Fluid Mechanics — Pressure, Buoyancy, and Flow

Fluids: Liquids and Gases Together

A fluid is any substance that can flow and takes the shape of its container. Both liquids and gases are fluids, but they behave differently: liquids are nearly incompressible (fixed volume), while gases are highly compressible.

Fluid mechanics studies forces and motion in fluids. It has two main branches:

Fluid statics: Fluids at rest (pressure, buoyancy)
Fluid dynamics: Fluids in motion (flow rate, Bernoulli’s principle)

Pressure in Fluids

Pressure is force per unit area: $P = \frac{F}{A}$ .

For a fluid, pressure acts equally in all directions at any point (Pascal’s law basis).

P = P_0 + \rho g h

$P_0$ = pressure at the surface (atmospheric pressure for open containers)
$\rho$ = density of the fluid (kg/m³)
$g$ = gravitational acceleration (9.8 m/s²)
$h$ = depth below the surface

Why does pressure increase with depth? Each layer of fluid supports the weight of all fluid above it. The deeper you go, the more fluid presses down.

Pressure depends only on depth — not on the shape of the container or the total volume of fluid. This is why a submarine at 1 km depth experiences the same pressure regardless of whether the ocean is wide or narrow above it.

Pascal’s Law

Pascal’s law: A change in pressure applied to an enclosed fluid is transmitted equally throughout the fluid.

This is the operating principle of hydraulic systems:

\frac{F_1}{A_1} = \frac{F_2}{A_2} \Rightarrow F_2 = F_1 \times \frac{A_2}{A_1}

If $A_2 >> A_1$ , a small force $F_1$ produces a large force $F_2$ — the basis of hydraulic jacks, brakes, and lifts.

Example: A hydraulic press has pistons of area 2 cm² and 100 cm². A force of 20 N on the small piston produces:

F_2 = 20 \times \frac{100}{2} = 1000 \text{ N}

The work done on both sides is equal (energy conservation): the large piston moves a smaller distance by the same factor.

Archimedes’ Principle and Buoyancy

When an object is submerged in a fluid, the fluid exerts an upward force — the buoyant force — equal to the weight of the fluid displaced.

F_b = \rho_f V_{submerged} g

$\rho_f$ = density of the fluid
$V_{submerged}$ = volume of the object that is submerged
$g$ = gravitational acceleration

Why buoyancy occurs: Pressure increases with depth. The pressure on the bottom face of a submerged object is greater than on the top face, creating a net upward force.

Floating, Sinking, and Hovering

An object:

Floats if $\rho_{object} < \rho_{fluid}$ (buoyancy > weight when fully submerged, so it rises until partially submerged)
Sinks if $\rho_{object} > \rho_{fluid}$ (weight > buoyancy even when fully submerged)
Hovers (neutral buoyancy) if $\rho_{object} = \rho_{fluid}$

For a floating object: Weight = Buoyant force

m g = \rho_f V_{displaced} g

\frac{V_{submerged}}{V_{total}} = \frac{\rho_{object}}{\rho_{fluid}}

Example: Ice (density 900 kg/m³) floats in water (1000 kg/m³) with 90% of its volume submerged.

A common JEE question: An ice cube with an iron ball inside it floats in water. What happens to the water level when the ice melts? When ice melts: the water from ice contributes exactly the volume the ice was displacing (since ice and water have the same mass). But the iron ball now sinks and displaces less water than it did when embedded in the floating ice. So the water level falls.

Equation of Continuity

For fluid flowing through a pipe, the equation of continuity states that mass is conserved:

A_1 v_1 = A_2 v_2 \quad (\text{for incompressible fluids})

$A$ = cross-sectional area of the pipe
$v$ = fluid velocity

Where the pipe is narrower, the fluid flows faster. You experience this when you partially block a garden hose nozzle — the stream goes farther because it’s moving faster.

Bernoulli’s Theorem

Bernoulli’s theorem is the energy conservation equation for fluid flow:

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

Along any streamline in steady, incompressible, non-viscous flow.

Physical meaning: The sum of pressure energy, kinetic energy, and potential energy per unit volume is constant along a streamline.

Key consequence: Where velocity is high, pressure is low (and vice versa).

Applications of Bernoulli’s Theorem

1. Venturimeter: Measures fluid flow rate. The pressure difference between the wide and narrow sections tells us the velocity.

2. Aircraft lift: Air moves faster over the curved top of a wing than the flat bottom, creating lower pressure above and higher pressure below — net upward lift force.

The “lift on aircraft wings” explanation by Bernoulli alone is actually incomplete — angle of attack also contributes significantly. But for exam purposes, the Bernoulli explanation is accepted and expected.

3. Atomizer/sprayer: Fast-moving air above a liquid tube creates low pressure, drawing the liquid up and atomizing it.

4. Magnus effect: A spinning ball has higher velocity on one side (where spin and air flow add) and lower on the other side, creating a pressure difference. This is why a cricket ball swings when bowled with the seam at an angle.

Viscosity and Stokes’ Law

Viscosity is internal friction in a fluid — it’s what makes honey harder to pour than water. Coefficient of viscosity $\eta$ (eta) is measured in Pa·s (Pascal-seconds).

Viscosity decreases with temperature for liquids (honey flows more easily when warm) and increases with temperature for gases.

Stokes’ law: The drag force on a small sphere of radius $r$ moving with velocity $v$ through a viscous fluid:

F_{drag} = 6\pi \eta r v

Terminal velocity: When a body falls through a viscous fluid, drag increases with velocity until $F_{drag} + F_{buoyancy} = mg$ . At this point, acceleration is zero and the body falls at constant velocity — the terminal velocity.

v_t = \frac{2r^2(\rho - \rho_f)g}{9\eta}

Solved Examples

Example 1 — CBSE Level

A hydraulic lift has a small piston of area 50 cm² and a large piston of area 2000 cm². What force on the small piston is needed to lift a 1500 kg car?

Weight of car $= 1500 \times 10 = 15000$ N

F_1 = F_2 \times \frac{A_1}{A_2} = 15000 \times \frac{50}{2000} = 375 \text{ N}

Example 2 — JEE Main Level

Water flows through a pipe of radius 4 cm at 2 m/s. It then passes through a section of radius 2 cm. Find the velocity in the narrower section.

A_1 v_1 = A_2 v_2

\pi (0.04)^2 \times 2 = \pi (0.02)^2 \times v_2

v_2 = 2 \times \frac{(0.04)^2}{(0.02)^2} = 2 \times 4 = 8 \text{ m/s}

Example 3 — JEE Advanced Level

Water flows horizontally in a pipe. At a certain section, the radius is 2 cm and the velocity is 5 m/s. At another section, the radius is 1 cm. Find the pressure difference between these two sections. ( $\rho = 1000$ kg/m³)

First, find velocity in narrower section using continuity:

v_2 = v_1 \times \frac{A_1}{A_2} = 5 \times \frac{\pi(0.02)^2}{\pi(0.01)^2} = 5 \times 4 = 20 \text{ m/s}

Apply Bernoulli’s equation (same height, so $\rho g h$ terms cancel):

P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2

P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{1}{2} \times 1000 \times (400 - 25) = 500 \times 375 = 187500 \text{ Pa}

Common Mistakes to Avoid

Mistake 1: In hydraulic systems, confusing force multiplication with energy multiplication. Pascal’s law multiplies force, but energy is conserved — the large piston moves a proportionally shorter distance.

Mistake 2: Applying Bernoulli’s equation between two points not on the same streamline, or in situations with significant viscosity. Bernoulli applies only along a streamline in ideal (non-viscous) flow.

Mistake 3: Forgetting that buoyancy force depends on the volume submerged, not the object’s total volume. A floating object has $V_{submerged} < V_{total}$ .

Mistake 4: In continuity equation, using diameters instead of areas. If diameter doubles, area quadruples ( $A = \pi r^2$ ). So velocity decreases to one-quarter, not one-half.

Mistake 5: Thinking apparent weight = actual weight for an object in air. Even in air, there’s a small buoyant force equal to the weight of air displaced. This is why precise mass measurements use vacuum balances. For most problems this is negligible, but the concept appears in theory questions.

Exam-Specific Tips

CBSE Class 11: Focus on Pascal’s law, Archimedes’ principle with examples (ship, submarine, hot air balloon), and the definition of viscosity. Bernoulli’s applications (venturimeter, sprayer) are common theory questions.

JEE Main: Numerical problems on continuity + Bernoulli combined. Given flow rate and pipe dimensions, find pressure. Also: terminal velocity derivation. Know the formula for speed of efflux from a hole in a tank ( $v = \sqrt{2gh}$ ).

NEET: Buoyancy and floating/sinking conditions are straightforward. Questions often involve comparing densities or finding what fraction of an object floats above the liquid.

Practice Questions

Q1. A cube of side 10 cm and density 600 kg/m³ floats in water. What fraction of it is above water?

$\frac{V_{above}}{V_{total}} = 1 - \frac{\rho_{cube}}{\rho_{water}} = 1 - \frac{600}{1000} = 0.4 = 40\%$

Q2. Find the terminal velocity of a steel ball (radius 1 mm, density 7800 kg/m³) in glycerine (density 1260 kg/m³, $\eta = 1.5$ Pa·s).

$v_t = \frac{2r^2(\rho - \rho_f)g}{9\eta} = \frac{2 \times (10^{-3})^2 \times (7800 - 1260) \times 10}{9 \times 1.5} = \frac{2 \times 10^{-6} \times 6540 \times 10}{13.5} = \frac{0.1308}{13.5} \approx 9.7 \times 10^{-3}$ m/s $\approx 9.7$ mm/s

Q3. A tank of water has a hole at the bottom. The water surface is 1.8 m above the hole. Find the speed of water emerging from the hole.

By Torricelli’s theorem (special case of Bernoulli): $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 1.8} = \sqrt{36} = 6$ m/s

FAQs

Q: Why do ships made of steel float despite steel being denser than water?

A ship is a hollow structure. The effective density of the ship (total mass ÷ total volume, including the air inside) is less than water. The large volume of displaced water creates a buoyant force equal to the ship’s weight.

Q: Why is it easier to float in salt water than fresh water?

Seawater has higher density ( $\approx$ 1025 kg/m³) than fresh water (1000 kg/m³). A higher density fluid provides more buoyant force for the same submerged volume, so you float higher with less volume submerged.

Q: Why does a spinning ball curve in cricket?

The Magnus effect. A spinning ball has higher air velocity on one side (where spin direction matches airflow direction) and lower on the other side. By Bernoulli’s principle, lower velocity = higher pressure. The pressure difference creates a sideways force, curving the ball’s path.

Q: What is gauge pressure?

Pressure measured above atmospheric pressure. Many gauges (tire pressure, blood pressure) read gauge pressure, not absolute pressure. Absolute pressure = gauge pressure + atmospheric pressure ( $1.013 \times 10^5$ Pa).