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# There is no change in the volume of a wire due to change in its length on stretching. The Poisson’s ratio of the material of the wire isA.0.05B.-0.05C.0.25D.-0.25

Last updated date: 13th Jun 2024
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Hint: In this question first we will use the relation between the volume, length and the side of the wire, and then we will deduce an expression for the Poisson’s ratio. Further, we will also study the basics of Poisson’s effect and Poisson’s ratio, for our better understanding.
Formula used:
$V = l \times {s^2}$
$\eta = - \dfrac{{\dfrac{{ds}}{s}}}{{\dfrac{{dl}}{l}}}$

Here, let length of the material l and side s. As the material maintains constant volume during stretching, this volume V is given as:
$V = l \times {s^2}$
Now, we differentiate the above equation with respect to small length dl:
$dV = dl \times 1.2s.ds$
Further, solving this we get:
\eqalign{& dl.s = 2l.ds \cr & \Rightarrow \dfrac{{ds}}{{dl}} = - \dfrac{s}{{2l}} \cr}
$\therefore \eta = - \dfrac{{\dfrac{{ds}}{s}}}{{\dfrac{{dl}}{l}}} = \dfrac{1}{2}$

Therefore, the correct option is A) i.e., the Poisson’s ratio of the material of the wire is 0.5, which is given by the above result.