The longitudinal stress in a rod (with rectangular cross-section)...

$∵$ Stress, $\dfrac{F}{∆ A} = 1$ of $Y = \dfrac{Y}{100}$ But $Young′s$ modulus, $Y = \dfrac{stress}{strain} = \dfrac{\dfrac{F}{∆ A}}{\dfrac{∆ l}{l}}$

∴ Y = \dfrac{\dfrac{Y}{100}}{\dfrac{∆ l}{l}} \left( putting\dfrac{F}{∆ A} = \dfrac{Y}{100}\right)<br /><br />∴ $\dfrac{∆ l}{l} = \dfrac{1}{100}<br /><br />${Poisson}^{′}s$ratio, \sigma = \dfrac{- \dfrac{∆ r}{r}}{\dfrac{∆ l}{l}}<br /><br />$∴ $\dfrac{∆ r}{r} = - \sigma$⋅ $\dfrac{∆ l}{l} = \dfrac{- 0.3}{100}<br /><br />\dfrac{∆ r}{r} = \dfrac{- 0.3}{100}<br /><br />$∴$Change in volume, \dfrac{∆ V}{V} = \dfrac{2 ∆ r}{r} + \dfrac{∆ l}{l}<br /><br />=\dfrac{2 \times \left(- 0.3\right)}{100} + \dfrac{1}{100}<br /><br />=\dfrac{1 - 0.6}{100} = \dfrac{0.4}{100}</p><p>So the percentage change in volume = 0.4$