First slide

Surface tension

Question

When the lower end of a vertical capillary tube of radius r is dipped in water, water rises in it due to capillarity action and change in gravitational potential energy of water is U. If the tube is replaced by another tube of radius 2r, charge in gravitational potential energy will be

Moderate

A

4U
B

2U
C

U/2
D

U

Solution

Capillary rise, $h = \frac{2 T}{ρ r g}$
Mass of water in the tube, $m = π r^{2} h ρ$
$∴$ Charge in gravitational potential energy
$U = m g \frac{h}{2} = \frac{1}{2} (π r^{2} h ρ) . g . h$
$= \frac{π}{2} r^{2} ρ g . {(\frac{2 T}{ρ r g})}^{2} = \frac{2 π T^{2}}{ρ g}$
So change in gravitational potential energy is independent of r
$∴ U^{1} = U$

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