Mechanical Properties of Fluids

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XI, Chapter 9.

Why is a dam thicker at the bottom?

Water pushes sideways on a wall, and it pushes harder the deeper you go, so engineers build the base of a dam far thicker than its top. The same chapter explains why a 70 tonne aircraft stays up on thin air, why a steel needle can rest on water, and why a small steel ball sinks through oil at a steady crawl instead of speeding up forever. Four ideas do all the work: pressure, flow, viscosity and surface tension. Learn why each one holds and the numericals become routine.

What is a fluid, and what is pressure?

A fluid is anything that flows: a liquid or a gas. A solid resists a sideways (shear) push and springs back. A fluid gives way and keeps flowing, which is why it takes the shape of its container. Liquids are almost incompressible, while gases are easy to squeeze.

Pressure is the normal force per unit area on a surface:

$$P = \dfrac{F_{\perp}}{A}$$

Only the component of force perpendicular to the surface counts. Pressure is a scalar. A fluid at rest pushes on any surface at right angles to it, but at a given point the pressure is the same whichever way the surface faces. Since no direction belongs to the pressure itself, it cannot be a vector. Density is $\rho = m/V$ in kg m⁻³, and relative density is the ratio of a substance's density to that of water at 4 °C ($1.0 \times 10^{3}$ kg m⁻³), so it has no unit.

Pressure grows with depth, and only with depth

Pick a vertical cylinder of fluid of area A and height h, with its top at point 1 and its bottom at point 2. The cylinder is at rest, so the upward push at the bottom must balance the downward push at the top plus the cylinder's weight:

$P_{2}A = P_{1}A + \rho(Ah)g$, so $P_{2} - P_{1} = \rho g h$

Put point 1 at the open surface, where the pressure is atmospheric $P_{a}$, and you get the absolute pressure at depth h:

$$P = P_{a} + \rho g h$$

The part $\rho g h$ is the gauge pressure, the pressure over and above the atmosphere. A tyre gauge, a blood pressure monitor and most pressure gauges report gauge pressure.

Notice what is missing: the shape of the container and the area of its base. Take three vessels of very different shapes joined at the bottom and fill them with water. The water stands at the same level in all three. If the wide vessel had a higher bottom pressure, it would shove water into the narrow one until the levels matched. This is the hydrostatic paradox: the pressure at the bottom depends only on the vertical height of the liquid, though the total weight of liquid may differ enormously. Wide vessels press harder on their base only because the force is pressure times area.

Pause and predict: a narrow tube of base 1 cm² and a lake of base 1 km² are both filled to a depth of 5 m. Where is the pressure at the bottom greater?

Check your answer: it is the same in both, $\rho g h$. The lake's base feels a gigantic force, but force is pressure multiplied by area. Pressure itself depends only on depth and the liquid's density.

Worked example: a diver in a lake. A diver is 30 m below the surface of a freshwater lake. Find the gauge pressure and the absolute pressure on the diver. Take $\rho = 1000$ kg m⁻³, g = 9.8 m s⁻² and $P_{a} = 1.01 \times 10^{5}$ Pa.

The common wrong step is to report $2.94 \times 10^{5}$ Pa as the pressure on the diver's eardrum. That is only the extra pressure. The air above the lake also pushes down, and the liquid transmits it, so the total is nearly 4 atm.

Atmospheric pressure, barometer and manometer

The atmosphere is a sea of air, and the pressure at sea level is the weight of a unit-area column of air reaching the top of the atmosphere: $1.013 \times 10^{5}$ Pa.

In Torricelli's barometer, a tube full of mercury is inverted into a mercury dish. The mercury falls until the column's pressure equals the atmosphere's, leaving a vacuum above it. Then $P_{a} = \rho g h$, and with $\rho = 13.6 \times 10^{3}$ kg m⁻³ the height is about 0.76 m, since $13.6 \times 10^{3} \times 9.8 \times 0.76 \approx 1.013 \times 10^{5}$ Pa. Mercury is used because it is very dense, so the column is short enough to handle. Tilting the tube or using a wider one changes the length of mercury in the tube but not its vertical height.

An open-tube manometer is a U-tube with a liquid in it. One arm is connected to the gas whose pressure you want and the other is open to the air. The difference in the liquid levels gives the gauge pressure of the gas, $P - P_{a} = \rho g h$. Oil is used for small pressure differences and mercury for large ones.

Beyond the textbook. For the exam: 1 atm = 76 cm of Hg = $1.013 \times 10^{5}$ Pa, and the barometer reading depends only on vertical height. In reality: water would need a column about 10.3 m tall to balance the atmosphere, since $\rho$ is 13.6 times smaller than that of mercury. That is why a suction pump can never lift water higher than about 10 m. The pump does not pull the water up. It lowers the pressure above the water, and the atmosphere pushes the water up until its column weighs as much as the atmosphere's push can support. Why it matters: it tells you the atmosphere is the one doing the lifting in every straw and suction pump.

Pascal's law and the hydraulic lift

Pascal's law: a change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container. The reason is simple. A fluid at rest has the same pressure at the same level, and any extra push at one place has to be balanced everywhere, so the whole fluid's pressure rises by the same amount.

A hydraulic lift uses this. A small force $F_{1}$ on a small piston of area $A_{1}$ creates a pressure $F_{1}/A_{1}$. The same pressure acts on a large piston of area $A_{2}$, giving

$$F_{2} = F_{1} \times \dfrac{A_{2}}{A_{1}}$$

The area ratio is the force multiplier. Hydraulic brakes, car jacks and earth movers all work this way.

Worked example: lifting a car. A hydraulic lift has pistons of radius 1.0 cm and 10 cm. What force on the small piston lifts a 1500 kg car (g = 10 m s⁻²)? How far must the small piston move to raise the car by 10 cm?

Two wrong steps are common. One is to use the radius ratio (10) instead of the area ratio (100). The other is to think the lift gives you extra energy. It trades force for distance, and the work in equals the work out. This is why a real jack needs many pumps.

Fluids in motion: streamlines and continuity

In steady (streamline) flow, every particle that passes a given point has the same velocity at that point, and this does not change with time. A streamline is the path of such a particle. Its tangent at each point gives the fluid's velocity there, so two streamlines can never cross, because a particle at a crossing point would need two velocities at once.

The equation of continuity is conservation of mass applied to a pipe. In steady flow the same mass must enter and leave every section each second, so for an incompressible fluid

$$A_{1}v_{1} = A_{2}v_{2} = \text{constant}$$

The product $Av$ is the volume flow rate in m³ s⁻¹. When the pipe narrows, the fluid speeds up. A thumb over a garden hose does exactly this.

Beyond the textbook. For the exam: flow is either streamlined or turbulent, depending on whether the speed is below or above a critical value. In reality: engineers use a dimensionless number, the Reynolds number $\text{Re} = \rho v d/\eta$, where d is the pipe diameter and $\eta$ the viscosity. Flow in a long, smooth pipe stays laminar for Re up to roughly 2000 and becomes turbulent above roughly 4000, so a thick, sticky fluid is more likely to stay smooth. Why it matters: it shows that the critical speed is not a property of the fluid alone. It depends on the pipe size, the density and the viscosity together.

Bernoulli's principle: speed trades against pressure

For an ideal fluid (non-viscous and incompressible) in steady flow, along a streamline

$$P + \dfrac{1}{2}\rho v^{2} + \rho g h = \text{constant}$$

Each term is an energy per unit volume: pressure energy, kinetic energy and potential energy. The equation is the work–energy theorem for a small chunk of fluid. As the chunk moves from section 1 to section 2, the net work done on it by the pressure forces, $(P_{1} - P_{2})\,\Delta V$, equals the gain in its kinetic and potential energies. Divide by the volume $\Delta V$ and you get the equation above.

Here is the physical meaning. To make a fluid speed up as it enters a narrow section, something has to push it forward harder than it is pushed back. So the pressure behind it must exceed the pressure ahead, and where the fluid is faster, the pressure is lower. This is not a rule about fluids "preferring" low pressure. It is Newton's second law.

At rest ($v = 0$) the equation reduces to $P_{2} - P_{1} = \rho g(h_{1} - h_{2})$, the same depth relation as before. Bernoulli's equation fails for viscous fluids, compressible flow and turbulent flow, so it does not hold for rapids in a river.

Worked example: a pipe that narrows. Water flows through a horizontal pipe at 2.0 m s⁻¹ where its cross-section is 20 cm². The pipe narrows to 5.0 cm². Find the speed in the narrow part and the drop in pressure. Take $\rho = 1000$ kg m⁻³.

The pressure in the narrow part is 30 kPa (about 0.3 atm) lower. The common wrong step is to say the pressure is higher where the fluid is faster, because "more speed means more push". The faster part has the lower pressure.

Torricelli's law. A tank is open at the top, and there is a small hole at depth h below the surface. At the surface and at the hole the pressure is atmospheric. The surface drops so slowly that its speed is negligible, so Bernoulli gives $(1/2)\rho v^{2} = \rho g h$, which means

$$v = \sqrt{2gh}$$

This is the speed of a body falling freely through the same height h. If the air above the liquid is at a higher pressure P, then $v = \sqrt{2gh + 2(P - P_{a})/\rho}$. It also explains why a jet from a hole low on a bottle shoots further than one from near the top.

Pause and predict: you hang two sheets of paper close together and blow air between them. Do they swing apart or together?

Check your answer: they swing together. The moving air between them has lower pressure, and the still air outside, at atmospheric pressure, pushes the sheets inward.

Beyond the textbook. For the exam: the wing's shape makes air move faster above than below, so the pressure difference $\Delta P = (1/2)\rho(v_{2}^{2} - v_{1}^{2})$ gives lift. In reality: the popular line that air must take the same time over the top and bottom is wrong, because air above the wing actually reaches the rear edge first. A wing deflects air downward, and by Newton's third law the air pushes the wing up. The pressure difference and the downward deflection of air are two descriptions of one effect. Why it matters: never use "equal transit time" as a reason in an answer. Use "faster flow, lower pressure".

Viscosity: friction inside a fluid

When one layer of a fluid slides over another, they drag on each other. This internal friction is viscosity. Imagine a liquid between two plates of area A and gap l, with the top plate moving at speed v. The force needed to keep it moving is

$$F = \eta A \dfrac{v}{l}$$

so $\eta = (F/A)/(v/l)$: the ratio of shear stress to strain rate. Its SI unit is the pascal second (Pa s), also called the poiseuille, and its dimensions are [M L⁻¹ T⁻¹]. Water is about $1 \times 10^{-3}$ Pa s and air about $1.8 \times 10^{-5}$ Pa s at room temperature. In the CGS system, 1 poise = 0.1 Pa s.

Stokes' law and terminal velocity. A small sphere of radius a moving at speed v through a fluid of viscosity $\eta$ feels a viscous drag

$$F = 6\pi \eta a v$$

Drop a sphere of density $\rho$ into a fluid of density $\sigma$. At first it speeds up, but the drag grows with speed. At the terminal velocity the weight is balanced by drag plus buoyancy, so

$6\pi\eta a v_{t} = (4/3)\pi a^{3}(\rho - \sigma)g$, which gives $v_{t} = 2a^{2}(\rho - \sigma)g/(9\eta)$

Terminal velocity varies as the square of the radius and inversely as the viscosity. It also disappears when $\rho = \sigma$, since there is then no net force to drive the ball downward.

Pause and predict: you double the radius of a small steel ball falling through oil. By what factor does its terminal speed change?

Check your answer: it becomes four times larger, because $v_{t} \propto a^{2}$. A larger ball gains weight as $a^{3}$ but its drag grows only as a, so it must reach a higher speed before the forces balance.

Worked example: a ball in oil. A steel ball of radius 1.0 mm and density $8.0 \times 10^{3}$ kg m⁻³ falls through oil of density $0.8 \times 10^{3}$ kg m⁻³ and viscosity 0.40 Pa s. Find its terminal velocity (g = 10 m s⁻²).

The classic wrong step is to put in the diameter (2.0 mm) for a. Then $a^{2}$ is four times too big and you get 0.16 m s⁻¹. The other is to use $\rho$ alone and forget buoyancy.

Beyond the textbook. For the exam: $v_{t} = 2a^{2}(\rho - \sigma)g/(9\eta)$. In reality: Stokes' law holds only for slow, smooth flow around small spheres. Putting a 1 mm radius raindrop into it gives a speed above 100 m s⁻¹, yet real raindrops of that size fall at only about 6 to 7 m s⁻¹. At those speeds air flows turbulently around the drop and the drag grows roughly as $v^{2}$ instead of $v$. For the tiny droplets in a cloud, with a radius near 10 µm, the formula works well and gives about 1 cm s⁻¹, which is why clouds float. Why it matters: a formula is only as good as its assumptions.

Surface tension: the skin on a liquid

A molecule deep inside a liquid is pulled equally in all directions, so the forces cancel. A molecule at the surface has neighbours only below and beside it, so it is pulled inward. Bringing a molecule up to the surface therefore costs energy. A liquid minimises the number of molecules at its surface, so it behaves as if its surface were a stretched elastic skin.

Surface tension S can be read in two equivalent ways. As force per unit length acting along the surface, perpendicular to an imaginary line drawn on it, $S = F/l$. As energy per unit area, $S = W/\Delta A$. To see why they agree, pull a soap film on a wire frame outward by a distance d against a force F. The film has two surfaces, so the area added is $2ld$. The work $Fd$ equals the surface energy $S \times 2ld$, giving $S = F/2l$. The unit is N m⁻¹, which equals J m⁻². Water has $S \approx 0.072$ N m⁻¹ at room temperature, and mercury about 0.47 N m⁻¹. Surface tension decreases as temperature rises, and it falls sharply when soap or detergent is added.

A liquid drop is spherical because a sphere has the least surface area for a given volume.

To see where 2S/r comes from, imagine the drop's radius growing by dr. The extra pressure $\Delta P$ pushes out a surface of area $4\pi r^{2}$, so the work is $\Delta P \times 4\pi r^{2}\,dr$. The surface energy gained is $S \times d(4\pi r^{2}) = S \times 8\pi r\,dr$. Equating the two gives $\Delta P = 2S/r$. A soap bubble has two surfaces, so the surface energy gained is doubled and $\Delta P = 4S/r$. A smaller bubble has a higher internal pressure.

Pause and predict: two soap bubbles of different sizes are joined by a tube. Which one grows?

Check your answer: the larger one. The small bubble has the higher pressure ($4S/r$ is bigger for smaller $r$), so air flows from small to large. The small bubble shrinks and the large one swells.

Worked example: breaking up a drop. A water drop of radius 1.0 cm splits into 1000 identical droplets. Find the energy required, taking S = 0.072 N m⁻¹.

The common wrong step is to take $r = R/1000$, forgetting that volume, not radius, is shared out among the droplets. In general $r = R/n^{1/3}$. When droplets merge, this surface energy is released, which is a reason raindrops warm up very slightly as they coalesce.

Angle of contact. Where a liquid surface meets a solid, the angle of contact $\theta$ is the angle between the tangent to the liquid surface and the solid surface, measured inside the liquid. Balancing the three interface tensions gives $S_{la} \cos\theta = S_{sa} - S_{sl}$.

Soaps and detergents are wetting agents, since they lower the angle of contact so that water spreads into the cloth. Waterproofing agents raise it.

Capillary rise. In a narrow tube of radius a dipped in water, the meniscus is concave. Its radius of curvature is $R = a/\cos\theta$, so the pressure just below it is lower than atmospheric by $2S/R = 2S\cos\theta/a$. The atmosphere pushes the liquid up until the extra column $\rho g h$ balances this:

$$h = \dfrac{2S\cos\theta}{\rho g a}$$

The rise is larger in a thinner tube and for a lower density. For mercury, $\theta$ is obtuse, so $\cos\theta$ is negative and h comes out negative: the level is depressed.

Worked example: mercury in a glass tube. Mercury ($\rho = 13.6 \times 10^{3}$ kg m⁻³, S = 0.465 N m⁻¹, $\theta$ = 135°) is in a glass capillary of radius 1.0 mm. Find the level change. Take g = 9.8 m s⁻².

The mercury level inside the tube is 4.9 mm below the outside level. The common wrong step is to drop the minus sign and say mercury rises.

Pause and predict: water rises 3 cm in a capillary tube on Earth. How high would it rise in the same tube on the Moon, where g is about one sixth of Earth's?

Check your answer: about 18 cm. Since $h \propto 1/g$, a weaker g makes the column six times taller.

Beyond the textbook. For the exam: water climbs a capillary tube by $h = 2S\cos\theta/(\rho g a)$. In reality: capillarity alone cannot lift water up a tall tree. For a 1 µm wide xylem channel the formula gives only about 15 m, yet trees exceed 100 m. The main driver is evaporation from leaves, which pulls on a continuous, cohesive thread of water, a mechanism called the cohesion–tension theory. Why it matters: the formula is correct, but it is not the whole story of how plants drink.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Mechanical Properties of Solids deals with stress, strain and elastic moduli, and the bulk modulus tells you why liquids are nearly incompressible.
• Same ideas elsewhere: Work, Energy and Power supplies the work–energy theorem that Bernoulli's equation is built on, and Laws of Motion explains why a falling sphere stops accelerating at terminal velocity.
• Next: Thermal Properties of Matter moves from how fluids flow to how they carry and store heat.