A uniform glass tube of length $2 m$ is closed at the lower end. It is completely filled with water and rigidly fixed at upper end. Now the tube is stretched downwards. Although the length of tube increases by $0.12 c m,$ the length of water column increases only by $0.08 c m .$ Assume water is incompressible. What is poisson’s ratio $σ$ for glass ? (Neglect surface tension effects)

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answer is 0.33.

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Detailed Solution

$σ = - \frac{\frac{d r}{r}}{\frac{d l}{l}}$
Given $\frac{d l}{l} = \frac{0.12 \times 10^{- 2}}{2} = 0.06 \times 10^{- 2}$
Also, since water is incompressible $\Rightarrow$ volume of water remains same.
$∴ A_{0} \times 2 = A_{N} (2 + 0.08 \times 10^{- 2})$
$\Rightarrow A_{N} = \frac{A_{0} \times 2}{2 + 0.08 \times 10^{- 2}}$

$\Rightarrow \frac{A_{N}}{A_{0}} = \frac{2}{2 (1 + 0.04 \times 10^{- 2})} = {(1 + 0.04 \times 10^{- 2})}^{- 1} \Rightarrow \frac{π {(r_{0} + Δ r)}^{2}}{π r_{0}^{2}} = 1 - 0.04 \times 10^{- 2} \Rightarrow {(1 + \frac{d r}{r_{0}})}^{2} = 1 - 0.04 \times 10^{- 2} \Rightarrow \frac{d r}{r} = - 0.02 \times 10^{- 2}$
$∴ σ = \frac{- \frac{d r}{r}}{\frac{d l}{l}} = \frac{0.02 \times 10^{- 2}}{0.06 \times 10^{- 2}} = \frac{1}{3}$