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

Concept. Pressure in a fluid at rest increases with depth, P₂ − P₁ = ρgh (Pascal’s law and Eq. 9.6). Pressure is defined as normal force per unit area, so it has magnitude only.

Question. 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.

(a) Blood pressure at the feet and at the brain

The blood in the arteries and veins is a continuous liquid column. In a standing person the feet are about 1.7 m below the brain, so the vertical column of blood above the feet is longer. The hydrostatic pressure difference is P_feet − P_brain = ρgh, where ρ is the density of blood (1.06×10³ kg m⁻³, Table 9.1) and h the height difference. With h ≈ 1.7 m this is about 1.06×10³ × 9.8 × 1.7 ≈ 1.8×10⁴ Pa (about 130 mm of Hg), added to the pressure produced by the heart. Hence the pressure at the feet is greater than at the brain. This is also why a person may feel faint on standing up suddenly.

(b) Atmospheric pressure halves at 6 km

The formula P = P_a + ρgh is valid only for an incompressible fluid of constant density, i.e. for liquids. Air is a gas, highly compressible: the lower layers are compressed by the weight of the air above, so the density of air is greatest at sea level and decreases with height (and g also decreases slightly). Atmospheric pressure at a height equals the weight of the air column above it; since the air is densest near the ground, about half of the mass of the atmosphere lies within the lowest 5–6 km. The pressure at 6 km is therefore nearly half the sea-level value, and the remaining half is spread over the enormous height of more than 100 km where the air is extremely thin. If the density were constant (as Example 9.3 assumes) the atmosphere would end at 8 km.

(c) Why hydrostatic pressure is a scalar

In the definition P = F/A the force that appears is only the component of force normal to the area, and the direction of this force is fixed by the orientation of the area, not by the pressure. At a given point in a fluid at rest the pressure is the same in all directions (Pascal’s law, Fig. 9.2): rotating the area at that point changes the direction of the force but not the pressure. Since no single direction can be assigned to it, pressure has magnitude only and so is a scalar, like other types of stress. It adds as ordinary numbers (P = P_a + ρgh).

Extra depth. A numerical check for (b): treat the atmosphere as isothermal with T = 288 K. Then P = P₀ exp(−Mgh/RT) with scale height H = RT/(Mg) = (8.31 × 288)/(0.029 × 9.8) ≈ 8.4 km, so at h = 6 km P/P₀ = e^(−6/8.4) ≈ 0.49. This is the beyond-the-book model, but it reproduces the NCERT statement that the pressure halves at about 6 km.

Conclusion: (a) the blood column is longer above the feet (ΔP = ρgh); (b) air is compressible so its density falls with height; (c) pressure is the normal force per unit area, equal in all directions, hence a scalar.

Concept. Cohesion (attraction between like molecules) and adhesion (attraction between unlike molecules) decide the shape of a liquid surface at a solid wall: if adhesion is larger, the liquid wets the solid (acute angle); if cohesion is larger, it does not (obtuse angle).

Question. 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; (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.

(a) Obtuse for mercury, acute for water

The angle of contact is determined by the balance S_la cosθ + S_sl = S_sa (Eq. 9.22). For mercury on glass the mercury molecules attract one another (cohesion) much more strongly than they are attracted by glass molecules (adhesion), so S_sl is large (S_sl > S_la), cosθ is negative and θ is obtuse (the meniscus is convex). For water on glass the attraction between water and glass molecules is greater than that between water molecules, S_sl is small (S_sl < S_la), cosθ is positive and θ is acute (concave meniscus).

(b) Water spreads on glass, mercury forms drops

Because of the acute angle, the water film on clean glass lowers the total surface energy by spreading: water wets glass. Because of the obtuse angle, mercury molecules prefer to stay together; it costs a lot of energy to create a mercury-glass interface, so mercury pulls itself into drops (minimum surface area) and does not wet glass.

(c) Surface tension is independent of the area

Surface tension is the force per unit length (equivalently the surface energy per unit area) in the plane of the interface. It depends on the nature of the liquid and the medium in contact and on temperature, but not on how large the surface is. When the area of a film is increased, molecules from the interior come to the surface until the surface has the same energy per unit area as before; the force per unit length on the bar remains F/2l, whatever the area. (This differs from a stretched rubber sheet where the force increases with stretching.)

(d) Detergent solution has small angle of contact

Detergents are wetting agents: they reduce the liquid-solid and liquid-air interfacial energies, so cosθ increases and θ becomes small. A small angle of contact means the solution spreads and penetrates the fibres of a cloth and carries the dirt away. (Water-proofing agents do the opposite: they produce a large angle of contact.)

(e) A free drop is spherical

A liquid-air interface has surface energy proportional to its area, and a system under no external forces takes the configuration of least energy. For a given volume, the surface with the least area is a sphere. Therefore a drop under no external forces (gravity, air resistance negligible) is spherical, as seen in tiny drops in a spray or the weightless water in a spacecraft.

Extra depth. For exact language: a drop of mercury on glass has cosθ = (S_sa − S_sl)/S_la negative ⇒ obtuse. Values: water-glass θ ≈ 0°–8°, mercury-glass θ ≈ 140°. A large drop is flattened by gravity, which is why only small drops are really spherical.

Conclusion: (a, b) cohesion vs adhesion decides the angle (acute for water, obtuse for mercury); (c) S is force per unit length; (d) detergents make θ small; (e) a sphere has the least surface area for a given volume.

Concept. Use the facts: S falls with temperature; viscosity of liquids falls and that of gases rises with temperature; fluids resist the rate of shear strain; continuity is conservation of mass; critical speed ∝ 1/(size) through the Reynolds number.

Question. Fill in the blanks using the word(s) from the list appended with each statement.

Explanations. (a) Surface tension is the extra energy of molecules at the surface; heating weakens the attraction between molecules, so S decreases (it is 0.0727 N m⁻¹ for water at 20°C and smaller at 100°C). (b) The viscosity of a liquid falls on heating (honey flows faster when warm) but that of a gas rises (Points to Ponder 7). (c) In a solid the shear strain is determined by the force and G; in a fluid the strain keeps increasing with time under a constant stress, so the stress is related to the rate of change of strain. (d) The speed rises at a narrow part because the same volume per second must pass (A₁v₁ = A₂v₂); this is a mass-conservation statement. (e) Turbulence begins when the Reynolds number R = ρvD/η exceeds a critical value (the same for model and aeroplane for similar flows); the model has a much smaller size D, so a larger speed is needed to reach the critical Reynolds number.

Extra depth. In (e) the Reynolds number is the dimensionless combination ρvD/η that decides whether flow is laminar or turbulent: R < about 1000 laminar, R > about 2000 turbulent (for flow in a pipe). Since it contains the size D, flow around a small model at moderate speed corresponds to the flow around a big aeroplane only at a much larger speed — that is why wind tunnels use high speeds or pressurised air.

Conclusion: (a) decreases; (b) increases, decreases; (c) shear strain, rate of shear strain; (d) conservation of mass; (e) greater.

Concept. Link each situation to its law: pressure and speed (Bernoulli), flow rate and area (continuity), flow through a narrow tube (viscosity), momentum conservation (thrust), and spin (Magnus effect).

Question. 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.

(a) Blow over, not under, the paper

When you blow over the paper the air above moves fast, so by Bernoulli’s principle (P + ½ρv² = constant) the pressure above falls below the atmospheric pressure below the paper. The greater pressure beneath pushes the paper up and keeps it horizontal. If you blow under it, the pressure below falls and the larger atmospheric pressure above pushes the paper down.

(b) Jets of water between the fingers

Water leaves the tap at a fixed volume flow rate Q = Av. When we close the opening partly with fingers, the area A of the openings becomes much smaller. By the equation of continuity (Av = constant) the speed v through the small gaps increases greatly, so fast narrow jets gush out.

(c) Needle size controls the flow rate

The flow of a viscous liquid through a narrow tube of radius r and length L under pressure difference ΔP is given by Poiseuille’s formula Q = πr⁴ΔP/(8ηL) (a result of viscous flow; not derived in the text). The flow rate depends on the fourth power of the radius of the needle but only linearly on the pressure. A change of needle radius by a factor of two changes Q by 16 times, whereas the doctor’s thumb can change the pressure by only a small factor. Hence the size of the needle controls the flow rate far better than the thumb pressure.

(d) Backward thrust on the vessel

As the fluid leaves the hole with speed v it carries away momentum in the forward direction. By conservation of momentum (Newton’s third law, as in a rocket) the vessel gets an equal momentum in the opposite direction, i.e. a backward thrust F = (dm/dt)v = ρav². For an open tank v = √(2gh) so F = 2ρgha (a = area of the hole). Also, the pressure of the liquid acts on the walls except on the area of the hole, leaving a net unbalanced force on the opposite wall.

(e) A spinning cricket ball does not follow a parabola

A spinning ball drags air along with it. Relative to the ball, the air on one side moves faster and on the other side slower, so the pressure on the two sides differs (Bernoulli) and there is a net sideways or upward force, the Magnus effect (dynamic lift). This extra force, beyond gravity, makes the path deviate from a parabola; it is the basis of swing and spin bowling.

Extra depth. The thrust in (d) is the principle of jet propulsion, garden sprinklers and rockets; for a hole of area 1 cm² at 1 m depth of water the thrust is 2 × 10³ × 9.8 × 1 × 10⁻⁴ = 1.96 N. In (c), blood-vessel or tube blockage also changes r⁴: a 20% reduction in the radius of an artery cuts the flow to about 40% unless the pressure rises.

Conclusion: (a) lower pressure above; (b) smaller area, larger speed (continuity); (c) flow ∝ r⁴ so the needle size controls it; (d) momentum of the jet gives a backward thrust; (e) Magnus effect.

Concept. Pressure = force/area. The force is the weight of the girl, mg, and the area is the circular heel, πr² = πd²/4.

Question. 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?

Given

• Mass m = 50 kg, g = 9.8 m s⁻² ⇒ F = mg = 490 N (acting normal to the floor)
• Diameter d = 1.0 cm = 1.0×10⁻² m ⇒ r = 0.50 cm = 5.0×10⁻³ m
• Area A = πr² = 3.14 × (5.0×10⁻³)² = 7.85×10⁻⁵ m²

Solution

P = F/A = 490/(7.85×10⁻⁵) = 6.2×10⁶ Pa

The pressure exerted by the heel is about 6.2×10⁶ Pa, i.e. about 62 atmospheres (6.2×10⁶/1.013×10⁵ = 61.6 atm). This is why a stiletto heel can dent a wooden floor, while the same girl standing flat on both feet of area, say, 150 cm² each would exert only 490/(300×10⁻⁴) ≈ 1.6×10⁴ Pa, about 380 times smaller.

Extra depth. This is the pressure idea of Section 9.2: the same force over a smaller area gives a larger pressure (needle, knife, high heel), which is why a blunt object does not pierce the skin but a needle does, and why a wooden plank spreads the force on the performer’s chest. In real life the whole weight is not carried by one heel at once, so the pressure is smaller, but the question assumes balancing on a single heel.

More to know. Compare other everyday pressures: a sharp knife edge or needle tip (area of the order 10⁻⁹ m²) can produce pressures of 10⁹ Pa for an ordinary hand force; the atmosphere itself presses on the human body (area about 1.5 m²) with a total force of 1.5 × 10⁵ N — about 15 tonnes — without hurting us because the internal pressure of the body fluids is equal and the pressure is transmitted equally in all directions (Pascal’s law). Snow-shoes, skis, caterpillar tracks of tanks and wide tyres of tractors all increase the area to reduce the pressure.

Conclusion: P = mg/(πd²/4) = 6.2×10⁶ Pa, about 60 times atmospheric pressure, because the contact area is very small.

Concept. A liquid barometer balances atmospheric pressure by a column of liquid: P_a = ρgh. Hence h = P_a/(ρg), and the height is inversely proportional to the density.

Question. Torricelli’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.

Given

• Normal atmospheric pressure P_a = 1.013×10⁵ Pa
• Density of wine ρ = 984 kg m⁻³, g = 9.8 m s⁻²

Solution

h = P_a/(ρg) = (1.013×10⁵)/(984 × 9.8) = 1.013×10⁵/9643 = 10.5 m

So the wine column would be about 10.5 m high. For comparison, mercury (ρ = 13.6×10³ kg m⁻³) gives h = 1.013×10⁵/(13.6×10³ × 9.8) = 0.76 m = 76 cm. Since h ∝ 1/ρ, the ratio is h_wine/h_Hg = 13.6×10³/984 = 13.8, and 0.76 × 13.8 ≈ 10.5 m: a good cross-check.

Extra depth. The barometer height depends on the density of the liquid but not on the cross-section or the shape of the tube (Eq. 9.7: only the height of the column matters). Water would need a tube about 10.3 m long (h = 1.013×10⁵/(10³ × 9.8)), which is why water barometers are impractical and why a suction pump cannot lift water by more than about 10 m. Mercury is used because it is dense (short tube) and has a negligible vapour pressure at room temperature; wine, being a solution of alcohol and water, has a larger vapour pressure that would depress the column slightly.

More to know. Torricelli (1608–1647) invented the mercury barometer in 1643, and Blaise Pascal repeated it with water and with red wine to show that the height of the column is inversely proportional to the density of the liquid; the experiment also made people accept that the atmosphere has weight. Pascal later sent his brother-in-law up the Puy de Dôme mountain with a barometer and found that the mercury column was shorter at the top, proving that atmospheric pressure decreases with height (the key idea of Ex 9.1(b)). Units to remember: 760 mm of Hg = 1 atm = 1.013×10⁵ Pa = 10.3 m of water.

Quick table. Column height for P_a = 1.013×10⁵ Pa (g = 9.8 m s⁻²): mercury (13.6×10³) 0.760 m; sea water (1.03×10³) 10.0 m; fresh water (1.00×10³) 10.3 m; wine (984) 10.5 m; ethyl alcohol (0.806×10³) 12.8 m. The less dense the liquid the taller the barometer.

Conclusion: h = P_a/(ρg) = 10.5 m for wine (density 984 kg m⁻³), compared with 0.76 m for mercury.

Concept. The structure at the bottom is compressed by the hydrostatic pressure P = ρgh (plus P_a if absolute). Compare this stress with the maximum stress the structure can withstand, 10⁹ Pa.

Question. 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.

Given

• Depth h = 3 km = 3000 m, ρ = 10³ kg m⁻³ (water), g = 9.8 m s⁻²
• Maximum stress the structure can withstand = 10⁹ Pa

Solution

P = ρgh = 10³ × 9.8 × 3000 = 2.94×10⁷ Pa

If the atmospheric pressure at the surface (1.01×10⁵ Pa) is added, the absolute pressure at the sea bed is 2.95×10⁷ Pa; the difference is negligible. The compressive stress on the structure at the bottom is therefore about 2.9×10⁷ Pa, which is less than the design limit of 10⁹ Pa by a factor of about 34.

Conclusion: the structure is suitable. The greatest stress it will meet is the hydrostatic pressure at the sea bed, 2.9×10⁷ Pa, much below the 10⁹ Pa which it can withstand (ocean currents being ignored, as the question allows).

Extra depth. The pressure depends only on the depth and not on the shape of the structure (Fig. 9.4, hydrostatic paradox). In practice such structures must also be designed against the pressure of the oil in the well and against waves and currents, and the safety factor of 34 gives a big margin. Using g = 10 m s⁻² and ρ = 1.03×10³ kg m⁻³ for sea water would give 3.09×10⁷ Pa; the conclusion is unchanged.

More to know. At the sea bed, 3 km deep, the pressure is about 290 times atmospheric pressure (2.94×10⁷/1.013×10⁵ = 290 atm). A force of this order acts on every square metre: 2.94×10⁷ N, about 3000 tonnes wt. Submarines and deep-sea probes are therefore built as spherical or cylindrical pressure hulls so that the compressive stress is spread uniformly. In the problem, the structure is “vertical”, so the sideways pressure acts on its walls and the vertical hydrostatic force on its base; both are of order ρgh. The stress on its walls is the gauge pressure outside minus whatever pressure is maintained inside, so a hollow structure filled with air at 1 atm would face the full 2.9×10⁷ Pa difference.

Conclusion: The maximum pressure at 3 km is ρgh = 2.94×10⁷ Pa, far below 10⁹ Pa, so the structure is suitable.

Concept. By Pascal’s law the pressure in the liquid is the same everywhere in the lift, so the pressure on the small piston equals the pressure on the large piston, P = F₂/A₂.

Question. 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?

Given

• Maximum mass m = 3000 kg ⇒ load F₂ = mg = 3000 × 9.8 = 2.94×10⁴ N
• Area of the load piston A₂ = 425 cm² = 425×10⁻⁴ m² = 4.25×10⁻² m²

Solution

P = F₂/A₂ = (2.94×10⁴)/(4.25×10⁻²) = 6.92×10⁵ Pa

The pressure is transmitted undiminished through the liquid, so the smaller piston also has to bear a pressure of 6.92×10⁵ Pa (about 6.8 atm), regardless of its own area. The area of the smaller piston decides only the force F₁ needed: F₁ = P A₁, so a smaller A₁ needs a smaller force.

Extra depth. This answer does not need the area of the small piston; the pressure in the lift is the same. If, for example, A₁ were 10 cm² the force needed would be 6.92×10⁵ × 10⁻³ = 692 N (about 71 kg wt) to lift 3000 kg: a mechanical advantage of 425/10 = 42.5. Remember that the small piston must move 42.5 times farther than the load piston since A₁L₁ = A₂L₂.

More to know. A hydraulic lift is an application of the second form of Pascal’s law (pressure applied to an enclosed fluid is transmitted undiminished). It multiplies force, not energy or work. The designer chooses the area ratio A₂/A₁ so that an easily applied force F₁ balances the heavy load; typical car lifts in garages use compressed air of a few atmospheres (6.9 atm here). Safety valves limit the pressure. Hydraulic brakes, dentist’s chairs, hydraulic jacks, excavators and aircraft control surfaces are other examples of the same principle.

Check in atmospheres. 6.92×10⁵ Pa / 1.013×10⁵ Pa = 6.8 atm gauge, or about 7.8 atm absolute if the atmospheric pressure is included. The question asks for the pressure that the smaller piston has to bear, which is the pressure in the liquid; if the pressure is expressed as gauge pressure then the figure above is the answer, and if in absolute terms add 1.01×10⁵ Pa.

Conclusion: P = F₂/A₂ = 6.92×10⁵ Pa on both pistons (Pascal’s law).

Concept. At the same level in the same liquid (mercury) the pressure is equal in both arms. Hence the pressure from the water column equals the pressure from the spirit column: h_wρ_wg = h_sρ_sg.

Question. 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?

Solution

The mercury surfaces in the two arms are at the same level, so the pressure just above the mercury surface is the same on both sides (pressure is the same at the same level in the same liquid at rest):

P_a + h_wρ_wg = P_a + h_sρ_sg  ⇒  ρ_s/ρ_w = h_w/h_s

Specific gravity of spirit = 10.0/12.5 = 0.80

Specific gravity (relative density) is the ratio of the density to that of water, so it has no unit. The density of the spirit is 0.80 × 10³ = 800 kg m⁻³. A less dense liquid needs a taller column to give the same pressure, which is why the spirit column (12.5 cm) is taller than the water column (10.0 cm).

Extra depth. The atmospheric pressure acts on the top of both columns and cancels, and the mercury does not enter the calculation because its surfaces are at the same level; mercury only separates the two liquids. If the mercury levels were unequal (as in Ex 9.10), the difference in height times the density of mercury would also appear in the balance.

Conclusion: h_wρ_w = h_sρ_s gives the specific gravity of the spirit as 10.0/12.5 = 0.80.

Concept. Equate the pressures at the lower mercury surface (the surface at the lower level on one side). The column on the other side is the other liquid plus an extra mercury column of height h.

Question. 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.)

Given

• Water column now 10.0 + 15.0 = 25.0 cm; spirit column 12.5 + 15.0 = 27.5 cm
• Specific gravity of spirit = 0.80 (from Ex 9.9), of mercury = 13.6, of water = 1
• Let the mercury level in the spirit arm be higher than in the water arm by h cm.

Solution

Pressure of the water column (25.0 cm) at the level of the lower mercury surface (the water side). The same level on the other side lies inside the mercury of the spirit arm, below its surface by h. The pressure there is the spirit column plus the extra mercury column of height h. Equating (and cancelling g and P_a):

25.0 × 1 = 27.5 × 0.80 + h × 13.6  ⇒  25.0 = 22.0 + 13.6 h

h = 3.0/13.6 = 0.22 cm

The mercury is about 0.22 cm (2.2 mm) higher in the spirit arm than in the water arm. The difference is small because mercury is so dense: a pressure difference equivalent to 3.0 cm of water (294 Pa) needs only 0.22 cm of mercury.

Extra depth. The difference arises because the new columns are no longer in the ratio 1 : 0.8 × 1 to the original: the water column pressure, 25.0 gρ_w (in cm of water), is greater than the spirit column pressure, 27.5 × 0.8 = 22.0, by 3.0 cm of water. This difference pushes the mercury down in the water arm and up in the spirit arm until the extra mercury column of height h = 3.0/13.6 balances it (a U-tube manometer working in reverse).

Conclusion: Equating pressures: 25.0 = 27.5 × 0.80 + 13.6 h, so h = 0.22 cm, mercury higher in the spirit arm.

Concept. Bernoulli’s equation P + ½ρv² + ρgh = constant is derived for steady, streamline (laminar), non-viscous, incompressible flow. It can be applied only where these conditions hold.

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

Answer: No

The flow of water through a rapid is turbulent and not steady. The water meets rocks and a steep, uneven bed; it forms eddies and whirlpools (white water), and at any point the velocity and the pressure change rapidly with time in magnitude and in direction. Bernoulli’s equation does not hold for non-steady or turbulent flow, since it assumes that the velocity at a point does not change with time and that the streamlines are smooth and do not cross (Section 9.4 and Points to Ponder 5 and 6).

A large part of the mechanical energy of the water is converted into heat and sound in the turbulent region, so P + ½ρv² + ρgh is not constant along the flow; it decreases downstream. Hence the equation cannot be used to calculate the speed or pressure at a point in a rapid.

Extra depth. Where could one use it in a river? In the smooth part of the river far from the rapids the flow is steady and nearly laminar, so the continuity equation and Bernoulli’s equation give a good first estimate (for instance the speed of the water as the river narrows and its depth changes). For the rapid itself one must use experiments, models with dimensionless numbers such as the Reynolds number, or computer simulation.

More to know. Even in the rapid, one can speak of an average flow, and the continuity equation holds for the time-averaged flow (the mass entering equals the mass leaving), because mass is conserved in any flow. It is only the energy equation that cannot be used in this simple form, since energy is dissipated by eddies. In engineering the head loss in such flows is found from empirical formulas; the Reynolds number R = ρvD/η is far above 2000 in a river rapid (for water η = 10⁻³ Pa s, v = 2 m s⁻¹, D = 1 m, R = 2×10⁶), confirming that the flow is turbulent.

Conclusion: No: the flow through a rapid is turbulent and non-steady, with energy loss to viscosity, so P + ½ρv² + ρgh is not constant.

Concept. In Bernoulli’s equation the pressures at two points enter in the form P₁ − P₂. Gauge pressure is P − P_a, so replacing P by (P − P_a) at both points subtracts the same P_a from both sides if the atmospheric pressure is the same at both points.

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

Answer

Write Bernoulli’s equation for two points:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

With P₁ = P_g1 + P_a1 and P₂ = P_g2 + P_a2 (gauge + atmospheric), this becomes

P_g1 + ½ρv₁² + ρgh₁ + P_a1 = P_g2 + ½ρv₂² + ρgh₂ + P_a2

If the atmospheric pressure is the same at the two points (P_a1 = P_a2), the two P_a cancel and the equation has exactly the same form with gauge pressures. So it does not matter whether gauge or absolute pressure is used, as long as the same atmospheric pressure acts at both points. This is true in most laboratory and engineering situations (pipes, tanks, nozzles), and gauge pressure is often more convenient because it is what gauges read.

Exception: if the two points are at greatly different heights so that the atmospheric pressure differs appreciably between them (for example at the top and bottom of a tall mountain or a very high pipeline), the two P_a do not cancel and one must use absolute pressures, or add the difference of the atmospheric pressures. The same cancellation is used in Torricelli’s law, where P_a at the hole and at the top of an open tank is the same and only ρgh remains.

Extra depth. The “gauge pressure” form is what makes Ex 9.4(c) simple: the net force on the window of a submarine is P_gA, since atmospheric pressure acts on both sides. In Bernoulli problems with an open tank one writes P₁ = P₂ = P_a and finds v = √(2gh) with no pressure term at all — the same idea.

Conclusion: It does not matter, because the atmospheric pressure cancels from both sides when it is the same at the two points; it would matter only if the atmospheric pressures differ significantly.

Concept. For steady laminar flow of a viscous liquid in a horizontal tube of length L and radius r, the volume flow rate Q is given by Poiseuille’s formula Q = πr⁴ΔP/(8ηL), so ΔP = 8ηLQ/(πr⁴). (Poiseuille’s formula is a standard result of viscous flow; it is not derived in the text.)

Question. 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 = 0.83 Pa s.) [Also check if the assumption of laminar flow in the tube is correct.]

Given

• L = 1.5 m, r = 1.0 cm = 1.0×10⁻² m, η = 0.83 Pa s, ρ = 1.3×10³ kg m⁻³
• Mass flow rate = 4.0×10⁻³ kg s⁻¹

Step 1 — volume flow rate

Q = (mass per second)/ρ = (4.0×10⁻³)/(1.3×10³) = 3.08×10⁻⁶ m³ s⁻¹

Step 2 — pressure difference (Poiseuille)

ΔP = 8ηLQ/(πr⁴) = (8 × 0.83 × 1.5 × 3.08×10⁻⁶)/(3.14 × (10⁻²)⁴) = 3.07×10⁻⁵/(3.14×10⁻⁸) ≈ 9.8×10² Pa

Step 3 — is the flow laminar? (Reynolds number)

• Average speed v = Q/(πr²) = 3.08×10⁻⁶/(3.14×10⁻⁴) = 9.8×10⁻³ m s⁻¹ (about 1 cm per second)
• Reynolds number R = ρvd/η = (1.3×10³ × 9.8×10⁻³ × 2.0×10⁻²)/0.83 = 0.31

A flow in a tube is laminar if R is less than about 1000 and turbulent if R is more than about 2000 (a standard criterion; the textbook only mentions the critical speed). Here R = 0.31 << 1000, so the flow of the glycerine is laminar and the assumption is correct.

Extra depth. Because the tube is horizontal there is no height term in Bernoulli’s equation, but Bernoulli itself cannot be used because viscosity is significant: it would predict a pressure drop of zero (constant speed) in a uniform tube. The pressure drop of 9.8×10² Pa (about 7 mm of Hg) is entirely used to overcome viscous friction between the layers. The result ΔP ∝ 1/r⁴ is dramatic: halving the radius increases the required pressure difference sixteen times, which is why blocked arteries make the heart work much harder and why hypodermic needles control flow (Ex 9.4(c)).

Conclusion: ΔP = 8ηLQ/(πr⁴) ≈ 9.8×10² Pa and R ≈ 0.31, so the flow is laminar.

Concept. The lift equals the pressure difference between the lower and upper surfaces times the wing area. Bernoulli’s equation for the two surfaces (same height): ΔP = ½ρ(v_upper² − v_lower²) is the amount by which pressure below exceeds pressure above.

Question. 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.

Given

• v_upper = 70 m s⁻¹, v_lower = 63 m s⁻¹, A = 2.5 m², ρ = 1.3 kg m⁻³

Step 1 — Bernoulli’s equation for the two surfaces (negligible height difference)

P_lower + ½ρv_lower² = P_upper + ½ρv_upper²

ΔP = P_lower − P_upper = ½ρ(v_upper² − v_lower²) = ½ × 1.3 × (70² − 63²)

70² = 4900 and 63² = 3969, so v_upper² − v_lower² = 931 m² s⁻² and ΔP = 0.65 × 931 = 605 Pa.

Step 2 — lift = pressure difference × area

Lift = ΔP × A = 605 × 2.5 = 1.51×10³ N ≈ 1.5×10³ N

The upward lift on the wing is about 1.5×10³ N (the weight of about 154 kg), directed upward because the pressure below the wing is higher than above. The speed above is only 10% higher than below, yet the lift is large.

Extra depth. For comparison, in Example 9.7 a speed difference of only 8% produces the lift for a 3.3×10⁵ kg aeroplane because its wing area is 500 m² and the speeds are much larger (267 m s⁻¹). Doubling the speed of the airflow quadruples the lift for the same wing (ΔP ∝ v² difference), which is why aeroplanes need high speed on take-off.

Conclusion: Lift = ½ρ(v_upper² − v_lower²)A = 1.5×10³ N.

Concept. For steady flow of a non-viscous liquid through a horizontal tube, the pressure is lower where the speed is higher (Bernoulli) and the speed is higher where the tube is narrower (continuity). The height of liquid in a vertical manometer tube measures the pressure there.

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

Answer: Figure 9.20(a) is incorrect

1. By the equation of continuity A₁v₁ = A₂v₂, the speed of the liquid is larger in the narrow part (the constriction) than in the wider part.
2. By Bernoulli’s equation, for a horizontal tube P + ½ρv² = constant, so where the speed is larger the pressure is smaller. So the pressure at the constriction is less than the pressure in the wider part.
3. The height of the liquid in a vertical open tube is proportional to the (gauge) pressure at that point, P = ρgh. Thus the liquid level in the manometer at the constriction must be lower than the level in the manometer at the wide part.
4. In Fig. 9.20(a) the level at the constriction is higher, which would mean a higher pressure where the speed is higher — contradicting Bernoulli’s principle. In Fig. 9.20(b) the level at the constriction is lower: correct.

The measurement of the pressure drop at a constriction in this way is the principle of the Venturi meter, used to find the flow speed of liquids in pipes: from the two heights, v₂² − v₁² = 2g(h₁ − h₂), and v₁ follows from continuity.

Extra depth. The same argument explains why the two ships sailing closely side by side are drawn together (fast water in the gap, lower pressure), and why a fire hose or spray gun lifts liquid up a tube at the constriction (atomiser). In the diagram the pressure difference between the wide and narrow regions is ΔP = ½ρ(v₂² − v₁²) = ρgΔh.

Conclusion: (a) is incorrect: the pressure (and hence the liquid level) must be lower at the constriction where the speed is higher.

Concept. The volume of liquid flowing per second is the same through the tube and through all the holes (equation of continuity): A₁v₁ = A₂v₂, where A₂ is the total area of all the holes.

Question. 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?

Given

• Cross-section of the tube A₁ = 8.0 cm² = 8.0×10⁻⁴ m²
• Speed in the tube v₁ = 1.5 m min⁻¹ = 1.5/60 = 0.025 m s⁻¹
• Number of holes n = 40; diameter of each d = 1.0 mm ⇒ radius r = 0.5 mm = 5.0×10⁻⁴ m

Step 1 — total area of the holes

A₂ = n π r² = 40 × 3.14 × (5.0×10⁻⁴)² = 40 × 7.85×10⁻⁷ = 3.14×10⁻⁵ m²

Step 2 — equation of continuity

v₂ = A₁v₁/A₂ = (8.0×10⁻⁴ × 0.025)/(3.14×10⁻⁵) = 2.0×10⁻⁵/3.14×10⁻⁵ = 0.64 m s⁻¹

The liquid is ejected through the holes with a speed of 0.64 m s⁻¹ (about 25 times the speed 0.025 m s⁻¹ in the tube, because the total hole area is about 1/25 of the tube cross-section). The volume of liquid delivered per second is A₁v₁ = 2.0×10⁻⁵ m³ s⁻¹ = 20 cm³ s⁻¹.

Extra depth. Do not use the area of one hole: the same flow divides among 40 holes, and the continuity equation is written for the total cross-section A₂ = 40πr². An equivalent way: each hole delivers one-fortieth of the flow: v₂ = (A₁v₁/40)/(πr²) gives the same answer. This is why a fine rose on a watering can or a shower head gives a fast spray: many small holes have a total area much smaller than the pipe.

Conclusion: v₂ = A₁v₁/(40πr²) = 0.64 m s⁻¹, because the total area of the holes is much smaller than the tube cross-section.

Concept. The film has two surfaces, each of which pulls on the slider with a force S l. At equilibrium the total upward force 2Sl equals the weight W supported, so S = W/2l.

Question. 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?

Given

• Weight supported W = 1.5×10⁻² N
• Length of the slider l = 30 cm = 0.30 m

Solution

The film has a front and a back surface, so the total length of the contact line which pulls on the slider is 2l. In equilibrium (Eq. 9.20 and 9.21):

S × 2l = W  &Rightarrow;  S = W/2l = (1.5×10⁻²)/(2 × 0.30) = 2.5×10⁻² N m⁻¹

The surface tension of the soap film is 2.5×10⁻² N m⁻¹ (0.025 N m⁻¹), about one-third of the value for pure water (7.27×10⁻² N m⁻¹, Table 9.3); the soap lowers the surface tension. The surface energy per unit area is the same number in J m⁻²: 2.5×10⁻² J m⁻² per surface.

Extra depth. The idea is the same as the measurement with the glass plate, S = mg/2l (Eq. 9.21), where the liquid film pulling on the plate is also two-sided (liquid on each of the two faces of the plate). A beginner’s error is to use S = W/l, which doubles the answer. If the film is stretched slowly by an extra distance d, the work done W d goes into surface energy S (2 l d), consistent with Eq. 9.19; the film does not need to support more weight as it is stretched, because S does not depend on area (Ex 9.2(c)).

Conclusion: S = W/2l = 1.5×10⁻²/(2 × 0.30) = 2.5×10⁻² N m⁻¹.

Concept. The force of surface tension on a movable bar depends only on S and the length of the line of contact: F = 2Sl. It does not depend on the shape or area of the film.

Question. Figure 9.21(a) shows a thin liquid film supporting a small weight of 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.

Answer

The weight supported by the film in (b) and in (c) is also 4.5×10⁻² N.

Physical explanation

• The film pulls on the movable bar by surface tension, which is a force per unit length acting along the line of contact with the bar: F = S × (total length of the contact line) = S × 2l for a two-sided film.
• In all three diagrams the length of the bar is the same, 40 cm, and the liquid and temperature are the same, so S is the same. Hence the force of surface tension on the bar and the weight supported are the same in (a), (b) and (c).
• The surface area of the films is quite different (the film in (b) is longer and in (c) has a different shape), but surface tension, being a force per unit length (equivalently the energy per unit area), does not depend on the area of the film. Increasing the area does not increase the force, because fresh molecules come from the interior to the surface so that the surface energy per unit area stays the same.

Numerically, the surface tension of the liquid is S = W/2l = 4.5×10⁻²/(2 × 0.40) = 5.6×10⁻² N m⁻¹, the same in all three cases (a number comparable with soapy water).

Extra depth. This question tests the same idea as Ex 9.2(c): S is independent of the area. A film stretched like a rubber sheet would pull harder as it is stretched (Hooke’s law), but a liquid film does not — its restoring force per unit length remains constant as more surface is created.

Conclusion: In (b) and (c) the film supports the same weight 4.5×10⁻² N, because the force 2Sl depends only on S and the 40 cm length, not on the area or shape.

Concept. A liquid drop has one liquid-air surface; the pressure inside is greater than that outside by the excess pressure 2S/r.

Question. 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.

Given

• r = 3.00 mm = 3.00×10⁻³ m
• S = 4.65×10⁻¹ N m⁻¹
• P_o = P_a = 1.01×10⁵ Pa

Step 1 — excess pressure (Eq. 9.25)

P_i − P_o = 2S/r = (2 × 4.65×10⁻¹)/(3.00×10⁻³) = 0.93/(3.00×10⁻³) = 310 Pa

Step 2 — pressure inside

P_i = P_a + 2S/r = 1.01×10⁵ + 310 = 1.0131×10⁵ Pa ≈ 1.01×10⁵ Pa

The pressure inside the drop is 1.0131×10⁵ Pa (to three significant figures 1.01×10⁵ Pa) and the excess pressure is 310 Pa. The excess pressure is only 0.3% of the atmospheric pressure, but it is exactly this small pressure difference that holds the drop spherical.

Extra depth. The excess pressure depends inversely on the radius: a 0.3 mm mercury droplet would have an excess pressure ten times larger (3100 Pa). Mercury has a high S (0.465 N m⁻¹ compared with 0.073 for water), so its drops are very nearly spherical when small and are flattened by gravity when large (ρgh = 13.6×10³ × 9.8 × 3×10⁻³ = 400 Pa, the same order as 2S/r, so a 3 mm mercury drop is already slightly flattened).

More to know. Why do small drops tend to coalesce into bigger ones? The pressure inside a small drop is higher than inside a large one (2S/r is larger for small r), so if two drops of different size are connected by a tube, air or liquid flows from the smaller to the larger one — the small one shrinks and the large one grows. This is why a tiny drop of mercury runs into a larger drop when they touch. Worked comparison for mercury (S = 0.465 N m⁻¹): r = 3.00 mm gives 310 Pa, r = 1.00 mm gives 930 Pa, r = 0.100 mm gives 9300 Pa. Only the tiniest drops have an excess pressure that is a measurable fraction of the atmospheric pressure.

Conclusion: P_i = P_a + 2S/r = 1.0131×10⁵ Pa; excess pressure = 2S/r = 310 Pa.

Concept. A soap bubble in air has two liquid-air surfaces: excess pressure 4S/r. An air bubble formed inside a liquid has only one surface: excess pressure 2S/r, and the pressure outside it is P_a + ρgh.

Question. What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of the soap solution at 20°C is 2.50×10⁻² N m⁻¹? If an air bubble of the same dimension were formed at a 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 atmosphere = 1.01×10⁵ Pa.)

Part 1 — soap bubble in air (two surfaces)

P_i − P_o = 4S/r = (4 × 2.50×10⁻²)/(5.00×10⁻³) = 0.10/(5.00×10⁻³) = 20 Pa

Part 2 — air bubble at 40.0 cm depth (one surface)

• Density of the soap solution ρ = 1.20 × 10³ kg m⁻³; depth h = 0.400 m
• Pressure outside the bubble P_o = P_a + ρgh = 1.01×10⁵ + (1.20×10³)(9.8)(0.400) = 1.01×10⁵ + 4704 = 1.0570×10⁵ Pa
• Excess pressure for a single surface: 2S/r = (2 × 2.50×10⁻²)/(5.00×10⁻³) = 10 Pa

P_i = P_o + 2S/r = 1.0570×10⁵ + 10 = 1.06×10⁵ Pa

So the excess pressure inside the soap bubble in air is 20 Pa, and the pressure inside the air bubble at 40.0 cm depth in the soap solution is 1.06×10⁵ Pa (1.0571×10⁵ Pa), of which only 10 Pa is due to surface tension and about 4.7×10³ Pa due to the depth of the liquid.

Extra depth. Note how small the surface-tension contribution (10 Pa) is compared with the hydrostatic part (4700 Pa) for a bubble of 5 mm radius. The bubble of Example 9.10 (r = 1 mm at 8 cm depth) is similar: 146 Pa from surface tension, 784 Pa from depth. A soap bubble has two surfaces (the inner and the outer of the thin film), so its excess pressure is double that of an air bubble in liquid of the same radius and S.

Conclusion: Soap bubble in air: 4S/r = 20 Pa. Air bubble at 40 cm depth in the solution: P_i = P_a + ρgh + 2S/r = 1.06×10⁵ Pa.