NCERT Solutions Class 11 Physics Chapter 9 — Mechanical Properties of Fluids

NCERT Class 11 Physics — Mechanical Properties of Fluids, Chapter 9 Exercises. All 20 questions solved.

The exercises at the end of Chapter 9 fall into three groups. Questions 9.1 and 9.5 to 9.10 are about pressure in a fluid at rest: how it grows with depth, how a barometer or a U-tube balances columns of liquid, and how a hydraulic lift passes it on. Questions 9.4 and 9.11 to 9.16 are about fluids in motion — the equation of continuity, Bernoulli’s principle and viscosity. Questions 9.2, 9.3 and 9.17 to 9.20 are about surface tension. Almost all of it runs on three sets of results:

• Pressure in a fluid at rest — $P = P_a + \rho g h$ at depth $h$, the same at every point at the same level; and by Pascal’s law, a pressure applied to an enclosed liquid is transmitted undiminished to every part of it.
• Fluids in motion — the equation of continuity $A_1v_1 = A_2v_2$ for an incompressible fluid, and Bernoulli’s principle, $P + \tfrac12\rho v^2 + \rho g h = \text{constant}$ along a streamline in steady, non-viscous flow.
• Surface tension — $S$ is the force per unit length along any line in a liquid surface; the pressure inside a drop or an air bubble exceeds that outside by $2S/r$, and inside a soap bubble, which has two surfaces, by $4S/r$.

Throughout, $g = 9.8\ \text{m s}^{-2}$ and the density of water is $1.00\times10^3\ \text{kg m}^{-3}$.

Key insight. Nearly every question here is a balance. In a fluid at rest, pressure depends only on depth, so two points at the same level in one connected liquid are at the same pressure — that one fact solves the barometer (9.6), the U-tubes (9.9, 9.10) and the hydraulic lift (9.8). In a moving fluid, Bernoulli’s principle trades pressure for speed: where the fluid moves faster, its pressure is lower (9.4, 9.14, 9.15). And a liquid surface pulls along every edge with a force $S$ per unit length, counted once for each surface — twice for a film (9.17, 9.18) or a soap bubble (9.20).

Question 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 $6\ \text{km}$ decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than $100\ \text{km}$ (c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

Question 9.1 (a)

Solution. Blood is a liquid, and the pressure in a liquid increases with depth by $\rho g h$. In a standing person the feet are roughly $1.5\ \text{m}$ below the brain, so the blood in the feet carries the extra weight of a column of blood about that tall. With $\rho \approx 1.06\times10^3\ \text{kg m}^{-3}$ for blood, $\rho g h \approx 1.06\times10^3\times9.8\times1.5 \approx 1.6\times10^4\ \text{Pa}$, about $120\ \text{mm}$ of mercury — a large part of the pressure itself. Lying down removes the difference, which is why blood pressure is measured on the arm, level with the heart.

Question 9.1 (b)

Solution. Air is compressible. The pressure at any height is the weight, per unit area, of all the air above that height, and that weight squeezes the lower layers: the air is densest at the ground and thins out rapidly with height. So the mass of the atmosphere is not spread evenly over its $100\ \text{km}$; about half of it is packed into the lowest $6\ \text{km}$ or so. Above that height there is only half the air left, and so only half the pressure. The formula $P = P_a + \rho g h$, with constant $\rho$, works for liquids but not for a gas whose density changes with height.

Question 9.1 (c)

Solution. At a point in a fluid at rest, the pressure is the same in every direction. Place a small surface there in any orientation and the fluid pushes on it with a force normal to it, of the same size per unit area whichever way it faces. The direction of that force is set entirely by the surface you choose, not by the pressure: in $\vec F = P\,\vec A$, where $\vec A$ is the area vector along the normal, the pressure is simply the number that turns the area vector into the force vector. Pressure itself has a magnitude but no direction of its own, so it is a scalar.

Question 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. (Put differently, 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.

Question 9.2 (a)

Solution. Where a liquid meets a solid and air, three surface tensions meet along the edge of the liquid: liquid–air $S_{la}$, solid–liquid $S_{sl}$ and solid–air $S_{sa}$. Balancing them along the solid surface gives $S_{la}\cos\theta + S_{sl} = S_{sa}$, so

$$\cos\theta = \frac{S_{sa} – S_{sl}}{S_{la}}$$

For water on glass, the attraction between water and glass molecules (adhesion) is strong, $S_{sl}$ is less than $S_{sa}$, $\cos\theta$ is positive and $\theta$ is acute. For mercury, the attraction between mercury atoms (cohesion) is far stronger than their attraction to glass, $S_{sl}$ exceeds $S_{sa}$, $\cos\theta$ is negative and $\theta$ is obtuse, about $140^\circ$.

Question 9.2 (b)

Solution. This is the same fact seen from the side. Water molecules are attracted to glass more strongly than to one another, so a water–glass interface costs less energy than the glass–air surface it replaces, and water lowers its energy by spreading to cover more glass. Mercury atoms attract one another much more strongly than they attract glass, so mercury keeps its contact with the glass as small as it can and pulls itself into drops.

Question 9.2 (c)

Solution. Surface tension is the extra energy per unit area of surface, or equally the force per unit length along a line in the surface. It arises because each molecule in the surface has fewer neighbours than one inside, and so a little more energy. Enlarging the surface brings more molecules up from the interior, each carrying the same extra energy, so the surface energy grows in proportion to the area and the energy per unit area stays the same. This is unlike a stretched rubber sheet, whose tension rises as it is stretched, because there the molecules themselves are pulled further apart.

Question 9.2 (d)

Solution. Detergents are added so that water wets the things it is meant to clean — cloth fibres and greasy surfaces. Detergent molecules gather at the surfaces and reduce the surface tension of water and of the water–solid interface, which raises $\cos\theta = (S_{sa} – S_{sl})/S_{la}$ and so makes $\theta$ small. With a small angle of contact the water spreads over the surface and soaks into the narrow spaces between fibres, where it can reach and lift away the dirt.

Question 9.2 (e)

Solution. Surface tension means a surface has energy in proportion to its area, so a liquid left to itself takes the shape of least surface area. For a given volume, the shape with the least surface area is a sphere. Normally gravity flattens a drop against whatever it rests on; with no external force there is nothing to oppose surface tension, and the drop is spherical.

Question 9.3

Fill in the blanks using the word(s) from the list appended with each statement: (a) Surface tension of liquids generally … with temperatures (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 for turbulence for an actual plane (greater / smaller)

Question 9.3 (a)

Solution. As the temperature rises, the molecules move more vigorously and the cohesive pull that holds the surface together becomes less effective, so surface tension falls; it reaches zero at the critical temperature.

Question 9.3 (b)

Solution. In a liquid, viscosity comes from the attraction between neighbouring molecules; heating makes the molecules more mobile and the liquid flows more easily. In a gas, viscosity comes from molecules carrying momentum between layers moving at different speeds, and heating makes that random motion faster, so the gas becomes more viscous.

Question 9.3 (c)

Solution. A solid resists being sheared by an amount: its modulus of rigidity is shear stress divided by shear strain, so the force is proportional to the strain. A fluid cannot hold a shear at all; it keeps flowing, and its viscosity is shear stress divided by the rate at which the strain changes, so the force is proportional to the rate of shear strain.

Question 9.3 (d)

Solution. In steady flow of an incompressible fluid the same mass passes every cross-section each second, so $Av$ is constant and a smaller area forces a larger speed. That is conservation of mass, the equation of continuity. Bernoulli’s principle then tells us what happens to the pressure at the constriction — it falls — but the speed increase itself comes from mass conservation.

Note on the book’s answer. The key prints “conservation of mass, Bernoulli’s equation”, naming both. The statement has one blank, and the speed increase follows from conservation of mass; Bernoulli’s equation accounts for the accompanying drop in pressure, which is presumably why the key names it too.

Question 9.3 (e)

Solution. Whether a flow becomes turbulent depends on the dimensionless combination $\rho v l/\eta$ (the Reynolds number), where $l$ is a length that sets the scale of the flow. Turbulence sets in when this number passes a critical value. A model is smaller than the real plane, so $l$ is smaller, and in the same air the speed has to be greater to reach the same value.

Question 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

Question 9.4 (a)

Solution. Held at one edge, a sheet of paper droops under its own weight. Blowing over it makes the air above move fast, and by Bernoulli’s principle fast-moving air is at a lower pressure. The still air underneath is at atmospheric pressure, so the pressure difference pushes the paper up, holding it horizontal. Blowing underneath would lower the pressure below the paper and make it droop further.

Question 9.4 (b)

Solution. The volume of water flowing out each second is set by the supply, not by our fingers. By the equation of continuity, $Av$ stays the same, so when the fingers leave only narrow gaps for the water, its speed through them must rise in proportion. The water comes out as fast jets.

Question 9.4 (c)

Solution. By Bernoulli’s principle, a pressure difference $\Delta P$ drives the liquid out of the needle at a speed $v = \sqrt{2\Delta P/\rho}$, so the flow rate is $Q = Av = A\sqrt{2\Delta P/\rho}$. The flow rate is proportional to the needle’s cross-section but only to the square root of the pressure: doubling the thumb pressure increases the flow by a factor of only $\sqrt2$, while the choice of needle changes it directly. So the needle size is the effective control, and the thumb only fine-tunes it.

Question 9.4 (d)

Solution. The fluid leaves through the hole with a speed $v$, carrying momentum away with it at the rate $\rho Av^2$ each second. Something must push the fluid out to give it that momentum, and that push comes from the vessel, through the pressure of the fluid inside. In terms of pressure: everywhere else the pressure on opposite walls balances, but at the hole there is no wall for the fluid to press on, so the pressure on the opposite wall is unbalanced. By Newton’s third law the fluid pushes the vessel backwards with a force $\rho Av^2$.

Question 9.4 (e)

Solution. A parabola is the path of a body acted on by gravity alone. A spinning ball drags a thin layer of air round with it. On one side the ball’s surface moves in the same direction as the air streaming past it, and the air there moves faster; on the other side it moves against the stream, and the air is slowed. By Bernoulli’s principle the pressure is lower on the faster side, so there is a net sideways force on the ball — the Magnus effect. With a force other than gravity acting, the path is no longer a parabola, and the ball swings or dips.

Question 9.5

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

Solution. Balancing on one heel, the girl’s whole weight $mg = 50\times9.8 = 490\ \text{N}$ rests on the heel. Its radius is half the diameter, $0.5\ \text{cm} = 5.0\times10^{-3}\ \text{m}$, so its area is $A = \pi r^2 = \pi(5.0\times10^{-3})^2 = 7.85\times10^{-5}\ \text{m}^2$:

$$P = \frac{F}{A} = \frac{490}{7.85\times10^{-5}} = 6.2\times10^{6}\ \text{Pa}$$

That is about sixty times atmospheric pressure, which is why stiletto heels dent soft floors.

Question 9.6

Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density $984\ \text{kg m}^{-3}$. Determine the height of the wine column for normal atmospheric pressure.

Solution. In a barometer the space above the liquid column is empty, so the column is held up by the atmosphere alone: the pressure $\rho g h$ at its foot equals atmospheric pressure. With $P_a = 1.013\times10^5\ \text{Pa}$:

$$h = \frac{P_a}{\rho g} = \frac{1.013\times10^5}{984\times9.8} = 10.5\ \text{m}$$

Wine is about $14$ times less dense than mercury, so the column is about $14$ times taller than mercury’s $76\ \text{cm}$.

Question 9.7

A vertical off-shore structure is built to withstand a maximum stress of $10^9\ \text{Pa}$. 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 $3\ \text{km}$, and ignore ocean currents.

Solution. The greatest pressure the structure meets is at its base, on the sea floor. Taking the density of sea water as $1.03\times10^3\ \text{kg m}^{-3}$:

$$\rho g h = 1.03\times10^3\times9.8\times3000 = 3.0\times10^7\ \text{Pa}$$

Atmospheric pressure, $10^5\ \text{Pa}$, adds less than one per cent to this. The pressure at the sea floor is about $3\times10^7\ \text{Pa}$, some thirty times less than the $10^9\ \text{Pa}$ the structure can withstand.

Question 9.8

A hydraulic automobile lift is designed to lift cars with a maximum mass of $3000\ \text{kg}$. The area of cross-section of the piston carrying the load is $425\ \text{cm}^2$. What maximum pressure would the smaller piston have to bear?

Solution. The largest pressure under the load piston is the heaviest car’s weight spread over that piston’s area:

$$P = \frac{mg}{A} = \frac{3000\times9.8}{425\times10^{-4}} = 6.92\times10^5\ \text{Pa}$$

By Pascal’s law, pressure in the enclosed liquid is transmitted undiminished, so the smaller piston bears exactly this pressure. What is smaller at the small piston is the force, because its area is smaller; that is the whole point of a hydraulic lift.

Question 9.9

A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with $10.0\ \text{cm}$ of water in one arm and $12.5\ \text{cm}$ 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, and points at the same level in one connected liquid at rest are at the same pressure. So the pressure of the water column on one mercury surface equals the pressure of the spirit column on the other (the atmosphere presses equally on both open ends):

$$h_w\rho_w g = h_s\rho_s g \;\Rightarrow\; \frac{\rho_s}{\rho_w} = \frac{h_w}{h_s} = \frac{10.0}{12.5} = 0.800$$

The ratio $\rho_s/\rho_w$ is the specific gravity. The less dense liquid needs the taller column to exert the same pressure.

Question 9.10

In the previous problem, if $15.0\ \text{cm}$ 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 $= 13.6$)

Solution. After pouring, the water arm holds $10.0 + 15.0 = 25.0\ \text{cm}$ of water and the spirit arm $12.5 + 15.0 = 27.5\ \text{cm}$ of spirit. In centimetres of water, their pressures are $25.0$ and $27.5\times0.800 = 22.0$. The water column presses harder, so it pushes the mercury down in its own arm and up in the spirit arm, until the extra height $h$ of mercury on the spirit side makes up the difference.

Take the level of the mercury surface in the water arm. At that level in the spirit arm, the liquid above is $h$ of mercury topped by the $27.5\ \text{cm}$ of spirit. The two pressures at this level are equal, and dividing through by $\rho_w g$ leaves specific gravities:

$$25.0\times1 = 27.5\times0.800 + 13.6\,h \;\Rightarrow\; h = \frac{25.0 – 22.0}{13.6} = 0.221\ \text{cm}$$

Question 9.11

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

Solution. Bernoulli’s equation is derived for steady, streamline flow of a non-viscous, incompressible fluid, in which each particle follows a fixed streamline and no energy is lost. Water in a rapid is turbulent: the velocity at any point fluctuates wildly, eddies form and break up, and a great deal of energy is dissipated. There are no steady streamlines to apply the equation along.

Question 9.12

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

Solution. Bernoulli’s equation compares two points: $P_1 + \tfrac12\rho v_1^2 + \rho g h_1 = P_2 + \tfrac12\rho v_2^2 + \rho g h_2$. Gauge pressure is absolute pressure minus atmospheric pressure, so using gauge pressures subtracts $P_a$ from both sides. If the atmospheric pressure is the same at the two points, the same quantity comes off each side and the equation is unchanged. It would matter only if the atmospheric pressures at the two points were significantly different, as they might be for points at very different heights.

Question 9.13

Glycerine flows steadily through a horizontal tube of length $1.5\ \text{m}$ and radius $1.0\ \text{cm}$. If the amount of glycerine collected per second at one end is $4.0\times10^{-3}\ \text{kg s}^{-1}$, what is the pressure difference between the two ends of the tube? (Density of glycerine $= 1.3\times10^3\ \text{kg m}^{-3}$ and viscosity of glycerine $= 0.83\ \text{Pa s}$). [You may also like to check if the assumption of laminar flow in the tube is correct].

Solution. The pressure difference needed to drive a viscous liquid steadily through a tube is given by Poiseuille’s equation for laminar flow, a standard result that this edition of the book no longer derives. For volume flow rate $Q$ through a tube of radius $r$ and length $l$,

$$Q = \frac{\pi\,\Delta P\,r^4}{8\eta l} \quad\Rightarrow\quad \Delta P = \frac{8\eta l Q}{\pi r^4}$$

The question gives the mass flow rate, so first convert it to volume: $Q = \dfrac{4.0\times10^{-3}}{1.3\times10^3} = 3.08\times10^{-6}\ \text{m}^3\,\text{s}^{-1}$. With $r = 1.0\times10^{-2}\ \text{m}$:

$$\Delta P = \frac{8\times0.83\times1.5\times3.08\times10^{-6}}{\pi\times(1.0\times10^{-2})^4} = 9.8\times10^2\ \text{Pa}$$

Checking the flow is laminar. The average speed is $v = Q/\pi r^2 = 3.08\times10^{-6}/(3.14\times10^{-4}) = 9.8\times10^{-3}\ \text{m s}^{-1}$. The Reynolds number, based on the tube’s diameter $D = 2.0\ \text{cm}$, is

$$Re = \frac{\rho v D}{\eta} = \frac{1.3\times10^3\times9.8\times10^{-3}\times0.020}{0.83} \approx 0.3$$

Flow in a tube stays laminar for $Re$ up to about $2000$, so this very viscous, slow flow is laminar and Poiseuille’s equation applies.

Question 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 $70\ \text{m s}^{-1}$ and $63\ \text{m s}^{-1}$ respectively. What is the lift on the wing if its area is $2.5\ \text{m}^2$? Take the density of air to be $1.3\ \text{kg m}^{-3}$.

Solution. The upper and lower surfaces are at practically the same height, so Bernoulli’s principle gives $P + \tfrac12\rho v^2$ the same above and below the wing. The air moves faster over the top, so the pressure there is lower, and the pressure difference is

$$P_{\text{lower}} – P_{\text{upper}} = \tfrac12\rho\left(v_u^2 – v_l^2\right) = \tfrac12\times1.3\times(70^2 – 63^2) = 0.65\times931 = 605\ \text{Pa}$$

This difference acts upwards over the whole wing area:

$$\text{lift} = (605)(2.5) = 1.5\times10^3\ \text{N}$$

Question 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?

Solution. The liquid is incompressible, so by the equation of continuity it must move faster through the constriction, where the cross-section is smaller. The pipe is horizontal, so by Bernoulli’s principle, $P + \tfrac12\rho v^2 = \text{constant}$, the faster-moving liquid at the constriction is at a lower pressure. Each vertical tube shows the pressure beneath it by the height of its liquid column, so the column above the constriction must stand lower than the column above the wide part. Figure (b) shows exactly this. Figure (a) shows the column at the constriction standing higher, which would mean a higher pressure where the liquid moves faster.

Question 9.16

The cylindrical tube of a spray pump has a cross-section of $8.0\ \text{cm}^2$ one end of which has $40$ fine holes each of diameter $1.0\ \text{mm}$. If the liquid flow inside the tube is $1.5\ \text{m min}^{-1}$, what is the speed of ejection of the liquid through the holes?

Solution. All the liquid that flows along the tube leaves through the holes, so by the equation of continuity the volume per second is the same: $A_1v_1 = (40\,a)\,v_2$, where $a$ is the area of one hole. The speed in the tube is $v_1 = 1.5/60 = 0.025\ \text{m s}^{-1}$, the tube’s area is $A_1 = 8.0\times10^{-4}\ \text{m}^2$, and each hole has radius $0.5\ \text{mm}$, so the 40 holes together have area $40\times\pi(0.5\times10^{-3})^2 = 3.14\times10^{-5}\ \text{m}^2$:

$$v_2 = \frac{A_1v_1}{40\,a} = \frac{(8.0\times10^{-4})(0.025)}{3.14\times10^{-5}} = 0.64\ \text{m s}^{-1}$$

Question 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 $1.5\times10^{-2}\ \text{N}$ (which includes the small weight of the slider). The length of the slider is $30\ \text{cm}$. What is the surface tension of the film?

Solution. The film pulls on the slider along its whole length, with a force $S$ per unit length for each surface. A soap film has two surfaces, front and back, so the total upward pull on the slider is $2SL$, and in equilibrium this balances the weight:

$$2SL = W \;\Rightarrow\; S = \frac{W}{2L} = \frac{1.5\times10^{-2}}{2\times0.30} = 2.5\times10^{-2}\ \text{N m}^{-1}$$

Question 9.18

Figure 9.21 (a) shows a thin liquid film supporting a small weight $= 4.5\times10^{-2}\ \text{N}$. 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.

Solution. As in Question 9.17, the film holds up the weight by pulling on the slider along its length, and the pull is $2SL$: $S$ per unit length for each of the film’s two surfaces, along the slider’s length $L$. The weight supported therefore depends only on $S$ and $L$. The liquid and temperature are the same, so $S$ is the same, and the slider is $40\ \text{cm}$ long in all three frames. The height of the film in (b) and the shape of its top in (c) change the area of the film but not the force on the slider.

Physically, this is because surface tension does not depend on the area of the surface (Question 9.2(c)). A liquid film is not like a rubber sheet, whose tension grows as it is stretched; enlarging a film only brings more molecules into its surface, and the pull along each unit length of its edge stays $S$. (For this liquid, $S = 4.5\times10^{-2}/(2\times0.40) = 5.6\times10^{-2}\ \text{N m}^{-1}$.)

Question 9.19

What is the pressure inside the drop of mercury of radius $3.00\ \text{mm}$ at room temperature? Surface tension of mercury at that temperature ($20\ ^\circ\text{C}$) is $4.65\times10^{-1}\ \text{N m}^{-1}$. The atmospheric pressure is $1.01\times10^5\ \text{Pa}$. Also give the excess pressure inside the drop.

Solution. A drop has a single surface, and the curved surface pulled tight by surface tension squeezes the liquid inside, raising its pressure above the outside by $2S/r$:

$$P_{\text{excess}} = \frac{2S}{r} = \frac{2\times4.65\times10^{-1}}{3.00\times10^{-3}} = 310\ \text{Pa}$$

The pressure inside is atmospheric pressure plus this:

$$P = 1.01\times10^5 + 310 = 1.0131\times10^5\ \text{Pa}$$

The data are given to three significant figures, so the total pressure is properly quoted as $1.01\times10^5\ \text{Pa}$: the excess is too small to show at that precision.

Question 9.20

What is the excess pressure inside a bubble of soap solution of radius $5.00\ \text{mm}$, given that the surface tension of soap solution at the temperature ($20\ ^\circ\text{C}$) is $2.50\times10^{-2}\ \text{N m}^{-1}$? If an air bubble of the same dimension were formed at depth of $40.0\ \text{cm}$ inside a container containing the soap solution (of relative density $1.20$), what would be the pressure inside the bubble? ($1$ atmospheric pressure is $1.01\times10^5\ \text{Pa}$).

Solution. The soap bubble in air is a thin film with two surfaces, inner and outer, each contributing $2S/r$, so the excess pressure inside is $4S/r$:

$$\frac{4S}{r} = \frac{4\times2.50\times10^{-2}}{5.00\times10^{-3}} = 20.0\ \text{Pa}$$

The air bubble inside the solution has only one surface, the boundary between the air and the liquid, so its excess pressure is half as much:

$$\frac{2S}{r} = \frac{2\times2.50\times10^{-2}}{5.00\times10^{-3}} = 10.0\ \text{Pa}$$

The pressure just outside the air bubble is the pressure in the solution at a depth of $40.0\ \text{cm}$. The solution’s density is $1.20\times10^3\ \text{kg m}^{-3}$:

$$P_{\text{out}} = P_a + \rho g h = 1.01\times10^5 + 1.20\times10^3\times9.8\times0.400 = 1.01\times10^5 + 4.70\times10^3 = 1.057\times10^5\ \text{Pa}$$

The pressure inside is $10.0\ \text{Pa}$ more, $1.0571\times10^5\ \text{Pa}$. To three significant figures the excess makes no difference.

Common mistakes

• Question 9.5: using the diameter as the radius. The heel’s $1.0\ \text{cm}$ is its diameter, so $r = 0.5\ \text{cm}$. Using $1.0\ \text{cm}$ makes the area four times too large and the pressure four times too small, $1.6\times10^6\ \text{Pa}$.
• Question 9.8: expecting the smaller piston to bear a larger pressure. Pascal’s law transmits the pressure undiminished, so both pistons feel the same $6.92\times10^5\ \text{Pa}$. It is the force that is smaller on the small piston, and students who carry over the idea of “force multiplication” apply it to pressure instead.
• Question 9.10: making mercury rise in the water arm. The spirit column is taller, so it looks heavier. But $27.5\ \text{cm}$ of spirit exerts the pressure of only $22.0\ \text{cm}$ of water, less than the $25.0\ \text{cm}$ of water opposite, so the mercury is pushed down in the water arm and rises in the spirit arm.
• Question 9.13: putting the mass flow rate into Poiseuille’s equation. The equation needs the volume flow rate, so $4.0\times10^{-3}\ \text{kg s}^{-1}$ must first be divided by the density. And the radius enters as $r^4$: using the diameter instead makes the answer sixteen times too small.
• Question 9.14: squaring the difference of speeds. Bernoulli’s principle needs $v_u^2 – v_l^2 = 931\ \text{m}^2\,\text{s}^{-2}$, not $(v_u – v_l)^2 = 49\ \text{m}^2\,\text{s}^{-2}$. The shortcut gives a lift of only about $80\ \text{N}$.
• Question 9.16: forgetting that there are 40 holes. The liquid leaves through all of them at once, so the outlet area is 40 times one hole’s area; using a single hole gives an ejection speed 40 times too high.
• Questions 9.17 and 9.18: counting only one surface of the film. A film has a front and a back surface, and each pulls on the slider, so the force is $2SL$. With $SL$ the surface tension in 9.17 comes out twice too large, $5.0\times10^{-2}\ \text{N m}^{-1}$.
• Question 9.20: mixing up $4S/r$ and $2S/r$, or forgetting the depth. The soap bubble has two surfaces and the air bubble one, so their excess pressures differ by a factor of two. And the air bubble sits $40.0\ \text{cm}$ down, where the pressure outside it is already $4.70\times10^3\ \text{Pa}$ above atmospheric.

Practise next

• Chapter 10, Thermal Properties of Matter — the next chapter, where temperature, which changes surface tension and viscosity as 9.3 shows, becomes the subject.
• Chapter 8, Mechanical Properties of Solids — stress, strain and the modulus of rigidity that 9.3(c) contrasts with viscosity.
• Chapter 5, Work, Energy and Power — Bernoulli’s principle is the work–energy theorem applied to a flowing fluid.