Latest
  • Admissions openClass 11 Science, 2027‑28: JEE, NEET and MHT‑CET with junior college and hostel under one roofApply now
  • IMGSAT 2027Free scholarship and admission test for Class 10 students, every Saturday and Sunday at our Nashik campusRegister
  • Free Foundation 2026‑27Evening classes for Class 10 in Physics, Chemistry, Maths and Biology, taught by our IITian and doctor facultyJoin free

+91 70303 00666

IIT Medico Guide In collaboration with Ashoka Group of Schools

Physics · Class 11 · Chapter 9

Mechanical Properties of Fluids

Liquids and gases flow, and they cannot hold a shape of their own. This chapter covers fluids at rest (pressure), fluids in motion (Bernoulli), internal friction (viscosity) and the skin-like behaviour of liquid surfaces (surface tension).

In this chapter: pressure and its variation with depth, atmospheric and gauge pressure, Pascal's law and hydraulic machines, streamline flow and the equation of continuity, Bernoulli's principle, speed of efflux, dynamic lift, viscosity, Stokes' law and terminal velocity, surface tension, surface energy, excess pressure in drops and bubbles, angle of contact and capillary rise.

Pressure

A fluid at rest exerts a force normal to any surface in contact with it. Pressure is the normal force per unit area, P = F/A. It is a scalar, with SI unit pascal (1 Pa = 1 N m−2). One atmosphere is 1.013 × 105 Pa, the pressure of a 76 cm column of mercury at sea level.

Variation with depth

P = Pa + ρghPa is atmospheric pressure at the surface. The excess P − Pa = ρgh is the gauge pressure, which is what a tyre gauge reads.

The pressure at a point depends only on the depth, not on the shape of the container or the amount of liquid. Three vessels of different shapes filled to the same height have the same pressure at the base. This surprising result is called the hydrostatic paradox. All points at the same depth in a connected liquid at rest are at the same pressure.

A mercury barometer measures atmospheric pressure as Pa = ρgh, where h is the height of the mercury column. An open-tube manometer measures the gauge pressure of a gas as the difference in the liquid levels in its two arms.

Pascal's law and hydraulic machines

Pascal's law: a change in pressure applied to an enclosed incompressible fluid is transmitted undiminished to every point of the fluid and to the walls of the container. In a hydraulic lift, a small force F1 on a small piston of area A1 creates a pressure that acts on a large piston of area A2:

F2 = F1 × A2A1The force is multiplied, but the large piston moves through a correspondingly smaller distance, so no energy is gained.
Worked example: In a hydraulic lift, the small piston has an area of 10 cm2 and the large piston an area of 1000 cm2. What force on the small piston will lift a car of mass 1200 kg? (g = 10 m s−2)
Solution: F2 = 1200 × 10 = 12 000 N. F1 = F2 × A1/A2 = 12 000 × 10/1000 = 120 N. Hydraulic brakes in cars work on the same principle.

Streamline flow and continuity

In steady (streamline) flow, the velocity of the fluid at each point does not change with time. The path of a fluid particle is a streamline, and no two streamlines cross. Above a critical speed the flow becomes turbulent, with eddies and irregular motion.

For an incompressible fluid, the mass entering one end of a tube per second must leave the other end. This gives the equation of continuity:

A1v1 = A2v2 = constantThe volume flow rate Av stays the same, so the fluid moves faster where the tube is narrower. This is why water spurts faster when you partly close the end of a hose with your thumb.

Bernoulli's principle

For steady, non-viscous, incompressible flow, applying the work-energy theorem to a moving element of fluid gives

P + ½ρv2 + ρgh = constantAlong a streamline. The three terms are pressure energy, kinetic energy and potential energy per unit volume.
Bernoulli’s principle: flow through a pipe of varying area and heightwww.iitmedicoguide.comv1v2A1, P1A2, P2reference levelh1h2P + ½ρv² + ρgh = constant along a streamlineA1v1 = A2v2, so the fluid speeds up where the pipe narrowswww.iitmedicoguide.com
Where the pipe narrows, continuity forces the fluid to speed up. By Bernoulli's principle, the pressure there must then be lower (after allowing for the change in height).

For a horizontal pipe (h constant), a higher speed goes with a lower pressure. For fluid at rest (v = 0), Bernoulli's equation reduces to P = Pa + ρgh.

Speed of efflux: Torricelli's law

A liquid flowing out of a small hole at depth h below the open surface of a wide tank leaves with speed v = √2gh, the same speed as a body falling freely through h. For a hole 5.0 m below the surface, v = √(2 × 9.8 × 5.0) ≈ 9.9 m s−1.

Dynamic lift

An aeroplane wing (aerofoil) is shaped so that air flows faster over its upper surface than below it. The pressure above is lower, giving an upward force called dynamic lift. A spinning ball drags air around with it, so the air speed is different on its two sides; the pressure difference gives it a curved path (the Magnus effect). This is why a spinning cricket or tennis ball curves in the air.

Viscosity

Real fluids resist relative motion between their layers. This internal friction is viscosity. When a liquid flows over a fixed surface, the layer in contact is at rest and the velocity increases layer by layer. The coefficient of viscosity η is defined by

F = ηA dvdxSI unit: Pa s (also called poiseuille, Pl); dimensions [ML−1T−1].

The viscosity of liquids decreases as temperature rises, which is why honey pours more easily when warm. The viscosity of gases increases with temperature.

Stokes' law and terminal velocity

A small sphere of radius r moving with speed v through a viscous fluid feels a drag force F = 6πηrv. A sphere falling through the fluid accelerates until the drag and buoyancy balance its weight, after which it falls at a constant terminal velocity:

vt = 2r2(ρ − σ)g9ηρ is the density of the sphere and σ the density of the fluid. Note vt ∝ r2. If σ > ρ, the sphere rises, like an air bubble in water.
Worked example: A copper ball of radius 2.0 mm falls through oil at a terminal velocity of 6.5 cm s−1. Densities: copper 8.9 × 103 kg m−3, oil 1.5 × 103 kg m−3. Find the viscosity of the oil. (g = 9.8 m s−2)
Solution: η = 2r2(ρ − σ)g/(9vt) = 2 × (2 × 10−3)2 × (7.4 × 103) × 9.8/(9 × 6.5 × 10−2) = 0.580/0.585 ≈ 0.99 Pa s.

Surface tension

A molecule inside a liquid is pulled equally in all directions by its neighbours, but a molecule at the surface has neighbours only on one side. Surface molecules therefore have extra energy, and the liquid tends to reduce its surface area. This is why small drops are spherical.

S = F/L = surface energy per unit areaSI unit N m−1 or J m−2. For water at 20 °C, S ≈ 0.073 N m−1. S decreases as temperature rises, and detergents lower it.

Excess pressure in drops and bubbles

Because a curved surface pulls inward, the pressure on the concave side is higher than on the convex side.

liquid drop or air bubble inside a liquid: ΔP = 2S/rsoap bubble (two surfaces): ΔP = 4S/r

Angle of contact and capillary rise

The angle of contact θ is the angle between the tangent to the liquid surface and the solid surface, measured inside the liquid. If θ < 90° the liquid wets the solid (water on clean glass, θ ≈ 0); if θ > 90° it does not (mercury on glass, water on a waxy leaf). Soaps and detergents reduce θ so that water can wet greasy cloth.

In a narrow tube (capillary) dipped in a wetting liquid, the liquid rises to a height

h = 2S cos θρ g r
Capillary rise of water in a glass tubewww.iitmedicoguide.comhrconcavemeniscuswaterCapillary riseh = 2S cos θ / (ρ g r)Water wets glass (θ ≈ 0°),so it rises. Mercury (θ > 90°)is pushed down instead.Narrower tube: higher rise(h ∝ 1/r).www.iitmedicoguide.com
Water rises in a clean glass capillary and forms a concave meniscus. In mercury the meniscus is convex and the level inside the tube falls below the outside level.

For water in a clean glass tube of radius 0.5 mm, h = 2 × 0.073/(1000 × 9.8 × 5 × 10−4) ≈ 0.030 m, or about 3 cm.

Common mistakes: (1) Using 2S/r for a soap bubble, which has two surfaces and needs 4S/r. (2) Mixing up gauge pressure and absolute pressure. (3) Using Bernoulli's equation for viscous or turbulent flow. (4) Forgetting the buoyancy term (ρ − σ) in the terminal velocity. (5) Writing that capillary rise is larger in a wider tube; h is inversely proportional to r.

JEE and NEET focus

  • Pressure at depth, gauge versus absolute pressure, and the hydraulic lift.
  • Continuity and Bernoulli together: pressure difference in a pipe, speed of efflux and range of the jet.
  • Terminal velocity and its dependence on r2; drag from Stokes' law.
  • Excess pressure in drops and bubbles, and work done in blowing a bubble (surface energy × area, counting both surfaces).
  • Capillary rise, including a tube shorter than h (the liquid does not overflow; the meniscus radius adjusts).

Practice questions

The gauge pressure at a depth of 10 m in water (ρ = 1000 kg m−3, g = 10 m s−2) is:

  1. 103 Pa
  2. 104 Pa
  3. 105 Pa
  4. 106 Pa
Show answer
C. ρgh = 1000 × 10 × 10 = 105 Pa, about one atmosphere.

Water flows through a pipe whose radius halves at one section. The speed there becomes:

  1. Half
  2. Double
  3. Four times
  4. The same
Show answer
C. Av is constant and A ∝ r2.

The radius of a small sphere falling through a viscous liquid is doubled. Its terminal velocity becomes:

  1. Half
  2. Double
  3. Four times
  4. Eight times
Show answer
C. vt ∝ r2.

The excess pressure inside a soap bubble of radius r and surface tension S is:

  1. S/r
  2. 2S/r
  3. 4S/r
  4. 8S/r
Show answer
C. A soap film has two surfaces.

Water rises to a height h in a capillary tube. In a tube of the same glass with half the radius, it rises to:

  1. h/2
  2. h
  3. 2h
  4. 4h
Show answer
C. h ∝ 1/r.

When a liquid is heated, its coefficient of viscosity:

  1. Increases
  2. Decreases
  3. Stays the same
  4. First increases, then decreases
Show answer
B. For gases it increases with temperature.

Water flows out of a small hole 20 m below the surface of a wide open tank (g = 10 m s−2). The speed of efflux is:

  1. 10 m s−1
  2. 14 m s−1
  3. 20 m s−1
  4. 40 m s−1
Show answer
C. v = √(2gh) = √400 = 20 m s−1.

Mercury in a glass capillary tube shows:

  1. A rise, with a concave meniscus
  2. A fall, with a convex meniscus
  3. A rise, with a convex meniscus
  4. No change in level
Show answer
B. The angle of contact is obtuse, so cos θ is negative and h is negative.
Call WhatsApp Apply
Chat with us on WhatsApp