Capillarity , Ascent Formula

When a piece of chalk is dipped into water, it is observed that water rises through the pores of chalk and wets it.

Consider a glass capillary of radius R dipped in water as shown in the figure.

The pressure below the meniscus will be $ \displaystyle P_0 – \frac{2 T}{r} $

To compensate for this pressure difference, water in the capillary rises so that

$ \displaystyle \frac{2 T}{r} = \rho g h $

$ \displaystyle h = \frac{2 T}{\rho g r} $

where r is the radius of meniscus, and

$ \displaystyle r = \frac{R}{cos\theta} $     , where θ is the angle of contact

$ \displaystyle h = \frac{2 T cos\theta}{\rho g R} $

If θ < 90° , the meniscus will be concave, for Illustration: at a water-glass interface.

If θ > 90° , the meniscus will be convex, for Illustration: at a mercury-glass interface.

Illustration : A meniscus drop of radius 1 cm is sprayed into 106 droplets of equal size. Calculate the energy expended if surface tension of mercury is 435 × 10-3 N/m.

Solution : Energy expended will be the work done again the increase in surface area i.e.

n(4πr2) – 4 πR2

E = W = TΔS

= T. 4π (nr2 – R2)

But the total volume remains constant.

$ \displaystyle \frac{4}{3}\pi R^3 = n \frac{4}{3}\pi r^3 $

$ \displaystyle r = \frac{R}{n^{1/3}} $

Hence, E = 4 πR2T (n1/3 – 1)

= 4 × 3.14 × (1 × 10-2)2 × 435 × 10-3 (102 – 1)

= 54.1 × 10-3 J

Also Read:

Surface Tension
Surface Energy
Excess Pressure inside a soap bubble
Angle of contact
Viscosity , Stoke’s Law & Terminal Velocity
Poiseuille’s formula

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