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CalibratePhysics

Chapter 2

Mechanical Properties of Fluids

Maharashtra HSC · 9 subtopics · textbook pages 26–55

Formula SheetEvery formula this chapter can be examined on, in one table.

Pressure

FormulaSymbols, units and conditions
p=FAp = \dfrac{F}{A}
  • FF = normal force exerted by the fluid, AA = area of the surface of contact. Pressure is a scalar. SI unit N m−2^{-2} = pascal (Pa); dimensions [L−1M1T−2][L^{-1}M^{1}T^{-2}].
  • Other units: 1 bar=1051\ \text{bar} = 10^{5} N m−2^{-2}, 1 hPa=1001\ \text{hPa} = 100 Pa, 1 torr=11\ \text{torr} = 1 mm of Hg.
p=hρgp = h\rho g
  • Pressure due to a liquid column of height hh and density ρ\rho at depth hh below the free surface. Independent of the shape and size of the vessel and of the area of the column. Valid for liquids and for gases.
p=p0+hρgp = p_{0} + h\rho g
  • Absolute (total) pressure at depth hh; p0p_{0} = atmospheric pressure acting on the free surface.
p−p0=hρgp - p_{0} = h\rho g
  • Gauge pressure = absolute pressure −- atmospheric pressure.
p2=p1+ρg (x1−x2)p_{2} = p_{1} + \rho g\,(x_{1} - x_{2})
  • General result for two levels x1x_{1} (upper) and x2x_{2} (lower) in a fluid, heights measured upwards from a reference level.
p=p0−dρair gp = p_{0} - d\rho_{\text{air}}\,g
  • Atmospheric pressure at a height dd above the liquid surface (atmospheric density taken uniform up to that height).
  • Standard values the board expects:
  • Normal atmospheric pressure p0=1.013×105p_{0} = 1.013\times10^{5} N m−2^{-2} = 76 cm of Hg = 760 mm of Hg = 760 torr ≈1.013\approx 1.013 bar.
  • ρwater=1000\rho_{\text{water}} = 1000 kg m−3^{-3}, ρmercury=13.6×103\rho_{\text{mercury}} = 13.6\times10^{3} kg m−3^{-3}, g=9.8g = 9.8 m s−2^{-2}.
  • Specific gravity of mercury =ρHg/ρw=13.6= \rho_{\text{Hg}}/\rho_{w} = 13.6.

Pascal's Law and Hydraulic Machines

FormulaSymbols, units and conditions
F1A1=F2A2⇒F2=(A2A1)F1\dfrac{F_{1}}{A_{1}} = \dfrac{F_{2}}{A_{2}} \quad\Rightarrow\quad F_{2} = \left(\dfrac{A_{2}}{A_{1}}\right)F_{1}
  • Hydraulic lift and hydraulic brakes. A1A_{1} = area of the small (input / master) piston, A2A_{2} = area of the large (output / slave) piston. Since A2≫A1A_{2}\gg A_{1}, a small F1F_{1} gives a large F2F_{2}.
  • Maximum pressure a lift piston must bear for a load of mass MM: p=Mg/Ap = Mg/A.

Measurement of Pressure

FormulaSymbols, units and conditions
p0=hρgp_{0} = h\rho g
  • Mercury barometer: hh = length of the mercury column, ρ\rho = density of mercury. Pressure above the column (Torricelli's vacuum) is zero.
p=p0+hρg,p−p0=hρgp = p_{0} + h\rho g, \qquad p - p_{0} = h\rho g
  • Open tube manometer: hh = difference of levels of the manometric liquid, ρ\rho = its density. The manometer reads the gauge pressure of the gas.

Surface Tension and Surface Energy

FormulaSymbols, units and conditions
T=FLT = \dfrac{F}{L}
  • FF = tangential force acting on both sides of an imaginary line of length LL drawn on the free surface. SI unit N m−1^{-1}; dimensions [L0M1T−2][L^{0}M^{1}T^{-2}]. Equivalent SI unit J m−2^{-2}.
  • Force on a diametrical line of length LL on a liquid surface in a beaker: F=TLF = TL (one surface).
  • Movable wire of length ll on a frame carrying a film with two surfaces: F=2TlF = 2Tl.
  • Horizontal circular loop of radius rr pulled out of a liquid, film on both sides: F=T×2(2πr)=4πrTF = T\times 2(2\pi r) = 4\pi r T.
W=T ΔAandT=EΔAW = T\,\Delta A \qquad\text{and}\qquad T = \dfrac{E}{\Delta A}
  • Work done isothermally against surface tension is stored as surface energy. Hence surface tension is numerically equal to the surface energy per unit area.

Work Done in Changing the Surface Area

FormulaSymbols, units and conditions
W=4πT (nr2−R2)=4πTR2 (n1/3−1)W = 4\pi T\,(n r^{2} - R^{2}) = 4\pi T R^{2}\,(n^{1/3} - 1)
  • nn droplets of radius rr coalescing into one drop of radius R=n1/3rR = n^{1/3}r: energy released =4πT (nr2−R2)= 4\pi T\,(n r^{2} - R^{2}); decrease in area =4π(nr2−R2)= 4\pi (n r^{2} - R^{2}).
  • Ratio of work done in blowing two bubbles: W1/W2=(r1/r2)2W_{1}/W_{2} = (r_{1}/r_{2})^{2}.
  • Free surface energy of a liquid drop of radius rr (one surface): E=4πr2TE = 4\pi r^{2}T.
  • Work done in blowing a soap bubble of radius rr from nothing (two surfaces): W=2×4πr2T=8πr2TW = 2\times 4\pi r^{2}T = 8\pi r^{2}T.
  • Increasing a soap bubble from r1r_{1} to r2r_{2}: W=8πT (r22−r12)W = 8\pi T\,(r_{2}^{2} - r_{1}^{2}).
  • Soap film on a rectangular frame changed from area A1A_{1} to A2A_{2}: W=2T (A2−A1)W = 2T\,(A_{2} - A_{1}).
  • One drop of radius RR breaking into nn identical droplets: R3=nr3R^{3} = n r^{3}, so r=R/n1/3r = R/n^{1/3} and

Angle of Contact and Shape of a Drop

FormulaSymbols, units and conditions
cos⁡θ=T2−T1T3\cos\theta = \dfrac{T_{2} - T_{1}}{T_{3}}
  • T1T_{1} = surface tension at the liquid–solid interface, T2T_{2} at the air–solid interface, T3T_{3} at the air–liquid interface, θ\theta = angle of contact.
  • Condition at θ=90∘\theta = 90^\circ: AP=AC/2AP = AC/\sqrt{2}. Acute angle: AP>AC/2AP > AC/\sqrt{2}; obtuse angle: AP<AC/2AP < AC/\sqrt{2}, where APAP is the net adhesive force and ACAC the net cohesive force.

Excess Pressure — Laplace's Law of a Spherical Membrane (2.4.5, 2.4.6)

FormulaSymbols, units and conditions
pi−po=2Trp_{i} - p_{o} = \dfrac{2T}{r}
  • Liquid drop, or an air bubble inside a liquid — one free surface of radius rr.
pi−po=4Trp_{i} - p_{o} = \dfrac{4T}{r}
  • Soap bubble in air — two free surfaces.
  • Absolute pressure inside a raindrop: pi=p0+2Trp_{i} = p_{0} + \dfrac{2T}{r}.
  • Plane surface: no excess pressure. Pressure on the concave side is always greater than that on the convex side.

Capillarity

FormulaSymbols, units and conditions
h=2Tcos⁡θrρg⟺T=hrρg2cos⁡θh = \dfrac{2T\cos\theta}{r\rho g} \qquad\Longleftrightarrow\qquad T = \dfrac{h r \rho g}{2\cos\theta}
  • hh = rise (or fall), rr = radius of the capillary bore, θ\theta = angle of contact, ρ\rho = density of the liquid.
  • Radius of curvature of the meniscus: R=rcos⁡θR = \dfrac{r}{\cos\theta}, and the pressure-difference form hρg=2TRh\rho g = \dfrac{2T}{R}.
  • h∝1rh \propto \dfrac{1}{r} for a given liquid: h1r1=h2r2h_{1}r_{1} = h_{2}r_{2}. If the area of cross section is halved, rr becomes r/2r/\sqrt{2} and hh becomes 2 h\sqrt{2}\,h.
  • If θ\theta is obtuse, cos⁡θ\cos\theta is negative and hh is negative — the liquid is depressed (capillary fall).

Viscosity and Coefficient of Viscosity (2.6.1, 2.6.2)

FormulaSymbols, units and conditions
f=ηA(dvdx)f = \eta A \left(\dfrac{dv}{dx}\right)
  • Newton's law of viscosity. AA = area of the layer parallel to the flow (not the cross section), dv/dxdv/dx = velocity gradient measured from a stationary layer.
η=fA(dvdx)\eta = \dfrac{f}{A\left(\dfrac{dv}{dx}\right)}
  • Coefficient of viscosity = viscous force per unit area per unit velocity gradient. SI unit N s m−2^{-2}, also called decapoise. CGS unit poise; 11 decapoise =10= 10 poise, i.e. 11 poise =0.1= 0.1 N s m−2^{-2}.
  • For a plate resting on a layer of oil of thickness dxdx moving with velocity dvdv: η=F dxA dv\eta = \dfrac{F\,dx}{A\,dv}.

Critical Velocity and Reynolds Number

FormulaSymbols, units and conditions
vc=Rnηρd⟺Rn=ρvcdηv_{c} = \dfrac{R_{n}\eta}{\rho d} \qquad\Longleftrightarrow\qquad R_{n} = \dfrac{\rho v_{c} d}{\eta}
  • vcv_{c} = critical velocity, RnR_{n} = Reynolds number, η\eta = coefficient of viscosity, ρ\rho = density of the fluid, dd = diameter of the tube.
  • RnR_{n} is a pure number — no unit, no dimensions. Rn<1000R_{n} < 1000: streamline flow. Rn>2000R_{n} > 2000: turbulent flow. 1000<Rn<20001000 < R_{n} < 2000: unsteady, changing from streamline to turbulent.

Stokes' Law and Terminal Velocity

FormulaSymbols, units and conditions
Fv=6πηrvF_{v} = 6\pi\eta r v
  • Viscous force on a small sphere of radius rr moving with velocity vv through a medium of coefficient of viscosity η\eta. Streamline flow only.
v=29 r2g(ρ−σ)η⟺η=29 r2g(ρ−σ)vv = \dfrac{2}{9}\,\dfrac{r^{2}g(\rho - \sigma)}{\eta} \qquad\Longleftrightarrow\qquad \eta = \dfrac{2}{9}\,\dfrac{r^{2}g(\rho - \sigma)}{v}
  • Terminal velocity. ρ\rho = density of the sphere, σ\sigma = density of the medium. v∝r2v \propto r^{2}, so two bodies of radii in ratio 1:41:4 have terminal velocities in ratio 1:161:16.
  • If ρ<σ\rho < \sigma (an air bubble in a liquid) the answer comes out negative — the negative sign means the body rises.
  • The three forces at terminal velocity: weight 43πr3ρg\dfrac{4}{3}\pi r^{3}\rho g downwards, buoyant force 43πr3σg\dfrac{4}{3}\pi r^{3}\sigma g upwards, viscous force 6πηrv6\pi\eta r v upwards.

Equation of Continuity

FormulaSymbols, units and conditions
A1v1=A2v2i.e.Av=constantA_{1}v_{1} = A_{2}v_{2} \qquad\text{i.e.}\qquad Av = \text{constant}
  • Steady flow of an incompressible fluid. Av=dV/dtAv = dV/dt = volume flux (volume rate of flow); ρ dV/dt=dm/dt\rho\,dV/dt = dm/dt = mass flux.
  • v∝1A∝1d2v \propto \dfrac{1}{A} \propto \dfrac{1}{d^{2}} for a circular pipe of diameter dd.

Bernoulli's Equation

FormulaSymbols, units and conditions
p+12ρv2+ρgh=constantp + \dfrac{1}{2}\rho v^{2} + \rho g h = \text{constant}—
p1+12ρv12+ρgh1=p2+12ρv22+ρgh2p_{1} + \dfrac{1}{2}\rho v_{1}^{2} + \rho g h_{1} = p_{2} + \dfrac{1}{2}\rho v_{2}^{2} + \rho g h_{2}—
p1−p2=12ρ(v22−v12)+ρg(h2−h1)p_{1} - p_{2} = \dfrac{1}{2}\rho\left(v_{2}^{2} - v_{1}^{2}\right) + \rho g (h_{2} - h_{1})
  • Ideal fluid (incompressible, non-viscous, steady, irrotational). pp = pressure head, 12ρv2\dfrac{1}{2}\rho v^{2} = velocity head, ρgh\rho g h = potential head; each has the dimensions of energy per unit volume.
  • Horizontal pipe (h1=h2h_{1} = h_{2}): p1−p2=12ρ(v22−v12)p_{1} - p_{2} = \dfrac{1}{2}\rho\left(v_{2}^{2} - v_{1}^{2}\right).

Applications of Bernoulli's Equation

FormulaSymbols, units and conditions
v2=2gh1−(A2/A1)2  ⟶  v2=2gh(A2≪A1)v_{2} = \sqrt{\dfrac{2gh}{1 - \left(A_{2}/A_{1}\right)^{2}}} \;\longrightarrow\; v_{2} = \sqrt{2gh} \quad (A_{2} \ll A_{1})
  • Speed of efflux (Torricelli's law) through an orifice at depth hh below the free surface — the same as a body falling freely through height hh.
v=2pρv = \sqrt{\dfrac{2p}{\rho}}
  • Speed of efflux when the liquid is driven by a gauge pressure pp (a closed tank, or the drop in pressure on opening a valve).
p1−p2=12ρ(v22−v12)=hρHg gp_{1} - p_{2} = \dfrac{1}{2}\rho\left(v_{2}^{2} - v_{1}^{2}\right) = h\rho_{\text{Hg}}\,g
  • Venturi tube with a mercury manometer of level difference hh; combine with A1v1=A2v2A_{1}v_{1} = A_{2}v_{2} to get the rate of flow. Rearranged: h=ρw(v22−v12)2ρHg gh = \dfrac{\rho_{w}\left(v_{2}^{2} - v_{1}^{2}\right)}{2\rho_{\text{Hg}}\,g}.
Short NotesThe chapter subtopic by subtopic, in the textbook's own order.9 subtopics
  1. 1Introduction
  2. 2Fluid
  3. 3Pressure
  4. 4Surface Tension32 board questions
  5. 5Fluids in Motion1 board question
  6. 6Critical Velocity and Reynold's Number4 board questions
  7. 7Stoke's Law1 board question
  8. 8Equation of Continuity
  9. 9Bernoulli's Equation
The whole chapter in brief

2.1–2.2 Fluid and Ideal Fluid

  • A fluid is any substance that can flow — it deforms continually under an external force. Liquids, gases and plasmas are fluids.
  • The shear modulus of a fluid is zero: a fluid cannot resist a shear (tangential) stress in static equilibrium. Ideal fluids can be subjected only to normal, compressive stress, which is pressure.
  • Properties of fluids: they do not oppose deformation (they get permanently deformed), they can flow, and they take the shape of the container.
  • An ideal fluid is (i) incompressible — constant density, (ii) irrotational — smooth flow, no turbulence, (iii) non-viscous — no internal friction, (iv) steady — velocity at each point constant in time.
  • Real fluids display viscosity and can therefore sustain low levels of shear stress. Liquid helium at about 2 K is the only almost non-viscous fluid.
  • Hydrostatics = study of fluids at rest. Hydrodynamics = study of fluids in motion.

2.3 Pressure

  • Pressure = normal force per unit area of contact, p=F/Ap = F/A. Scalar. SI unit pascal; [L−1M1T−2][L^{-1}M^{1}T^{-2}].
  • Pressure due to a liquid column: p=hρgp = h\rho g. It does not depend on the area of the imaginary column, nor on the shape or size of the container, and holds for gases as well.
  • Absolute pressure at depth hh: p=p0+hρgp = p_{0} + h\rho g. Gauge pressure =p−p0=hρg= p - p_{0} = h\rho g.
  • Atmospheric pressure = weight of a column of air of unit cross section extending from that point to the top of the atmosphere. Highest at sea level; the effective air column is about 8 km. Normal atmospheric pressure = 1.013×1051.013\times10^{5} Pa = 76 cm of Hg.
  • Vacuum = region where the gas pressure is much less than atmospheric. Perfect vacuum = no matter at all.
  • Hydrostatic paradox: in interconnected vessels of different shapes the liquid stands at the same level, because the pressure at a point depends only on the height of the liquid column above it. The walls of a slanted vessel push on the liquid with forces normal to the wall whose vertical components support the extra weight.
  • Pascal's law: pressure applied at any point of an enclosed fluid at rest is transmitted equally and undiminished to every point of the fluid and to the walls of the container, provided gravity is neglected.
  • Applications: hydraulic lift and hydraulic brakes, both using F2=(A2/A1)F1F_{2} = (A_{2}/A_{1})F_{1}.
  • Mercury barometer (Torricelli): p0=hρgp_{0} = h\rho g, with Torricelli's vacuum above the column. Open tube manometer: p=p0+hρgp = p_{0} + h\rho g, so it reads the gauge pressure hρgh\rho g.

2.4 Surface Tension — definitions

  • Cohesive force: force of attraction between molecules of the same substance. Adhesive force: between molecules of different substances.
  • Range of molecular force: the maximum distance up to which the molecular force is effective — of the order of a few nanometres, hence a short range force.
  • Sphere of influence: an imaginary sphere with a molecule at its centre and radius equal to the molecular range.
  • Surface film: the surface layer of a liquid of thickness equal to the range of intermolecular force.
  • Free surface: the surface of a fluid which experiences no shear stress.
  • Surface tension TT: the tangential force acting per unit length on both sides of an imaginary line drawn on the free surface of a liquid. T=F/LT = F/L; N m−1^{-1}; [L0M1T−2][L^{0}M^{1}T^{-2}].
  • Surface energy: the extra potential energy of the molecules in the surface layer of a liquid. Work done isothermally to increase the area by dAdA is dW=T dAdW = T\,dA, so surface tension = surface energy per unit area (N m−1^{-1} = J m−2^{-2}).
  • Angle of contact θ\theta: the angle between the tangent to the free liquid surface and the tangent to the solid surface at the point of contact, measured within the liquid.
  • Capillarity: the rise or fall of a liquid inside a capillary tube dipped in it.

2.4 Surface Tension — standard results

SituationResult
Liquid drop / air bubble in liquidpi−po=2Trp_{i} - p_{o} = \dfrac{2T}{r}
Soap bubble in airpi−po=4Trp_{i} - p_{o} = \dfrac{4T}{r}
Capillary rise or fallh=2Tcos⁡θrρgh = \dfrac{2T\cos\theta}{r\rho g}
Surface energy of a dropE=4πr2TE = 4\pi r^{2}T
Work to blow a soap bubbleW=8πr2TW = 8\pi r^{2}T
Drop RR into nn dropletsW=4πTR2(n1/3−1)W = 4\pi T R^{2}(n^{1/3} - 1)

2.4 Shape of meniscus and angle of contact

Angle of contactMeniscusCohesive vs adhesiveExampleCapillary
Acute (θ<90∘\theta < 90^\circ)ConcaveAdhesive force dominatesWater or kerosene in glassRise
ZeroConcave, tangent along the wallCohesion negligibleHighly pure water on clean glassMaximum rise
Obtuse (θ>90∘\theta > 90^\circ)ConvexCohesive force dominatesMercury in glassFall
  • Typical values: pure water on clean glass 0∘0^\circ; ether–glass 16∘16^\circ; kerosene–glass 26∘26^\circ; water–paraffin 107∘107^\circ; mercury–glass 140∘140^\circ.
  • Angle of contact is constant for a given solid–liquid pair at a given temperature; it changes with impurity and decreases with rise in temperature; a small contamination changes it greatly.
  • Drop on a solid surface: cos⁡θ=(T2−T1)/T3\cos\theta = (T_{2} - T_{1})/T_{3}. If (T2−T1)>T3(T_{2} - T_{1}) > T_{3} then cos⁡θ>1\cos\theta > 1, which is impossible, and the liquid spreads instead of forming a drop.

2.4 Effect of impurity and temperature on surface tension

  • Soluble impurity (common salt in water): surface tension increases.
  • Sparingly soluble impurity (phenol, detergent): surface tension decreases — that is why detergent is added while washing clothes.
  • Insoluble impurity (dust on mercury): surface tension decreases.
  • For most liquids surface tension decreases as temperature rises; for molten copper and molten cadmium it increases. Surface tension becomes zero at the critical temperature.

2.4 Excess pressure across a curved surface

  • Plane surface: the resultant force of surface tension on a molecule just below the surface is zero, so the pressure just above and just below is the same.
  • Upper convex surface: the resultant force of surface tension is vertically downwards, adds to the atmospheric force, so the pressure just below (concave side) is greater.
  • Upper concave surface: the resultant force of surface tension is vertically upwards, opposes the atmospheric force, so the pressure just below (convex side) is smaller.
  • In every case: pressure on the concave side is greater than the pressure on the convex side.

2.5 Fluids in Motion

  • Flow line: the path of an individual particle in a moving fluid.
  • Streamline: a curve whose tangent at any point is along the velocity of flow there. For a steady flow, streamlines and flow lines are identical.
  • Flow tube: an imaginary bundle of flow lines bounded by an imaginary wall; in steady flow the fluid cannot cross its walls and adjacent tubes cannot mix.
  • Steady flow: a measurable property such as pressure or velocity at a given point is constant in time.
Sr. No.Streamline (laminar) flowTurbulent flow
iSmooth flow with velocity smaller than the critical velocityIrregular, unsteady flow when the velocity exceeds the critical velocity
iiVelocity at a given point is always constantVelocity at a point does not remain constant
iiiTwo streamlines never intersect; they are always parallel, hence no eddiesThe fluid may have rotational motion at some points, giving rise to eddies
ivThe flow can be divided into plane layers; in a pipe of uniform cross section the streamlines are parallel to the axisThe flow tube loses its order and particles move in random directions

2.6 Viscosity, critical velocity and Reynolds number

  • Viscosity is that property of a fluid by virtue of which the relative motion between different layers of the fluid experiences a dragging force, called the viscous drag.
  • In liquids the viscous drag is due to short range molecular cohesive forces; in gases it is due to collisions between fast moving molecules.
  • Velocity gradient: the rate of change of velocity dvdv with distance dxdx measured from a stationary layer, dv/dxdv/dx.
  • Newton's law of viscosity: f=ηA(dv/dx)f = \eta A (dv/dx), with AA the area of the layer parallel to the flow.
  • Coefficient of viscosity = viscous force per unit area per unit velocity gradient. SI unit N s m−2^{-2} (decapoise); 11 poise =0.1= 0.1 N s m−2^{-2}.
  • With rise in temperature the coefficient of viscosity of a liquid decreases but that of a gas increases.
  • Critical velocity: the velocity beyond which a streamline flow becomes turbulent, vc=Rnη/ρdv_{c} = R_{n}\eta/\rho d.
  • Reynolds number Rn=ρvcd/ηR_{n} = \rho v_{c} d/\eta is a pure number: Rn<1000R_{n} < 1000 streamline, Rn>2000R_{n} > 2000 turbulent, in between unsteady.

2.7 Stokes' law and terminal velocity

  • Stokes' law: the viscous force on a small sphere falling through a viscous medium is directly proportional to the radius of the sphere, its velocity and the coefficient of viscosity — Fv=6πηrvF_{v} = 6\pi\eta r v, the constant 6π6\pi being obtained empirically.
  • Three forces act on a sphere falling through a viscous fluid: weight downwards, viscous force upwards (increasing with speed) and buoyant force upwards (constant).
  • Terminal velocity is the constant downward velocity reached when Fg=Fv+FuF_{g} = F_{v} + F_{u}; the whole discussion applies to streamline flow only.
  • v=29r2g(ρ−σ)ηv = \dfrac{2}{9}\dfrac{r^{2}g(\rho - \sigma)}{\eta}, so v∝r2v \propto r^{2}.
  • A body can never move through a viscous fluid faster than the terminal velocity for that body in that fluid.

2.8 Equation of continuity

  • Statement: for a steady flow of an incompressible fluid, A1v1=A2v2A_{1}v_{1} = A_{2}v_{2}, i.e. AvAv = constant. It expresses conservation of mass.
  • The volume rate of flow dV/dt=AvdV/dt = Av is the volume flux; ρ dV/dt=dm/dt\rho\,dV/dt = dm/dt is the mass flux.
  • A particle moves faster through a narrow section and slower through a wider section, because there is less space to accommodate the particles behind it; otherwise the incompressible fluid would have to be compressed.

2.9 Bernoulli's equation

  • Statement: the work done per unit volume of a fluid by the surrounding fluid is equal to the sum of the changes in kinetic energy and potential energy per unit volume during the flow. It is an application of the work–energy theorem, and is consistent with conservation of energy.
  • p+12ρv2+ρgh=p + \dfrac{1}{2}\rho v^{2} + \rho g h = constant, for an ideal fluid in steady flow.
  • Bernoulli's principle: where the speed of a fluid is greater, its internal pressure is smaller.
  • Applications: speed of efflux (Torricelli's law v=2ghv = \sqrt{2gh}), Venturi tube, dynamic lift on an aeroplane wing, atomizer / paint gun / carburettor, blowing off of roofs by stormy wind.
  • Aeroplane: streamlines crowd above the aerodynamically shaped wing, so the air moves faster and the pressure above falls while that below is unchanged; the resulting upward dynamic lift raises the plane when it exceeds its weight.
  • Roof in a storm: fast wind over the roof lowers the pressure above it while the air inside stays at atmospheric pressure, so the roof is lifted and blown off.

Answer-in-brief points the board repeats

  • Surface tension of paints and lubricating oils is kept low so that they spread easily over a large area.
  • Work done in forming a soap bubble of radius rr is W=8πr2TW = 8\pi r^{2}T (two surfaces).
  • The basis of Bernoulli's principle is the conservation of energy (the work–energy theorem applied to a fluid element).
  • A low density manometric liquid is used because it gives a larger difference of levels for the same pressure difference, so the reading is more sensitive.
  • An incompressible fluid is one whose density is constant, i.e. does not change with pressure.
  • Two or more mercury drops merge into one because a single drop has a smaller total surface area, and hence a smaller surface energy; the system moves to minimum potential energy and energy is released.
  • Water speeds up on entering a narrow pipe because of the equation of continuity, AvAv = constant.
  • At a constriction in a horizontal pipe the speed increases by continuity, and Bernoulli's equation then requires the pressure to fall.

Traps that cost marks

  • A soap bubble has two surfaces, a liquid drop or an air bubble inside a liquid has one. Use 4T/r4T/r only for a soap bubble, and 2T/r2T/r for a drop or for a bubble in liquid.
  • In work-done problems, remember the factor 2 for a soap film or bubble; and always convert dyne cm−1^{-1} to N m−1^{-1} (11 dyne cm−1=10−3^{-1} = 10^{-3} N m−1^{-1}) and erg to joule (11 erg =10−7= 10^{-7} J).
  • The excess pressure formula gives a pressure difference. If the question asks for the pressure inside a raindrop, add the atmospheric pressure.
  • p=hρgp = h\rho g is the gauge pressure. If the question says "find the pressure" and gives the pressure on the free surface, the answer is p0+hρgp_{0} + h\rho g.
  • In the terminal velocity formula ρ\rho is the density of the body and σ\sigma that of the medium. For a rising bubble the answer is negative and the sign is part of the answer.
  • In η=f/[A(dv/dx)]\eta = f/[A(dv/dx)], AA is the area of the layer parallel to the flow, not the cross section, and dxdx is the thickness of the fluid layer.
  • Radius, not diameter, goes into 2T/r2T/r, 6πηrv6\pi\eta r v and h=2Tcos⁡θ/(rρg)h = 2T\cos\theta/(r\rho g). Halve the given diameter first.
  • Reynolds number has no unit and no dimension — do not attach one.
  • Capillary rise depends on the radius, not the area. If the area is halved, rr falls by 2\sqrt{2} and hh rises by 2\sqrt{2}.
Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter. 3 have come up in two or more papers.4
  1. 1
    Terminal velocity of a spherical body falling through a viscous medium
    Derive an expression for terminal velocity of a spherical object falling under gravity through a viscous medium.

    Asked in: March 2022 (Q.21, 3 marks)

    Show the derivation
    Expression for terminal velocity:
    Step 1
    Consider a sphere of radius (r) and density (ρ) falling under gravity through a liquid of density (σ) and coefficient of viscosity (η) as shown in figure.
    Step 2
    Forces acting on the sphere during downward motion are,
    a. Viscous force = Fv=6πηrvF_v=6\pi\eta rv (directed upwards)
    b. Weight of the sphere, (Fg)(F_g)
    mg=43πr3ρgmg=\dfrac{4}{3}\pi r^3\rho g (directed downwards)
    c. Upward thrust as Buoyant force (Fu)(F_u)
    Fu=43πr3σgF_u=\dfrac{4}{3}\pi r^3\sigma g (directed upwards)
    Step 3
    As the downward velocity increases, the viscous force increases. A stage is reached, when sphere attains terminal velocity.
    Step 4
    When the sphere attains the terminal velocity, the total downward force acting on the sphere is balanced by the total upward force acting on the sphere.
    Total downward force = Total upward force
    ∴\therefore Weight of sphere (mg) = Viscous Force + Buoyant force due to medium
    ∴43πr3ρg=6πηrv+43πr3σg\therefore \dfrac{4}{3}\pi r^3\rho g=6\pi\eta rv+\dfrac{4}{3}\pi r^3\sigma g
    ∴6πηrv=(43πr3ρg)−(43πr3σg)\therefore 6\pi\eta rv=\left(\dfrac{4}{3}\pi r^3\rho g\right)-\left(\dfrac{4}{3}\pi r^3\sigma g\right)
    ∴6πηrv=(43)πr3g(ρ−σ)\therefore 6\pi\eta rv=\left(\dfrac{4}{3}\right)\pi r^3g(\rho-\sigma)
    ∴v=(43)πr3g(ρ−σ)×16πηr\therefore v=\left(\dfrac{4}{3}\right)\pi r^3g(\rho-\sigma)\times\dfrac{1}{6\pi\eta r}
    ∴v=29r2g(ρ−σ)η....(1)\therefore v=\dfrac{2}{9}\dfrac{r^2g(\rho-\sigma)}{\eta}\qquad....(1)
    This is the expression for terminal velocity of the sphere.
  2. 2
    Relation between surface energy and surface tension Repeats
    Obtain the relation between surface energy and surface tension.

    Asked in: March 2013; October 2013; February 2023 (Q.30(i), 4 marks); February 2024 (Q.18, 3 marks); July 2024 (Q.15, 3 marks)

    Show the derivation
    Relation between surface tension and surface energy per unit area:
    Step 1
    Let ABCD be a rectangular frame of wire, fitted with a movable arm PQ.
    Step 2
    The frame held in horizontal position is dipped into soap solution and taken out so that a soap film APQB is formed. Due to surface tension of soap solution, a force 'F' will act on each arm of the frame. Under the action of this force, the movable arm PQ moves towards AB.
    Step 3
    Magnitude of force due to surface tension is,
    F=2Tl....[∵T=F/l]F=2Tl\qquad....[\because T=F/l]
    (A factor of 2 appears because soap film has two surfaces which are in contact with wire.)
    Step 4
    Let the wire PQ be pulled outwards through a small distance 'dx' to the position P'Q', by applying an external force F' isothermally, which is equal and opposite to F. Work done by this force, dW=F′dx=2TldxdW=F'dx=2Tldx.
    Step 5
    But, 2ldx=dA=2ldx=dA= increase in area of two surfaces of film.
    ∴dW=T dA\therefore dW=T\,dA
    Step 6
    This work done in stretching the film is stored in the area dA in the form of potential energy (surface energy).
    ∴\therefore Surface energy, E=T dAE=T\,dA
    ∴EdA=T\therefore \dfrac{E}{dA}=T
    Hence, surface tension = surface energy per unit area.
    Step 7
    Thus, surface tension is equal to the mechanical work done per unit surface area of the liquid, which is also called as surface energy.
  3. 3
    Laplace's law for a spherical membrane / bubble Repeats
    Derive Laplace's law for a spherical membrane of a bubble due to surface tension.

    Asked in: March 2015; March 2016; March 2018; February 2026 (Q.27, 4 marks)

    Show the derivation
    Expression for excess pressure inside a drop:
    Step 1
    Free surface of drops or bubbles are spherical in shape.
    Let, PiP_i = inside pressure of a drop or air bubble, PoP_o = outside pressure of bubble, r = radius of drop or bubble.
    Step 2
    As drop is spherical, Pi>PoP_i>P_o
    ∴\therefore excess pressure inside drop =Pi−Po=P_i-P_o
    Step 3
    Let the radius of drop increases from r to r+Δrr+\Delta r so that inside pressure remains constant.
    Step 4
    Initial area of drop A1=4πr2A_1=4\pi r^2, Final surface area of drop A2=4π(r+Δr)2A_2=4\pi(r+\Delta r)^2
    Increase in surface area,
    ΔA=A2−A1=4π[(r+Δr)2−r2]=4π[r2+2rΔr+Δr2−r2]=8πrΔr+4πΔr2\Delta A=A_2-A_1=4\pi[(r+\Delta r)^2-r^2]=4\pi[r^2+2r\Delta r+\Delta r^2-r^2]=8\pi r\Delta r+4\pi\Delta r^2
    Step 5
    As Δr is very small, the term containing Δr2\Delta r^2 can be neglected.
    ∴dA=8πrΔr\therefore dA=8\pi r\Delta r
    Step 6
    Work done by force of surface tension, dW=TdA=(8πrΔr)T....(1)dW=TdA=(8\pi r\Delta r)T\qquad....(1)
    This work done is also equal to the product of the force F which causes increase in the area of the bubble and the displacement Δr which is the increase in the radius of the bubble.
    ∴dW=FΔr\therefore dW=F\Delta r
    The excess force is given by, (Excess pressure) × (Surface area)
    ∴F=(Pi−Po)×4πr2\therefore F=(P_i-P_o)\times4\pi r^2
    ∴dF=(Pi−Po)A\therefore dF=(P_i-P_o)A
    dW=FΔr=(Pi−Po)AΔrdW=F\Delta r=(P_i-P_o)A\Delta r
    From equation (1), (Pi−Po)AΔr=(8πrΔr)T(P_i-P_o)A\Delta r=(8\pi r\Delta r)T
    ∴Pi−Po=8πrΔrT4πr2Δr(∵A=4πr2)\therefore P_i-P_o=\dfrac{8\pi r\Delta rT}{4\pi r^2\Delta r}\qquad(\because A=4\pi r^2)
    ∴Pi−Po=2Tr....(2)\therefore P_i-P_o=\dfrac{2T}{r}\qquad....(2)
    Equation (2) represents excess pressure inside a drop or air bubble. It is also called Laplace's law of spherical membrane.
    Step 7
    In case of soap bubble, there are two free surfaces in contact with air.
    ∴\therefore For a bubble, Total increase in the surface area of a soap bubble, while increasing its radius by Δr, is 2(8πrΔr)2(8\pi r\Delta r)
    The work done by this excess pressure is dW=(Pi−Po)4πr2Δr=T(16πrΔr)dW=(P_i-P_o)4\pi r^2\Delta r=T(16\pi r\Delta r)
    ∴Excess pressure, Pi−Po=4Tr\therefore \text{Excess pressure, }P_i-P_o=\dfrac{4T}{r}
  4. 4
    Rise of liquid in a capillary tube (capillarity) Repeats
    Derive an expression for the height to which liquid rises in a capillary tube dipped vertically in the liquid.

    Asked in: October 2009; March 2019

    Show the derivation
    Expression for capillary rise:
    Step 1
    When glass capillary tube is dipped into a liquid, then the liquid rises in the capillary against gravity. Hence, the weight of the liquid column must be equal and opposite to the component of force due to surface tension at the point of contact.
    Step 2
    The length of liquid in contact inside the capillary is the circumference 2πr.
    Let, r = radius of capillary tube, h = height of liquid level in the tube, T = surface tension of liquid, ρ = density of liquid, g = acceleration due to gravity
    Step 3
    The force of magnitude fTf_T acts tangentially on unit length of liquid surface which is in contact with wall of capillary tube and is given as, fT=T×2πrf_T=T\times2\pi r
    This force can be resolved into two components: a. fTcos⁡θf_T\cos\theta - vertically upward and b. fTsin⁡θf_T\sin\theta - along horizontal
    Step 4
    Vertical component is effective. Horizontal component is not responsible for capillary rise.
    Step 5
    Vertical component of force acting on liquid column (fT)v=(f_T)_v= force per unit length × circumference =Tcos⁡θ×2πr=T\cos\theta\times2\pi r
    Step 6
    Upward force balances weight of liquid in the capillary.
    W=mg=Vρg=πr2hρgW=mg=V\rho g=\pi r^2h\rho g
    where, V = volume of liquid rise in the tube (ignoring the liquid in the concave meniscus.) m = mass of the liquid in the capillary rise.
    Step 7
    This must be equal and opposite to the vertical component of the force due to surface tension. If liquid in meniscus is neglected, then for equilibrium,
    2πrTcos⁡θ=πr2hρg2\pi rT\cos\theta=\pi r^2h\rho g
    ∴h=2Tcos⁡θrρg....(1)\therefore h=\dfrac{2T\cos\theta}{r\rho g}\qquad....(1)
    This is the required expression for rise of liquid in capillary tube

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