Class 11 Physics Chapter 9: Mechanical Properties of Fluids | NCERT Solutions

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Study pressure, buoyancy, fluid flow, viscosity and surface tension through worked examples.

9.1

Fluids and how they differ from solids

Liquids and gases flow and are called fluids. They have no fixed shape; they take the container’s shape.
A liquid has a nearly fixed volume and a free surface. A gas expands to fill the container. Liquids and solids have much lower compressibility than gases; their volumes still change slightly with pressure.
A solid resists shear and can hold a shape. A fluid offers very little resistance to static shear: even a small sustained shear makes it flow and its strain keeps growing with time.
In this chapter, pressure concerns fluids at rest; streamline flow, viscosity and surface tension describe important fluid behaviour.

9.2

Pressure, density and units

A fluid at rest exerts force normal to every surface in contact with it. A tangential force component would make the fluid flow.
Average pressure is normal force per area; local pressure is its limit as the area shrinks: $P=\lim_{\Delta A\to0}\Delta F_\perp/\Delta A$. Pressure is a positive scalar, not a force vector.
SI unit: pascal, $1\,\mathrm{Pa}=1\,\mathrm{N\,m^{-2}}$; dimensions $[ML^{-1}T^{-2}]$. $1\,\mathrm{atm}=1.013\times10^5\,\mathrm{Pa}$; $1\,\mathrm{bar}=10^5\,\mathrm{Pa}$; $1\,\mathrm{torr}=1\,\mathrm{mmHg}\approx133\,\mathrm{Pa}$.
Density is $\rho=m/V$ (kg m⁻³, dimensions $[ML^{-3}]$); relative density is density divided by water’s density at 4 °C and is dimensionless. Water at 4 °C has $\rho\approx10^3\,\mathrm{kg\,m^{-3}}$. Liquids are nearly incompressible; gas density varies substantially with pressure and temperature.

Table 9.1 · Representative fluid densities (kg m⁻³)
Fluid	Density
Air (0 °C, 1 atm)	$1.29\times10^0$
Carbon dioxide	$1.98\times10^0$
Oxygen	$1.43\times10^0$
Hydrogen	$0.09\times10^0$
Water (4 °C)	$1.00\times10^3$
Sea water	$1.03\times10^3$
Mercury	$13.6\times10^3$

Fig. 9.1 · A fluid at rest pushes perpendicular to every boundary.

A small contact area raises pressure for the same force; that is why a sharp needle pierces more readily than a spoon’s broad back.

9.2.1

Pascal’s law: pressure acts equally in every direction

At one point in a fluid at rest, pressure is the same on surfaces of every orientation. At equal height, pressure is equal throughout a connected static fluid.
Apply force balance to a tiny right-angled prism of fluid. The weight is negligible compared with face forces as its size tends to zero. The sloping-face force components balance the two perpendicular-face forces.
For the prism geometry, $F_b\sin\theta=F_c$, $F_b\cos\theta=F_a$ and $A_b\sin\theta=A_c$, $A_b\cos\theta=A_a$. Dividing corresponding equations gives $F_a/A_a=F_b/A_b=F_c/A_c$, hence $P_a=P_b=P_c$.
A horizontal fluid element is also in equilibrium only if pressures at its ends match. A pressure difference would create flow.

Prism proof · Resolve normal-pressure forces and use the matching projected face areas.

9.2.2

Pressure increases with depth

Take a fluid cylinder of area $A$ and height $h$. Vertical equilibrium gives $(P_2-P_1)A=mg=\rho hAg$, so $P_2-P_1=\rho gh$.
For a liquid open to air, absolute pressure at depth $h$ is $P=P_a+\rho gh$. Gauge pressure is $P_g=P-P_a=\rho gh$. The result depends on vertical depth, density and $g$, not vessel shape or base area.
Points at the same horizontal level in a connected static liquid have equal pressure. This explains communicating vessels and the hydrostatic paradox: vessels with different shapes and liquid amounts can have the same bottom pressure if the liquid heights match.
The model assumes uniform density and approximately constant $g$. It works well for liquids; gas density can change strongly with height.

Hydrostatic paradox · Different liquid weights are balanced by forces from the differently shaped vessel walls.

NCERT Example 9.2

Pressure on a swimmer at depth

What is the pressure on a swimmer 10 m below the surface of a lake?

Step 1. $P=P_a+\rho gh=1.01\times10^5+1000(10)(10)=2.01\times10^5\,\mathrm{Pa}$.

Step 2. This is about 2 atm absolute; the gauge pressure alone is $1.00\times10^5\,\mathrm{Pa}$.

9.2.3

Atmospheric pressure, barometers and manometers

Atmospheric pressure is the weight per unit area of the air column above a point. At sea level it is approximately $1.013\times10^5$ Pa.
In Torricelli’s mercury barometer, the nearly empty space above the tube’s mercury has negligible pressure. At the same level in the reservoir and tube, $P_a=\rho_{Hg}gh$, giving $h\approx0.76$ m at sea level.
A U-tube manometer measures pressure relative to atmosphere: $P-P_a=\rho_mgh$ when the manometer liquid columns differ in height by $h$. Choose a low-density fluid for small pressure differences and a denser fluid for larger ones. Step through each connected fluid column, adding $\rho gh$ downward and subtracting it upward; pressure is continuous at interfaces.
Atmospheric pressure changes with weather and altitude; assuming air density constant gives only a rough height estimate.

Column-height differences translate to pressure differences using $\Delta P=\rho gh$.

NCERT Example 9.1

Pressure supported by the femurs

The two thigh bones (femurs), each of cross-sectional area 10 cm², support the upper part of a human body of mass 40 kg. Estimate the average pressure sustained by the femurs.

Step 1. Total area $A=2(10\,\mathrm{cm^2})=2.0\times10^{-3}\,\mathrm{m^2}$; weight $F=mg=400\,\mathrm N$.

Step 2. $P=F/A=400/(2.0\times10^{-3})=2.0\times10^5\,\mathrm{Pa}$.

NCERT Example 9.3

Rough height of a constant-density atmosphere

The density of the atmosphere at sea level is 1.29 kg m⁻³. Assume that it does not change with altitude. Then how high would the atmosphere extend?

Step 1. Set the pressure of the air column equal to sea-level pressure: $P_a=\rho gh$.

Step 2. $h=1.01\times10^5/(1.29\times9.8)=7.99\times10^3\,\mathrm m\approx8.0\,\mathrm{km}$.

Step 3. This is only a constant-density estimate; real air density and $g$ decrease with altitude, and the atmosphere extends far higher.

NCERT Example 9.4

Submarine window at ocean depth

At a depth of 1000 m in an ocean (a) what is the absolute pressure? (b) What is the gauge pressure? (c) Find the force acting on the window of area 20 cm × 20 cm of a submarine at this depth, the interior of which is maintained at sea-level atmospheric pressure. (The density of sea water is 1.03 × 10³ kg m⁻³, g = 10 m s⁻².)

Step 1. Gauge pressure: $\rho gh=1.03\times10^7\,\mathrm{Pa}$.

Step 2. Absolute pressure: $P_a+\rho gh=1.0401\times10^7\,\mathrm{Pa}\approx104$ atm.

Step 3. Area $A=0.040\,\mathrm{m^2}$. Since the cabin is at $P_a$, net force is gauge pressure times area: $F=1.03\times10^7(0.040)=4.12\times10^5\,\mathrm N$.

9.2.4

Pascal’s law and hydraulic machines

Pascal’s law: a pressure change applied to an enclosed fluid is transmitted undiminished to every point and to the container walls.
For pistons of areas $A_1,A_2$, ideal pressure transmission gives $F_1/A_1=F_2/A_2$. Thus force is multiplied by $A_2/A_1$. Mechanical advantage trades force for distance: incompressibility gives $A_1x_1=A_2x_2$.
Hydraulic lifts and brakes use a liquid. Brakes distribute pressure to wheel cylinders; the frictional brake pads then retard the wheels. A sealed car lift may use compressed air to press on the hydraulic fluid.

Pressure is shared; piston area sets the available force and travel.

NCERT Example 9.5

Two connected syringes

Two syringes of different cross-sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively. (a) Find the force exerted on the larger piston when a force of 10 N is applied to the smaller piston. (b) If the smaller piston is pushed in through 6.0 cm, how much does the larger piston move out?

Step 1. Area ratio is $(d_2/d_1)^2=3^2=9$, so $F_2=9F_1=90\,\mathrm N$.

Step 2. Conservation of displaced volume: $A_1x_1=A_2x_2$, hence $x_2=x_1/9=6.0/9=0.67\,\mathrm{cm}$.

NCERT Example 9.6

Force and pressure in a car lift

In a car lift compressed air exerts a force F₁ on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm. If the mass of the car to be lifted is 1350 kg, calculate F₁. What is the pressure necessary to accomplish this task? (g = 9.8 m s⁻².)

Step 1. Load $F_2=mg=1350(9.8)=13230\,\mathrm N$. Area ratio $A_1/A_2=(5/15)^2=1/9$, so $F_1=1470\,\mathrm N$.

Step 2. Required gauge pressure $P=F_1/(\pi(0.050)^2)=1.87\times10^5\,\mathrm{Pa}$, about 1.85 atm above atmosphere.

9.3

Streamlines, continuity and turbulence

A flow is steady when the velocity at each fixed point does not change with time. A particle can accelerate as it moves to another point. A streamline is tangent to the instantaneous velocity everywhere; streamlines cannot intersect in steady flow.
Conserve mass through a stream tube: $\rho_1A_1v_1=\rho_2A_2v_2$. For an incompressible liquid, $A_1v_1=A_2v_2=Q$, the volume-flow rate. Narrower cross-section means greater speed.
Laminar flow: smooth layers slide without large mixing; in a pipe, speed is greatest at the axis and zero at the stationary wall. Turbulent flow: irregular fluctuations and eddies arise beyond a critical speed.
Reynolds number $Re=\rho vD/\eta$ compares inertial with viscous effects. Small Re favours laminar flow; large Re favours turbulence. The transition depends on geometry and disturbances, so no single cutoff is universal. At equal Re, a smaller model scale needs a larger speed.

Continuity · the tube carries equal mass flow through every cross-section.

Trace flow through a constriction

Change the throat area. The tracer speeds up as the stream tube narrows.

Interactive 3D model

Cross-sectional area ratio A₂/A₁0.35–1.00

Adjust the control to explore the model.

The particle animation is schematic; the speed ratio follows A₁v₁=A₂v₂.

9.4

Bernoulli’s principle: proof and uses

For steady, incompressible, non-viscous flow along a streamline, pressure work changes kinetic and gravitational potential energy. Between points 1 and 2, equal volume $\Delta V$ passes both. Net pressure work is $(P_1-P_2)\Delta V$.
Work–energy theorem: $(P_1-P_2)\Delta V=\frac12\rho\Delta V(v_2^2-v_1^2)+\rho g\Delta V(h_2-h_1)$. Divide by $\Delta V$ and rearrange: $P+\frac12\rho v^2+\rho gh=\mathrm{constant}$.
Each term is energy per unit volume (Pa). At the same height, faster flow has lower pressure. At rest it reduces to hydrostatics. The relation is approximate for real fluids and fails for strongly viscous, compressible, unsteady or turbulent flow.
Torricelli efflux: for a large open tank, apply Bernoulli from the free surface to a small hole; both points are at atmospheric pressure and surface speed is negligible. Thus $v=\sqrt{2gh}$.
Venturi/atomiser: a constriction raises speed and lowers static pressure, drawing in a second fluid. Lift: air speed differences above and below an aerofoil create pressure differences. A spinning ball also changes the surrounding flow and pressure, causing a Magnus force; spinning alone does not prove the simple textbook Bernoulli model.

For equal height, Bernoulli gives $P_1+\frac12\rho v_1^2=P_2+\frac12\rho v_2^2$.

Compare speed and pressure

Adjust the throat area to compare the flow speed and pressure at the narrow section.

Interactive 3D model

Throat area ratio A₂/A₁0.35–1.00

Adjust the control to explore the model.

Assumes ideal horizontal steady flow; real viscosity causes pressure losses.

NCERT Example 9.7

Lift on a Boeing aircraft

A fully loaded Boeing aircraft has a mass of 3.3 × 10⁵ kg. Its total wing area is 500 m². It is in level flight with a speed of 960 km h⁻¹. (a) Estimate the pressure difference between the lower and upper surfaces of the wings. (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is ρ = 1.2 kg m⁻³.]

Step 1. Level flight: $\Delta P A=mg$, so $\Delta P=(3.3\times10^5)(9.8)/500=6.47\times10^3\,\mathrm{Pa}\approx6.5\,\mathrm{kPa}$.

Step 2. Convert average speed: $v_{av}=960/3.6=267\,\mathrm{m\,s^{-1}}$. For small difference, Bernoulli gives $\Delta P\approx\rho v_{av}\Delta v$.

Step 3. $\Delta v/v_{av}\approx\Delta P/(\rho v_{av}^2)=6.47\times10^3/[1.2(267)^2]=0.075\approx8\%$.

9.5

Viscosity and Poiseuille flow

Viscosity is internal friction: adjacent fluid layers moving at different speeds exert tangential drag on one another. The fluid at a solid boundary matches that boundary’s speed.
For a thin film between plates of area $A$ separated by $\ell$, if the top plate moves at speed $v$ and the bottom is fixed, shear stress $F/A=\eta(v/\ell)$. The velocity gradient is the rate of shear strain. SI unit of dynamic viscosity $\eta$: Pa s (poiseuille); dimensions $[ML^{-1}T^{-1}]$.
In pipe flow, speed is zero at the wall, rises toward a maximum on the axis, and the laminar profile is parabolic. For a circular tube, Poiseuille’s relation is $Q=\pi r^4\Delta P/(8\eta L)$. It requires steady laminar flow of a Newtonian fluid in a cylindrical tube. Since $Q\propto r^4$, small radius changes strongly affect flow.
Liquid viscosity generally falls as temperature rises; gas viscosity generally rises. The table values below are in mPa s (numerically cP).

The no-slip boundary condition gives zero fluid speed at a stationary wall.

Table 9.2 · Dynamic viscosities of selected fluids (mPa s)
Fluid	Temperature (°C)	η (mPa s)
Water	20	1.0
Water	100	0.3
Blood	37	2.7
Machine oil	16	113
Machine oil	38	34
Glycerine	20	830
Honey	—	200
Air	0	0.017
Air	40	0.019

NCERT Example 9.8

Viscosity from a moving plate

A metal block of area 0.10 m² is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in Fig. 9.13. A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.085 m s⁻¹. Find the coefficient of viscosity of the liquid.

Step 1. At constant speed, string tension balances the hanging weight: $F=mg=0.010(9.8)=0.098\,\mathrm N$.

Step 2. Shear stress $F/A=0.98\,\mathrm{Pa}$. Strain rate $v/\ell=0.085/(3.0\times10^{-4})=2.83\times10^2\,\mathrm{s^{-1}}$.

Step 3. $\eta=(F/A)/(v/\ell)=3.46\times10^{-3}\,\mathrm{Pa\,s}$.

Poiseuille profile and flow-rate derivation. In a tube of radius $R$ and length $L$, balance the pressure force on a cylindrical fluid core of radius $r$ with viscous shear on its curved surface: $\pi r^2\Delta P=-\eta(du/dr)(2\pi rL)$. Hence $du/dr=-r\Delta P/(2\eta L)$. Integrate and apply no-slip $u(R)=0$: $u(r)=\Delta P(R^2-r^2)/(4\eta L)$. Then $Q=\int_0^R2\pi r u(r)dr=\pi R^4\Delta P/(8\eta L)$. This derivation assumes steady, fully developed laminar flow of a Newtonian fluid in a cylindrical tube.

9.5.1

Stokes’ law and terminal velocity

A sphere moving slowly through a viscous fluid experiences opposing Stokes drag $F_d=6\pi\eta av$, where $a$ is sphere radius. NCERT states Stokes’ law experimentally; it does not derive the law here.
A falling sphere initially accelerates. Drag grows with speed until weight equals buoyancy plus drag; net force and acceleration then vanish and it falls at terminal speed.
At terminal speed: $6\pi\eta a v_t=(4\pi/3)a^3(\rho_s-\rho_f)g$. Therefore $v_t=2a^2(\rho_s-\rho_f)g/(9\eta)$. It scales as $a^2$ and density contrast, and inversely as viscosity.
This result presumes laminar creeping flow around a sphere; do not use Stokes’ law when its low-Reynolds-number conditions fail.

Force balance makes the terminal speed constant.

NCERT Example 9.9

Viscosity from a copper ball’s terminal speed

The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20 °C is 6.5 cm s⁻¹. Compute the viscosity of the oil at 20 °C. Density of oil is 1.5 × 10³ kg m⁻³, density of copper is 8.9 × 10³ kg m⁻³.

Step 1. Convert $a=2.0\times10^{-3}$ m and $v_t=6.5\times10^{-2}$ m s⁻¹; density difference is $7.4\times10^3$ kg m⁻³.

Step 2. Rearrange Stokes terminal speed: $\eta=2a^2(\rho_s-\rho_f)g/(9v_t)$.

Step 3. $\eta=2(2.0\times10^{-3})^2(7.4\times10^3)(9.8)/[9(6.5\times10^{-2})]=0.99\,\mathrm{Pa\,s}$.

9.6.1–9.6.2

Molecular picture, surface energy and surface tension

A molecule deep inside a liquid is attracted in all directions; its mean potential energy is lower than that of a surface molecule, which lacks neighbours above. Creating surface area therefore requires energy. A liquid tends to minimise its free surface area.
Surface tension $S$ is tangential force per unit length along a liquid interface: $S=F/\ell$ (N m⁻¹). Equivalently, it is the reversible work per unit increase in surface area (J m⁻²).
For a soap film, there are two surfaces. Pull a movable edge of length $\ell$ through $dx$: force $F=2S\ell$, work $dW=Fdx=2S\ell dx=S dA$, because total film area grows by $dA=2\ell dx$.
$S$ depends on the pair of media and temperature; it is not the property of a liquid alone. It generally decreases as a liquid warms. Values below include heats of vaporisation from the supplied table.

The energetic cost of making surface explains why a free drop tends to be spherical.

Table 9.3 · Surface tension and heat of vaporisation
Liquid	Temperature (°C)	S (N m⁻¹)	Heat of vaporisation (kJ mol⁻¹)
Helium	−270	0.000239	0.115
Oxygen	−183	0.0132	7.1
Ethanol	20	0.0227	40.6
Water	20	0.0727	44.16
Mercury	20	0.4355	63.2

Raise or depress a liquid in a capillary

Change tube radius to see the water rise; the model uses a wetting contact angle near zero.

Interactive 3D model

Tube radius0.2–2.0 mm

Adjust the control to explore the model.

The capillary-height display is scaled for visibility; the readout uses h=2S cosθ/(ρga).

9.6.3

Wetting and the angle of contact

The angle of contact $\theta$ is measured through the liquid between the solid surface and the tangent to the liquid meniscus at the contact point. It depends on both contacting materials.
Balance interfacial tensions at the contact line: $S_{la}\cos\theta+S_{sl}=S_{sa}$.
Small acute $\theta$: liquid wets the solid (water on clean glass). Large obtuse $\theta$: poor wetting (mercury on glass, water on a waxy leaf). At $\theta=90^circ$, the ideal vertical surface-tension component vanishes.
Detergent changes interfacial tension and helps water spread; capillary rise depends on $\cos\theta$, so non-wetting liquids can be depressed below the outside level.

The angle is measured through the liquid at its contact with the solid.

9.6.4

Drops, bubbles and excess pressure

If gravity is negligible, surface energy favours the least area for a given volume; a sphere minimises area, so small free drops and bubbles are spherical.
For a liquid drop of radius $r$, a small increase $dr$ creates area $dA=8pi r,dr$ and costs $dU=S(8pi r,dr)$. Expansion work is $(P_i-P_o)4pi r^2dr$. Equating gives $P_i-P_o=2S/r$.
For a gas bubble in a liquid there is one liquid–gas surface, so excess pressure is $2S/r$. A soap bubble has two surfaces, inner and outer; its excess pressure is $4S/r$. Smaller radius means larger excess pressure.
For either curved interface, the concave side has higher pressure than the convex side. Distinguish carefully: a liquid drop is one surface; a soap bubble has two.

Young–Laplace pressure follows from balancing surface-energy increase and pressure work.

9.6.5

Capillary rise: derivation

A narrow tube raises a wetting liquid or depresses a non-wetting one. The curved meniscus has a pressure difference: $P_i-P_o=-2S\cos\theta/a$ for tube radius $a$ (sign depends on which side is chosen).
At the meniscus, pressure inside the liquid is below atmospheric by $2S\cos\theta/a$. At the outside free surface it is atmospheric. Equating pressure at the same level inside and outside gives $\rho gh=2S\cos\theta/a$.
Thus $h=2S\cos\theta/(\rho ga)$. For water on glass, $\theta$ is acute and the liquid rises; for mercury on glass, $\theta$ is obtuse and it is depressed. Magnitude grows as tube radius shrinks.
Example scale: water with $S=0.073$ N m⁻¹, $a=0.50$ mm, $\theta\approx0$, $\rho=1000$ kg m⁻³ and $g=9.8$ m s⁻² rises about 3.0 cm.

The rise balances the pressure jump from the curved meniscus against the liquid-column weight.

NCERT Example 9.10

Pressure needed to blow a hemispherical bubble underwater

The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the surface of water. What pressure is required in the tube to blow a hemispherical bubble at its end? Take surface tension $S=7.30\times10^{-2}$ N m⁻¹, atmospheric pressure $1.01\times10^5$ Pa, water density $1000$ kg m⁻³ and $g=9.80$ m s⁻². Also calculate the excess pressure.

Step 1. Pressure outside at 8.00 cm depth: $P_o=P_a+\rho gh=1.01\times10^5+1000(9.80)(0.080)=1.01784\times10^5$ Pa.

Step 2. The hemispherical bubble radius is the capillary radius $r=1.00\times10^{-3}$ m. A gas bubble in liquid has one interface: $\Delta P=2S/r=2(7.30\times10^{-2})/(1.00\times10^{-3})=146$ Pa.

Step 3. Required inside pressure $P_i=P_o+\Delta P=1.01930\times10^5$ Pa $\approx1.02\times10^5$ Pa (3 s.f.); excess pressure is 146 Pa.

Quiz

Concept quiz

Choose an answer to see the reasoning and key idea.

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9.1 Fluids and static shear

1. What distinguishes a fluid from a solid under a sustained shear stress?

A fluid at rest cannot support a sustained tangential stress without beginning to flow; a solid can hold a static shape under shear.

Remember

Sustained shear causes continuing fluid deformation.

2. Which statement best compares liquid and gas compressibility?

A liquid has a nearly fixed volume and a free surface; a gas expands to fill its container and is much more compressible.

Remember

Liquids have nearly fixed volume; gases expand to fill their container.

9.2 Pressure, density and units

3. Which description of pressure is correct?

Pressure is the limiting normal force per unit area. One pascal is one newton per square metre.

Remember

Pressure is scalar; the force it produces on a surface acts normally.

4. Relative density of a substance is:

Relative density is a ratio of densities, so the units cancel. Water at 4 °C is the stated reference.

Remember

Density has units kg m⁻³; relative density has no units.

9.2.1 Pressure in every direction

5. In the small-fluid-prism proof of Pascal’s law, why can the prism’s weight be neglected as its size tends to zero?

Face areas scale as length squared while the prism’s weight scales as volume. In the infinitesimal limit, weight is negligible in the balance.

Remember

Resolve pressure forces on the prism faces and use projected areas to obtain Pₐ = Pᵦ = P꜀.

6. For a connected fluid at rest, pressure at two points at the same height is:

Hydrostatic pressure depends on vertical depth, not the vessel’s shape or horizontal position at a shared level.

Remember

Equal height in a connected static fluid means equal pressure.

9.2.2 Pressure with depth

7. A liquid of uniform density ρ is open to air. At depth h, which pair is correct?

The liquid column adds ρgh above the surface pressure. Gauge pressure excludes atmospheric pressure; absolute pressure includes it.

Remember

P_g = ρgh and P_abs = Pₐ + ρgh.

8. Two differently shaped vessels contain the same liquid to the same height. Ignoring density changes, their bottom pressures are:

The column balance gives ΔP = ρgh; vessel shape and total liquid volume do not enter the result.

Remember

Hydrostatic paradox: same fluid height gives the same bottom pressure.

9.2.3 Barometers and manometers

9. In a mercury barometer, the pressure above the mercury column is negligible. The atmospheric pressure is approximately:

At the same level in the reservoir and column, pressure balance gives Pₐ = ρ_Hggh. The mercury height is about 0.76 m at sea level.

Remember

A barometer converts atmospheric pressure to a liquid-column height.

10. When stepping down through a connected liquid column in a manometer, the pressure change is:

Pressure increases on moving down a static column and decreases on moving up. Match pressure continuously across interfaces.

Remember

Track each column: down adds ρgΔh; up subtracts it.

9.2.4 Hydraulic machines

11. A hydraulic lift has input and output piston areas A₁ and A₂. Ideal Pascal pressure transmission gives:

The pressure increments are equal: F₁/A₁ = F₂/A₂. A larger output area therefore produces a larger force.

Remember

Force multiplication comes from piston area ratio.

12. For ideal hydraulic pistons filled with an incompressible liquid, which relation expresses volume conservation?

The volume pushed by one piston equals the volume displaced by the other. Ideal work F₁x₁ = F₂x₂ is conserved.

Remember

A force gain trades against distance; it does not create work.

9.3 Streamlines, continuity and turbulence

13. In steady flow, which statement about streamlines is correct?

A streamline is tangent to the instantaneous velocity. Two streamlines cannot intersect in steady flow because that would assign two velocities to one point.

Remember

Steady means velocity at a fixed point is time-independent; a moving particle may still accelerate.

14. For steady flow through a stream tube, the general continuity relation is:

Mass flow rate is conserved. For an incompressible liquid of constant density this reduces to A₁v₁ = A₂v₂ = Q.

Remember

Narrower area means higher speed for an incompressible flow.

15. The Reynolds number Re = ρvD/η compares:

Re is dimensionless and indicates the relative importance of inertia and viscosity. A transition value is not universal for every geometry and disturbance level.

Remember

Small Re favours laminar flow; sufficiently large Re favours turbulence.

9.4 Bernoulli principle and applications

16. Bernoulli’s equation applies most directly to flow that is:

The ideal derivation assumes steady, incompressible, non-viscous flow along a streamline. Real-fluid losses can make it approximate.

Remember

Check Bernoulli’s assumptions before applying the equation.

17. In Bernoulli’s proof between two points, the net pressure work on an equal volume ΔV is:

Use work–energy: (P₁−P₂)ΔV = ½ρΔV(v₂²−v₁²) + ρgΔV(h₂−h₁). Divide by ΔV and rearrange.

Remember

The result is P + ½ρv² + ρgh = constant along a streamline.

18. For a large open tank with a small hole at depth h, Torricelli’s ideal efflux speed is:

Apply Bernoulli from the nearly stationary free surface to the hole. Both points are at atmospheric pressure, giving v²/2 = gh.

Remember

Torricelli’s result is v = √(2gh), under the ideal-flow assumptions.

9.5 Viscosity and Poiseuille flow

19. For a Newtonian liquid layer between plates, viscous shear stress is proportional to:

Dynamic viscosity η relates tangential stress to velocity gradient. Its SI unit is Pa·s and dimensions are [ML⁻¹T⁻¹].

Remember

Viscosity measures internal resistance to relative motion between fluid layers.

20. For laminar flow of a Newtonian fluid through a cylindrical tube, Poiseuille’s law is:

The relation assumes steady, fully developed laminar flow in a cylindrical tube. The velocity profile is parabolic, zero at the wall and greatest at the axis.

Remember

At fixed ΔP, η and L, flow rate varies as R⁴.

21. As temperature rises, the usual trend is:

Heating generally lowers a liquid’s viscosity but raises a gas’s viscosity; these are distinct molecular effects.

Remember

Remember the opposite temperature trends for liquids and gases.

9.5.1 Stokes drag and terminal speed

22. For a sphere of radius a moving slowly at speed v through a viscous fluid, Stokes’ drag magnitude is:

Stokes’ law gives F_d = 6πηav for slow creeping flow around a sphere; the force opposes relative motion.

Remember

Use Stokes’ law only when its low-Reynolds-number conditions hold.

23. For a sphere denser than its fluid, Stokes-regime terminal speed is:

At terminal speed the net force is zero. Balancing weight minus buoyancy with 6πηav gives the stated result, so v_t ∝ a².

Remember

Terminal speed requires zero resultant force, not zero individual forces.

9.6.1–9.6.2 Surface energy and surface tension

24. Why does creating more liquid surface generally require energy?

Molecules inside the liquid experience attractions in all directions; surface molecules lack neighbours above, so increasing area costs energy.

Remember

A free liquid surface tends to minimise area when other effects are negligible.

25. Surface tension S is equivalently:

S = F/ℓ and S = dW/dA. Their units N m⁻¹ and J m⁻² are equivalent.

Remember

Surface tension is both an interfacial force measure and a surface-energy measure.

26. A soap film has two surfaces. A movable edge of length ℓ is pulled slowly. The surface-tension force is:

Both film surfaces pull on the slider, so their forces add. The work F dx equals S dA with total area increase dA = 2ℓ dx.

Remember

The factor 2 comes from the two faces of a soap film.

9.6.3 Contact angle and wetting

27. The contact angle θ is measured:

The angle is defined at the contact line and measured through the liquid. An acute angle usually indicates wetting; an obtuse angle indicates poor wetting.

Remember

Water on clean glass is acute and wets; mercury on glass is obtuse and does not.

28. Young’s balance of interfacial tensions at a contact line is:

Resolve the liquid–air tension along the solid surface and balance it with the solid–liquid and solid–air interfacial tensions.

Remember

Contact angle depends on both materials; detergent changes interfacial tensions and helps water spread.

9.6.4 Drops, bubbles and excess pressure

29. Which excess-pressure pair is correct for radius r?

A drop has one liquid–gas interface. A soap bubble has inner and outer interfaces, doubling the excess pressure.

Remember

Count the interfaces before choosing the Laplace pressure.

30. For a liquid drop, equating pressure work to the increase in surface energy gives:

A radius increase dr changes volume by 4πr²dr and area by 8πrdr. Setting ΔP dV = S dA yields ΔP = 2S/r.

Remember

The pressure excess is larger for smaller radius; negligible-gravity drops are spherical because a sphere minimises area.

9.6.5 Capillary rise

31. For a capillary of radius a, the magnitude of rise or depression is:

Balance the vertical surface-tension force around the rim, 2πaS cosθ, against the liquid column’s weight, ρgπa²h.

Remember

h = 2S cosθ/(ρga); the sign of cosθ determines rise or depression.

32. A non-wetting liquid with an obtuse contact angle is:

For an obtuse θ, cosθ < 0, so the signed capillary height is negative. Smaller tube radius increases the magnitude.

Remember

Acute contact angle gives rise; obtuse contact angle gives depression.

Exercises

NCERT Exercises

Try each question before opening its solution.

9.1

Explain why (a) the blood pressure in humans is greater at the feet than at the brain; (b) atmospheric pressure at a height of about 6 km decreases to nearly half of its value at sea level, though the height of the atmosphere is more than 100 km; (c) hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

Step 1. At lower feet, the vertical blood-fluid column adds hydrostatic pressure $\rho gh$ relative to the brain.

Step 2. Air density decreases with altitude, so the pressure integral $P(h)=\int_h^\infty\rho(z)g\,dz$ falls rapidly near Earth; atmospheric height alone does not set the pressure scale.

Step 3. Pressure is the normal-force magnitude per unit area at a point and has no direction. The associated force on a chosen surface is normal and vector-valued.

9.2

Explain why (a) the angle of contact of mercury with glass is obtuse, while that of water with glass is acute; (b) water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops (water wets glass while mercury does not); (c) surface tension of a liquid is independent of the area of the surface; (d) water with detergent dissolved in it should have small angles of contact; (e) a drop of liquid under no external forces is always spherical in shape.

Step 1. (a) Water molecules adhere to clean glass strongly relative to water cohesion, giving acute $\theta$; mercury cohesion dominates adhesion, giving obtuse $\theta$.

Step 2. (b) Acute contact wets and spreads; obtuse contact does not wet and surface energy favours a compact drop.

Step 3. (c) Surface tension is an intensive interfacial property fixed by materials and temperature, not the total area.

Step 4. (d) Detergent alters and lowers interfacial tension, improving wetting and reducing contact angle on many surfaces.

Step 5. (e) At fixed volume, a sphere has minimum surface area and hence minimum surface energy when gravity is negligible.

9.3

Fill in the blanks using the word(s) from the list appended with each statement: (a) surface tension of liquids generally ___ with temperature (increases / decreases); (b) viscosity of gases ___ with temperature, whereas viscosity of liquids ___ with temperature (increases / decreases); (c) for solids with elastic modulus of rigidity, the shearing force is proportional to ___, while for fluids it is proportional to ___ (shear strain / rate of shear strain); (d) for a fluid in a steady flow, the increase in flow speed at a constriction follows ___ (conservation of mass / Bernoulli’s principle); (e) for the model of a plane in a wind tunnel, turbulence occurs at a ___ speed than for an actual plane (greater / smaller).

Step 1. (a) decreases; (b) increases, decreases; (c) shear strain, rate of shear strain; (d) conservation of mass (continuity); (e) greater.

Step 2. At equal critical Reynolds number, $v_c\propto1/D$. A smaller model length scale therefore needs a larger speed to match the full aircraft’s Reynolds number.

9.4

Explain why (a) to keep a piece of paper horizontal, you should blow over, not under, it; (b) when we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers; (c) the size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection; (d) a fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel; (e) a spinning cricket ball in air does not follow a parabolic trajectory.

Step 1. (a) Faster air over the paper has lower pressure; greater pressure underneath pushes it upward.

Step 2. (b) Flow accelerates through each narrow opening; local pressure falls, and the jets emerge at high speed.

Step 3. (c) For laminar tube flow $Q\propto r^4\Delta P$, so needle radius has a strong effect on flow rate.

Step 4. (d) The outgoing liquid carries forward momentum; the vessel gets opposite momentum (Newton’s third law/conservation of momentum).

Step 5. (e) Spin changes the airflow and creates a sideways Magnus force; gravity is not the only force, so the path is not parabolic.

9.5

A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?

Step 1. $F=mg=490$ N; radius $r=0.005$ m and area $A=\pi r^2=7.85\times10^{-5}$ m².

Step 2. $P=F/A=6.24\times10^6$ Pa (about 6.2 MPa).

9.6

Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m⁻³. Determine the height of the wine column for normal atmospheric pressure.

Step 1. Set $P_a=\rho gh$: $h=1.013\times10^5/(984\times9.8)=10.5$ m.

9.7

A vertical off-shore structure is built to withstand a maximum stress of 10⁹ Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.

Step 1. Gauge pressure is $\rho gh=1.03\times10^3(10)(3000)=3.09\times10^7$ Pa. Absolute pressure is about $3.10\times10^7$ Pa.

Step 2. This is well below $10^9$ Pa, so the structure meets the stated pressure-stress limit; this simplified check omits currents and other loads.

9.8

A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm². What maximum pressure would the smaller piston have to bear?

Step 1. $A=425\times10^{-4}=0.0425$ m² and $F=mg=29400$ N.

Step 2. By Pascal’s law, $P=F/A=6.92\times10^5$ Pa gauge (about 6.8 atm). The same pressure increment acts throughout the fluid.

9.9

A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit?

Step 1. Equal pressures at the mercury interfaces give $\rho_wg(0.100)=\rho_sg(0.125)$.

Step 2. $\rho_s/\rho_w=0.100/0.125=0.80$. Spirit specific gravity is 0.80.

9.10

In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6)

Step 1. The added liquid columns create pressure imbalance $\Delta P=g(\rho_w-0.80\rho_w)(0.150)=0.030\rho_wg$.

Step 2. Mercury balances it: $\rho_{Hg}g\Delta h=\Delta P$, so $\Delta h=0.030/13.6=0.00221$ m = 2.21 mm.

9.11

Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.

Step 1. Generally not across a turbulent rapid: the flow is unsteady and strongly mixed, with appreciable viscous energy loss. Bernoulli’s ideal steady, non-viscous relation does not apply there.

9.12

Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.

Step 1. No, if the same atmospheric reference is added to or subtracted from every pressure along the streamline: the same constant cancels from both sides. Do not mix gauge and absolute values, or use different reference pressures at the two points.

9.13

Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10⁻³ kg s⁻¹, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 × 10³ kg m⁻³ and viscosity of glycerine = 0.83 Pa s.) [You may also like to check if the assumption of laminar flow in the tube is correct.]

Step 1. Volume flow $Q=\dot m/\rho=4.0\times10^{-3}/(1.3\times10^3)=3.08\times10^{-6}$ m³ s⁻¹.

Step 2. Poiseuille’s relation: $\Delta P=8\eta LQ/(\pi r^4)=9.78\times10^2$ Pa.

Step 3. Mean speed $v=Q/(\pi r^2)=9.80\times10^{-3}$ m s⁻¹. With diameter 0.020 m, $Re=\rho vD/\eta\approx0.31$, safely laminar.

9.14

In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s⁻¹ and 63 m s⁻¹ respectively. What is the lift on the wing if its area is 2.5 m²? Take the density of air to be 1.3 kg m⁻³.

Step 1. At equal height, $P_{lower}-P_{upper}=\frac12\rho(v_{upper}^2-v_{lower}^2)=0.65(70^2-63^2)=604.5$ Pa.

Step 2. Lift $=\Delta P A=604.5(2.5)=1.51\times10^3$ N upward.

9.15

Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?

Fig. 9.20 · Reconstructed schematic for the referenced exercise. In the narrow section, continuity gives higher speed and Bernoulli gives lower pressure.

Step 1. Figure (a) is incorrect: at the narrower throat, continuity requires greater speed. Bernoulli at equal height then requires lower pressure there. Its indicated pressure trend contradicts those two equations; figure (b) shows the consistent trend.

9.16

The cylindrical tube of a spray pump has a cross-section of 8.0 cm², one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min⁻¹, what is the speed of ejection of the liquid through the holes?

Step 1. Inlet area $8.0\times10^{-4}$ m², speed $0.025$ m s⁻¹; flow $Q=Av=2.0\times10^{-5}$ m³ s⁻¹.

Step 2. Total hole area $40\pi(0.5\times10^{-3})^2=3.14\times10^{-5}$ m².

Step 3. Continuity gives $v_{out}=Q/A_{holes}=0.637$ m s⁻¹.

9.17

A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10⁻² N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?

Step 1. $F=2S\ell$ because the film has two surfaces.

Step 2. $S=F/(2\ell)=1.5\times10^{-2}/0.60=2.5\times10^{-2}$ N m⁻¹.

9.18

Figure 9.21(a) shows a thin liquid film supporting a small weight = 4.5 × 10⁻² N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c)? Explain your answer physically.

Fig. 9.21 · Schematic films with the same pulled edge length; frame shape and film area do not set the force.

Step 1. Both support $4.5\times10^{-2}$ N. For the same liquid and temperature, $S$ is unchanged; the same slider length gives the same pull $F=2S\ell$. Frame shape and total film area do not change the force on that edge.

9.19

What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20 °C) is 4.65 × 10⁻¹ N m⁻¹. The atmospheric pressure is 1.01 × 10⁵ Pa. Also give the excess pressure inside the drop.

Step 1. A drop has one interface: $\Delta P=2S/r=2(0.465)/(0.00300)=310$ Pa.

Step 2. $P_i=P_a+\Delta P=1.0131\times10^5$ Pa; excess pressure is 310 Pa.

9.20

What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10⁻² N m⁻¹? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 10⁵ Pa).

Step 1. A soap bubble has two surfaces: $\Delta P=4S/r=4(0.0250)/(0.00500)=20.0$ Pa.

Step 2. At depth, surrounding liquid pressure is $P_o=P_a+\rho gh=1.01\times10^5+(1.20\times10^3)(9.8)(0.40)=1.05704\times10^5$ Pa.

Step 3. The air bubble has one interface, so add $2S/r=10.0$ Pa: $P_i\approx1.05714\times10^5$ Pa.

Revision

Review the chapter’s definitions, equations and proofs.

9.1 · Fluids and static shear

Review concept and proof

Liquids and gases flow and take the shape of their container; a liquid has nearly fixed volume, while a gas fills the container.
Fluids offer little resistance to static shear: sustained shear makes them keep deforming and flowing.
Solids can sustain static shear and retain a shape; liquids and solids are much less compressible than gases.

9.2 · Pressure, density and Pascal’s law

Review concept and proof

Pressure is normal force per area: P = F⊥/A; local pressure is the limiting value as area shrinks.
Pressure is a scalar (Pa = N m⁻²; dimensions [ML⁻¹T⁻²]); the force exerted on a chosen surface is normal.
Density ρ = m/V has units kg m⁻³; relative density compares with water at 4 °C and is dimensionless.

9.2.1 · Pressure in every direction

Review concept and proof

Take a tiny right-angled fluid prism at rest; its weight becomes negligible relative to face forces in the infinitesimal limit.
Resolve the sloping-face pressure force and use projected areas: F_b sinθ = F_c, A_b sinθ = A_c (and corresponding cosine relations).
Dividing force balances by face areas gives P_a = P_b = P_c: pressure at a point is isotropic.

9.2.2 · Pressure with depth

Review concept and proof

For a column of area A and height h, equilibrium gives (P₂ − P₁)A = ρhAg, so P₂ − P₁ = ρgh.
Open liquid: gauge pressure P_g = ρgh; absolute pressure P_abs = Pₐ + ρgh.
Pressure depends on vertical depth, density and g, not vessel shape; uniform ρ and nearly constant g are assumed.

9.2.3 · Barometers and manometers

Review concept and proof

Mercury barometer: pressure balance gives Pₐ = ρ_Hggh; the space above the column is nearly a vacuum and h ≈ 0.76 m at sea level.
Manometer: equal-level pressures match; descending a column adds ρgΔh and ascending subtracts it.
Air density varies with altitude and weather, so a constant-density atmosphere gives only a rough height estimate.

9.2.4 · Hydraulic machines

Review concept and proof

Pascal’s law: a pressure change applied to a confined fluid is transmitted undiminished throughout it.
Equal pressure increments give F₁/A₁ = F₂/A₂, so the ideal force gain is A₂/A₁.
Incompressibility gives A₁x₁ = A₂x₂; the force gain trades against distance and ideal work is conserved.

9.3 · Streamlines and continuity

Review concept and proof

Steady flow has time-independent velocity at each fixed point; particles may still accelerate as they move. Streamlines are tangent to velocity and cannot intersect.
Mass conservation gives ρAv = constant; for an incompressible fluid Av = Q. Narrower area means greater speed.
Laminar layers move smoothly; turbulence contains fluctuations and eddies. Re = ρvD/η compares inertia with viscosity; transition depends on geometry and disturbances.

9.4 · Bernoulli principle

Review concept and proof

Bernoulli assumes steady, incompressible, non-viscous flow along a streamline.
Pressure work on equal volume ΔV is (P₁−P₂)ΔV = ½ρΔV(v₂²−v₁²) + ρgΔV(h₂−h₁).
Divide by ΔV: P + ½ρv² + ρgh = constant. At equal height faster flow has lower pressure; Torricelli efflux from a large tank is v = √(2gh).

9.5 · Viscosity and Poiseuille flow

Review concept and proof

Viscosity is internal friction between layers; Newtonian shear stress obeys F/A = η(dv/dy). η has units Pa·s and dimensions [ML⁻¹T⁻¹].
Pipe flow has zero speed at the wall and a parabolic laminar profile. Poiseuille: Q = πR⁴ΔP/(8ηL).
Poiseuille’s law requires steady, fully developed laminar flow of a Newtonian fluid in a cylindrical tube; liquid viscosity usually falls with temperature, gas viscosity rises.

9.5.1 · Stokes drag and terminal speed

Review concept and proof

Stokes drag on a slowly moving sphere is F_d = 6πηav, opposite to its motion; the law requires low-Reynolds-number creeping flow.
At terminal speed, weight = buoyancy + drag: 6πηav_t = (4πa³/3)(ρ_s−ρ_f)g.
Thus v_t = 2a²(ρ_s−ρ_f)g/(9η): it scales with a² and density contrast and inversely with viscosity.

9.6.1 · Surface energy and tension

Review concept and proof

Bulk molecules are attracted in all directions; surface molecules have fewer neighbours and higher potential energy. Creating surface area costs energy.
Surface tension S is force per length and reversible work per area: S = F/ℓ = dW/dA; N m⁻¹ = J m⁻².
A soap film has two surfaces; moving an edge of length ℓ gives F = 2Sℓ and dA = 2ℓdx.

9.6.3 · Contact angle and wetting

Review concept and proof

Contact angle θ is measured through the liquid between the solid and the meniscus tangent at the contact line.
Young’s balance: S_la cosθ + S_sl = S_sa; θ depends on both contacting materials.
Acute θ generally wets (water on clean glass); obtuse θ resists wetting (mercury on glass). Detergent changes interfacial tensions and helps water spread.

9.6.4 · Drops and bubbles

Review concept and proof

With gravity negligible, a sphere minimises area for a given volume, so small free drops and bubbles are spherical.
For a drop, dA = 8πrdr and dV = 4πr²dr; equating surface-energy increase S dA to pressure work ΔP dV gives ΔP = 2S/r.
One interface (drop or gas bubble in liquid): ΔP = 2S/r. Soap bubble has two interfaces: ΔP = 4S/r; smaller radius means greater excess pressure.

9.6.5 · Capillary rise

Review concept and proof

The curved meniscus creates a pressure difference whose vertical surface-tension component around the rim is 2πaS cosθ.
Balance it with column weight ρgπa²h to derive h = 2S cosθ/(ρga).
Acute contact gives cosθ > 0 and rise; obtuse contact gives cosθ < 0 and depression. Magnitude grows as tube radius a shrinks.

Tests

Practice tests

Choose a level, answer the questions and submit to check your score.

1. A pressure of 2×10⁵ Pa acts on area 0.02 m². Force is:

2. A 0.5 m² piston supports 5000 N. Pressure is:

3. If pipe area halves for incompressible flow, speed:

4. In Bernoulli flow, P+½ρv²+ρgh is:

5. A liquid drop has excess pressure:

1. If capillary radius doubles, rise becomes:

2. Poiseuille flow rate varies with tube radius as:

3. At terminal speed, resultant force on sphere is:

4. The Reynolds number compares:

5. For the same liquid film and slider length, force depends on:

1. At depth h, gauge pressure is:

2. For an ideal fluid, viscosity is assumed:

3. Stokes-regime terminal speed varies with sphere radius as:

4. A Venturi meter uses:

5. A wetting liquid generally has contact angle: