies \frac{\Delta l}{l} = \frac{1}{100} = 0.01$.Poisson's ratio $
u = -\frac{\Delta w / w}{\Delta l / l} = -\frac{\Delta t / t}{\Delta l / l} = 0.3$.Thus, the lateral strain $\frac{\Delta w}{w} = \frac{\Delta t}{t} = -
u \left( \frac{\Delta l}{l} ight) = -0.3 	imes 0.01 = -0.003$.The volume of a rectangular rod is $V = l 	imes w 	imes t$.The fractional change in volume is $\frac{\Delta V}{V} \approx \frac{\Delta l}{l} + \frac{\Delta w}{w} + \frac{\Delta t}{t}$.Substituting the values: $\frac{\Delta V}{V} = 0.01 + (-0.003) + (-0.003) = 0.01 - 0.006 = 0.004$.Therefore, the percentage change in volume is $0.004 	imes 100 \% = 0.4 \%$.

The stress along the length of a rod (with rectangular cross-section) is $1 \%$ of the Young's modulus of its material. What is the approximate percentage of change of its volume (in $\%$)? (Poisson's ratio of the material of the rod is $0.3$.)

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