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Q.a) Water rises up in a narrow tube in spite of gravity. This phenomenon is called .......

(1)
b) Derive an expression for the height of water in the tube in terms of radius of the tube and surface tension of the liquid.
(3)
c) Water with detergent dissolved in it should have ......... angle of contact. (small/large) (1)
Kerala DhseKerala DHSE Plus One Board 2018Subjective· 5mImportance★★★★★
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Capillary rise happens because surface tension curves the liquid surface inside a narrow tube into a meniscus, which creates an extra (Laplace) pressure that pushes liquid up until it is balanced by the weight of the raised column: h=2Tcos⁡θρgrh = \dfrac{2T\cos\theta}{\rho g r}. A detergent needs a small angle of contact so it wets cloth fibres thoroughly.

a) Name of the phenomenon

When a narrow (capillary) tube is dipped into a wetting liquid such as water, the liquid climbs up inside the tube well above the level in the surrounding vessel, even though gravity acts downward on it. This rise of liquid in a narrow tube is called capillarity or capillary rise / capillary action. (For a liquid that does not wet the tube, such as mercury in glass, the level is instead depressed — capillary depression — but here water is the wetting case being asked about.)

b) Derivation of the height of rise

Consider a capillary tube of radius rr dipped vertically into water of surface tension TT and density ρ\rho, and let the water make an angle of contact θ\theta with the glass wall of the tube (for water and clean glass, θ≈0\theta \approx 0).

Step 1 — shape of the meniscus. Because water wets glass, the liquid surface inside the tube curves upward into a concave meniscus. If rr is the radius of the tube and θ\theta is the angle of contact, simple geometry (the radius of the tube, the line of contact, and the radius of curvature RR of the meniscus form a right triangle) gives

r=Rcos⁡θ⇒R=rcos⁡θr = R\cos\theta \quad\Rightarrow\quad R = \frac{r}{\cos\theta}

Step 2 — excess pressure across the curved surface. A curved liquid surface (a spherical meniscus here) has an excess pressure on its concave side compared to its convex side, given by Laplace's formula for a spherical liquid surface:

ΔP=2TR=2Tcos⁡θr\Delta P = \frac{2T}{R} = \frac{2T\cos\theta}{r}

This excess pressure exists just below the meniscus, and it is lower than atmospheric pressure just under the concave meniscus (the concave side faces down into the liquid), which is exactly why liquid is pushed/pulled up the tube.

Step 3 — pressure balance. Consider a point A just below the meniscus inside the tube (at the top of the risen column) and a point B at the same level, but outside the tube, in the free liquid surface, and a point C directly below A at the level of the free liquid surface in the vessel. The pressure at B (open to atmosphere, flat surface) is atmospheric pressure P0P_0. The pressure at C must equal P0P_0 too, since C is on the free surface outside. But by hydrostatics, the pressure at C equals the pressure at A (top of column, just below meniscus, where it is P0−ΔPP_0 - \Delta P by Step 2) plus the weight of the column of height hh above it:

P0=(P0−ΔP)+hρgP_0 = (P_0 - \Delta P) + h\rho g

⇒ΔP=hρg\Rightarrow \Delta P = h \rho g

Step 4 — combine. Equating the two expressions for ΔP\Delta P: …

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