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Angle of contact is the angle which is tangent to a liquid surface at the point of contact makes with the solid surface inside the liquid.

It's value is acute (90^{o}) for liquid which do notwet the walls of container.

Capillary Tube

A glass tube of very fine bare (hair like) is known as capillary tube. Liquid having concave meniscus (θ ^{o}) depresses below the free surface inside a capillary tube. When capillary tube open at both ends when dipped in a liquid.

This rise or fall of a liquid in a capillary tube is known as capillary action or capillarity.

**Measurement of Surface Tension of a liquid by Capillarity**

Let us consider a capillary tube opened at both ends of radius is dipped in a liquid of density (Ρ) having concave meniscus.

Let h be the height of liquid rises in the tube above free surface as shown in the figure.Suppose the liquid molecules at the point of contact with the tube. The force due to the surface tension (T) acts on it along the tangent to the meniscus inside the liquid.

According to Newton's third law of motion, the tube also exerts on equal and opposite reaction force (R) can be resolved into two compounds i.e Tcosθ which acts vertically upward and Tsinθ which acts horizontally. The horizontal components Tsinθ acting on molecules at the point of contact cancel each other whereas vertical component added up. Hence a liquid experiences a net upward force on the liquid $$F = T\cos\theta \times 2\pi r \dots (i) $$

In equilibrium,

\begin{align*} \text {Net force on the surface of the liquid} (F) &= \text {height of the liquid in the tube} (w) \\ \text {Now volume of liquid in the tube (v)} &= \text {volume of cylinder of radius 'r' and height} (h) -\text {volume of radius} (v) \\ V &= \pi r^2r + \pi r^2r + \frac 23 \pi r^3 \dots (iii) \\ \end{align*}

Now, equation (i) becomes

\begin{align*} T\cos\theta \times 2\pi r &= V\rho g\\ T\cos\theta \times 2\pi r &= (\pi r^2 h +\frac 13 \pi r^3)\rho g \\ 2T\cos\theta &= r^2 \left ( h + \frac r3 \right ) \rho g \\ T &= \frac {r \left ( h + \frac r3\right)\rho g}{2\cos\theta}\end{align*}

Here \(\frac r3 << h\) so neglect \(\frac r3\)

\begin{align*} T &= \frac {rh\rho g}{2\cos \theta} \\ h &= \frac {2\cos\theta}{r\rho g} \dots (vi) \\ h \propto \frac 1r \\ \end{align*}

Equation (vi) is known ascent formula

For liquid having \(\theta > 90^o, \cos \theta\) becomes negative such liquid depresses below the free surface the corresponding formula is known as descent formula.

**Rise of Liquid in a Tube of Insufficient Length**

Let us consider a capillary tube opened at both ends of radius is dipped in a liquid of density (Ρ) having concave meniscus.

Let h be the height of liquid rises in the tube above free surface as shown in the figure.Suppose the liquid molecules at the point of contact with the tube. The force due to the surface tension (T) acts on it along the tangent to the meniscus inside the liquid.

As we know that the height of a liquid that rises in a capillary tube of radius 'r' is given by

$$h =\frac {2T\cos \theta}{r \rho g} \dots (i)$$

From figure

\begin{align*} \cos\theta = \frac rR \\ \therefore r = R\cos\theta \dots (ii) \\ \end{align*}

Where R is the radius of curvature of the concave meniscus. Equation (i) becomes

\begin{align*} h = \frac{2T\cos \theta}{R\cos \theta \rho g}\\ hR = \frac {2T}{\rho g} =\text {constant} \dots (iii) \end{align*}

If a capillary tube if insufficient height 'h' is dipped in the same liquid then equation (iii) becomes

$$h'R' = \frac {2T}{\rho g} \dots (iv)$$

Fro, equation (iii) and (iv)

$$ hR = h'R' $$

As \( h' < h \: \text {so} \: R' > R\)

liquid will not spill out from the tube but the radius of curvature of its meniscus will increase.

1In equilibrium,

\begin{align*} \text {Net force on the surface of the liquid} (F) &= \text {height of the liquid in the tube} (w) \\ \text {Now volume of liquid in the tube (v)} &= \text {volume of cylinder of radius 'r' and height} (h) -\text {volume of radius} (v) \\ V &= \pi r^2r + \pi r^2r + \frac 23 \pi

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