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Properties of Bulk Matter - Surface Tension

Grade 11ICSEPhysics

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

🔑Concepts

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Surface Tension (TT): The property of a liquid surface at rest to behave like a stretched elastic membrane. It is measured as the force per unit length acting normally on either side of an imaginary line drawn on the liquid 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. The relation between work done (WW) and surface tension is W=TΔAW = T \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. For pure water and clean glass, θ≈0∘\theta \approx 0^\circ; for mercury and glass, θ≈135∘\theta \approx 135^\circ.

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Excess Pressure: Due to surface tension, the pressure on the concave side of a liquid surface is greater than on the convex side. For a liquid drop, Pexcess=2TRP_{excess} = \frac{2T}{R}. For a soap bubble, Pexcess=4TRP_{excess} = \frac{4T}{R} because it has two surfaces.

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Capillarity: The phenomenon of rise or fall of a liquid in a narrow tube (capillary). A liquid that wets the glass (like water) rises, while a liquid that does not wet the glass (like mercury) falls.

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Effect of Temperature: Surface tension generally decreases with an increase in temperature. It becomes zero at the critical temperature. The variation is given by Tt=T0(1−αt)T_t = T_0(1 - \alpha t).

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Effect of Impurities: Highly soluble substances (like salt in water) increase surface tension, while sparingly soluble substances (like soap or phenol) decrease it.

📐Formulae

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

W=T⋅ΔAW = T \cdot \Delta A

Pexcess=2TR (for a liquid drop)P_{excess} = \frac{2T}{R} \text{ (for a liquid drop)}

Pexcess=4TR (for a soap bubble)P_{excess} = \frac{4T}{R} \text{ (for a soap bubble)}

h=2Tcos⁡θrρg (Ascent Formula)h = \frac{2T \cos \theta}{r \rho g} \text{ (Ascent Formula)}

T=rhρg2cos⁡θT = \frac{rh\rho g}{2 \cos \theta}

💡Examples

Problem 1:

Calculate the work done in blowing a soap bubble of radius 0.1 m0.1 \text{ m}. The surface tension of the soap solution is 30×10−3 N/m30 \times 10^{-3} \text{ N/m}.

Solution:

W=T⋅ΔAW = T \cdot \Delta A Since a soap bubble has two free surfaces (inner and outer), ΔA=2×4πR2\Delta A = 2 \times 4\pi R^2. W=T×8πR2W = T \times 8\pi R^2 W=30×10−3×8×3.14×(0.1)2W = 30 \times 10^{-3} \times 8 \times 3.14 \times (0.1)^2 W=30×10−3×8×3.14×0.01W = 30 \times 10^{-3} \times 8 \times 3.14 \times 0.01 W=7.536×10−3 JW = 7.536 \times 10^{-3} \text{ J}

Explanation:

The work done to increase the surface area of a soap bubble is twice that of a solid drop because the soap film has two surfaces in contact with air.

Problem 2:

Water rises to a height of 20 cm20 \text{ cm} in a capillary tube of radius rr. If the radius of the capillary tube is reduced to r3\frac{r}{3}, 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}. Since TT, θ\theta, ρ\rho, and gg are constant, we have: h∝1r  ⟹  h1r1=h2r2h \propto \frac{1}{r} \implies h_1 r_1 = h_2 r_2 Given h1=20 cmh_1 = 20 \text{ cm} and r2=r13r_2 = \frac{r_1}{3}: 20×r1=h2×r1320 \times r_1 = h_2 \times \frac{r_1}{3} h2=20×3=60 cmh_2 = 20 \times 3 = 60 \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 - Revision Notes & Key Formulas | ICSE Class 11 Physics