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Physics · MECHANICAL PROPERTIES OF SOLIDS
हिन्दीA wire of original diameter \( d \) and length \( L \) has Poisson's ratio \( \nu \). When it is stretched longitudinally so that the longitudinal strain is \( \varepsilon = \frac{\Delta L}{L} \), the fractional change in its volume (\( \frac{\Delta V}{V} \)) for small deformations is given by:
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Correct answer · Option B
\( \varepsilon (1 - 2\nu) \)
Explanation
Volume \( V = \frac{\pi}{4} d^2 L \). Differentiating gives \( \frac{\Delta V}{V} = \frac{\Delta L}{L} + 2\frac{\Delta d}{d} \). Since lateral strain is \( \frac{\Delta d}{d} = -\nu \frac{\Delta L}{L} = -\nu\varepsilon \), we have \( \frac{\Delta V}{V} = \varepsilon(1 - 2\nu) \).
