A compressive force is applied to a uniform r | vedclass

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(C) Given: The fractional change in length is $\frac{\Delta l}{l} = -0.01$ (since it decreases by $1 \%$).Poisson's ratio, $\sigma = 0.2$.For a rod, t

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decreases by $0.6 \%$

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(C) Given: The fractional change in length is $\frac{\Delta l}{l} = -0.01$ (since it decreases by $1 \%$).Poisson's ratio, $\sigma = 0.2$.For a rod, the volume $V = A 	imes l$, where $A$ is the cross-sectional area.The fractional change in volume is given by $\frac{\Delta V}{V} = \frac{\Delta A}{A} + \frac{\Delta l}{l}$.Since $A = w 	imes t$ (width $	imes$ thickness), $\frac{\Delta A}{A} = \frac{\Delta w}{w} + \frac{\Delta t}{t}$.By definition of Poisson's ratio, $\sigma = -\frac{\Delta w / w}{\Delta l / l} = -\frac{\Delta t / t}{\Delta l / l}$.Thus, $\frac{\Delta w}{w} = -\sigma \frac{\Delta l}{l}$ and $\frac{\Delta t}{t} = -\sigma \frac{\Delta l}{l}$.Substituting these into the volume equation:$\frac{\Delta V}{V} = -\sigma \frac{\Delta l}{l} - \sigma \frac{\Delta l}{l} + \frac{\Delta l}{l} = \frac{\Delta l}{l} (1 - 2\sigma)$.Substituting the values: $\frac{\Delta V}{V} = -0.01 	imes (1 - 2 	imes 0.2) = -0.01 	imes (1 - 0.4) = -0.01 	imes 0.6 = -0.006$.Therefore, the volume decreases by $0.6 \%$.

A compressive force is applied to a uniform rod of rectangular cross-section so that its length decreases by $1 \%$. If the Poisson's ratio for the material of the rod is $0.2,$ which of the following statements is correct? "The volume approximately ..........."

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