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Mechanical Properties of Fluids - Surface Tension and Capillary Rise

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Surface Tension (TT): It is the property of a liquid at rest by virtue of which its free surface behaves like a stretched elastic membrane and tends to occupy the minimum surface area. It is defined as the force per unit length acting perpendicular to an imaginary line drawn on the surface: T=FlT = \frac{F}{l}.

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Surface Energy: The extra potential energy possessed by the molecules at the surface of a liquid is called surface energy. The relation between surface tension and work done is W=T×ΔAW = T \times \Delta A, where ΔA\Delta A is the increase in surface area.

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Angle of Contact (θ\theta): The angle between the tangent to the liquid surface at the point of contact and the solid surface inside the liquid. It is acute for liquids that wet the glass (like H2OH_2O) and obtuse for liquids that do not wet the glass (like HgHg).

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Excess Pressure: Due to surface tension, the pressure inside a curved liquid surface is greater than the pressure outside. For a liquid drop, Pexcess=2TRP_{excess} = \frac{2T}{R}. For a soap bubble (which has two surfaces), Pexcess=4TRP_{excess} = \frac{4T}{R}.

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Capillarity: The phenomenon of rise or fall of a liquid in a tube of very fine bore (capillary tube) is called capillarity. The height hh to which a liquid rises is given by the Ascent Formula: h=2Tcos⁡θrρgh = \frac{2T \cos \theta}{r \rho g}.

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Factors affecting Surface Tension: It decreases with the increase in temperature. The addition of highly soluble impurities (like NaClNaCl) increases surface tension, while sparingly soluble impurities (like phenol or soap) decrease it.

📐Formulae

T=FlT = \frac{F}{l}

W=TΔAW = T \Delta A

Pdrop=2TRP_{drop} = \frac{2T}{R}

Pbubble=4TRP_{bubble} = \frac{4T}{R}

h=2Tcos⁡θrρgh = \frac{2T \cos \theta}{r \rho g}

R=rcos⁡θR = \frac{r}{\cos \theta}

💡Examples

Problem 1:

Calculate the work done in blowing a soap bubble from a radius of 2 cm2\text{ cm} to 5 cm5\text{ cm}. The surface tension of the soap solution is 3×10−2 N/m3 \times 10^{-2}\text{ N/m}.

Solution:

Given: r1=2×10−2 mr_1 = 2 \times 10^{-2}\text{ m}, r2=5×10−2 mr_2 = 5 \times 10^{-2}\text{ m}, T=3×10−2 N/mT = 3 \times 10^{-2}\text{ N/m}. A soap bubble has two free surfaces, so ΔA=2×(4πr22−4πr12)\Delta A = 2 \times (4\pi r_2^2 - 4\pi r_1^2). W=T×8π(r22−r12)W = T \times 8\pi(r_2^2 - r_1^2). W=3×10−2×8×3.14×[(25−4)×10−4]W = 3 \times 10^{-2} \times 8 \times 3.14 \times [(25 - 4) \times 10^{-4}]. W=0.7536×21×10−6≈1.58×10−3 JW = 0.7536 \times 21 \times 10^{-6} \approx 1.58 \times 10^{-3}\text{ J}.

Explanation:

Since a soap bubble is hollow, it has both an inner and an outer surface in contact with air, hence the area change is doubled.

Problem 2:

Water rises to a height of 10 cm10\text{ cm} in a capillary tube. If the radius of the tube is made half of its previous value, what will be the new height of the water column?

Solution:

From the ascent formula, h=2Tcos⁡θrρgh = \frac{2T \cos \theta}{r \rho g}, we see that h∝1rh \propto \frac{1}{r} (Jurin's Law). Therefore, h1r1=h2r2h_1 r_1 = h_2 r_2. Given h1=10 cmh_1 = 10\text{ cm} and r2=r12r_2 = \frac{r_1}{2}. 10×r1=h2×r12  ⟹  h2=20 cm10 \times r_1 = h_2 \times \frac{r_1}{2} \implies h_2 = 20\text{ cm}.

Explanation:

According to Jurin's Law, the height of the liquid column in a capillary tube is inversely proportional to the radius of the tube.

Surface Tension and Capillary Rise - Revision Notes & Key Formulas | CBSE Class 11 Physics