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)

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 0.33.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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.
$∴ A_{0} \times 2 = A_{N} (2 + 0.08 \times 10^{- 2})$
$\Rightarrow A_{N} = \frac{A_{0} \times 2}{2 + 0.08 \times 10^{- 2}}$

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