First slide

Surface tension and surface energy

Question

Water rises to a height h in a capillary tube of cross-sectional area A. The height to which water will rise in a capillary tube of cross-sectional area 4A will be

Easy

A

h
B

h/2
C

h/4
D

4h

Solution

$h = \frac{2 S \cos θ}{r ρ g}$
Cross-sectional area increases four times, which means radius gets doubled
$So, h' = \frac{2 S \cos θ}{2 (r ρ g)} = \frac{h}{2}$

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