A capillary tube made of glass of radius $0.15 mm$ is dipped vertically in a beaker filled with methylene iodide (surface tension = $0.05 N m^{- 1},$ density = $667 k g m^{- 3}$ ) which rises to height ‘h’ in the tube. It is observed that the two tangents drawn from liquid-glass interfaces (from opp. sides of the capillary) make an angle of $60^{0}$ with one another. Then ‘h’ is close to (g = $10 m s^{- 2}$ ).

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0.049 m

0.087 m

0.137 m

0.172 m

answer is B.

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Detailed Solution

Angle of contact $θ = 30^{0}$
Surface tension, $T = 0.05 N m^{- 1}$
Radius of capillary tube, r = 0.15 mm = $0.15 \times 10^{- 3} m$
Density of methylene iodide, $ρ = 667 k g m^{- 3}$
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$c a p i l l a r y r i s e, h = \frac{2 T \cos θ}{ρ g r}$ $= \frac{2 \times 0.05 \times \frac{\sqrt{3}}{2}}{667 \times 10 \times 0.15 \times 10^{- 3}} = 0.087 m$