There is no change in the volume of a wire du | vedclass

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(B) The volume $V$ of a wire of radius $r$ and length $l$ is given by $V = \pi r^2 l$.Since the volume remains unchanged during stretching,$dV = 0$.Di

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(B) The volume $V$ of a wire of radius $r$ and length $l$ is given by $V = \pi r^2 l$.Since the volume remains unchanged during stretching,$dV = 0$.Differentiating the volume equation,we get $dV = 2\pi rl dr + \pi r^2 dl = 0$.Rearranging the terms,$2\pi rl dr = -\pi r^2 dl$.This simplifies to $\frac{dr}{r} = -\frac{1}{2} \frac{dl}{l}$.Poisson's ratio $\sigma$ is defined as the ratio of lateral strain to longitudinal strain: $\sigma = -\frac{dr/r}{dl/l}$.Substituting the value from our equation,$\sigma = -(-\frac{1}{2}) = 0.5$.

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 is

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