This MCQ module is based on: Viscosity Surface Tension
TOPIC 7 OF 33
Viscosity Surface Tension
🎓 Class 11
Physics
CBSE
Theory
Ch 9 – Mechanical Properties of Fluids
⏱ ~14 min
🌐 Language: [gtranslate]
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Viscosity Surface Tension
9.5 Viscosity
Water pours easily; honey oozes; pitch takes years to flow. The property that makes one fluid "thicker" than another is viscosity — internal friction between adjacent layers of a fluid moving at different velocities.
Viscosity is the resistance a fluid offers to relative motion between its layers. When one layer slides past another, tangential forces act to drag the faster layer back and pull the slower layer along.
9.5.1 Newton's Law of Viscous Flow
Consider a fluid confined between two horizontal plates separated by a distance \(dx\) (Fig 9.15). The bottom plate is stationary; the top plate is pulled with velocity \(v\). Fluid layers near the top move fast, those near the bottom are at rest — a velocity gradient \(dv/dx\) is established.
Newton found that the tangential viscous force \(F\) needed to maintain the motion is proportional to the plate area \(A\) and the velocity gradient:
\[F = \eta\,A\,\frac{dv}{dx}\]
where \(\eta\) (eta) is the coefficient of viscosity. SI unit: Pa·s (= N·s/m²). In CGS: poise (P), with 1 Pa·s = 10 P. Dimensions: \([ML^{-1}T^{-1}]\).
9.5.2 Typical Values of Viscosity
| Fluid | η (Pa·s) at ~20 °C | Remarks |
|---|---|---|
| Air | 1.8 × 10⁻⁵ | Very low — gases slip freely |
| Water | 1.0 × 10⁻³ | Reference fluid |
| Blood (whole) | ~4 × 10⁻³ | Non-Newtonian; varies with shear |
| Machine oil (light) | 0.1 | Lubricant |
| Glycerine | 0.83 | ~830× water |
| Honey | 2 – 10 | Strongly temperature-dependent |
In general, viscosity of a liquid decreases as temperature increases (think honey on a warm day), while viscosity of a gas increases with temperature (more thermal agitation → more momentum transfer between layers).
9.5.3 Stokes' Law
A small sphere of radius \(r\) moving slowly with speed \(v\) through a viscous fluid of viscosity \(\eta\) experiences a viscous drag force given by Stokes' law:
\[\boxed{\,F = 6\pi\eta r v\,}\]
This is valid for streamline flow (low speeds) and for a smooth rigid sphere.
9.5.4 Terminal Velocity
When a sphere (say a raindrop, or a ball bearing falling through oil) descends in a fluid, it experiences three forces: gravity (downward), buoyancy (upward), and viscous drag (upward, opposing motion). As the sphere speeds up, the drag increases. At a certain velocity the three forces balance and the sphere falls with a constant terminal velocity \(v_t\).
Let the density of the sphere be \(\rho\) and that of the fluid be \(\sigma\). At terminal velocity:
\[\tfrac{4}{3}\pi r^3 \rho g \;=\; \tfrac{4}{3}\pi r^3 \sigma g \;+\; 6\pi\eta r v_t\] \[\Longrightarrow\quad \boxed{\,v_t = \frac{2 r^2 (\rho - \sigma) g}{9\eta}\,}\]Key observations: \(v_t\propto r^2\) — a drop of radius 2 mm falls 4× faster than one of radius 1 mm. Larger raindrops hit the ground harder. Dust particles have such small \(r\) that \(v_t\) is tiny — they stay suspended in air for hours. Parachutes, by contrast, use a very high-drag shape to keep a human's terminal velocity low enough for safe landing.
9.5.5 Poiseuille's Equation (brief)
For laminar flow of a liquid through a cylindrical tube of length \(L\) and radius \(r\) under a pressure difference \(\Delta P\), the volume flow rate is:
\[Q = \frac{\pi\,\Delta P\,r^4}{8\eta L}\]The astonishing \(r^4\) dependence means halving the radius of a pipe cuts the flow to 1/16. This explains why even minor narrowing of blood vessels (atherosclerosis) causes dramatic reductions in blood supply.
9.6 Surface Tension
Water forms drops, insects walk on ponds, narrow tubes suck water up against gravity — these are all manifestations of surface tension, the tendency of a liquid surface to behave like a stretched elastic membrane.
9.6.1 Molecular Origin
A molecule deep in the bulk of a liquid is surrounded on all sides by identical neighbours; the attractive intermolecular forces cancel out. A molecule at the surface has neighbours below and beside, but none above — so it experiences a net inward force. The liquid surface tries to minimise its area, hence the spherical shape of free droplets (a sphere has the smallest surface-area-to-volume ratio).
9.6.2 Definition — Surface Tension & Surface Energy
Surface tension (S): the force per unit length acting tangent to the liquid surface, perpendicular to any line drawn on it.
\[S = \frac{F}{L}\]
Unit: N/m (or J/m² — the two are dimensionally identical).
Surface energy: the extra energy stored per unit area of liquid surface. Numerically equal to surface tension: \[\text{Work done in extending the surface by area } \Delta A = S\,\Delta A\]
Surface energy: the extra energy stored per unit area of liquid surface. Numerically equal to surface tension: \[\text{Work done in extending the surface by area } \Delta A = S\,\Delta A\]
A soap bubble has two surfaces (inner and outer). To increase its total surface area by \(\Delta A\), the work done is \(W = S\times 2\Delta A\).
9.6.3 Angle of Contact
Where a liquid meets a solid, the liquid surface makes a definite angle of contact (θ) with the solid, measured through the liquid. This angle depends on the relative strengths of the cohesive forces (liquid–liquid) and the adhesive forces (liquid–solid).
| Case | Angle θ | Behaviour | Example |
|---|---|---|---|
| Adhesion > Cohesion | θ < 90° (acute) | Liquid wets surface; concave meniscus; rises in capillary | Water on clean glass |
| Cohesion > Adhesion | θ > 90° (obtuse) | Liquid does not wet; convex meniscus; depressed in capillary | Mercury on glass |
| Perfectly wetting | θ = 0° | Liquid spreads as a thin film | Water on very clean glass |
9.6.4 Excess Pressure Inside a Drop / Bubble
Surface tension tries to shrink a curved surface; so the pressure inside a curved drop must be greater than outside to balance it.
Liquid drop (one surface): \[P_{\text{in}} - P_{\text{out}} = \frac{2S}{r}\]
Soap bubble (two surfaces — inner and outer): \[P_{\text{in}} - P_{\text{out}} = \frac{4S}{r}\]
Air bubble inside a liquid (one surface): \(\dfrac{2S}{r}\)
Because \(\Delta P \propto 1/r\), a small bubble has a higher excess pressure than a big one. If you connect a small bubble to a large one, air flows from the small to the large — the small bubble shrinks to nothing while the large one grows. A counter-intuitive but easily verified result.
9.6.5 Capillarity — The Jurin Formula
Dip a narrow glass tube in water; water rises up the tube, against gravity. This is capillary rise. Dip it in mercury, and the mercury is depressed below the free surface.
Consider a tube of radius \(r\) with a liquid of density \(\rho\), surface tension \(S\), angle of contact \(\theta\). Balance the upward vertical component of surface tension around the contact circle (\(2\pi r\cdot S\cos\theta\)) against the weight of the raised column (\(\pi r^2 h\,\rho g\)):
\[\boxed{\,h = \frac{2S\cos\theta}{\rho g r}\,}\]
For water on glass (\(\theta \approx 0\)), \(h = 2S/(\rho g r)\) — narrower tube means higher rise. For mercury (\(\theta \approx 140°\)), \(\cos\theta\) is negative so \(h\) is negative — the liquid is depressed.
Capillarity is how plants draw sap up from the roots through thin xylem vessels, how a blotting paper soaks ink, and how a wick draws oil up a lamp.
9.6.6 Detergent Action
Water alone cannot wet an oily surface — the angle of contact is obtuse, so water beads up and rolls off. Detergents and soaps contain molecules with a hydrophilic (water-loving) head and a hydrophobic (oil-loving) tail. Dissolving them in water drops the surface tension to about 1/3 of pure water, reduces the angle of contact, and allows water to wet oil and dirt. The detergent tails stick to the oil, wrap it in tiny droplets (micelles), and these are washed away.
Worked Examples
Example 9.9: Terminal velocity of a raindrop
Find the terminal velocity of a spherical raindrop of radius 0.20 mm falling through air. Density of water = 1000 kg/m³, density of air = 1.2 kg/m³, viscosity of air = 1.8 × 10⁻⁵ Pa·s, g = 9.8 m/s².
\(r = 0.20 \times 10^{-3} = 2.0\times 10^{-4}\) m.
Since \(\sigma \ll \rho\), approximate \((\rho - \sigma) \approx \rho = 1000\) kg/m³. \[v_t = \frac{2r^2(\rho - \sigma)g}{9\eta} = \frac{2\times(2.0\times 10^{-4})^2 \times 1000\times 9.8}{9\times 1.8\times 10^{-5}}\] \[= \frac{2\times 4\times 10^{-8}\times 9800}{1.62\times 10^{-4}} = \frac{7.84\times 10^{-4}}{1.62\times 10^{-4}} = 4.84\text{ m/s}\] Answer: \(v_t \approx \boxed{4.8\text{ m/s}}\). In reality large raindrops deform and have a slightly lower terminal speed.
Since \(\sigma \ll \rho\), approximate \((\rho - \sigma) \approx \rho = 1000\) kg/m³. \[v_t = \frac{2r^2(\rho - \sigma)g}{9\eta} = \frac{2\times(2.0\times 10^{-4})^2 \times 1000\times 9.8}{9\times 1.8\times 10^{-5}}\] \[= \frac{2\times 4\times 10^{-8}\times 9800}{1.62\times 10^{-4}} = \frac{7.84\times 10^{-4}}{1.62\times 10^{-4}} = 4.84\text{ m/s}\] Answer: \(v_t \approx \boxed{4.8\text{ m/s}}\). In reality large raindrops deform and have a slightly lower terminal speed.
Example 9.10: Capillary rise of water
A clean glass capillary of internal radius 0.20 mm is dipped vertically in water. How high does the water rise? Surface tension of water = 0.073 N/m, angle of contact = 0°, ρ = 1000 kg/m³, g = 9.8 m/s².
\[h = \frac{2S\cos\theta}{\rho g r} = \frac{2\times 0.073\times 1}{1000\times 9.8\times 2.0\times 10^{-4}}\]
\[= \frac{0.146}{1.96} = 0.0745\text{ m} \approx 7.45\text{ cm}\]
Answer: \(h \approx \boxed{7.5\text{ cm}}\).
Example 9.11: Excess pressure in a soap bubble
Calculate the excess pressure inside a soap bubble of radius 1.0 cm. Surface tension of soap solution = 0.030 N/m.
A soap bubble has two surfaces, so
\[\Delta P = \frac{4S}{r} = \frac{4\times 0.030}{0.010} = 12.0\text{ Pa}\]
Answer: Excess pressure \(= \boxed{12\text{ Pa}}\).
Example 9.12: Work done in blowing a soap bubble
How much work must be done to blow a soap bubble from zero size to a radius of 2 cm in air? Surface tension = 0.030 N/m.
Total surface area created (two surfaces) = \(2\times 4\pi r^2 = 8\pi r^2\).
\[W = S\times 8\pi r^2 = 0.030 \times 8\pi \times (0.02)^2\]
\[= 0.030 \times 8\pi \times 4\times 10^{-4} = 3.02\times 10^{-4}\text{ J}\]
Answer: \(W \approx \boxed{3.0\times 10^{-4}\text{ J}}\).
Example 9.13: Viscous force between plates
A metal plate of area 0.10 m² slides horizontally over a thin film of oil (η = 0.80 Pa·s) of thickness 0.50 mm on a fixed table. What horizontal force is required to keep the plate moving at a constant 0.20 m/s?
Velocity gradient \(dv/dx = 0.20/(5\times 10^{-4}) = 400\) s⁻¹.
\[F = \eta A \frac{dv}{dx} = 0.80\times 0.10\times 400 = 32\text{ N}\]
Answer: \(F = \boxed{32\text{ N}}\).
Example 9.14: Two drops merging
Eight identical water drops each of radius 1 mm coalesce to form a single large drop. How much energy is released? (Surface tension of water = 0.073 N/m.)
Volume conservation: \(8\times \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi R^3 \Rightarrow R = 2r = 2\) mm.
Initial surface area: \(A_i = 8\times 4\pi r^2 = 32\pi r^2\).
Final surface area: \(A_f = 4\pi R^2 = 4\pi(2r)^2 = 16\pi r^2\).
Decrease in area: \(\Delta A = 32\pi r^2 - 16\pi r^2 = 16\pi r^2 = 16\pi\times (10^{-3})^2 = 5.03\times 10^{-5}\) m².
Energy released: \(E = S\,\Delta A = 0.073\times 5.03\times 10^{-5} = 3.67\times 10^{-6}\) J.
Answer: \(E \approx \boxed{3.7\times 10^{-6}\text{ J}}\). The energy shows up as a slight rise in temperature of the merged drop.
Initial surface area: \(A_i = 8\times 4\pi r^2 = 32\pi r^2\).
Final surface area: \(A_f = 4\pi R^2 = 4\pi(2r)^2 = 16\pi r^2\).
Decrease in area: \(\Delta A = 32\pi r^2 - 16\pi r^2 = 16\pi r^2 = 16\pi\times (10^{-3})^2 = 5.03\times 10^{-5}\) m².
Energy released: \(E = S\,\Delta A = 0.073\times 5.03\times 10^{-5} = 3.67\times 10^{-6}\) J.
Answer: \(E \approx \boxed{3.7\times 10^{-6}\text{ J}}\). The energy shows up as a slight rise in temperature of the merged drop.
Interactive: Terminal Velocity Calculator L3 Apply
Adjust the sphere's radius, its density and the fluid viscosity — watch the terminal velocity change.
Activity — A Needle on Water
L3 Apply
Predict: Can you float a steel sewing needle on the surface of water? Steel is 8× denser than water — yet water striders do it with legs, not wings. How?
- Fill a clean bowl with water to the brim.
- Rest a small piece of tissue paper on the surface.
- Place a dry sewing needle carefully on the tissue paper.
- Gently poke the tissue with a pencil until it sinks, leaving the needle behind.
Observation: The needle floats on the surface, supported by a slight depression in the water.
Explanation: The water surface behaves like a stretched elastic membrane because of surface tension. The vertical components of the surface-tension force around the needle (F = S × contact length) support the needle's weight. Add a drop of detergent and the surface tension falls — the needle sinks at once.
Explanation: The water surface behaves like a stretched elastic membrane because of surface tension. The vertical components of the surface-tension force around the needle (F = S × contact length) support the needle's weight. Add a drop of detergent and the surface tension falls — the needle sinks at once.
Competency-Based Questions
An aerosol paint-spray produces droplets of radius 0.05 mm. A physics student investigates how the drops travel and settle. Density of paint = 900 kg/m³; viscosity of air = 1.8 × 10⁻⁵ Pa·s; density of air = 1.2 kg/m³. Surface tension of paint = 0.035 N/m.
Q1. L1 Remember State Stokes' law and name the quantity denoted by η.
\(F = 6\pi\eta r v\) is the viscous drag on a small sphere of radius r moving at speed v through a fluid of viscosity η (coefficient of viscosity).
Q2. L3 Apply Find the terminal velocity of the paint droplets.
\(r = 5\times 10^{-5}\) m. \(v_t = \frac{2r^2(\rho-\sigma)g}{9\eta} \approx \frac{2\times 25\times 10^{-10}\times 900\times 9.8}{9\times 1.8\times 10^{-5}} = \frac{4.41\times 10^{-5}}{1.62\times 10^{-4}} = \boxed{0.27\text{ m/s}}\).
Q3. L3 Apply Find the excess pressure inside one such droplet.
Single surface: \(\Delta P = 2S/r = 2\times 0.035/(5\times 10^{-5}) = \boxed{1400\text{ Pa}}\).
Q4. L4 Analyse Explain why the spray forms a fine mist that hangs in the air rather than falling straight down.
Because \(v_t \propto r^2\), these very small droplets have a terminal speed of only ~0.3 m/s. Any slight air current easily overpowers gravity, so the droplets drift and hover. They appear as a "mist" and can be carried across a room.
Q5. L5 Evaluate The manufacturer claims that reducing droplet radius improves paint coverage. Evaluate this claim in terms of both benefits and drawbacks.
Benefit: smaller drops → larger total surface area per unit mass → more uniform coverage and less drip. Drawback: smaller drops have higher excess pressure (\(\propto 1/r\)) and tend to re-evaporate rapidly; they also have very low terminal velocity, so a lot of paint is wasted as airborne mist (overspray) that never reaches the target. There is an engineering trade-off — typical commercial droplets are 20–100 µm.
Assertion-Reason Questions
Assertion (A): Small raindrops are nearly spherical but larger drops are flattened.
Reason (R): For small drops, surface tension dominates and minimises surface area giving a sphere; for larger drops, air drag and gravity deform the shape significantly.
Answer: A. Standard result from the balance of surface tension, drag and gravity.
Assertion (A): Narrower capillaries produce higher capillary rise.
Reason (R): The height of rise \(h = 2S\cos\theta/(\rho g r)\) is inversely proportional to the radius r.
Answer: A. Direct consequence of Jurin's formula.
Assertion (A): When two soap bubbles of different sizes are connected by a tube, the smaller bubble shrinks and the larger one grows.
Reason (R): Excess pressure inside a soap bubble (\(\Delta P = 4S/r\)) is larger for smaller bubbles.
Answer: A. The higher pressure of the smaller bubble drives air into the larger bubble until the smaller collapses.
Frequently Asked Questions - Viscosity Surface Tension
What is the main concept covered in Viscosity Surface Tension?
In NCERT Class 11 Physics Chapter 9 (Mechanical Properties of Fluids), "Viscosity Surface Tension" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Viscosity Surface Tension useful in real-life applications?
Real-life applications of Viscosity Surface Tension from NCERT Class 11 Physics Chapter 9 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Viscosity Surface Tension?
Key formulas in Viscosity Surface Tension (NCERT Class 11 Physics Chapter 9 Mechanical Properties of Fluids) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 9?
NCERT Class 11 Physics Chapter 9 (Mechanical Properties of Fluids) is structured so each part builds on the previous one. Viscosity Surface Tension connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Viscosity Surface Tension?
CBSE board questions from Viscosity Surface Tension typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Viscosity Surface Tension lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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