9.1 Introduction
Chapter 9 — Mechanical Properties of Fluids
In the 1640s, the French scientist Blaise Pascal performed a demonstration that looked almost impossible. He took a strong wooden barrel, filled it completely with water, and sealed it shut. Then he fixed a very thin, tall pipe into the top of the barrel and climbed up to pour water into that pipe from a height.
The barrel did not hold a lot of extra water — the thin pipe could take only a small amount. Yet as the water in the pipe rose higher and higher, something dramatic happened: the sturdy barrel burst apart.
How could a thin column of water, weighing far less than the barrel itself, split open a solid barrel? The secret was not the amount of water, but the height of the water column. The tall pipe created a large pressure at the bottom, and — as Pascal realised — that pressure was passed on equally to every part of the sealed liquid, pushing on the barrel walls from inside until they gave way.
This single idea, that a fluid at rest transmits pressure throughout itself, quietly runs the modern world. It lifts cars in a garage, stops a speeding truck, and lets a submarine dive to crushing depths. The behaviour of fluids — liquids and gases that can flow — shapes weather, flight, blood circulation, and much more.
Figure to come
Fig. 9.0 – Pascal’s barrel experiment: a sealed water-filled barrel with a tall thin vertical pipe rising from its top, water poured into the pipe, and cracks bursting outward on the barrel walls.
By the end of this chapter, you will be able to answer each of these questions using the physics of fluids at rest and in motion.
In this chapter, we study some common physical properties of liquids and gases. The most striking thing that liquids and gases share is that both of them can flow from one place to another and take up the shape of whatever holds them. Because of this ability to flow, liquids and gases are together called fluids. This single property is what separates them, in a basic way, from solids.
Fluids are all around us, at every moment. The Earth is wrapped in a thick blanket of air, and nearly two-thirds of its surface is covered with water. Water is not just something we drink to stay alive — the body of almost every mammal, including our own, is made mostly of water.
In fact, nearly every life process, in animals as well as in plants, is carried out with the help of fluids. Blood carries oxygen through our body, sap rises through a tree, and air moves in and out of our lungs. For all these reasons, understanding how fluids behave is genuinely important.
Now let us ask the two questions that shape this whole chapter: How is a fluid different from a solid, and what do liquids and gases have in common with each other?
The first difference is about shape. Unlike a solid, a fluid has no definite shape of its own — it simply takes the shape of its container. A solid, on the other hand, holds its own shape.
The next point is about volume. Solids and liquids have a fixed volume, but a gas has no fixed volume — it spreads out and fills the entire space of whatever container it is placed in.
Figure to come
Fig. 9.1a – Three identical containers: one holding a solid block keeping its own shape, one holding a liquid that takes the container’s base shape but keeps a level top surface, and one filled completely by a gas.
To go further, we need one idea carried over from the previous chapter: stress. Stress is simply the force acting on each unit of area of a material. When this force acts along the surface, trying to slide one layer over another, it is called shear stress. When it acts squeezing the material inward from all sides, it changes the material’s volume.
We learnt earlier that the volume of a solid can be changed by applying such stress. The same is true for liquids and gases: the volume of a solid, liquid, or gas depends on the stress or pressure acting on it.
So when we say a solid or a liquid has a “fixed” volume, we really mean its volume under normal atmospheric pressure — the ordinary pressure of the air around us. The full meaning of pressure is developed in the next section; for now, think of it as the push the surroundings apply on the material.
The real difference lies in how much the volume changes when this outside pressure changes. For solids and liquids, squeezing them harder changes their volume only very slightly. For gases, the same change in pressure changes the volume a great deal.
We describe this using compressibility — how easily a material’s volume can be reduced by pressure. Solids and liquids have very low compressibility, while gases have high compressibility.
There is one more, deeper property that truly defines a fluid — its response to shear stress. Recall that a shear stress tries to slide one layer of a material past the next.
A solid resists this strongly: a shear stress can change the shape of a solid while keeping its volume fixed, but the solid pushes back and holds a definite form. A fluid behaves very differently.
The key property of fluids is that they offer very little resistance to shear stress. Even a very small shear stress is enough to make a fluid change its shape and keep flowing. Put in numbers, the shearing stress a fluid can bear is about a million times smaller than that of a solid.
This is the deep reason a fluid “flows” while a solid does not. It is also the starting point for everything that follows in this chapter — from fluids sitting at rest under pressure, to fluids moving through pipes, to the thin films and droplets held together at a liquid’s surface.
9.2 Pressure
A sharp needle pressed against our skin pierces it easily. Yet if we press the back of a spoon against our skin with the same force, the skin stays unbroken. The force is identical in both cases — so what makes the difference?
The answer is the area over which the force acts. The needle touches the skin over a tiny area, while the spoon spreads the same force over a much wider area. A smaller area concentrates the force, and that is what breaks the skin.
Everyday life is full of such examples. If an elephant were to step on a person’s chest, the ribs would crack. But a circus performer can survive a much heavier load if a large, strong wooden plank is first placed across the chest, so that the weight is spread over a wide area.
These experiences tell us something clear: the effect of a force depends not only on how big the force is, but also on the area over which it is applied. The smaller the area on which a given force acts, the greater is its effect. This effect is what we call pressure.
Now let us look at pressure inside a fluid. When an object is submerged in a fluid that is at rest, the fluid pushes on the object’s surface with a force. An important fact is that this force always acts normal — that is, perpendicular — to the object’s surface, as shown in Fig. 9.1(a).
Figure to come
Fig. 9.1a – A solid object submerged in a beaker of liquid, with small arrows on its surface and on the beaker walls all pointing perpendicular (normal) to those surfaces.
Why must the force be perpendicular? Suppose the fluid pushed on the surface at a slant, so that part of the force ran along the surface (a sideways, or parallel, component). By Newton’s third law — every action has an equal and opposite reaction — the object would then push back on the fluid sideways too. That sideways push would make the fluid flow along the surface. But the fluid is at rest and is not flowing. So no sideways force can exist. Only the perpendicular part is allowed, which is why the fluid’s force at rest is always normal to the surface it touches.
This normal force can actually be measured. Fig. 9.1(b) shows an idealised device for doing so. It is a sealed, evacuated (air-removed) chamber fitted with a piston attached to a spring that has been calibrated to read off force. When this device is placed at a point inside the fluid, the fluid pushes the piston inward, and the spring pushes back outward. When the two balance, the spring reading gives the force the fluid exerts on the piston.
Figure to come
Fig. 9.1b – A small evacuated cylindrical chamber with a piston of area A on one face, connected to a calibrated spring inside; the fluid outside pushes the piston in.
Let \(F\) be the size of this normal force acting on a piston of area \(A\). The average pressure \(P_{av}\) is then defined as the normal force acting per unit area:
\[P_{av} = \frac{F}{A} \tag{9.1}\]
Here \(F\) is the normal force in newtons (N), \(A\) is the area in square metres (m\(^2\)), and \(P_{av}\) is the average pressure.
In principle, we can make the piston area smaller and smaller to find the pressure at a single point rather than an average over a large area. As the area shrinks towards zero, we write the pressure in a limiting sense:
\[P = \lim_{\Delta A \to 0} \frac{\Delta F}{\Delta A} \tag{9.2}\]
Here \(\Delta F\) is the small normal force acting on a very small patch of area \(\Delta A\), and \(P\) is the pressure right at that point.
An important point about pressure: pressure is a scalar quantity. It has magnitude but no direction. This may seem surprising, because force — which appears in the numerator — is a vector. The key is that the numerator is only the component of force perpendicular to the area, not the full vector force. So no single direction can be assigned to pressure itself; it acts on any area regardless of how that area is turned.
The dimensions of pressure are \([ML^{-1}T^{-2}]\). Its SI unit is newton per square metre, N m\(^{-2}\), which has been given the special name pascal (Pa) in honour of the French scientist Blaise Pascal (1623–1662), who carried out pioneering studies on fluid pressure.
A common larger unit is the atmosphere (atm) — the pressure the atmosphere exerts at sea level:
\[1 \text{ atm} = 1.013 \times 10^{5} \text{ Pa}\]
Another quantity that is indispensable in describing fluids is density, written with the symbol \(\rho\) (the Greek letter “rho”). For a fluid of mass \(m\) occupying a volume \(V\), the density is:
\[\rho = \frac{m}{V} \tag{9.3}\]
Here \(m\) is the mass in kilograms (kg) and \(V\) is the volume in cubic metres (m\(^3\)).
The dimensions of density are \([ML^{-3}]\), and its SI unit is kg m\(^{-3}\). Density is a positive scalar quantity.
Recall from the introduction that a liquid is largely incompressible. Because its volume hardly changes when pressure changes, its density stays nearly constant at all pressures. A gas behaves very differently — since a gas is easily compressed, its density varies a great deal as the pressure on it changes.
As a reference value, the density of water at \(4\,^\circ\)C (\(277\) K) is: \[\rho_{water} = 1.0 \times 10^{3} \text{ kg m}^{-3}\]
Using water as a standard, we define the relative density of a substance. It compares how heavy a material is, volume for volume, against water.
For example, the relative density of aluminium is \(2.7\). This means aluminium is \(2.7\) times as dense as water, so its density is: \[\rho_{aluminium} = 2.7 \times 10^{3} \text{ kg m}^{-3}\]
The densities of some common fluids are listed in Table 9.1.
Table 9.1 — Densities of some common fluids at STP*
| Fluid | \(\rho\) (kg m\(^{-3}\)) |
|---|---|
| Water | \(1.00 \times 10^{3}\) |
| Sea water | \(1.03 \times 10^{3}\) |
| Mercury | \(13.6 \times 10^{3}\) |
| Ethyl alcohol | \(0.806 \times 10^{3}\) |
| Whole blood | \(1.06 \times 10^{3}\) |
| Air | \(1.29\) |
| Oxygen | \(1.43\) |
| Hydrogen | \(9.0 \times 10^{-2}\) |
| Interstellar space | \(\approx 10^{-20}\) |
* STP means standard temperature (\(0\,^\circ\)C) and 1 atm pressure.
9.2.1 Pascal’s Law
The French scientist Blaise Pascal made a simple but powerful observation about fluids at rest. He found that in a fluid at rest, the pressure is the same at all points that lie at the same height (the same depth below the surface).
Let us see why this is true, and along the way discover a second important fact: that the pressure at a point acts equally in all directions.
Consider a very small element taken from somewhere inside a fluid at rest, shaped like a right-angled prism and labelled ABC–DEF, as shown in Fig. 9.2. The triangular faces ABC and DEF form the two ends, and the three flat rectangular faces run between them.
Figure to come
Fig. 9.2 – A small right-angled triangular prism ABC–DEF inside a fluid, with normal forces \(F_a\), \(F_b\), \(F_c\) pushing inward on its three rectangular faces, and the angle \(\theta\) marked in the triangular cross-section.
We deliberately choose this element to be extremely small. Because it is so tiny, every part of it lies at practically the same depth below the liquid surface, so gravity affects all parts of it equally and its effect can be ignored in comparing the pressures. (In the figure the prism has been drawn enlarged only so that we can see it clearly.)
The forces acting on this element come from the surrounding fluid pressing on its faces. As we established earlier, in a fluid at rest these forces are always normal (perpendicular) to each face. So the fluid exerts pressures \(P_a\), \(P_b\) and \(P_c\) on the three rectangular faces BEFC, ADFC and ADEB, with corresponding normal forces \(F_a\), \(F_b\) and \(F_c\). Let the areas of these three faces be \(A_a\), \(A_b\) and \(A_c\) respectively.
Since the element is at rest, it is in equilibrium — the forces on it must balance. Resolving the slanted force \(F_b\) into its horizontal and vertical parts and setting them against \(F_c\) and \(F_a\) gives the two balance conditions:
\[F_b \sin\theta = F_c, \qquad F_b \cos\theta = F_a \quad \text{(by equilibrium)}\]
Next we use the geometry of the prism. The areas of the three rectangular faces are proportional to the three sides of the right-angled triangle, so the same angle \(\theta\) relates them by simple trigonometry:
\[A_b \sin\theta = A_c, \qquad A_b \cos\theta = A_a \quad \text{(by geometry)}\]
Now divide each force equation by the matching area equation. The \(\sin\theta\) and \(\cos\theta\) factors cancel, leaving:
\[\frac{F_b}{A_b} = \frac{F_c}{A_c} = \frac{F_a}{A_a}; \qquad P_b = P_c = P_a \tag{9.4}\]
Since force divided by area is just pressure, this says the pressures on all three faces are equal, even though the faces point in different directions.
Hence the pressure exerted at a point in a fluid at rest is the same in all directions. This reminds us once again that pressure, unlike a force, has no direction of its own — it is a scalar. The force on any area inside or bounding a fluid at rest is normal to that area, no matter how the area is oriented.
Now let us prove the first part of Pascal’s observation — that pressure is the same at all points at the same height. Imagine a fluid element in the shape of a horizontal bar of uniform cross-section, lying inside the fluid at rest, as in Fig. 9.2b.
Figure to come
Fig. 9.2b – A horizontal cylindrical bar of fluid inside a container, with equal and opposite horizontal forces from the surrounding fluid pushing on its two flat end faces.
This bar is in equilibrium, so the horizontal forces on it must balance. The only horizontal forces are those pushing on its two end faces. For these to balance, the pressures at the two ends must be equal. Since the bar can be placed anywhere along a horizontal line, this shows that in a liquid in equilibrium the pressure is the same at all points in a horizontal plane.
We can also see this by reasoning about flow. Suppose the pressure were not the same across a horizontal level — say higher on one side than the other. Then there would be a net sideways force on the fluid, and the fluid would flow from the high-pressure region to the low-pressure region. But the fluid is at rest and is not flowing. Therefore the pressure must be the same everywhere in a horizontal plane.
9.2.2 Variation of Pressure with Depth
We have just seen that in a fluid at rest the pressure is the same everywhere at one horizontal level. But what happens as we move down, deeper into the fluid? Everyday experience — the growing pressure on your ears as you swim to the bottom of a pool — suggests the pressure increases with depth. Let us find the exact rule.
Consider a fluid at rest in a container. In Fig. 9.3, point 1 is at a height \(h\) above point 2, so point 2 is deeper. Let the pressures at these points be \(P_1\) and \(P_2\) respectively.
To relate them, imagine a small cylindrical column of the fluid itself, standing vertically with its top face at point 1 and its bottom face at point 2. Let this cylinder have base area \(A\) and height \(h\).
Figure to come
Fig. 9.3 – A vertical cylindrical column of fluid of base area \(A\) and height \(h\) inside a larger body of fluid; point 1 at the top face, point 2 at the bottom face, pressure forces \(P_1A\) pushing down on top, \(P_2A\) pushing up on the bottom, and weight \(mg\) acting downward.
Since the fluid is at rest, this column is in equilibrium. That means two things: the horizontal forces on it cancel out, and the vertical forces must balance the column’s own weight.
Look at the vertical forces. The fluid above presses down on the top face with a force \(P_1A\) (pressure \(\times\) area), acting downward. The fluid below presses up on the bottom face with a force \(P_2A\), acting upward. And the weight of the fluid in the column, \(mg\), pulls it down.
For balance, the upward force must equal the total downward force: \[P_2 A = P_1 A + mg\]
Rearranging gives: \[(P_2 - P_1)\, A = mg \tag{9.5}\]
Now we express the weight in terms of density. If \(\rho\) is the mass density of the fluid, then the mass of fluid in the cylinder is its density times its volume, \(m = \rho V = \rho (h A) = \rho h A\). Substituting this into Eq. (9.5): \[(P_2 - P_1)\, A = (\rho h A)\, g\]
The base area \(A\) cancels from both sides, leaving the central result: \[P_2 - P_1 = \rho g h \tag{9.6}\]
Here \(P_2 - P_1\) is the pressure difference (in Pa), \(\rho\) is the fluid’s density (in kg m\(^{-3}\)), \(g\) is the acceleration due to gravity (in m s\(^{-2}\)), and \(h\) is the vertical height difference between the two points (in m).
Notice what this equation tells us: the pressure difference depends only on the vertical distance \(h\) between the two points, the density \(\rho\) of the fluid, and \(g\). The deeper you go (larger \(h\)), the greater the pressure — which matches our experience underwater.
Now shift point 1 all the way up to the top surface of the liquid (say water), where the liquid is open to the air. At that surface the pressure is simply the atmospheric pressure \(P_a\). So we replace \(P_1\) by \(P_a\), and write the pressure at the lower point (depth \(h\)) simply as \(P\). Equation (9.6) then becomes: \[P = P_a + \rho g h \tag{9.7}\]
So the pressure at a depth below a liquid’s open surface is greater than atmospheric pressure by exactly \(\rho g h\). This total pressure \(P\) is called the absolute pressure.
The extra pressure over and above the atmosphere, that is \(P - P_a = \rho g h\), is given a special name — the gauge pressure at that point.
A striking feature of Eq. (9.7) is that the base area \(A\) has cancelled out — it does not appear at all. This means the pressure depends only on the depth \(h\), not on the cross-sectional area or the shape of the container.
The liquid pressure is therefore the same at all points that lie at the same depth (same horizontal level), whatever the shape of the vessel above them. This surprising result is beautifully shown by the hydrostatic paradox.
Consider three vessels A, B and C of completely different shapes, connected at the bottom by a horizontal pipe, as in Fig. 9.4. When water is poured in, the level rises to exactly the same height in all three — even though the three vessels hold very different amounts of water.
Figure to come
Fig. 9.4 – Three vessels A, B and C of different shapes (e.g. narrow, wide, slanted) connected at the base by a horizontal pipe, all filled with water standing at the same height.
The reason is exactly Eq. (9.7): the water at the bottom sits at the same depth below each surface, so it has the same pressure below every vessel. Equal pressure at the connecting base means the levels must match, regardless of how much water each shape contains.
9.2.3 Atmospheric Pressure and Gauge Pressure
We live at the bottom of a deep ocean of air. Just as water pressure grows with depth, the air above us presses down on everything below it. The pressure of the atmosphere at any point is equal to the weight of a column of air of unit cross-sectional area, stretching from that point all the way up to the top of the atmosphere.
At sea level this atmospheric pressure has the value: \[P_a = 1.013 \times 10^{5} \text{ Pa} = 1 \text{ atm}\]
But how can we actually measure the pressure of something as invisible as air? The first method was devised by the Italian scientist Evangelista Torricelli.
Let us see how the mercury barometer works. A long glass tube, closed at one end, is completely filled with mercury and then carefully inverted into an open trough of mercury, as shown in Fig. 9.5(a). The mercury falls a little, leaving a column standing in the tube.
Figure to come
Fig. 9.5a – A vertical glass tube closed at the top, inverted in a trough of mercury; the mercury column stands to height \(h\), with a vacuum (point A) at the top, point B at the tube bottom, and point C on the open mercury surface at the same level as B.
Above the mercury column, at point A, there is only a tiny amount of mercury vapour. Its pressure is so small that we can treat it as zero. So the pressure at point A is effectively \(P = 0\).
Now compare two points at the same horizontal level: point B, at the bottom of the mercury column inside the tube, and point C, on the open mercury surface in the trough. By Pascal’s law, points at the same level in a connected fluid at rest are at the same pressure. Point C is exposed to the air, so its pressure is atmospheric, \(P_a\). Therefore point B must also be at pressure \(P_a\).
Next, use the depth relation from the previous section. The pressure at B equals the pressure at A plus \(\rho g h\), where \(h\) is the height of the mercury column. Since the pressure at A is zero, the pressure at B is just \(\rho g h\). Setting this equal to the atmospheric pressure at the same level:
\[P_a = \rho g h \tag{9.8}\]
Here \(\rho\) is the density of mercury (in kg m\(^{-3}\)), \(g\) is the acceleration due to gravity (in m s\(^{-2}\)), and \(h\) is the height of the mercury column (in m).
Experiment shows that at sea level this mercury column stands about \(76\) cm tall, which corresponds to one atmosphere. (The same \(76\) cm can be worked out by putting the density of mercury into Eq. (9.8).)
Because the barometer reads pressure as a height of mercury, it is common to state pressures directly in centimetres or millimetres of mercury (Hg). A pressure equal to \(1\) mm of mercury is called one torr, again after Torricelli: \[1 \text{ torr} = 133 \text{ Pa}\]
The mm of Hg and the torr are widely used in medicine and physiology. In meteorology (the study of weather), a common unit is the bar and its smaller relative the millibar: \[1 \text{ bar} = 10^{5} \text{ Pa}\]
Another useful instrument is the open tube manometer, used to measure the pressure of a gas or a system. It is a U-shaped tube containing a suitable liquid, as shown in Fig. 9.5(b). A low-density liquid (such as oil) is used for measuring small pressure differences, and a high-density liquid (such as mercury) for large ones.
Figure to come
Fig. 9.5b – A U-tube manometer: one arm open to the atmosphere, the other arm connected to a vessel of gas at pressure \(P\); the liquid levels differ by height \(h\), with point A in the connected arm and point B in the open arm at the same level.
One arm of the U-tube is open to the atmosphere; the other is connected to the system whose pressure \(P\) we want to find. The pressure at point A (on the system side) equals the pressure at point B (on the open side at the same level), because the two points are at the same height in the connected fluid.
What the manometer directly gives us is the gauge pressure, \(P - P_a\), which by Eq. (9.8) is proportional to the difference in liquid levels, the manometer height \(h\).
An important reason all this works so cleanly for liquids is that liquid density stays almost constant. For liquids, the density varies very little over wide ranges of pressure and temperature, so we can safely treat it as a constant here. Gases behave differently — their densities change a great deal with pressure and temperature. Unlike gases, liquids are therefore largely treated as incompressible.
9.2.4 Hydraulic Machines
So far we have studied fluids sitting quietly at rest. Now let us ask a different question: what happens when we deliberately increase the pressure on a fluid trapped in a container?
Consider a horizontal cylinder fitted with a piston, and with three vertical tubes rising from different points along it, as in Fig. 9.6(a). The pressure inside the horizontal cylinder shows up as the height to which liquid rises in each vertical tube — and, importantly, that height is the same in all three tubes.
Figure to come
Fig. 9.6a – A horizontal cylinder with a piston at one end and three vertical open tubes at different points along the top; the liquid stands at the same level in all three tubes.
Now push the piston in. The liquid level rises in all three tubes at once, and once again the levels settle to the same height in each tube.
This simple observation tells us something powerful. When we increased the pressure at one place (by pushing the piston), that increase was not confined to that spot — it spread evenly to every part of the fluid. In other words, whenever an external pressure is applied on any part of a fluid enclosed in a vessel, it is transmitted undiminished and equally in all directions throughout the fluid.
This is another form of Pascal’s law — the transmission form — and it is the basis of many machines used in daily life.
A whole family of devices, such as the hydraulic lift and hydraulic brakes, are built on this law. In each of them, a fluid is used to carry (transmit) pressure from one place to another.
Let us work out the hydraulic lift, shown schematically in Fig. 9.6(b). Two pistons of different sizes are connected by a space filled with liquid. A small piston has cross-sectional area \(A_1\), and a large piston has a bigger area \(A_2\).
Figure to come
Fig. 9.6b – A hydraulic lift: a small piston of area \(A_1\) on the left pushed by force \(F_1\), connected through an enclosed liquid to a large piston of area \(A_2\) on the right supporting a car; the upward force on the large piston is \(F_2\).
We press down on the small piston with a force \(F_1\). This creates a pressure in the liquid directly beneath it: \[P = \frac{F_1}{A_1}\]
By Pascal’s law, this pressure \(P\) is transmitted undiminished through the liquid to the large piston. Acting over the larger area \(A_2\), the same pressure produces a much larger upward force: \[F_2 = P \times A_2 = \frac{F_1}{A_1} \times A_2 = \frac{F_1 A_2}{A_1}\]
Because \(A_2\) is larger than \(A_1\), the output force \(F_2\) is larger than the input force \(F_1\). This is why a small push on the small piston can support the large weight of a car or a truck resting on the large piston.
By varying the small force \(F_1\), we can raise or lower the platform. Comparing input and output, the force has been multiplied by the factor \(A_2 / A_1\). This factor is called the mechanical advantage of the device.
9.3 Streamline Flow
Up to now we have studied fluids sitting at rest. We now turn to fluids in motion. The study of fluids in motion is called fluid dynamics.
Think of a water tap turned on gently. At low speed, the water comes out as a smooth, glassy stream. But open the tap fully, and the stream becomes rough, broken and noisy. Clearly, how a fluid flows depends on how fast it moves. To describe flow carefully, we watch what happens to the fluid particles passing through each point in space.
We call a flow steady when, at any chosen point in space, every fluid particle arriving there has the same velocity as the particle that passed that point just before it. In other words, the velocity at a fixed point does not change with time.
This does not mean the velocity is the same everywhere. A particle can speed up, slow down, or change direction as it travels from one point to another. What “steady” requires is only that every particle reaching a given point behaves exactly like the one before it at that same point.
Because each particle then follows a smooth, well-defined path that does not cross the path of any other particle, we can map the flow using streamlines.
The path taken by a fluid particle in steady flow is called a streamline. Fig. 9.7(a) shows such a path, describing how one fluid particle moves with time from P to Q.
Figure to come
Fig. 9.7a – A smooth curved path PQ traced by a single fluid particle, with velocity arrows drawn tangent to the curve at several points.
The streamline PQ acts like a permanent map of the flow, showing the direction the fluid moves at each place. A key rule is that no two streamlines can ever cross. If they did, a fluid particle arriving at the crossing point could go off in either of two directions — which would mean two possible velocities at one point, and the flow would no longer be steady. Hence, in steady flow, this map of streamlines stays fixed in time.
Now suppose we want to picture many streamlines together. Drawing a line for every single particle would fill the region with a continuous blur of lines. Instead, we look at flat cross-sections (planes) drawn perpendicular to the flow — say at three points P, R and Q along a tube, as in Fig. 9.7(b).
Figure to come
Fig. 9.7b – A tube of flow bounded by streamlines, wider at P and R and narrower at Q, with cross-sectional areas \(A_P\), \(A_R\), \(A_Q\) marked and flow arrows through each.
We choose these cross-sections so that their edges are bounded by the same bundle of streamlines. Because streamlines never cross, no fluid leaks out sideways through this bundle. So the number of fluid particles crossing each cross-section in a given time must be the same at P, R and Q.
Let us turn this into an equation. Let the cross-sectional areas at the three points be \(A_P\), \(A_R\), \(A_Q\), and the fluid speeds there be \(v_P\), \(v_R\), \(v_Q\).
Consider a small time interval \(\Delta t\). In that time, the fluid at P advances a distance \(v_P \Delta t\). So the volume of fluid crossing P is (area \(\times\) distance) \(= A_P (v_P \Delta t)\), and its mass is density \(\times\) volume \(= \rho_P A_P v_P \Delta t\).
By the same reasoning, the mass crossing at R in time \(\Delta t\) is \(\rho_R A_R v_R \Delta t\), and the mass crossing at Q is \(\rho_Q A_Q v_Q \Delta t\).
Since no fluid is created or destroyed, the mass flowing out must equal the mass flowing in. Therefore:
\[\rho_P A_P v_P \Delta t = \rho_R A_R v_R \Delta t = \rho_Q A_Q v_Q \Delta t \tag{9.9}\]
For an incompressible fluid (such as a liquid), the density is the same everywhere: \[\rho_P = \rho_R = \rho_Q\]
Cancelling the equal densities and the common \(\Delta t\) from Eq. (9.9) leaves:
\[A_P v_P = A_R v_R = A_Q v_Q \tag{9.10}\]
This is the equation of continuity. It is simply the statement that mass is conserved in the flow of an incompressible fluid. In general we write it as:
\[A v = \text{constant} \tag{9.11}\]
The quantity \(Av\) is called the volume flux or flow rate — the volume of fluid passing a cross-section per unit time. The equation tells us this flow rate stays the same all along the pipe.
Here is the physically important consequence: since \(Av\) is constant, where the pipe is narrow (small \(A\)) the fluid must move fast (large \(v\)), and where it is wide (large \(A\)) it moves slowly. On a streamline map, this shows up as streamlines crowding close together where the fluid is fast, and spreading apart where it is slow.
Looking at Fig. 9.7(b), the tube is wider at R than at Q, so \(A_R > A_Q\), which means \(v_R < v_Q\). The fluid therefore speeds up (accelerates) as it passes from R to Q. As we will see in the next section, this change in speed is linked to a change in pressure — a fact that leads directly to Bernoulli’s principle.
So far we have described smooth, steady flow. But this only holds at low flow speeds. Beyond a certain limiting speed, called the critical speed, the flow loses its steadiness and becomes turbulent — full of irregular swirls and eddies. You can see this where a fast-flowing stream hits rocks and breaks into foamy, whirlpool-like “white-water rapids.”
Figure 9.8 shows streamlines for two typical kinds of flow. In Fig. 9.8(a), the flow is laminar: the fluid moves in smooth layers, and although the velocity may differ in magnitude at different points, the directions of flow are neatly parallel. Fig. 9.8(b) sketches turbulent flow, where a jet of air strikes a flat plate placed perpendicular to it and breaks into chaotic motion.
Figure to come
Fig. 9.8a – Smooth parallel streamlines representing laminar flow, arrows all pointing in the same general direction.
Figure to come
Fig. 9.8b – A jet of air striking a flat plate placed perpendicular to the flow, streamlines breaking into irregular swirls behind the plate, representing turbulent flow.
9.4 Bernoulli’s Principle
Fluid flow is a complicated phenomenon. But for steady, streamline flow we can extract some very useful results by applying one of the most reliable ideas in physics — the conservation of energy.
Consider an incompressible fluid flowing steadily through a pipe whose cross-sectional area changes from place to place, and which is also raised to different heights, as shown in Fig. 9.9.
Figure to come
Fig. 9.9 – A pipe of varying cross-section rising from a lower, wider region 1 (with area \(A_1\), speed \(v_1\), pressure \(P_1\), height \(h_1\)) to a higher, narrower region 2 (area \(A_2\), speed \(v_2\), pressure \(P_2\), height \(h_2\)), with the fluid element BC advancing to DE in time \(\Delta t\).
By the equation of continuity, the fluid’s speed must change as the pipe widens or narrows. A change in speed means acceleration, and acceleration requires a net force. That force comes from the surrounding fluid, which means the pressure must be different in different regions of the pipe.
Bernoulli’s equation is a general relation that connects the pressure difference between two points in a pipe to the changes in the fluid’s speed (kinetic energy) and its height (potential energy).
Let us derive it. Look at the flow at two regions of the pipe: region 1 (labelled BC) and region 2 (labelled DE). Focus on the portion of fluid that initially lies between B and D. In a very small time interval \(\Delta t\), this whole portion shifts forward.
Let \(v_1\) be the fluid speed at B and \(v_2\) the speed at D. In time \(\Delta t\), the fluid at B advances a distance \(v_1 \Delta t\) to reach C. (Since \(\Delta t\) is tiny, we can treat the cross-section along BC as constant.) In the same time, the fluid at D advances a distance \(v_2 \Delta t\) to reach E.
At the two ends, the surrounding fluid pushes with pressures \(P_1\) and \(P_2\) on the plane faces of areas \(A_1\) and \(A_2\).
Now calculate the work done. At the left end (BC), the pressure \(P_1\) pushes the fluid forward, doing positive work: \[W_1 = P_1 A_1 (v_1 \Delta t) = P_1 \Delta V\]
Here \(\Delta V = A_1 v_1 \Delta t\) is the small volume of fluid that moves. By the equation of continuity, exactly the same volume \(\Delta V\) emerges at the other end. There, the fluid must push against the outside pressure \(P_2\), so the work done on the fluid at the right end is negative: \[W_2 = P_2 A_2 (v_2 \Delta t) = P_2 \Delta V\]
So the total work done on the fluid is: \[W_1 - W_2 = (P_1 - P_2)\, \Delta V\]
This net work goes into two things: changing the fluid’s kinetic energy, and changing its gravitational potential energy. To track these, note that the mass of fluid that passes through in time \(\Delta t\) is: \[\Delta m = \rho A_1 v_1 \Delta t = \rho \Delta V\] where \(\rho\) is the fluid’s density.
The change in gravitational potential energy as this mass rises from height \(h_1\) to \(h_2\) is: \[\Delta U = \rho g \Delta V (h_2 - h_1)\]
The change in its kinetic energy, as its speed changes from \(v_1\) to \(v_2\), is: \[\Delta K = \frac{1}{2} \rho \Delta V (v_2^2 - v_1^2)\]
We now apply the work–energy theorem (from Chapter 6) to this volume of fluid: the net work done on it by the pressure forces equals the total change in its kinetic and potential energy. This gives: \[(P_1 - P_2)\, \Delta V = \frac{1}{2} \rho \Delta V (v_2^2 - v_1^2) + \rho g \Delta V (h_2 - h_1)\]
Every term contains \(\Delta V\), so we divide through by \(\Delta V\): \[(P_1 - P_2) = \frac{1}{2} \rho (v_2^2 - v_1^2) + \rho g (h_2 - h_1)\]
Rearranging so that everything about point 1 is on one side and point 2 on the other: \[P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \tag{9.12}\]
This is Bernoulli’s equation. Since points 1 and 2 can be any two locations along the same streamline, we can write it in general form: \[P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant} \tag{9.13}\]
In words: as we move along a streamline, the sum of three quantities stays constant — the pressure \(P\), the kinetic energy per unit volume \(\left(\dfrac{\rho v^2}{2}\right)\), and the potential energy per unit volume \((\rho g h)\).
Here \(P\) is the pressure (Pa), \(\rho\) is the fluid density (kg m\(^{-3}\)), \(v\) is the flow speed (m s\(^{-1}\)), \(g\) is the acceleration due to gravity (m s\(^{-2}\)), and \(h\) is the height above a reference level (m).
A very important consequence appears when the pipe is horizontal, so \(h_1 = h_2\) and the height terms cancel. Then wherever the fluid moves faster (higher \(v\)), the pressure \(P\) must be lower, and wherever it moves slower, the pressure is higher. This “fast flow means low pressure” idea explains a surprising number of everyday effects.
We must be careful about the conditions under which Bernoulli’s equation holds, since we made some idealisations in deriving it.
First, we assumed that no energy is lost to friction. In reality, when a fluid flows, its layers move at different speeds and rub against one another, exerting frictional forces. This internal friction is called viscosity (studied in detail in the next section), and it converts some of the fluid’s kinetic energy into heat. So Bernoulli’s equation applies ideally only to non-viscous fluids — fluids with zero viscosity.
Second, the fluid must be incompressible, because we did not account for any elastic energy stored by compressing the fluid.
Third, Bernoulli’s equation does not hold for non-steady or turbulent flows, because in those cases the velocity and pressure keep fluctuating in time and there is no single steady value along a streamline.
Despite these restrictions, Bernoulli’s equation works very well in practice for low-viscosity, incompressible fluids, and it explains a wide variety of natural and technological phenomena.
Finally, a useful consistency check. If the fluid is at rest — its velocity is zero everywhere — then both \(v_1\) and \(v_2\) are zero, and Bernoulli’s equation reduces to: \[P_1 + \rho g h_1 = P_2 + \rho g h_2\] \[(P_1 - P_2) = \rho g (h_2 - h_1)\]
This is exactly the pressure–depth relation of Eq. (9.6) that we found earlier for a fluid at rest. Bernoulli’s principle therefore contains our earlier hydrostatic result as a special case.
9.4.1 Speed of Efflux: Torricelli’s Law
Bernoulli’s equation lets us answer a very practical question: if a tank of liquid has a hole in its side, how fast does the liquid squirt out? The outflow of a fluid through an opening is called efflux, and the speed at which it leaves is the speed of efflux.
Torricelli found a beautifully simple result: the speed of efflux from an open tank is exactly the same as the speed a body would reach by falling freely through the same height. Let us derive this from Bernoulli’s equation.
Consider a large tank holding a liquid of density \(\rho\), with a small hole in its side at a height \(y_1\) above the bottom, as shown in Fig. 9.10. The liquid surface is at a height \(y_2\), and the air above the surface presses down with pressure \(P\).
Figure to come
Fig. 9.10 – A tank of liquid with its top surface at height \(y_2\) (area \(A_2\), pressure \(P\) above it) and a small side hole at height \(y_1\) (area \(A_1\)) from which liquid squirts out horizontally with speed \(v_1\); the depth of the hole below the surface is marked \(h = y_2 - y_1\).
Take point 1 at the hole and point 2 at the top surface. By the equation of continuity: \[v_1 A_1 = v_2 A_2 \quad \Rightarrow \quad v_2 = \frac{A_1}{A_2} v_1\]
Now, the tank is wide but the hole is tiny, so the cross-sectional area of the tank \(A_2\) is much larger than that of the hole \(A_1\) (that is, \(A_2 \gg A_1\)). From the relation above, \(v_2\) is then much smaller than \(v_1\), so we may treat the top surface as approximately at rest and set \(v_2 = 0\).
Apply Bernoulli’s equation between point 1 (the hole) and point 2 (the surface). At the hole, the liquid emerges into the open air, so its pressure there is atmospheric, \(P_1 = P_a\). At the surface, the pressure is \(P\). Using Eq. (9.12): \[P_a + \frac{1}{2}\rho v_1^2 + \rho g y_1 = P + \frac{1}{2}\rho v_2^2 + \rho g y_2\]
Putting \(v_2 = 0\) and letting \(h = y_2 - y_1\) be the depth of the hole below the surface, this rearranges to: \[\frac{1}{2}\rho v_1^2 = (P - P_a) + \rho g h\]
Solving for the efflux speed \(v_1\): \[v_1 = \sqrt{2gh + \frac{2(P - P_a)}{\rho}} \tag{9.14}\]
Here \(v_1\) is the speed of efflux (m s\(^{-1}\)), \(h\) is the depth of the hole below the surface (m), \(P\) is the pressure of the gas above the liquid and \(P_a\) the atmospheric pressure (both Pa), \(\rho\) is the liquid density (kg m\(^{-3}\)), and \(g\) is the acceleration due to gravity (m s\(^{-2}\)).
Let us look at two important special cases.
First, suppose the gas above the liquid is at a very high pressure, so that \(P \gg P_a\) and the \(2gh\) term is negligible in comparison. Then the efflux speed is set almost entirely by the container pressure. This is essentially what happens in rocket propulsion, where high-pressure gas drives the exhaust out at great speed.
Second, suppose the tank is simply open to the atmosphere. Then the pressure above the liquid is just atmospheric, \(P = P_a\), so the pressure term vanishes and Eq. (9.14) becomes: \[v_1 = \sqrt{2gh} \tag{9.15}\]
This is Torricelli’s law.
Notice that \(\sqrt{2gh}\) is exactly the speed a body reaches after falling freely from rest through a height \(h\) (a result from kinematics). So the liquid leaves the hole as though it had simply dropped through the depth \(h\) — which is why Torricelli’s law is often stated as “the speed of efflux equals that of a freely falling body.”
9.4.2 Dynamic Lift
One of the most striking uses of Bernoulli’s principle is in explaining dynamic lift — the force that pushes up on a body moving through a fluid, purely because of that motion.
We see hints of it in sport. In cricket, tennis, baseball or golf, a spinning ball is often observed to swerve away from the simple parabolic path it would otherwise follow through the air. This deviation can be explained, at least in part, using Bernoulli’s principle. Let us build up to it in two steps.
(i) A ball moving without spin. Fig. 9.11(a) shows the streamlines of air around a smooth ball that moves through the air but does not spin.
Figure to come
Fig. 9.11a – Symmetric streamlines of air flowing past a non-spinning ball, identical above and below the ball.
Because the streamlines are symmetric — the same above and below — the air moves at the same speed above and below the ball at matching points. By Bernoulli’s principle, equal speeds mean equal pressures, so there is no pressure difference between the top and bottom. The air therefore exerts no net upward or downward force on the ball.
(ii) A ball moving with spin. Now suppose the ball is also spinning. A spinning ball drags the layer of air next to its surface along with it (a rough surface drags even more air). Fig. 9.11(b) shows the streamlines for a ball that is both moving forward and spinning.
Figure to come
Fig. 9.11b – Streamlines around a ball spinning clockwise while moving forward; streamlines crowded (closer together) above the ball and spread out (rarefied) below it.
Relative to the ball, the air streams backward. On the side where the surface spin runs the same way as this airflow, the dragged air adds to the flow, so the air moves faster there; on the opposite side it moves slower. In the case drawn, the air above the ball is faster and the air below is slower. As a result, the streamlines get crowded above and spread out (rarefied) below.
By Bernoulli’s principle, faster air above means lower pressure above, and slower air below means higher pressure below. This pressure difference produces a net upward force on the ball.
This dynamic lift produced by spinning is called the Magnus effect.
Aerofoil — lift on an aircraft wing. The same idea, in a steadier form, keeps aeroplanes in the air. Fig. 9.11(c) shows an aerofoil: a solid shape designed to produce an upward dynamic lift when it moves horizontally through air. The cross-section of an aeroplane’s wing looks much like this aerofoil.
Figure to come
Fig. 9.11c – Air flowing past an aerofoil (wing cross-section); streamlines crowded together above the wing and more widely spaced below, with an upward lift force marked.
When the aerofoil moves against the oncoming air, its shape and tilt (orientation relative to the flow) make the streamlines crowd together more above the wing than below it. Crowded streamlines mean the air flows faster over the top than underneath. By Bernoulli’s principle, the faster air on top has lower pressure, and the slower air below has higher pressure. This pressure difference gives a net upward force — the dynamic lift — which balances the weight of the plane and keeps it flying.
9.5 Viscosity
When we derived Bernoulli’s equation, we assumed an “ideal” fluid that loses no energy to friction. Real fluids are not like that. Most real fluids resist being made to flow, offering an internal resistance to motion. This internal resistance behaves like a kind of friction inside the fluid — much like the friction a solid feels when it slides over a surface. We call this property viscosity.
Viscosity only shows up when different layers of the fluid move relative to each other. To see it clearly, imagine a fluid such as oil held between two flat glass plates, as in Fig. 9.12(a). The bottom plate is kept fixed, while the top plate is dragged sideways at a constant velocity \(v\).
Figure to come
Fig. 9.12a – A thin layer of liquid sandwiched between two parallel horizontal glass plates; the lower plate fixed, the upper plate pulled to the right with velocity \(v\) by a force \(F\), and the liquid layers in between shown with arrows increasing in length from bottom to top.
If we replace the oil with honey, we find we must pull much harder to move the top plate at the same speed. This tells us honey resists relative motion more strongly than oil — honey is more viscous than oil.
A key experimental fact makes this work: the fluid layer touching a surface moves with the same velocity as that surface. This is often called the “no-slip” condition. So the layer of liquid touching the moving top plate travels with velocity \(v\), while the layer touching the fixed bottom plate stays at rest.
Between these two, the velocity of the layers increases smoothly and uniformly, from zero at the bottom up to \(v\) at the top. Each layer is pulled forward by the faster layer just above it and dragged back by the slower layer just below it. This tug between neighbouring layers is the viscous force.
This smooth, layer-over-layer motion is exactly the laminar flow we met in Section 9.3. The layers slide over one another like the pages of a book sliding when you push the top cover sideways while the book lies flat on a table.
The same layered pattern appears when a fluid flows through a pipe. There, the liquid moves fastest along the central axis of the tube, and its speed decreases steadily toward the walls, becoming zero right at the wall, as shown in Fig. 9.12(b). All points on a given cylindrical surface (at the same distance from the axis) move at the same speed.
Figure to come
Fig. 9.12b – Velocity distribution for viscous flow in a pipe: a set of horizontal arrows across the pipe cross-section, longest along the central axis and shortening to zero at the walls, forming a curved (parabolic) profile.
Now let us make the idea quantitative. Because of the shearing motion, a portion of the liquid that at one instant has the rectangular shape ABCD becomes slanted into the shape AEFD after a short time interval \(\Delta t\). In that time the liquid has undergone a shear strain of \(\Delta x / l\), where \(\Delta x\) is the sideways shift of the top and \(l\) is the thickness of the layer.
Here is the crucial difference from a solid. In a solid, a fixed shear stress produces a fixed shear strain, and the solid then settles. In a flowing fluid, the strain keeps growing continuously with time — the fluid never stops deforming while the force acts. So the stress in a fluid is found, by experiment, to depend not on the strain itself, but on the rate of change of strain (the “strain rate”): \[\text{strain rate} = \frac{\Delta x}{l\,\Delta t} = \frac{v}{l}\]
The coefficient of viscosity, written \(\eta\) (the Greek letter “eta”), is defined as the ratio of the shearing stress to the strain rate:
\[\eta = \frac{F/A}{v/l} = \frac{F l}{v A} \tag{9.16}\]
Here \(F\) is the tangential (shearing) force applied (N), \(A\) is the area of the layer over which it acts (m\(^2\)), \(v\) is the relative velocity between the layers (m s\(^{-1}\)), and \(l\) is the perpendicular separation between them (m). The quantity \(\eta\) measures how viscous the fluid is.
The SI unit of viscosity is the poiseuille (Pl). Equivalent forms of this unit are N s m\(^{-2}\) or Pa s. The dimensions of viscosity are \([ML^{-1}T^{-1}]\).
In general, thin liquids like water and alcohol are less viscous, while thick liquids like coal tar, blood, and glycerine are more viscous. The coefficients of viscosity for several common fluids are listed in Table 9.2.
Two facts about blood and water are worth noting. As Table 9.2 shows, blood is “thicker” — more viscous — than water. Also, the relative viscosity of blood (its viscosity compared with that of water, \(\eta / \eta_{water}\)) stays nearly constant between \(0\,^\circ\)C and \(37\,^\circ\)C.
Table 9.2 — The viscosities of some fluids
| Fluid | T (\(^\circ\)C) | Viscosity (mPl) |
|---|---|---|
| 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 |
An important temperature effect appears in this table: the viscosity of liquids decreases with rising temperature, whereas the viscosity of gases increases with temperature.
9.5.1 Stokes’ Law
We have seen that viscosity produces a resisting force between layers of a moving fluid. This same viscosity also resists any solid body that tries to move through a fluid.
When a body falls through a fluid, it drags along the layer of fluid touching its surface. This sets up relative motion between the fluid layers, and as a result the body feels a backward, retarding force opposing its motion. Everyday examples include a raindrop falling through air and a pendulum bob swinging through it.
Experiment shows that this viscous retarding force is proportional to the body’s velocity and acts opposite to the direction of motion. For a sphere, the force \(F\) also depends on the viscosity \(\eta\) of the fluid and the radius \(a\) of the sphere.
Sir George G. Stokes (1819–1903) stated this viscous drag force clearly: \[F = 6\pi \eta a v \tag{9.17}\]
Here \(F\) is the viscous drag force (N), \(\eta\) is the coefficient of viscosity of the fluid (Pa s), \(a\) is the radius of the sphere (m), and \(v\) is the speed of the sphere relative to the fluid (m s\(^{-1}\)). This relation is known as Stokes’ law. (We shall not derive it here.)
Stokes’ law is a fine example of a retarding force that grows with velocity. Let us use it to study a body falling through a viscous medium — a raindrop in air.
At first, the raindrop accelerates under gravity, so its speed increases. But as the speed grows, the viscous retarding force (which is proportional to speed) grows too. Meanwhile the fluid also pushes up on the drop with a buoyant force — the upthrust equal to the weight of the fluid displaced by the drop.
Eventually a balance is reached: the upward viscous force plus the upward buoyant force become equal to the downward force of gravity. The net force is then zero, so the acceleration is zero, and the drop falls at a steady, constant speed. This steady speed is called the terminal velocity, \(v_t\).
Figure to come
Fig. 9.13b – A sphere of radius \(a\) falling through a fluid at terminal velocity, with the downward weight, the upward buoyant force, and the upward viscous drag force \(6\pi\eta a v_t\) all marked.
Let us set up this balance for a sphere of radius \(a\). Let \(\rho\) be the density of the sphere and \(\sigma\) the density of the fluid. The sphere’s volume is \(\frac{4}{3}\pi a^3\), so its weight (downward) is \(\frac{4}{3}\pi a^3 \rho g\), and the buoyant force (upward, equal to the weight of displaced fluid) is \(\frac{4}{3}\pi a^3 \sigma g\). At terminal velocity, gravity is balanced by buoyancy plus viscous drag:
\[6\pi \eta a v_t = \frac{4}{3}\pi a^3 (\rho - \sigma) g\]
Solving this for the terminal velocity, the factors of \(\pi a\) cancel and the arithmetic \(\frac{4}{18} = \frac{2}{9}\) gives:
\[v_t = \frac{2 a^2 (\rho - \sigma) g}{9 \eta} \tag{9.18}\]
So the terminal velocity depends on the square of the sphere’s radius, and inversely on the viscosity of the medium. (You may like to refer back to Example 6.2 for a related discussion of motion under a velocity-dependent force.)
9.6 Surface Tension
We now turn to a set of familiar but puzzling everyday observations. Oil and water do not mix. Water wets our skin but rolls off a duck’s feathers. Mercury does not wet glass, yet water sticks to it. Oil creeps up a cotton wick against gravity, and sap and water rise to the top leaves of a tall tree.
There are more such examples. The hairs of a paint brush stay apart when dry, and even when the brush is dipped in water, but they draw together into a fine tip the moment the wet brush is lifted out.
What do all these effects have in common? They all involve the free surface of a liquid.
A liquid has a fixed volume but no fixed shape, so when poured into a container it settles with a definite free surface at the top. Careful study shows that this surface is special: it stores some extra energy compared with the liquid deep inside.
This extra energy at the surface makes the surface behave as though it were a stretched, elastic sheet always trying to shrink. This phenomenon is called surface tension.
Because it depends on having a free surface, surface tension is a property of liquids only — gases, which spread out to fill their container, have no free surface and therefore show no surface tension. In the sections that follow, we build up this idea carefully, starting from what happens to the individual molecules at a liquid’s surface.
9.6.1 Surface Energy
To understand surface tension, we look at the forces between the molecules of a liquid. A liquid stays together as a body because its molecules attract one another.
Consider a molecule sitting well inside the liquid, deep below the surface, as in Fig. 9.14(a). It is surrounded on all sides by other molecules within attracting distance, so it is pulled equally in every direction by its neighbours.
Figure to come
Fig. 9.14a – A molecule deep inside a liquid, surrounded on all sides by other molecules, with attractive-force arrows pointing outward toward neighbours in every direction.
This all-round attraction gives the interior molecule a negative potential energy — “negative” because energy would have to be supplied to pull it away from its attracting neighbours. The exact value depends on how many molecules surround it and how they are arranged, but on average every interior molecule has the same potential energy.
We can see the effect of this attraction in a familiar quantity: the heat of evaporation. To evaporate a liquid, we must pull its molecules far apart against their mutual attraction, and this needs a large amount of energy. For water, the heat of evaporation is of the order of \(40\) kJ/mol — a large value, confirming how strongly the molecules attract one another.
Now consider a molecule right at the surface, as in Fig. 9.14(b). Only its lower half is surrounded by liquid molecules; above it there is only air, with far fewer molecules to attract it.
Figure to come
Fig. 9.14b – A molecule at the liquid surface, with attractive-force arrows only from below and to the sides (from the liquid), and none from above (only air/vapour above).
So a surface molecule is attracted by fewer neighbours than an interior one. Its negative potential energy is therefore smaller in size — roughly half that of a fully surrounded interior molecule.
A smaller amount of negative energy means the surface molecule sits at a higher energy than an interior molecule. In other words, molecules on the surface carry some extra energy compared with molecules in the interior.
This single fact explains a great deal. Since every bit of surface carries extra energy, a larger surface means more energy. Nature tends toward the lowest energy state, so a liquid tends to take on the least surface area that external conditions allow. Making the surface bigger costs energy; shrinking it releases energy.
Most surface phenomena — from droplets pulling into beads to a wet brush forming a tip — can be understood from this one principle: increasing surface area requires energy, so a liquid resists it.
How much energy does it take to bring one molecule to the surface? As we reasoned above, a surface molecule has lost about half of its binding, so the energy needed to place a molecule at the surface is roughly half the energy needed to remove it entirely from the liquid — that is, about half the heat of evaporation.
Finally, what exactly is a “surface”? Since a liquid is made of molecules in constant motion, there is no perfectly sharp boundary. Instead, as we move outward across the surface (the direction marked in Fig. 9.14c), the density of liquid molecules drops rapidly to zero over a distance of just a few molecular sizes.
Figure to come
Fig. 9.14c – A graph/schematic showing the liquid molecule density falling rapidly to zero across the surface region (around z = 0) over a distance of a few molecular sizes, with the balance of attractive (A) and repulsive (R) forces indicated.
9.6.2 Surface Energy and Surface Tension
We have seen that extra energy is associated with the surface of a liquid. Now we connect this surface energy to a measurable force — the surface tension — using a simple experiment.
Take a horizontal liquid film (like a soap film) whose one edge is a light bar of length \(l\) that can slide freely along two parallel guides, as in Fig. 9.15. Creating more surface, while keeping the volume of liquid fixed, means sliding this bar outward to stretch the film.
Figure to come
Fig. 9.15 – (a) A liquid film held on a rectangular frame with one side a movable bar of length \(l\), the film in equilibrium; (b) the same film with the bar pulled out a small extra distance \(d\) by a force \(F\), increasing the film area.
Suppose we pull the bar out by a small distance \(d\). The area of the surface increases, so the system now stores more surface energy. That extra energy must have come from work done against an internal pulling force.
Let this internal force be \(F\). The work done by the applied force in moving the bar a distance \(d\) is: \[W = F \cdot d = Fd\]
By conservation of energy, this work is stored as the additional surface energy of the film.
Here is a subtle but important point: a film has two surfaces — a front face and a back face — both in contact with the air, with the liquid sandwiched in between. So when the bar of length \(l\) moves out by \(d\), new area is created on both faces. The total extra area is: \[\text{extra area} = 2 \, l \, d = 2dl\]
Let \(S\) be the surface energy per unit area of the film. Then the extra surface energy created is \(S\) times the extra area, \(S(2dl)\). Setting this equal to the work done: \[S\,(2dl) = Fd \tag{9.19}\]
Cancelling \(d\) from both sides and solving for \(S\): \[S = \frac{Fd}{2dl} = \frac{F}{2l} \tag{9.20}\]
This quantity \(S\) is the magnitude of the surface tension. Notice that Eq. (9.20) gives it two equivalent meanings at once: \(S\) is the surface energy per unit area of the liquid interface, and it is also the force per unit length that the liquid exerts on the movable bar.
Here \(F\) is the force needed to hold the bar (N), \(l\) is the length of the bar (m), and the factor \(2\) accounts for the film’s two surfaces.
So far we have discussed the surface of a single liquid exposed to air. More generally, a fluid surface can be in contact with another fluid or with a solid. In that case, the surface energy depends on the materials on both sides of the surface. If the molecules on the two sides attract each other, the surface energy is reduced; if they repel, it is increased.
More correctly, then, the surface energy is really the energy of the interface between two materials, and it depends on both of them.
From all of this, we can make two key observations.
- Surface tension is a force per unit length (equivalently, a surface energy per unit area) acting in the plane of the interface between the liquid and any other substance. It is also the extra energy that the molecules at the interface have compared with those in the interior.
- Consider any point on the interface, away from its boundary. Draw a small line through it. Equal and opposite surface tension forces, of magnitude \(S\) per unit length, act on the two sides of this line, perpendicular to it and lying in the plane of the interface. To picture this, imagine a line of molecules at the surface: the molecules to the left pull the line toward them, and those to the right pull it toward them, so the line stays in equilibrium under tension. But if the line lies at the true edge of the interface, there is liquid on one side only, so there is a single unbalanced force \(S\) per unit length pulling inward.
Table 9.3 lists the surface tension of several liquids, along with their heats of vaporisation at the stated temperatures.
Table 9.3 — Surface tension of some liquids at the temperatures indicated, with heats of vaporisation
| Liquid | Temp (\(^\circ\)C) | Surface Tension (N/m) | Heat of vaporisation (kJ/mol) |
|---|---|---|---|
| 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 |
The value of surface tension depends on temperature. Just like viscosity, the surface tension of a liquid usually falls as its temperature rises.
Finally, we can describe a direct way to measure surface tension, which also brings in the idea of a liquid sticking to a solid.
A fluid will stick to (wet) a solid surface if the surface energy between the fluid and the solid is smaller than the sum of the surface energies of the solid–air and fluid–air interfaces. There is then a net attraction between the solid surface and the liquid, and this attraction can be measured directly.
The arrangement is shown schematically in Fig. 9.16. A flat vertical glass plate forms one arm of a balance, with a vessel of liquid kept below it. The plate is balanced by weights on the other side, with its horizontal lower edge held just above the water. The vessel is then raised slightly until the liquid just touches the plate, and surface tension pulls the plate down a little. Weights are added on the other side until the plate just clears the water.
Figure to come
Fig. 9.16 – A vertical glass plate hanging from one arm of a balance, its lower edge touching the water surface in a vessel below, with weights \(W\) on the other pan to balance the downward pull of surface tension on the plate.
Suppose the additional weight required to just pull the plate free is \(W\). Then, using Eq. (9.20) and the reasoning behind it, the surface tension of the liquid–air interface is: \[S_{la} = \frac{W}{2l} = \frac{mg}{2l} \tag{9.21}\] where \(m\) is the extra mass added, \(l\) is the length of the plate’s edge, and the factor \(2\) again accounts for the two liquid surfaces clinging to the plate. The subscript “la” emphasises that this is the liquid–air interface tension.
9.6.3 Angle of Contact
In the previous section we noted that surface energy really belongs to the interface between two materials. Let us use this idea to understand how a liquid behaves when it meets a solid.
A fluid will stick to (spread on) a solid surface when the surface energy between the fluid and the solid is smaller than the sum of the surface energies of the solid–air and fluid–air interfaces. In that case there is a net attraction between the solid and the liquid.
Before studying that attraction, note that the liquid–air surface tension itself can be measured directly, as sketched in Fig. 9.16. A flat vertical glass plate hangs from one arm of a balance, with its lower horizontal edge held just above the liquid. The vessel of liquid is raised until the liquid just touches the plate and pulls it down slightly (because of surface tension). Weights are then added on the other pan until the plate just breaks free of the liquid.
Figure to come
Fig. 9.16 – A vertical glass plate suspended from one arm of a balance, its lower edge touching the liquid surface below, balanced by weights \(W\) on the other arm.
If the extra weight needed is \(W\), then using Eq. (9.20) and the reasoning behind it, the surface tension of the liquid–air interface is: \[S_{la} = \frac{W}{2l} = \frac{mg}{2l} \tag{9.21}\]
Here \(m\) is the extra mass added (kg), \(g\) is the acceleration due to gravity (m s\(^{-2}\)), and \(l\) is the length of the plate’s edge (m). The subscript “\(la\)” reminds us this is the liquid–air interface tension. (The factor \(2\) appears because the liquid wets both faces of the plate’s edge.)
Now to the shape of the liquid near a solid. Where a liquid surface meets a solid, the surface is generally curved. The angle this curve makes with the solid is a key quantity called the angle of contact.
Figure to come
Fig. 9.17 – Water drops showing the angle of contact \(\theta\): (a) a rounded droplet beading up on a lotus leaf (large \(\theta\)); (b) a flattened drop spreading over a clean plastic plate (small \(\theta\)), with the three interfacial tensions marked at the contact line.
The value of \(\theta\) is different for each pair of liquid and solid, and it decides whether a liquid spreads out over a solid or gathers into droplets. For example, water forms rounded droplets on a lotus leaf (Fig. 9.17a), but spreads out into a thin film over a clean plastic plate (Fig. 9.17b).
To understand what sets \(\theta\), we consider the three interfacial tensions acting where the liquid, solid and air all meet: the liquid–air tension \(S_{la}\), the solid–air tension \(S_{sa}\), and the solid–liquid tension \(S_{sl}\).
At the line of contact, these three surface forces must be in equilibrium. Balancing the forces along the solid surface (as seen in Fig. 9.17b) gives the relation: \[S_{la} \cos\theta + S_{sl} = S_{sa} \tag{9.22}\]
From this balance we can read off two cases.
The angle of contact is obtuse (\(\theta > 90^\circ\)) when \(S_{sl} > S_{la}\), as for water on a lotus leaf. Here the liquid’s molecules are attracted strongly to one another but only weakly to the solid. Creating a liquid–solid contact then costs a lot of energy, so the liquid does not wet the solid. This is what happens with water on a waxy or oily surface, and with mercury on almost any surface.
The angle of contact is acute (\(\theta < 90^\circ\)) when \(S_{sl} < S_{la}\), as for water on clean plastic. Here the liquid’s molecules are strongly attracted to the solid, which reduces \(S_{sl}\) and makes \(\theta\) small. The liquid then spreads and wets the solid. This is what happens for water on glass or plastic, and for kerosene oil on virtually anything (it just spreads).
This is also why we can control wetting chemically. Wetting agents such as soaps, detergents and dyeing substances lower the angle of contact, so the liquid penetrates and spreads well — which is exactly what a detergent needs to do to clean a fabric. Waterproofing agents, on the other hand, are added to increase the angle of contact between water and the fibres, so water beads up and runs off instead of soaking in.
9.6.4 Drops and Bubbles
One striking consequence of surface tension is that free liquid drops and bubbles are spherical, provided the effects of gravity can be neglected. You can see this clearly in the tiny drops formed in a high-speed spray or jet, and in the soap bubbles most of us blew as children.
Why are drops and bubbles spherical, and what keeps a soap bubble stable?
The reason follows directly from surface energy. A liquid–air interface has energy proportional to its area, so for a given volume of liquid, the shape with the least energy is the shape with the least surface area. Among all shapes enclosing a fixed volume, the sphere has the smallest surface area. (Proving this fully is beyond our scope, but you can check that a sphere beats even a cube for the same volume.)
So, if gravity and other forces such as air resistance were absent, a free liquid drop would settle into a perfect sphere — the shape of least surface energy.
A second important consequence of surface tension is that the pressure inside a spherical drop is greater than the pressure outside it. Let us find by how much.
Figure to come
Fig. 9.18a – A spherical liquid drop of radius \(r\) with higher pressure \(P_i\) inside and lower pressure \(P_o\) outside, and an inward-pulling surface.
Consider a spherical drop of radius \(r\) in equilibrium, and imagine its radius increasing slightly by \(\Delta r\). The surface area grows, so extra surface energy is created. This extra energy is the surface tension times the increase in area: \[\left[4\pi (r + \Delta r)^2 - 4\pi r^2\right] S_{la} = 8\pi r\,\Delta r\, S_{la} \tag{9.23}\]
(Here we expanded \((r+\Delta r)^2\) and dropped the very small \((\Delta r)^2\) term.)
If the drop is in equilibrium, this energy cost must be paid for by the work done in the expansion by the pressure difference \((P_i - P_o)\) between the inside and the outside. That work is the pressure difference times the increase in volume (\(4\pi r^2 \Delta r\)): \[W = (P_i - P_o)\, 4\pi r^2 \Delta r \tag{9.24}\]
Setting the work equal to the extra surface energy and cancelling the common factor \(4\pi r \Delta r\): \[(P_i - P_o) = \frac{2 S_{la}}{r} \tag{9.25}\]
So the excess pressure inside a spherical drop is \(2S_{la}/r\). Notice it is larger for a smaller drop — tiny drops have surprisingly high internal pressure.
In general, for any curved liquid–gas interface, the concave side (the inside) has higher pressure than the convex side (the outside). For instance, an air bubble (a cavity) formed inside a liquid, shown in Fig. 9.18(b), has higher pressure inside it than in the surrounding liquid.
Figure to come
Fig. 9.18b – An air cavity (bubble) of radius \(r\) inside a liquid, with higher pressure inside and the liquid surface curving around it.
A soap bubble, shown in Fig. 9.18(c), is different from a solid drop or an air cavity: it has two liquid surfaces — an inner one and an outer one — with a thin film of liquid between them. Applying the same argument to both surfaces doubles the effect, so for a soap bubble: \[(P_i - P_o) = \frac{4 S_{la}}{r} \tag{9.26}\]
Figure to come
Fig. 9.18c – A soap bubble of radius \(r\) showing its two interfaces (inner and outer liquid surfaces) with the enclosed air at higher pressure \(P_i\).
This is probably why you have to blow hard, but not too hard, to form a soap bubble: a little extra air pressure is needed inside to overcome the surface tension of both surfaces and hold the bubble open.
9.6.5 Capillary Rise
A direct consequence of the pressure difference across a curved liquid–air surface is a very familiar effect: water climbs up a narrow tube on its own, rising against gravity. This is called capillary rise.
The word capilla means “hair” in Latin. If the tube were as thin as a hair, the rise would be very large. To study it, consider a vertical capillary tube of circular cross-section (radius \(a\)) dipped into an open vessel of water, as shown in Fig. 9.19.
Figure to come
Fig. 9.19 – (a) A narrow vertical tube standing in a vessel of water, with the water risen to a height \(h\) inside the tube and a concave meniscus at the top; points A (inside, just below the meniscus) and B (in the vessel at the same level) marked. (b) An enlarged view of the concave meniscus showing the contact angle \(\theta\), tube radius \(a\), and radius of curvature \(r\).
The contact angle between water and glass is acute, so the water surface inside the tube curves into a concave shape (a meniscus that dips in the middle). Because this surface is curved, there is a pressure difference across it. Using the excess-pressure idea from the last section, and relating the meniscus’s radius of curvature \(r\) to the tube radius \(a\) by \(r = a\sec\theta\): \[(P_i - P_o) = \frac{2S}{r} = \frac{2S}{a\sec\theta} = \frac{2S}{a}\cos\theta \tag{9.27}\]
Here \(S\) is the surface tension (N m\(^{-1}\)), \(a\) is the tube radius (m), \(\theta\) is the contact angle, \(P_i\) is the pressure just below the meniscus and \(P_o\) the pressure just above it (atmospheric).
Because the surface is concave, this means the pressure of the water just below the meniscus is less than the atmospheric pressure above it.
Now compare two points at the same horizontal level: point A, inside the tube just below the meniscus, and point B, out in the open vessel at the same height. Since they are at the same level in the connected liquid, they must be at the same pressure. Writing the pressure at B (atmospheric \(P_0\) plus the column of height \(h\)) equal to the pressure at A: \[P_0 + h\rho g = P_i = P_A \tag{9.28}\]
where \(\rho\) is the density of water and \(h\) is the height the liquid has risen — the capillary rise.
Combining Eqs. (9.27) and (9.28), the \(P_0\) terms cancel and we get: \[h\rho g = (P_i - P_0) = \frac{2S\cos\theta}{a} \tag{9.29}\]
Rearranging gives the capillary rise directly: \[h = \frac{2S\cos\theta}{\rho g a}\]
The equation makes it clear that capillary rise is caused by surface tension, and that it is larger for a smaller tube radius \(a\). For fine capillaries, the rise is typically a few centimetres.
As a worked figure, take a tube of radius \(a = 0.05\) cm \(= 5 \times 10^{-4}\) m, with water’s surface tension from Table 9.3 (and treating \(\cos\theta \approx 1\) for water on glass): \[h = \frac{2S}{\rho g a} = \frac{2 \times (0.073 \text{ N m}^{-1})}{(10^{3} \text{ kg m}^{-3})(9.8 \text{ m s}^{-2})(5 \times 10^{-4} \text{ m})}\] \[h = 2.98 \times 10^{-2} \text{ m} = 2.98 \text{ cm}\]
There is an important opposite case. If the meniscus is convex instead of concave — as it is for mercury, whose contact angle is obtuse so that \(\cos\theta\) is negative — then Eq. (9.29) gives a negative \(h\). The liquid is then pushed down, so it stands lower inside the capillary than outside. This is called capillary depression.
9.7 Summary
The basic property of a fluid is that it can flow. A fluid offers no resistance to a change of its shape, so the shape of a fluid is governed by the shape of its container.
A liquid is incompressible and has a free surface of its own. A gas is compressible and expands to occupy all the space available to it.
If \(F\) is the normal force exerted by a fluid on an area \(A\), then the average pressure \(P_{av}\) is defined as the ratio of the force to the area: \[P_{av} = \frac{F}{A}\]
The SI unit of pressure is the pascal (Pa), which is the same as N m\(^{-2}\). Other common units of pressure are: \[1 \text{ atm} = 1.01 \times 10^{5} \text{ Pa}\] \[1 \text{ bar} = 10^{5} \text{ Pa}\] \[1 \text{ torr} = 133 \text{ Pa} = 0.133 \text{ kPa}\] \[1 \text{ mm of Hg} = 1 \text{ torr} = 133 \text{ Pa}\]
Pascal’s law states that: the pressure in a fluid at rest is the same at all points which are at the same height. A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the containing vessel.
The pressure in a fluid varies with depth \(h\) according to the expression: \[P = P_a + \rho g h\] where \(\rho\) is the density of the fluid, assumed uniform.
The volume of an incompressible fluid passing any point every second in a pipe of non-uniform cross-section is the same in steady flow: \[vA = \text{constant}\] where \(v\) is the velocity and \(A\) is the area of cross-section. This equation is due to the conservation of mass in incompressible fluid flow.
Bernoulli’s principle states that as we move along a streamline, the sum of the pressure (\(P\)), the kinetic energy per unit volume (\(\rho v^2 / 2\)), and the potential energy per unit volume (\(\rho g y\)) remains constant: \[P + \frac{\rho v^2}{2} + \rho g y = \text{constant}\] This equation is basically the conservation of energy applied to non-viscous fluid motion in the steady state. No real fluid has zero viscosity, so the statement is true only approximately. Viscosity acts like friction and converts kinetic energy into heat.
Although shear strain in a fluid does not require a shear stress, when a shear stress is applied to a fluid, motion is generated which causes a shear strain that grows with time. The ratio of the shear stress to the time rate of shearing strain is known as the coefficient of viscosity, \(\eta\): \[\eta = \frac{F/A}{v/l}\] where the symbols have their usual meaning as defined in the text.
Stokes’ law states that the viscous drag force \(F\) on a sphere of radius \(a\) moving with velocity \(v\) through a fluid of viscosity \(\eta\) is: \[F = 6\pi \eta a v\]
Surface tension is a force per unit length (or surface energy per unit area) acting in the plane of the interface between the liquid and the bounding surface. It is the extra energy that the molecules at the interface have compared with those in the interior.
9.8 Points to Ponder
Pressure is a scalar quantity. Defining pressure as “force per unit area” may give the false impression that pressure is a vector. But the “force” in the numerator is the component of the force normal to the area on which it acts. In describing fluids, we shift from particle and rigid-body mechanics, because we are now concerned with properties that vary from point to point within the fluid.
One should not think of the pressure of a fluid as being exerted only on a solid — like the walls of a container or a solid object immersed in the fluid. Pressure exists at all points within a fluid. An element of fluid (such as the one shown in Fig. 9.4) is in equilibrium because the pressures exerted on its various faces are equal.
The expression \(P = P_a + \rho g h\) holds true only if the fluid is incompressible. In practice it holds for liquids, which are largely incompressible, so their density is essentially constant with height.
The gauge pressure is the difference between the actual pressure and the atmospheric pressure: \[P - P_a = P_g\] Many pressure-measuring devices measure the gauge pressure. These include the tyre pressure gauge and the blood pressure gauge (sphygmomanometer).
A streamline is a map of fluid flow. In steady flow, two streamlines do not intersect, because an intersection would mean the fluid particle there had two possible velocities at once.
Bernoulli’s principle does not hold in the presence of viscous drag on the fluid. In that case the work done by this dissipative viscous force must be taken into account, and \(P_2\) (Fig. 9.9) will be lower than the value given by Eq. (9.12).
As the temperature rises, the atoms of a liquid become more mobile, and the coefficient of viscosity \(\eta\) falls. In a gas, a temperature rise increases the random motion of atoms, and \(\eta\) increases.
Surface tension arises due to the excess potential energy of the molecules at the surface compared with their potential energy in the interior. Such surface energy is present at the interface separating two substances, at least one of which is a fluid. It is not the property of a single fluid alone.
9.9 Table of Physical Quantities
| Physical Quantity | Symbol | Dimensions | Unit | Remarks |
|---|---|---|---|---|
| Pressure | \(P\) | \([M\,L^{-1}\,T^{-2}]\) | pascal (Pa) | \(1 \text{ atm} = 1.013 \times 10^{5}\) Pa; Scalar |
| Density | \(\rho\) | \([M\,L^{-3}]\) | kg m\(^{-3}\) | Scalar |
| Specific Gravity | — | dimensionless | no unit | \(\dfrac{\rho_{substance}}{\rho_{water}}\); Scalar |
| Coefficient of viscosity | \(\eta\) | \([M\,L^{-1}\,T^{-1}]\) | Pa s or poiseuille (Pl) | Scalar |
| Surface Tension | \(S\) | \([M\,T^{-2}]\) | N m\(^{-1}\) | Scalar |
9.10 NCERT Questions
(NCERT 9.1) Explain why:
- The blood pressure in humans is greater at the feet than at the brain.
- 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.
- Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.
(NCERT 9.2) Explain why:
- The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
- 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.)
- Surface tension of a liquid is independent of the area of the surface.
- Water with detergent dissolved in it should have small angles of contact.
- A drop of liquid under no external forces is always spherical in shape.
(NCERT 9.3) Fill in the blanks using the word(s) from the list appended with each statement:
- Surface tension of liquids generally … with temperatures (increases / decreases).
- Viscosity of gases … with temperature, whereas viscosity of liquids … with temperature (increases / decreases).
- 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).
- For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle).
- For the model of a plane in a wind tunnel, turbulence occurs at a … speed than for turbulence for an actual plane (greater / smaller).
(NCERT 9.4) Explain why:
- To keep a piece of paper horizontal, you should blow over, not under, it.
- When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.
- The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection.
- A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel.
- A spinning cricket ball in air does not follow a parabolic trajectory.
(NCERT 9.5) A 50 kg girl wearing high-heel shoes balances on a single heel. The heel is circular with a diameter of 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
(NCERT 9.6) 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.
(NCERT 9.7) A vertical off-shore structure is built to withstand a maximum stress of \(10^9\) 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.
(NCERT 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?
(NCERT 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?
(NCERT 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)
(NCERT 9.11) Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.
(NCERT 9.12) Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.
(NCERT 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 \(4.0 \times 10^{-3}\) kg s⁻¹, what is the pressure difference between the two ends of the tube? (Density of glycerine \(= 1.3 \times 10^3\) kg m⁻³ and viscosity of glycerine \(= 0.83\) Pa s.) [You may also like to check whether the assumption of laminar flow in the tube is correct.]
(NCERT 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.
(NCERT 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?
(NCERT 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?
(NCERT 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 \(1.5 \times 10^{-2}\) 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?
(NCERT 9.18) Figure 9.21(a) shows a thin liquid film supporting a small weight \(= 4.5 \times 10^{-2}\) 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.
(NCERT 9.19) What is the pressure inside a drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20 °C) is \(4.65 \times 10^{-1}\) N m⁻¹. The atmospheric pressure is \(1.01 \times 10^5\) Pa. Also give the excess pressure inside the drop.
(NCERT 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 \(2.50 \times 10^{-2}\) 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 atmospheric pressure is \(1.01 \times 10^5\) Pa.)
9.11 Check Your Concepts
Solids and liquids are said to have much lower compressibility than gases. Using the idea of molecular spacing, explain why a gas can be compressed easily but a liquid cannot.
Pressure is defined using force, which is a vector, yet pressure is a scalar. Explain how this is consistent, referring to what the “force” in the definition actually represents.
Three vessels of very different shapes are connected at the base and filled with water to the same height. The pressure at the base is found to be the same in all three, even though they hold different amounts of water. Explain this hydrostatic paradox.
The wall of a dam is built much thicker at its base than near its top. Justify this design using the variation of fluid pressure with depth.
A tyre pressure gauge reads zero for a completely flat tyre, even though air at atmospheric pressure is still present inside. Explain what the gauge is actually measuring and why the reading is zero.
In a hydraulic lift, a small applied force supports a very large load. Show, using the idea of the distances moved by the two pistons, that this does not violate the conservation of energy.
Using the equation of continuity, explain why a smooth stream of water falling from a tap becomes narrower as it descends.
State the assumptions under which Bernoulli’s equation is derived, and explain why it does not apply to turbulent flow.
A person standing close to the edge of a platform feels a pull toward a fast train as it passes. Explain this effect using Bernoulli’s principle.
Explain, in terms of the speed of air above and below the wing, how the shape and tilt of an aircraft wing generate an upward lift.
The viscosity of a liquid decreases as its temperature rises, whereas the viscosity of a gas increases. Explain the different molecular reasons behind these two opposite behaviours.
A raindrop falling from a great height reaches the ground at a moderate, constant speed rather than an ever-increasing one. Explain, using the forces acting on it, why it attains a terminal velocity.
Molecules at the surface of a liquid possess more energy than those deep inside. Explain why, and connect this to the tendency of a liquid to minimise its surface area.
The excess pressure inside a soap bubble is twice that inside a liquid drop of the same radius. Explain the reason for this factor of two.
A fine steel needle, though denser than water, can be made to float on a water surface, while it sinks at once if the surface is disturbed with detergent. Explain both observations.
Water rises in a fine glass capillary tube but mercury is depressed in the same tube. Explain both effects in terms of the angle of contact and the shape of the meniscus.
9.12 Practice with Numericals
A rectangular water tank has water standing to a depth of 5.0 m. Taking \(\rho_{water} = 1000\) kg m⁻³, \(g = 10\) m s⁻², and atmospheric pressure \(P_a = 1.01 \times 10^5\) Pa, find the gauge pressure and the absolute pressure at the bottom of the tank.
The density of a certain oil is \(8.5 \times 10^2\) kg m⁻³. Find its relative density, and calculate the mass of \(2.0\) litres of this oil. (Take \(\rho_{water} = 1.0 \times 10^3\) kg m⁻³.)
A man of mass 70 kg stands on both feet, each shoe sole having a contact area of \(180\) cm². Calculate the average pressure he exerts on the ground. (Take \(g = 10\) m s⁻². Be careful to convert the area into m².)
In a hydraulic lift, the smaller piston has a diameter of 5.0 cm and the larger piston a diameter of 25 cm. If a force of 200 N is applied to the smaller piston, find (a) the maximum load that can be raised on the larger piston, and (b) the mechanical advantage of the lift.
One arm of a mercury manometer is open to the atmosphere, and its mercury level stands 22.0 cm higher than the arm connected to a gas cylinder. Taking the density of mercury as \(13.6 \times 10^3\) kg m⁻³, \(g = 9.8\) m s⁻², and \(P_a = 1.01 \times 10^5\) Pa, find the gauge pressure and the absolute pressure of the gas.
Water flows steadily through a horizontal pipe whose cross-sectional radius narrows from 5.0 cm to 2.5 cm. If the speed of water in the wider section is 1.2 m s⁻¹, find its speed in the narrower section using the equation of continuity.
Water flows through a horizontal pipe. At a wide section the speed is 3.0 m s⁻¹ and the pressure is \(2.0 \times 10^5\) Pa; at a narrow section the speed rises to 9.0 m s⁻¹. Using Bernoulli’s equation with \(\rho_{water} = 1000\) kg m⁻³, find the pressure at the narrow section.
A large open water tank has a small hole in its side at a depth of 2.5 m below the water surface. Using Torricelli’s law (take \(g = 10\) m s⁻²), find the speed with which water leaves the hole.
In a wind-tunnel test, air flows over the top of a model wing at 90 m s⁻¹ and under it at 84 m s⁻¹. If the wing area is 1.5 m² and the density of air is 1.2 kg m⁻³, estimate the lift force on the wing (ignore the height difference across the wing).
A flat plate of area \(0.30\) m² rests on a layer of oil \(0.40\) mm thick, of viscosity \(1.0\) Pa s, spread on a table. What horizontal force is needed to move the plate over the oil at a steady speed of \(0.15\) m s⁻¹?
A small steel ball of radius \(1.0\) mm falls through a tall column of oil of viscosity \(0.90\) Pa s. The density of steel is \(7.8 \times 10^3\) kg m⁻³ and that of the oil is \(1.2 \times 10^3\) kg m⁻³. Taking \(g = 9.8\) m s⁻², find the terminal velocity of the ball.
A soap film is formed on a rectangular frame whose movable wire is \(10\) cm long. A force of \(5.0 \times 10^{-3}\) N is needed to hold this wire against the pull of the film. Treating the film as having two surfaces, find the surface tension of the soap solution.
Calculate the excess pressure inside (a) a water drop of radius \(1.0\) mm, and (b) a soap bubble of radius \(1.0\) mm. Take the surface tension of water as \(0.073\) N m⁻¹ and of the soap solution as \(0.030\) N m⁻¹.
A clean glass capillary tube of internal radius \(0.25\) mm is dipped vertically into water. Taking the surface tension of water as \(0.073\) N m⁻¹, the contact angle as \(0^\circ\), density of water \(1000\) kg m⁻³, and \(g = 9.8\) m s⁻², find the height to which water rises in the tube.