JEE MainPhysicsModerateMechanical Properties of Fluids

A vertical capillary tube with inner radius 0.5 mm, is submerged
into water so that the length of its part above the water surface is h =
25 mm. Radius of curvature of meniscuses formed is
$\frac{\alpha}{10}mm$ . Find $\alpha$ . (surface tension of water = 0.075
N/m. Density of water 1000kgm^-3, angle of contact = 0°)

Show answer and solution

h_{0} = \frac{2{Scos}\theta}{r_{0}\rho g} = \frac{2 \times 7.5 \times 10^{- 2}}{5 \times 10^{- 4} \times 10^{3} \times 10}

= \frac{15 \times 10^{- 2}}{5} = 3 \times 10^{- 2}m = 30mm

h_{0}r_{0} = hr

\Rightarrow r = \frac{h_{0}}{h}\ r_{0} = \left( \frac{30}{25} \right)(0.5) = 0.6mm

Source: Grass-G / P-24 Physcis Paper 31 Dec.docx

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